Product of quotient topology not homeomorphic to the quotient of product topology How can I construct a counter-example for the following statement: if $\sim_X$ is an equivalence relation on $X$, $\sim_Y$ is an equivalence relation on $Y$, then $X/{\sim_X} \times Y/{\sim_Y}$ is homeomorphic to $(X \times Y)/{\sim_{XY}}$ where $(x_1,y_1) \sim_{XY} (x_2,y_2)$ iff $x_1 \sim_X x_2$ and $y_1 \sim_Y y_2$?

There is a related question on this, Products of quotient topology same as quotient of product topology , which says that it is false that if $p : X \to Z_1$ and $q : Y \to Z_2$ are quotient maps, then $p \times q$ is necessarily a quotient map. However, that this particular map is not a quotient map doesn't imply that an homeomorphism cannot be constructed.

This answer is incorrect, and does not contribute to answering the posted question. Don't read it, and if you do, keep in mind it only illustrates how one could make a mistake.
Let $X=Y=[0,1]$ with $0\sim1$. Then $X/\sim$ and $Y/\sim$ each is homeomorphic to a circle, so their product is a torus. On the other hand $(X\times Y)/\sim$
is a square with the four corners identified, which is not homeomorphic to a torus.
Edit to address the comment below.
Take a handkerchief and lift the four corners, bending it, so they touch. (But, apart from the corners meeting at a single point, the edges do not touch.) This is $(X\times Y)/\sim$.
Here is a more formal justification that there is no homeomorphism between $(X/\sim)\times(Y/\sim)$ (a torus) and $(X\times Y)/\sim$.
Removing one point from $(X\times Y)/\sim$ - the point where the corners of the handkerchief touch - leaves a simply connected space (a closed square with the four corners removed). There is no point with this property on the torus.


Possible combinations of two sets of three into a set of two I'm trying to find the mathematical term for a certain phenomenon and hopefully a more general way to solve such problems. Say there are two sets, each $= \{A, B, C\}.$ I would like to know how many different possible ways I can combine these sets into a set of two $\{...,...\}.$ The combination $\{A, B\}$ is considered the same as $\{B, A\}.$ The only way I can think to do this, is:

$$ 3*3 - \binom{3}{2} = 6 $$

Basically, I am counting up the total number of possible combinations (3*3) and subtracting out the combinations I overcounted. Is there a name for this kind of a calculation? Is there anyway to describe this more mathematically, as opposed to this intuitive approach?

Thanks!

You are actually counting the number of subsets of $\{A,B,C\}$ of a fixed size ($2$, in your example), without regarding order, and allowing duplicates. A more general way to look at it, is taking some set $A=\{1,\ldots,n\}$, and trying to assign a number of occurrences to each of $1\ldots n$, that is, "$1$ appears once, $2$ appears five times, etc.", while demanding that the sum of these numbers be some constant $k$ - that is, you're only selecting a set of $k$ elements.

This is equivalent to the following situation:

You have $n$ buckets and $k$ balls. You wish to toss the balls into the buckets, putting as many balls in one bucket as you like. In order to calculate the number of ways to do that, it's more convenient to look at the $n$ buckets as $n-1$ partitions - defining $n$ different spaces, that you can place among the balls, like in the following illustration:

$$\cdot|\cdot\cdot\cdot||\cdot|\cdot\cdot||$$

This has $7$ buckets, with $1,3,0,1,2,0,0$ balls respectively.

Now we can give a general equation.

Sort the partitions and the balls together - there are $(n-1+k)!$ ways to do that.

Disregard the internal ordering of the partitions and the internal ordering of the balls, as you're only interested in the number of balls in each "bucket", and the "buckets" are defined by their position regardless of which partitions surround them.

Therefore, the number of ways to select a set of size $k$, with duplicates allowed and without regarding order, from a set of size $n$ is $\frac{(n+k-1)!}{(n-1)!k!}=\binom{n+k-1}{k}$.

In your particular case, $n=3$ and $k=2$, so this equals $\binom{4}{2}=6$.

I assume that pairs like $\text{ {A,A}, {B,B}, {C,C}}$ are allowed.

I'll get sets of 5 exactly same numbers to make things clearer. Let

$$a = {A,B,C,D,E} \quad b = {A,B,C,D,E}$$

First we chose A there are $5$ other option to make a pair.

After that we count possibilities with B. There are $4$ possibilites to make a pair that doesn't exist.

For C there are $3$.

Can you spot the pattern.

$$\text{Number of possibilities} = n(n-1)(n-2)...2 \cdot 2$$

Or you can denote it as:

$$\text{Number of possibilities} = n!$$


Solving $\sin(x^2)=\sin(x)^2$ (without transcendence of $\pi$) Is it possible to find all solutions by analytic means? Maybe even without using the fact that $\pi$ does not satisfy a quadratic polynomial with integer coefficients?

We are in a case of relationship $(f \circ g - g \circ f)(x)=0$ with $f = \sin$ and $g:x \rightarrow x^2$. That may or not have in interest. In order to have an idea a general position of roots, we have plotted curves of $f$ and $g$ (we have limited the plot to the first quadrant because both functions are even). There is an infinite number of roots for which a general formula probably does not exist.

First way

Maybe one method could be the usage of Taylor Series.

Indeed let's try to a low order:

$$\sin x^2 \approx x^2$$

$$\sin^2(x)\approx x^2$$

Which means $$x^2 = x^2$$ not very useful since every real number could be solution of that identity and it won't obviously work. We need more terms.

With more terms, you'll find

$$\sin x^2 \approx x^2 - \frac{x^6}{6}$$

$$\sin^2(x) \approx x^2 - \frac{x^4}{3}$$

thence you have to solve

$$-\frac{x^6}{6} + \frac{x^4}{3} = 0$$

$$-\frac{x^4}{3}\left(\frac{x^2}{2} - 1\right) = 0$$

thence besides the trivial solution $x = 0$ you have $$x^2 = 2$$

More terms more solutions. But surely you won't get them all I guess..

Also this is a numerical method so it's really approximative. But it's a try like everyotherelse


If, for every $n$, $f_n\to0$ at infinity and $|f_n(x_n)|\geq C$, can one deduce that $(x_n)$ is bounded? Let $f_n:\mathbb R \to \mathbb C$ such that $f$ is continuous on $\mathbb R$ and $\|f_n\|_{L^{\infty}(\mathbb R)} \geq C >0$ for all $n\in \mathbb N$ (where $C>0$ is fixed constant)

Therefore, we have

(1) There exists $\{x_n\}\subset \mathbb R$ such that  $$|f_n(x_n)| \geq C $$ for all $n\in \mathbb N.$

Assume that $|f_n(x)| \to 0$ as $|x| \to \infty.$

  Question: Can we say $\{x_n\}$  (as obtained in (1) above) is bounded? In other words, Dose there exists $M>0$ such that $|x_n|\leq M$ for all $n$, and $|f_n(x_n)| \geq C$?

This is false in general.

This can happen: Take something like a piecewise affine function on $\mathbb R$ which is $0$ outside $(0,1/n)$, such that $f(1/2n) = C$ and extends linearly on $(0,1/n)$. This is continuous and there is pointwise convergence to $0$ everywhere. The sequence $1/2n$ is bounded.

Now, notice that the same construction with the interval being $(n,n+1)$ gives you a counter-example.


What is the maximum number of triangles we can draw in a planar graph using at most 30 nodes and 50 edges I'm trying to solve this problem because I need to implement it into my program for calculating some values related to this.

Namely, I need to find what is the maximum number of triangles we can draw if we can use only up to 30 nodes and up to 50 edges. The triangles can share point, points or sides.

I started thinking if we can put one point in the center and then we put all the other points around it and then connect with the point in the center, but it looks like this wont be the maximum number.

Take the complete graph on 10 nodes. It needs ${10 \choose 2}=45$ edges. You get ${10 \choose 3}=120$ triangles. And you have 5 more edges.


Simply normal sequence of $-1$'s and $1$'s as coefficients of harmonic series Suppose $s_{n}$ is either $1$ or $-1$ for $n=1,2, 3,\ldots$ and that half the $s_{n}$'s are 1; i.e.
$$
\lim_{n\to \infty} \frac{\#\{i\leq n: s_{i}=1\}}{n}=\frac{1}{2}.
$$
Does then the series $\sum_{i=1}^{\infty} \frac{s_{n}}{n}$ converge?

The point here is that the alternating harmonic series converges, and if the $s_n$'s are the result of a fair coin flip ($-1$ for heads and $1$ for tails), then the series converges almost surely. So I'm asking about convergence for a simply normal sequence of $1$'s and $-1$'s. The sequence $1,-1,1,-1,1,-1,\ldots$ is simply normal, as is the result of the coin flip (almost surely).

I just found this paper in which there's an exercise at the end claiming that "Balancing the plus and
minus signs does not guarantee convergence."

Well I pretty sure that your series converges :)

So... as far as i understand, $s_n$ is the sequence which for each $n$ gives you either $1$ or $-1$, but not neccesarily in the order $1,-1,1,-1,1...$ Is that right?

If the sequence is $s_n=(-1)^n$ then of course the series $\sum_{k=1}^n\frac{s_n}{n}$ converges. Here is a link to the alterating series test: http://www.math.com/tables/expansion/tests.htm
whether it converges to $\frac{1}{2}$ i can't say for sure but you could try to make the deference between $\frac{1}{2}$ and your series up til $n$ less then an arbitrary number, $\epsilon$. In that way you would have proved that the series equals $\frac{1}{2}$

But I'm not completely sure if we're just looking at random series with mixed up $1$ and $-1$...


How to calculate the intersection of 2 circles? I am working on making inverse kinematics for a game I am making so I decided that the best way to start would be to do it in 2D.

After asking some people how to do it, they said you need to calculate the intersection of 2 circles.

How would I go about doing this without using any trigonometric formula? I searched online, however the answers and solutions to get the answer are not very well worded and easy to understand.

If you know the two positions of the circles, the distance between them and the radius of the circles, how would you calculate the intersection? And if you are answering can you clearly explain each step and what each thing means please and it would help me very much, e.g if you are using $x_1^2$, $x_2^2$, $y_1^2$, $y_2^2$, $r_1$, $r_2$ etc as such they do not show much meaning to me as I do not understand what they are showing? If you could explain this in the very best way possible for me to understand that would be great! thank you very much!

I have searched far and wide, but anywhere I found the wording is too hard to understand and follow - I am asking for someone to answer in a way that is understandable.

https://i.stack.imgur.com/EN6nS.png

WLOG, $x_1=y_1=0$ (if not, you translate the two circles so that the first goes to the origin).

By subtracting the two implicit equations,

$$\begin{cases}x^2+y^2=r_1^2,\\(x-x_2)^2+(y-y_2)^2=r_2^2\end{cases}$$
you get
$$\begin{cases}(x-x_2)x_2+(y-y_2)y_2=r_2^2-r_1^2\\x^2+y^2=r_1^2\end{cases},$$

which combines the equation of a straight line and a circle.

You can draw $y$ from the first equation and plug into the second, to obtain a quadratic equation in $x$.

You must plot your circles on an $xy$-coordinate plane. Since you already have the known radii and distances of both centers from each other, you can easily find their equivalent coordinates in the $xy$-plane.
If I were you, it's best that I position one circle on the origin (i.e. one circle's center must be at $(0,0)$), so that it would be a lot easier.

You can just use the general formula for a circle:
$$(x-h)^2+(y-k)^2=r^2 \quad{\text{where $(h,k)$ is the center of the circle}}$$
Solving for $y$ to make things easier, you get:
$$y=k\pm\sqrt{-h^2+2 h x+r^2-x^2}$$
For your two circles, you'll get:
$$(x-x_1)^2+(y-y_1)^2=r^2$$
$$(x-x_2)^2+(y-y_2)^2=r^2$$
You can just solve for $y$ for both equations, set them equal to each other and solve for $x$. After getting $x$, you then find $y$.
Notice that you have to solve for both cases $\pm$ of $(x,y)$ since two circles intersect at two points.


Squares with squares Are there a group of numbers whose squares are made up of squares? For example, $7$ would be one because $7^2$ is $49$ which has $2^2$ and $3^2$. $20$ would be another example.

What are these numbers called?

Please help me find good tags for this question.

OEIS calls them Numbers whose square is a nontrivial concatenation of other squares, and calls the sequence of their squares Squares which are a decimal concatenation of two or more squares. One of the references from that latter sequence calls the squares Smarandache square-partial-digital [numbers].

I am not agree with taking $20$ as a number of that kind ( or trivial in whose case $10, 30,40,50,60,70,80,90$ are too).

We show here that the only two-digit number with four-digit square of this kind is $41$. Looking at the algorithm of extraction of the square root in the figure below we can write

$$ab=x^2\Rightarrow 4\le x\le 9$$
$$y(20x+y)=cd=z^2\Rightarrow y(20x+y)\le 81\Rightarrow x=4$$

Hence $$y(80+y)\le81\Rightarrow y=1$$
No other possibility. In fact $41^2=1681$ as we wished.


Prove that the recurrence is true I am working on an assignment question, and am having trouble moving ahead. The question is as follows:

  Let the total number of bit strings with three consecutive zeros be $t_n$. Prove for $n \ge 4$ that $t_n= t_{n-1} + t_{n-2} + t_{n-3} + 2^{n-3}$

So, I started to check if the base case is true with n=4
$t_4 = t_3 + t_2 + t_1 + 2^1 = 1 + 0 + 0 + 2 = 3$

But isn't the total number of combinations for n =4 only 2?
1000 or 0001

I might just be going about this the wrong way. Any guidance would be appreciated.

Base case is fine if you include the option of more than 3 consecutive zeros:

$$  t4: 1000, 0000, 0001 $$

Breaking the problem down, you can consider the first bit as either 1 or 0. For one, there are no extra results and you get t3. For 0, you get a bonus set of cases where the first 3 are zero, which occurs 2^(n-3) times. Repeating the process for the second and third digit - minus the already counted cases in 2^(n-3) - and you get the rest of the argument you were requested to prove.


Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$

This inequality is strengthening of the following Vasile Cirtoaje's one, which he  created in 2005.

  Let $a$, $b$ and $c$ be positive numbers. Prove that:
  $$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$

My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220

But this way does not help for the starting inequality.

A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is  $0.0000158...$.

I tried also to use Cauchy-Schwarz, but without success.

Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.

@Maxim Gilula asked a good question: What is the point of presenting a solution via computer?

Let me address it from two aspects.

First, from my experience (the problems from MSE, AoPS, academic research etc.), the inequalities with elegant solutions (e.g. by hand)
are usually those designed e.g. for contest. Except for the designed inequalities, the inequalities require usually the help of computer.
Many inequalities in MSE, AoPS are open for many years without any solution by hand.

Second, although also with the help of computer, the solutions are different.

1) Some solutions e.g. Buffalo Way (BW) or Sum of Squares (SOS) provide step-by-step rigorous complete analytical solutions with detailed explanation.
Moreover, these solutions are often not very long. Usually one or several A4 pages are enough to write down the step-by-step rigorous complete analytical solutions with detailed explanation.

2) In contrast, some solutions does not provided step-by-step solutions. For example, the step-by-step solutions are in Mathematica's innards,
and for some cases, Mathematica run several hours to output 'true' which means that the step-by-step solutions may be very, very long,
probably 10,000 A4 pages are required to write down the step-by-step solutions.
Sometimes the method of Lagrange Multiplier works well, however, sometimes it results in very complicated equations which requires Mathematica etc.

3) Often, although the thinking process is very complicated and requires the help of computer,
the solution is elementary and easy to check even by hand. For example, recently, I used computer to solve an inequality in AoPS to get the identity
$a^3 + b^3 + c^3 - a^2b - b^2c - c^2a
= \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2
+ \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}$. In contrast, Mathematica only outputs 'true'. Which solution is better?

EDIT

Remark: Actually, the Buffalo Way works though the solution is ugly. I do not put it here. With computer, here is an SOS (Sum of Squares) solution:

WLOG, assume that $c = \min(a, b, c)$.
We have
\begin{align*}
&\frac{a^3}{13a^2+5b^2} + \frac{b^3}{13b^2+5c^2} + \frac{c^3}{13c^2+5a^2} - \frac{a+b+c}{18}\\
=\ & \frac{1}{18(13a^2+5b^2)(13b^2+5c^2)(13c^2+5a^2)}\\
&\quad \cdot \frac{1}{2223}\Big(cz_1^TA_1 z_1 + c(a-c)(b-c)z_2^TA_2z_2 + (b-c)z_3^TA_3z_3 + (a-c)z_4^TA_4z_4\Big)
\end{align*}
(Remark: $A_1, A_2, A_3, A_4$ are all positive definite. So the inequality is true.)
where
$$z_1 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right),
\quad z_2 = \left(\begin{array}{c} {\left(a - c\right)}^2\\ \left(a - c\right)\, \left(b - c\right)\\ {\left(b - c\right)}^2\\ c\, \left(a - c\right)\\ c^2\\ c\, \left(b - c\right) \end{array}\right),
$$
$$z_3 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right)
, \quad z_4 = \left(\begin{array}{c} {\left(a - c\right)}^2\, \left(b - c\right)\\ \left(a - c\right)\, {\left(b - c\right)}^2\\ c\, {\left(a - c\right)}^2\\ c\, \left(a - c\right)\, \left(b - c\right)\\ c\, {\left(b - c\right)}^2\\ c^2\, \left(b - c\right)\\ c^2\, \left(a - c\right) \end{array}\right),$$
and
$$A_1 = \left(\begin{array}{ccccccc} 453245 & -100035 & 111397 & 113607 & -146718 & 24687 & 6498\\ -100035 & 166231 & -19773 & 24453 & -82004 & -11286 & 31356\\ 111397 & -19773 & 444600 & 86526 & -144261 & 13585 & -6669\\ 113607 & 24453 & 86526 & 760500 & -344736 & -126711 & 149188\\ -146718 & -82004 & -144261 & -344736 & 297882 & -26676 & -53352\\ 24687 & -11286 & 13585 & -126711 & -26676 & 160056 & -111150\\ 6498 & 31356 & -6669 & 149188 & -53352 & -111150 & 160056 \end{array}\right),
$$
$$A_2 = \left(\begin{array}{cccccc} 281580 & 33098 & -64467 & 40261 & 4275 & -24219\\ 33098 & 329004 & -124982 & 13572 & 20241 & -51129\\ -64467 & -124982 & 106704 & -51129 & -6840 & -30875\\ 40261 & 13572 & -51129 & 326610 & -8645 & -79794\\ 4275 & 20241 & -6840 & -8645 & 62244 & -46683\\ -24219 & -51129 & -30875 & -79794 & -46683 & 153216 \end{array}\right)
,$$
$$A_3 = \left(\begin{array}{ccccccc} 258609 & -100035 & 102011 & 75582 & -56069 & -57798 & 126945\\ -100035 & 55575 & -37791 & -29393 & 2223 & 26923 & -26676\\ 102011 & -37791 & 589342 & 224757 & -195624 & -155376 & 313272\\ 75582 & -29393 & 224757 & 693576 & -279851 & -320283 & 535977\\ -56069 & 2223 & -195624 & -279851 & 200070 & 91143 & -306945\\ -57798 & 26923 & -155376 & -320283 & 91143 & 293436 & -435708\\ 126945 & -26676 & 313272 & 535977 & -306945 & -435708 & 1016158 \end{array}\right)
,$$
$$A_4 = \left(\begin{array}{ccccccc} 144495 & -57057 & 3705 & 24206 & -22230 & 2457 & 26923\\ -57057 & 55575 & 4199 & -15561 & -22724 & 13338 & -1989\\ 3705 & 4199 & 200070 & 18031 & -46449 & -9063 & 17784\\ 24206 & -15561 & 18031 & 351468 & -140049 & -6435 & -13509\\ -22230 & -22724 & -46449 & -140049 & 189202 & -117990 & -46449\\ 2457 & 13338 & -9063 & -6435 & -117990 & 177840 & -48659\\ 26923 & -1989 & 17784 & -13509 & -46449 & -48659 & 253422 \end{array}\right).
$$

I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

  $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5a^2}\geq \frac{a+b}{18}$$

Proof:

We have with $x=\frac{a}{b}$ :
$$\frac{x^3}{13x^2+5}+\frac{1}{13+5x^2}\geq \frac{1+x}{18}$$
Or
$$5(x+1)(x-1)^2(5x^2-8x+5)\geq 0$$

So we have (if we permute the variables $a,b,c$ and addition the three inequalities ) :

$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}+\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{9}$$

If we have $\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq\sum_{cyc}\frac{a^3}{13a^2+5c^2}$

We have :
$$\sum_{cyc}\frac{a^3}{13a^2+5b^2}\geq \frac{a+b+c}{18}$$
But also
$$\frac{(a-\epsilon)^3}{13(a-\epsilon)^2+5b^2}+\frac{(b)^3}{13(b)^2+5(c+\epsilon)^2}+\frac{(c+\epsilon)^3}{13(c+\epsilon)^2+5(a-\epsilon)^2}\geq \frac{a+b+c}{18}$$
If we put $a\geq c $ and $\epsilon=a-c$

We finally obtain :
$$\sum_{cyc}\frac{a^3}{13a^2+5c^2}\geq \frac{a+b+c}{18}$$

So all the cases are here so it's proved !


Special probability distribution If $p(x)$ is a probability distribution with non-zero values on $[0,+\infty)$, what conditions on $p(x)$ yield $\int_0^{\infty}p(x)\log{\frac{ p(x)}{(1+\epsilon)p({x}(1+\epsilon))}}dx \leq c \epsilon^2$, where $c$ is a constant and $0<\epsilon<1$.

The inequality above is actually a Kullback-Leibler Divergence between distribution $p(x)$ and stretched version of it ${(1+\epsilon)}p({x}{(1+\epsilon)})$. I have found out that this inequality holds for Exponential, Gamma, and Weibull distributions and I am interested to know if that works for a larger class of probability distributions.

Any idea what does that inequality mean?

Before, I thought this inequality holds for all distributions and I was looking for proofs, but I realized this is not true in general by counter examples such as the one introduced as an answer bellow in which lets
$1/\beta =1+\epsilon$ and equivalently considers $\int_0^{\infty}p(x)\log{\frac{\beta p(x)}{p({x}/{\beta})}}dx \leq c (1-\frac{1}{\beta})^2$.

I don't think thus is true in general.

Let's construct a case where the relative entropy is infinity. Let's consider a special case where $\beta = 2$. For each set $A$, let

$\bar A = \{2x|x\in A\}$.

Now, we can pick a sequence of segments on $R^+$, $A_1, A_2, ..., $ such that $A_1, \bar A_1, A_2, \bar A_2 ...$ do not intersect each other.

Now assign constant density on these sets, so that,

$P(\bar A_i) = \frac{1}{4^i}$ and

$P(A_i) =  \frac{1}{2}e^{-4^i}P(\bar A_i)$

Note that the density $p(x) = \frac{P(\bar A_i)}{\mu(\bar A_i)}$ and $p(x/2) = \frac{P(A_i)}{\mu(A_i)}$ for $x\in \bar A_i$, where $\mu(A)$ is the standard borel measure of $A$.

Also note that $\mu(\bar A_i) = 2\mu(A_i)$ by construction, so we have,

$log(\frac{p(x)}{\frac{1}{2}p(x/2)}) = 4^i$

on $\bar A_i$ and

$\int_{(\cup \bar A_i)}p(x)log(\frac{p(x)}{\frac{1}{2}p(x/2)})dx = \infty  $

On the remaining set, just assign, say, the exponential distribution, scaled by some constant to make $p$ a probability.  By renormalization, you can prove that

$\int_{(\cup \bar A_i)^c}p(x)log(\frac{p(x)}{\frac{1}{2}p(x/2)})dx >-C $

for some constant $C$.

So, overall, the relative entropy is $\infty$.


how to determine a matrix has a single eigenvalue Find the jordan form of the matrix
$$A = \begin{pmatrix}
 1 & 1 & 2 & 2\\
 1 & -2 & -1 & -1\\
 -2 & 1 & -1 & -1\\
1 & 1 & 2 & 2 \\
\end{pmatrix}$$

Hint: check that the matrix A has a single eigenvalue, and $trace(A) = 0$.

How can I check that the matrix has a single eigenvalue without using the determinant?

That's what I managed to do so far:

Let $G$ the Jordan form of $A$, using the hint that there is a single eigenvalue:
$$trace(A) = 0\implies 0 = trace(A) = trace(G) = 4a \implies a = 0$$

meaning that theres a single eigenvalue which is zero.

Then the characteristic polynomial is $A^4$  and the minimal polynomial is $A^3$.

Then:
$G = diag\{J_3(0), J_1(0)\}$

But how do I prove that there is a single eigenvalue?

The rank is 2 (the first two columns span the image) and trace is zero. If there is one non-zero eigenvalue $a$, there is another non-zero eigenvalue $-a$. Hence the image would be spanned by the two eigenvectors of the non-zero eigenvalues and the same would automatically hold for all powers of $A$. But the first column is contained in the kernel and in the image of the matrix, hence the rank of $A^2$ will be at most $1$ (since $e_1$ is in the kernel of $A^2$, but not in the kernel of $A$), contradiction to the conclusion above!

Since the characteristic polynomial is $\chi_A (x) = x^4 $, this tells you that all eigenvalues are zero.

A matrix satisfies its characteristic polynomial by the Cayley Hamilton theorem, hence $A^4 = 0$. If $v$ is a non zero eigenvector with eigenvalue $\lambda$ then $0 = A^4 v = \lambda^4 v$ and so $\lambda = 0$.


Let S be a set of six positive integers who maximum is at most 14. Let $\mathcal S$ be a set of six positive integers who maximum is at most 14. Show that the sums of the elements in all the nonempty subsets of $\mathcal S$ cannot all be distinct

For each non-empty subset, $\mathcal A$ of $\mathcal S$ the sum of the elements in $\mathcal A$ denoted as $S_\mathcal A$ satisfies
$$1\leq S_\mathcal A \leq 9 + 10 + \cdots + 14 = 69$$
and there are $2^6−1=63$ non-empty subsets.

How do you know that the sums start from $9$ and goes to $14$? $(9+10+...+14= 69)$. Why can't it start from $1$?

Suppose $S$ contains two equal elements. Then the subsets of $S$ comprising only these elements have equal sums, so the sums are not all distinct.

Now suppose $S$ does not contain two equal elements, i.e. the elements of $S$ are all distinct.

Let $x$ be the smallest element of $S$.

Then the smallest possible value of $S_A$ is $x$.

And the largest possible value of $S_A$ is $x+10+11+12+13+14=x+60$.

Since $x<S_A<x+60$, there are $61$ possible values of $S_A$.

And since there are $2^6-1=63$ non-empty subsets, these cannot all have distinct sums since there are only $61$ possible values available for the sum.

import Java.util.Scanner;
public static void main(String [] args){
Scanner userInput=new Scanner(System.in);
int numberOfPigeons=0;
int numberOfPigeonholes=0;
System.out.println("Please enter the number of pigeons: ");
numberOfPigeons=userInput.nextInt();
System.out.println("Please enter the number of pigeonholes: );
numberOfPigeonholes=userInput.nextInt();
if(checkValidity(numberOfPigeons,numberOfPigeonholes)==true)
System.out.println("The statement is correct");
else
System.err.println("The statement is not correct");

}
public static boolean checkValidity(int numberOfPigeons,int numberOfPigeonholes){
if(numberOfPigeons<=numberOfPigeonholes){
return false;
}
else
return true;
}


Find $xy+yz+zx$, given quadratic form of equations. Given that $(x,y,z) \in \mathbb R^+$ and the following equations:

$$x^2 + y^2 + xy = 1,$$
$$y^2 + z^2 + yz = 2,$$
$$z^2 + x^2 + xz = 3.$$

How to find $xy + yz + zx$? Please help.

Consider a triangle ABC.
         take a point inside triangle say O. such that sides subtend angle of 2π/3 each.
let OA = x, OB = y, OC = z.
So sides will come out 1, √2, √3. ( cosine rule ).
as triangle is right angled, area  = 1/2*√2
Also add up area of triangle formed by smaller ones we get,
       xy + yz + zx = √(8/3).


Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3)(r+4)}=\frac{3}{(r+1)(r+3)(r+4)}$$ Please provide an alternate approach to form telescopic series.

The partial fraction expansion of your summand is
$$-\frac{1}{6}\frac{1}{r+1} + \frac{3}{2}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
Then notice that $3/2 = 4/3+1/6$ so you have
$$-\frac{1}{6}\frac{1}{r+1} + \frac{1}{6}\frac{1}{r+3}+\frac{4}{3}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
And now things telescope like crazy.

WLOG
$$\dfrac r{(r+1)(r+3)(r+4)}=a\left(\dfrac1{r+1}-\dfrac1{r+3}\right)+b\left(\dfrac1{r+3}-\dfrac1{r+4}\right)\  \ \ \  (1)$$
so that
$$f(n)=\sum_{r=1}^n\dfrac r{(r+1)(r+3)(r+4)}$$
$$=a\sum_{r=1}^n\left(\dfrac1{r+1}-\dfrac1{r+3}\right)+b\sum_{r=1}^n\left(\dfrac1{r+3}-\dfrac1{r+4}\right)$$
$$=a\left(\dfrac12-\dfrac14+\dfrac13-\dfrac15+\cdots+?\right)+b\left(\dfrac14-\dfrac1{n+4}\right)$$
$$\lim_{n\to\infty}f(n)=a\left(\dfrac12-\dfrac14+\dfrac13-\dfrac15\right)+b\left(\dfrac14\right)$$
Now from $(1),$  $$r=2a(r+4)+b(r+1)=r(2a+b)+8a+b$$
Now compare the coefficients of $r$ and constant to find $$8a+b=0\ \ \ \ (2), 2a+b=1\ \ \ \ (3)$$
Finally solve the two simultaneous solutions to find $a,b$


Find $\int\frac{x-1}{x^2-5x+6}dx$. Why my solution is different from book? I'm learning single variable calculus right now. Right now trying to understand integration with partial fraction. I'm confused in a problem from sometime. I think I'm doing right but answer in my book is something else.
Please have a look at the images.

The Solution given in my Book
I know there's is a difference in finding value of A and B. But in previous exercise I was applying the same method and was getting correct answer. Please help. thankyou in advance.

HINT: we get $$\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}$$
can you show this?
now we calculate the zeros of $x^2-5x+6$, these are:$$x_{1,2}=\frac{5}{2}\pm\sqrt{\frac{25}{4}-\frac{24}{4}}$$ thus we get $$x_1=3$$ or $$x_2=2$$
and we get $$x^2-5x+6=(x-2)(x-3)$$ and we can make the ansatz:
$$\frac{5x-5}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}$$
multiplying by the denominators we obtain:
$$\frac{5x-5}{x^2-5x+6}=\frac{A(x-3]+B(x-2)}{x^2-5x+6}$$
and we get $$5x-5=x(A+B)-3A-2B$$
from here you will get $$5=A+B$$ and $$-5=-3A-2B$$ you must solve this system

If your way is right, then the answer doesn't matter, because it is the representation of  the value, if you put the value and plot it on a garph, you will see you will get the same answer.


Proof of non-strictly convexity of $l_1$ and $l_{\infty}$ Given the Banach space $l_p = \left\{f\in R^n : \|f\| \leq \infty\right\}$ for $1\leq p \leq \infty$, we define the following norms

$\|f\|_p = \left(\displaystyle\sum_{j=0}^{\infty}
   |f_j|^p\,w_j\right)^{\frac{1}{p}},\quad p\geq 1,\,w_j>0,\,\forall
   j\in\mathbb N$
$\|f\|_{\infty} = \underset{j\in\mathbb N}{\sup} |f_j|$

I want to prove that the spaces $l_1$ and $l_{\infty}$ are not strictly convex.

For $l_1$, I'm trying to find two different sequences $\left\{a_n\right\}_{n=1}^{\infty}$ and $\left\{b_n\right\}_{n=1}^{\infty}$ such that

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

Respectively, for $l_{\infty}$, I have to find two sequences such that

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

are these argument correct?

If they are, then I think that I can prove it for $l_{\infty}$. One just has to take the sequences:

$$\left\{a_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2^{n}}\right\}_{n=1}^{\infty} $$
$$\left\{b_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2n}\right\}_{n=1}^{\infty}$$

Since then we have:

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \frac{1}{2} + \frac{1}{2} = 1 = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

I like the geometric interpretation of "non-strictly convex": the unit sphere $\{x:\|x\|=1\}$ contains a nondegenerate line segment (line segment is degenerate if it has only one point). If one keeps this in mind and follows julien's suggestion

  It is a good idea to draw the unit balls of $\ell_1$ and $\ell_\infty$ in dimension $2$ to get a visual idea of what is going on

then the answer comes out naturally: the line segment with endpoints $e_1$ and $e_2$ (or any two vectors of the standard basis) lies on the unit sphere of $\ell_1$; also, the line segment with endpoints $e_1$ and $e_1+e_2$ lies on the unit sphere of $\ell_\infty$. This is where the examples given by David Mitra come from.

I think

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

is not correct for all situations.

Consider the following
$$\left\{a_n\right\}_{n=0}^{\infty}=\left\{-\frac{1}{2^{n}}\right\}_{n=0}^{\infty} $$
$$\left\{b_n\right\}_{n=0}^{\infty}=\left\{\frac{1}{2^{n+1}}\right\}_{n=0}^{\infty}$$

$$\sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}|+\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}| > \sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}+\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|-\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}|$$


Sandwiching Limsups & liminfs of expectations Why is it that if we sandwich a liminf of an expectation between two equal quantities we get that the limit exists? Can we somehow deduce the limsup from that and conclude that it's the same or am I missing something else. A book I'm reading used a rule called Fatous lemma to give the below, given that $X_n \rightarrow X$ a.s. and $\mathbb{E}|X_n| \to \mathbb{E}|X|$.
\begin{equation}
\mathbb{E}|X| = \lim \mathbb{E}|X_n| \ge \liminf \mathbb{E}X^+ + \liminf\mathbb{E}X_n^- \ge \mathbb{E}X^+ + \mathbb{E}X^- = \mathbb{E}|X|
\end{equation}

The problem is that at this point it was concluded that $\lim \mathbb{E}X_n^+ \to \mathbb{E}X^+$ and $\lim \mathbb{E}X_n^- \to \mathbb{E}X^-$. I don't understand how can we conclude this just from the above equality. Thanks in advance.

I don't see why the mentioned equality yields the existence of the limits, but here is an alternative way to prove the (original) statement. It actually even shows $X_n \to X$ in $L^1$.

Since $$\mathbb{E}|X_n| \to \mathbb{E}|X| \tag{1}$$ and $X \in L^1$, we have $X_n \in L^1$ for sufficiently large $n$ and

$$|X_n-X| \leq |X_n|+ |X| \in L^1.$$

This means that we can apply Fatou's lemma to the non-negative sequence $|X_n|+|X|-|X_n-X|$:

$$\begin{align*} 2 \mathbb{E}|X| &\stackrel{(1)}{=} \mathbb{E}\left( \lim_{n \to \infty} |X_n|+|X|-|X_n-X| \right) \\ &= \mathbb{E}\left( \liminf_{n \to \infty} |X_n|+|X|-|X_n-X| \right)\\ &\leq \liminf_{n \to \infty} \left( \mathbb{E}|X_n|+\mathbb{E}|X| - \mathbb{E}|X_n-X| \right) \\ &\stackrel{(1)}{=} 2 \mathbb{E}|X| - \limsup_{n \to \infty} \mathbb{E}|X_n-X|. \tag{2}\end{align*}$$

Since $|X_n-X| \geq 0$ is non-negative, we conclude

$$0 \leq \liminf_{n \to \infty} \mathbb{E}|X_n-X| \leq \limsup_{n \to \infty} \mathbb{E}|X_n-X| \stackrel{(2)}{\leq} 0.$$

Hence,

$$\lim_{n \to \infty} \mathbb{E}|X_n-X| = 0,$$

i.e. $X_n \to X$ in $L^1$. Now note that the mapping $f(x) := x \mapsto \max\{0,x\}$ is contractive (i.e. $|f(x)-f(y)| \leq |x-y|$). We therefore find

$$|\mathbb{E}(X_n^+-X^+)| = |\mathbb{E}(f(X_n)-f(X))| \leq \mathbb{E}|f(X_n)-f(X)| \leq \mathbb{E}|X_n-X| \to 0.$$

Similarly, we get $\mathbb{E}X_n^- \to \mathbb{E}X^-$.

Fatou's lemma is an important result in measure theory. Suppose that $(X,\mathcal{B},\mu)$ is a measure space (just think $(\Omega, \mathcal{F},\mathbb{P}))$. Fatou's lemma states that, for any measurable positive functions $f_n \geq 0$,
$$\liminf_{n\rightarrow \infty} \int f_n d\mu \geq \int \liminf_{n\rightarrow\infty} f_n d\mu.$$
[One way to remember this is to think of a sequence of functions whose limit is $0$ but which all integrate to 1. Then the left-hand side above is always 1 while the right-hand side is 0.] Note that $X_n \rightarrow X$ a.s. does not guarantee $E|X_n| \rightarrow E|X|$.

Next, for fixed $\omega$, $X_n(\omega)$ is a sequence of real numbers. If $\lim_{n\rightarrow\infty} X_n(\omega)$ exists for almost all $x$, then $\lim_{n\rightarrow\infty} X_n(\omega) = \liminf_{n\rightarrow\infty} X_n(\omega)$ for all such $x$. This and $E|X_n| \rightarrow E|X|$ imply
$$\lim_{n\rightarrow\infty} E X_n^\pm = \liminf_{n\rightarrow\infty} E X_n^\pm \geq E\liminf_{n\rightarrow\infty} X_n^\pm = E\lim_{n\rightarrow\infty} X_n^\pm.$$ Note that $X^\pm \geq 0$, so we may apply Fatou's lemma.

The last part you bring up, and good to notice, is that
$$\liminf_{n\rightarrow\infty} EX_n^+ + \liminf_{n\rightarrow\infty}EX_n^- = EX^+ + EX^-$$
may not immediately imply equality of the two individual limits.

Write the above as $$a+b = c+d.$$ By Fatou's lemma, we have $a \geq c$ and $b \geq d$. If $a \neq c$, then $c = a-\epsilon > 0$ for $\epsilon > 0$. Hence $d = b+\epsilon > b$, a contradiction. It follows that $a = c$ and $b = d$.


Property of Mobius transformations which send $\mathbb{R}$ in $\mathbb{R}$ and that $ad-bc=1$ I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers.

So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see if I get some information about $a,b,c,d$ considering that $T(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ but this doesn't really work. Also I've considered $SL(2,\mathbb{C}) / \{ \pm I\}$ which is isomorphic to the group of Mobius transformations but this doesn't really help either.

Hints: if we assume $c\neq 0$, $T(z)=\frac{a}{c}-\frac{1}{c(cz+d)}$. Letting $z$ go to infinity shows $\frac{a}{c}$ is real, hence $c(cz+d)$ is real for all $z\in\mathbb R$. From there conclude $cd$ and $c^2$ are real. The rest should be easy. The case $c=0$ is also not difficult.

Suppose $T$ is as asked for.

Suppose $c\neq 0$. Then if $\frac{-d}{c} \in \mathbb{R}$ and we know that $T(\frac{-d}{c}) = \infty \notin \mathbb{R}$ unless the numerator is $0$ as well, in which $a(\frac{-d}{c}) = b = 0$ which means $\frac{-ad}{c} = -b$ or $-ad = -bc$ or $ad=bc$ and we have a contradiction with $ad-bc = 1$.

So $c=0$ and so $ad=1$. What more can you do now?


Exponential random variables with average number of beeps A device randomly beeps, with an average of 2 beeps per second.

If we assume that the time $T$ between any two beeps is an exponential random variable, then what is the probability that no beep is emitted in the next 5 seconds? Now, if we let $Y_i$ be exponential random variables modeling the time between the $i$th and the $i$ + 1st beeps, what is the probability that there are exactly 3 beeps in the next 4 seconds? I know we can parametrize this with a parameter $\lambda$ which is 2 beeps per second, but I am unsure of how to proceed in each of these situations.

Preliminary
Gamma distribution to the rescue! If a R.V. $T\sim Gamma(a,\mu)$, its pdf is...
$$
P[T = t] =
\begin{cases}
\frac{t^{a-1}e^{-\frac{t}{\mu}}}{\mu^a\Gamma(a)},  & t \gt 0 \\
0, & t \le 0
\end{cases}
$$
And its cdf is...
$$
P[T < t] =
\begin{cases}
\frac{\gamma(a,\frac{t}{\mu})}{\Gamma(a)},  & t \gt 0 \\
0, & t \le 0
\end{cases}
$$
Where $\mu$ is the average per trial and $a$ is the # inter-arrival times.
Answer 1
For your problem, $a = 1$ (The gamma reduces to the exponential distribution) and $\mu = 0.5$ [s/beep]...
$$
P[T>5] = 1-P[T<5] = 1 - \frac{\gamma(1,2\cdot 5)}{\Gamma(1)} = 1 - (1 - e^{-10})
\\\implies P[T>5] =e^{-10}
$$
Answer 2
Let $Y = Y_1 + Y_2 + Y_3$ so $a = 3$ this time...
$$
P[Y=4] = \frac{4^{3 - 1}e^{-2\cdot 4}}{0.5^3\Gamma(3)}
\\\implies P[Y=4] = 64e^{-8}
$$


Are Hamiltonian trajectories geodesics on the cotangent bundle? Suppose we have a Hamiltonian dynamics on a phase space, whose base space is also a Riemannian manifold. I was wondering if the Hamiltonian trajectories are whether geodesics or only locally geodesics on the cotangent bundle? Thanks!

A short answer is yes, but we need some more restriction on the Hamiltonian we are choosing.  See: https://en.wikipedia.org/wiki/Geodesics_as_Hamiltonian_flows .


How long will it take two clocks to show the same time again? 2 clocks A and B are now showing 3'o clock. Clock B is 2 mins/hr faster than clock A. How long will it take them to show the same time again

My solution:
hour hand has speed 1/2 degree per minute so Clock B will run 1 degree faster then clock B ,
for showing same time clock B have to move 360 degree so time = 360hrs

Am I correct?

I am interpreting this question as follows: the hour hand of A rotates through a full 30 degrees in an hour, and the minute hand of B rotates through a full 30 degrees plus the equivalent of two minutes (which is 1 degree) in an hour. If this is how you've interpreted it too, your answer's right: my thoughts are detailed below.

So, A goes at 30 deg/h and B goes at 31 deg/h. I'm safe to ignore the minute and second hands here: if the hour hands are in the same position, they show the same time. So the question is: how many hours have to elapse before A and B show the same time? If the number of hours that elapse is $x$, then the hands of A and B have rotated $30x$ and $31x$ degrees respectively. We want these to "show the same time". By this we mean that they should be equal, up to adding or subtracting integer multiples of 360 (because 360 degrees = a full revolution). So we want a solution to $30x = 31x + 360k$, for some integer k, and some x > 0 - in fact, we want $x$ to be as small as possible, because $x$ is in hours, and the question says "how long will it take?".

(However, clock A might not run correctly. But that's fine. If it goes at 53 deg/h, then B goes at 54 deg/h. This doesn't affect the solution.)

I think answer 360 hours is not correct. Two options are there

Assume Clock A (run correctly) travels at 60 minutes/ Hr & B travel at 62 minutes per Hour. B travels at 2 minutes faster than A per hour. So Answer should be LCM of 62 Minutes and (12 Hrs x 60 minutes = 720 minutes) which is equal to 22320 minutes which is equal 372 Hours

Assume Clock A (run correctly) travels at 60 minutes/ Hr & B travel at 58 minutes per Hour.A travels at 2 minutes faster than B per hour. So Answer should be LCM of 58 Minutes and (12 Hrs x 60 minutes = 720 minutes) which is equal to 20880 minutes which is equal 348 Hours

Answer can be either 348 hours or 372 hours


How do I make my TI-89 evaluate a recursive function? On my TI-89 I can assign variables recursively such as:
$1\to x$ returns 1
$x+1 \to x$ returns 2
$x+1 \to x$ returns 3
etc.
How could I do functions the same way:
$x \to f(x)$ returns Done
$2\cdot f(x)+x \to f(x)$ returns Done
but
$f(x)$ now returns Error: Memory< br />

I think that it is because it is trying to define $f(x)$ in terms of itself but each iteration I just want it to evaluate $f(x)$ and return the result to the function in terms of $x$ therefore $f(x)$ would return $3x$.
Someone might wonder why I don't just evaluate this one by hand but I am trying to understand the concept so I can apply it to more complex recursive formulas such as $$p_{k+1}(x)=2p_k(x-1)+\frac{1-2p_k(-1)}{k!}(1-x)(2-x)\cdots(k-x) $$

CORRECTED

I rebooted my old Ti92 and wrote you a little program for the symbolic evaluation of :
$$p_{k+1}(x)=2p_k(x-1)+\frac{1-2p_k(-1)}{k!}(1-x)(2-x)\cdots(k-x)$$

I had too to shift the $k$ index ($k\to k-1$) and add the indispensable initial condition (arbitrary $1$ for $k\le 1$).

The resulting function is (use  [APPS] [7] [3] to create it!) :

p(k,xx)
Func
   Local a,i
   If k<= 1 Then
      Return 1
   Else
      1->a
      For i, 1, k-1
         a*(i-xx)/i->a
      EndFor
      Return 2*p(k-1,xx-1)+(1-2*p(k-1,-1))*a
   EndIf
EndFunc

and you will get (use the keys $\,[\diamond]$ [Home]  for the input) :

$\displaystyle p(3,x)\qquad\qquad\qquad \frac{x^2}2+\frac x2+1$

Note that in my first version I obtained "circular definition" errors and could call only $\,p_k(x)$ instead of the wished $\,p_k(x-1)$. The simple replacement of the function parameter $x$ by $xx$ solved both problems as explained in the TI Knowledge Base.

Hoping this made all this useful,

As far as I know, the TI-89 cannot find closed forms for recursive functions. Try Wolfram|Alpha for that. However, we can easily graph recursively defined sequences.

Click MODE and switch Graph... to SEQUENCE. Then go to the plotter, and input your recursive function. For example, u1 = u1(n-1) + u1(n-2) with initial conditions of ui1 = {0,1}.

From there you may graph your sequence and extract a table of values, which can be used as data to find the best fitting curve, for example.


Concept about $\binom{ \infty}k$ concept about function $\binom{ \infty}k$
For
$ k \in \mathbb{N}$

The idea of this function is derived from my power sum formula

link for my power sum formula

Formula is as

$$\sum_{k=1}^{n} k^{m}=\sum_{b=1}^{m+1} \binom{n}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$

This formula helps to derive $\binom{ \infty}k$ function and to calculate it's value.  we know the negative values of zeta function.if

$$ \zeta(-m)=\lim_{n\to \infty}\sum_{k=1}^{n} k^{m}$$

So can we construct it as
 $$\zeta(-m)=\sum_{b=1}^{m+1} \binom{\infty}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$
Then we can calculate, if we substitute value $\zeta (0)=-1/2$ then$\binom{\infty}1=-1/2$

Again we can calculate next value using or substituting previous values of $\binom{\infty}k$.

Other values of $\binom{\infty}k$are

$$\binom{\infty}2=5/12$$
$$\binom{\infty}3=-3/8$$
$$\binom{\infty}4=251/720$$
$$...$$
And so on.

Application

Definition

Let's us define a sequence as :

$$a=(a_{1},a_{2},a_{3},...)$$

Difference between two term is as follows

$$\triangle^{0}a_{n}=a_{n}$$
$$\triangle^{1}a_{n}=a_{n+1}-a_{n}$$
More generally
$$\triangle^{m}a_{n}=\triangle^{m-1}a_{n+1}-\triangle^{m-1}a_{n}$$
If there exist some $m$ for $\triangle^{m}a_{n}=0$ such that $\forall n \in \mathbb{N}$

Then $$\sum_{k=1}^{n} a_{k}=\sum_{b=1}^{m+1} \binom{n}b\triangle^{b-1}a_1$$

now if $\lim_{n \to \infty}$ put values $\binom{\infty}b$

$$\sum_{k=1}^{\infty} a_{k}=\sum_{b=1}^{m+1} \binom{\infty}b\triangle^{b-1}a_1$$

And Get the result

Example

To calculate
$1+3+5+...+(2n-1)+...=1/3$

Question

    Q1- how this function $\binom{\infty}k$ impact to understanding  and  analysis of mathematics?

    Q2- can we derive it's definition/algorithm to calculate  function $\binom{\infty}k$ for $k\in\mathbb{C}$  values by analysis in some field , i mean what is generalization for function $\binom{\infty}k$?

    Q3-Is it have some interesting properties?

Thank you very much for your suggestions comments and answer.

There is a unique polynomial $P_m(x)$ such that $P_m(0) = 0$ and $P_m(x)-P_m(x-1)= x^m$ so $P_m(n)=\sum_{k=1}^n k^m$ and you are setting $$P_m(\infty) = \zeta(-m)$$
For any polynomial $f(x)$ such that $f(0) = 0$ then $f(x)-f(x-1)$ is a polynomial $$f(x)-f(x-1) = \sum_{j=0}^d c_j x^j$$

You are setting $$f(\infty) = \sum_{j=0}^d c_j \zeta(-j)$$
With $g_l(x) = {x \choose l}$ then $$g_l(x)-g_l(x-1) =\sum_{j=0}^{l-1} b_{l,j} x^j$$
$${\infty \choose l} =\sum_{j=0}^{l-1} b_{l,j} \zeta(-j)$$
From $\sum_{n=1}^N (-1)^{n+1} = \frac{1+(-1)^{N+1}}{2}$ we have that $\eta(s)=(1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is entire and $$\eta(-j) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1} n^j x^n$$
Let $$T (\sum_{j=0}^d c_j x^j)(x) = \sum_{j=0}^d \frac{c_j}{1-2^{1+j}} x^j$$
you are setting $$f(\infty) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1}x^n T(f(x)-f(x-1))(n) $$
The problem is that operator $T$ which makes it unobvious how to generalize your summation method to any sequences, not only polynomials.


How did the rule of addition come to be and why does it give the correct answer when compared empirically? I'm still a high school student and very interested in maths but none of my school books describe these kind of things, and only say how to do it not the whys. My question is very simple, for example:

        19
       +25
      = 44

because the one from adding 9 and 5 goes on to add with 1 and two. How did this rule of addition come to be?

Here's a bit of explanation that can be useful(sorry if it is frustrating):
Suppose we are a 3 year old child and no one teaches us how to add and we recognize 1 is our index and 5 all palm fingers held up. Someone gives us the problem add: 1+5 so we hold 'em up, right? and again someone gives us to add 8564+2345 so we can't lift 'em up. So we try to device a rule but we don't recognize 6+4= 10 in which 0 stays and one jumps neither can we say that only the digits from rightmost positions are to be added. This is what i meant.

Well essentially what you're doing is this:

$$19+25 = (10 + 9) + (20 + 5) = \underbrace{(9 + 5)}_{\text{ones}} + \underbrace{(10 + 20)}_{\text{tens}}$$

Then you'll get:

$$14 + (10 + 20)$$

So for the ones digit you get $4$ so let's subtract that from $14$ to get $10$. For the tens digit, since you still have $10$ left over from adding the ones digits you have to "carry over" $10$ into $(10+20)$:

$$10\,\text{(carried over digit)}+(10+20) = 20+20 = 40$$

Adding the result from the ones digit gives:
$$40 + 4 = 44$$

The "rule" of carrying over from one place value into another is just that, except that you do it vertically.

Though my answer is not quite what I want as well, this is what I could conclude seeing all the views here and it's not an algebric verification. If it will be possible I'll come back here and post an answer delving deeper into the history:


Dual spaces: Roots and Cartan subalgebra Can someone show that the roots and the Cartan subalgebra are dual vector spaces?

I don't see how simple roots acting on non-corresponding indices of a Cartan basis produce 0 and a simple root evaluated on its corresponding Cartan basis element equals 1.

According to the definition of dual spaces: $\beta (t_\beta)=1$
and $\beta (t_\alpha)=0$, where ${\alpha,\beta } $ (simple roots) corresponds to ${t_\alpha, t_\beta} $ (Cartan basis) respectively.

If this was true then the inner product of two simple roots would be zero. Since the inner product of roots uses the restriction of the killing form to the Cartan, and so does the root evaluation on a Cartan element.

So
$(\alpha,\beta)=k (t_\alpha, t_\beta)_{H\times H}=
\alpha (t_\beta)=\beta (t_\alpha)$

Where $H$ is the max Cartan subalgebra.

Since $\beta$ is be being evaluated on an uneven indice ($ t_\alpha $), as above, $\beta (t_\alpha) $ is 0, which is not true for most root systems. In most root systems the inner product of simple roots is less than 0. Also $\beta (t_\beta)$ is not always 1.

So I am confused how the dual space axioms apply here.

In addition, I see how the Cartan basis is orthonormal or at least orthogonal under its own killing form, but not under the Lie algebra's killing form restricted to the Cartan, which is non-degenerate.

I am probably confusing a bunch of stuff.

Help is really appreciated

First you have to realize that the roots are linear functions ($V$) that take elements in some vector space $V$ and produce (bijectively) elements in $\mathbb C$ (complex plane, however we can restrict it to real plane). I encourage to you read the following notes of Robert N. Cahn. In particular pages 15 and 19. There you will find in detail a very clear answer to your question:

http://phyweb.lbl.gov/~rncahn/www/liealgebras/texall.pdf


Covariance zero does not imply independence - but what does it imply? As is well known, there are many examples of (pairs of) random variables which have covariance zero, but which are not independent.

However, I'm wondering whether there are any general theorems about what having covariance zero does imply.   Are there theorems that say something like, "Covariance zero, under additional hypothesis H, implies the following relationship between $X$ and $Y$," or is there essentially no possible general relationship?

By the way, although for certain types of distributions/families the implication "Covariance zero implies independent" is valid, that is not the kind of theorem I am asking about.  What relationship can there be that is short of full independence?  Any reasonable hypothesis H would be interesting to me, as would references to online resources or standard references.

The covariance is just one number computed from two random variables. You cannot expect it to capture much information.

Saying that the covariance is zero is one equation; by comparison, saying that two RVs are independent is a huge amount of equations: one for each couple of real numbers! It is very strict.

All you can say when the covariance is zero is that a linear regression on a sample large enough will give you a horizontal line: you cannot predict the value of one variable from the other based on a linear model.

I would be happy to learn, but I'm afraid that not much more can be said.

The opposite to lack of correlation is full correlation. Typically, partial or full correlation between two events without dependence occurs when there is a third intevening variable. For example suppose there is a partial correlation between my time of waking and high tide. this may be explained by the fact that I need to get to work early in the morning and at my longditude  high tide tends to occur in the morning.  In this case time is the intervening variable.


What is the Probability of A intersect A? What is the P(A $\cap$ A)? Would it still be P(A)?

I know this may seem childish or elementary but I needed some clarification and nothing online says anything about it.

We can assume they are not independent because they describe the same event.

Visualize what P(A ∩ A) actually mean. The probability of A occurring and A occurring is an event inside a larger sample space. These two events perfectly overlap. Therefore if A occurs it is the same as saying A and A i.e. P(A ∩ A) occurs.

Do not think about it as the probability of A intersecting A, which would be obviously 1.
Instead Define an event $B = A\cap A$, so we have the probability of event B, which is
$$
    P(B) = \frac{\# B}{\# S}
$$
where S is the Sample Space, i.e. the set of all possible outcomes.
And since B=A, then you have
$$P(B)=\frac{\# A}{\# S}$$


Graph of the Dirichlet Function we know, the Dirichlet function as: $$ f(x) = \begin{cases} 1, &\text{if } x \text{ is irrational; and } \\ 0, &\text{if }x \text{ is rational}. \end{cases} $$ R.A. Silverman in his book Modern Calculus says that this wild function cannot be plotted at all while M.R.Spiegel in the book Advanced Calculus constructed a graph of $f(x)$ as two parallel lines with $x$-axe, gone via $1$ and $0$. The second athour says:  The graph is shown in the adjoining Fig. 2-3. From its appearence it would seem that there are two functional values $0$ and $1$ corresponding to each value of $x$, i.e. that $f(x)$ is multiple-valued, whereas it is actually single-valued. What happens in these two point of views?

Silverman would deny that this picture counts as "plotting" the function.  What look like two solid lines are actually not solid at all: the top line contains only points whose $x$ coordinate is irrational, and the bottom line contains only points whose $x$ coordinate is rational.  These facts are impossible to show in a picture.

It is impossible to plot the graph of such a function because there are breaks (infinitely many discontinuities).


Why does the image of an interval being unbounded imply there exists a sequence with the image of the sequence greater than n? I am trying to understand a specific step in the proof of the Extreme Value Theorem.
The step is showing that the image of a continuous function over a compact interval, $f:[a,b] \rightarrow \mathbb{R}$, is bounded.
The proof assumes the image, $f([a,b])$ is unbounded to get a contradiction.
Why is the following true?
$$f([a,b]) \text{ unbounded } \implies \exists \ (x_n)_{n \in \mathbb{N}} \in [a,b] \text{ such that } f(x_n) \geq n$$

That is the definition of unbounded.  Intuitively, it says how ever far away you look, there is a point farther.  If there were not such a point, the image would be bounded above by the first $n$ where it fails.  You must be considering functions with positive range or you should have absolute value signs around $f(x_n)$ to consider the case where $f$ is unbounded negative.


Calculating the core of a game In a coalitional game n miners find equal blocks of gold. Two can carry one piece home. The payoff of a coalition S is $\nu(S)=|S|/2$ if $|S|$ even and $(|S|-1)/2$ for $|S|$ odd.

Determine the core if n is even, that is the set of imputations $x$ such that $\sum_{i\in S}x_i\ge \nu(S)$ for all S.

Could someone show how to do this systematically? What if three miners can carry one lump?

If the cardinal of the grand coalition is odd, the core is empty because it is impossible to share an amount $v(N)$ such that $\forall i \in N$, the sum of the payoff of the players belonging to $v(N\setminus i)$ receive more than $v(N)=v(N\setminus i)$.

If the cardinal of the grand coalition is even, then it is easy to see that $x_i=\frac{1}{2}$ $\forall i \in N$ belongs to the core and it is the only one possible payoff.


Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$. Question: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^2 - A + (1/2)I = 0$, where $I$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Prove that $A^n \to 0$ as $n \to \infty$.

My attempt: Here is my idea so far. Consider $A$ as a $2 \times 2$ matrix over the field of complex numbers. Now, the polynomial

$$g(t) = t^2 - t + \frac{1}{2} $$

factors as

$$g(t) = \left(\frac{1}{2} - \frac{i}{2}\right)\left(\frac{1}{2}+\frac{i}{2}\right)$$

over $C$. This means that the minimal polynomial of $A$ over $C$ is:

$$m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right), m(t) = \left(t-(\frac{1}{2} + \frac{i}{2})\right), \ \ \ \text{ or } \ \ \ m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right)\left(t-(\frac{1}{2} + \frac{i}{2})\right),$$

and thus we have $A = Q D Q^{-1}$, where

$$D =\left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 - i)}
\end{array}} \right],$$

$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 + i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right],\ \ \ \text{ or } $$

$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right].$$

Now, $A^n = QD^nQ^{-1}$.

Here is where I get stuck. $D^n$ doesn't seem to be converging to $0$ as $n$ approaches infinity. In addition, I am concerned that my strategy is bad, since we are talking about a real matrix and I'm using a minimal polynomial over $C$. Is it still true that $A$ must have one of the three forms above, even if $A$ is supposed to be real?

Thanks for any help/suggestions you may be able to provide.

The eigenvalues of $A$ (or $D$) satisfy $|\lambda | = { 1\over \sqrt{2}} <1$. Hence
$|\lambda|^n \to 0$.

For generalisation, assume $$
        A=\begin{bmatrix}
        a & b \\
        c & d \\
        \end{bmatrix}
$$ Now, for every $2X2$ matrix, this equation is always valid $A^2-trace(A)A+det(A)I=0$ (prove this by youself, this is very easy). Now you can compare this equation with your original equation, you will get $a+d=-1$ and $ad-bc=\frac {1}{2}$. Assume any values of $a,b,c,d$ satistfying these equations which I had done earlier.

Assume the values $a=d=-0.5, -b=c=0.5$, then matrix will look like $$2A=
        \begin{bmatrix}
        -1 & -1  \\
         1 & -1  \\
        \end{bmatrix}
$$ This matrix satisfy all the given condition.Now you can proceed. You can get very easily that $A^n=0$ as $n \to \infty$ . And also you can also use the eigen value logic to prove that $A^n=0$ as $n \to \infty$

Eigen Value Logic can also be applied. Let the eigen value of this matrix be $e_1$ and $e_2$. From given equation it is clear that $det(A)=0.5=e_1e_2$ and both eigen values are non real. So $e_1e_2=e^2=0.5$ where $|e_1|=|e_2|=e$

Now $\alpha=|A^n|=|A|^n=2^{-n}$ and as $n \to \infty$, $\alpha=0$ which clearly proves that $A^n$ has to be zero matrix because determinant of that matrix is zero, otherwise determinat of that matrix will never be zero.


Integration using trigonometric identities Can someone please help me to integrate this function
$$\int_0^1 \frac{1}{\sqrt{1-x^2}} \frac{s}{s^2+\alpha^2 \cdot x^2} dx = \frac{s\pi \sqrt{\frac{\alpha^2}{s^2}}}{2\alpha^2\sqrt{1+\frac{s^2}{\alpha^2}}}$$
  My professor to me to substitute $x = \sin u$ to get
  $$s \int_0^{\pi/2} \frac{1}{s^2+\alpha^2 \cdot \sin^2 u} du$$
But I can't seem to get where the $\frac{1}{\sqrt{1-x^2}}$ go

Isn't it supposed to be $\frac{1}{\sqrt{1-\sin^2 u}}$. Or $\sec u$

Please help

Notice that we have the pythagorean identity $\sin^2u+\cos^2u=1$, or $\cos^2u=1-\sin^2u$.  That is how the first fraction was handled.  The remaining can be dealt with $\sin^2u=\frac{1-\cos(2u)}2$ and a tangent half-angle substitution, as described in this post.

Remember also the substitution law. You need to change $x$ to $\sin u$, but it is the only thing to change?

SPOILER:

Remember also to change $dx$ to $du$, that is:
$$
dx=\sin'udu.
$$


Applications of the Fibonacci sequence The Fibonacci sequence is very well known, and is often explained with a story about how many rabbits there are after $n$ generations if they each produce a new pair every generation. Is there any other reason you would care about the Fibonacci sequence?

Perhaps it's not an entirely practical application, but Fibonacci numbers can be used to convert from miles to kilometers and vice versa:

  Take two consecutive Fibonacci
  numbers, for example 5 and 8. And
  you're done converting. No kidding –
  there are 8 kilometers in 5 miles. To
  convert back just read the result from
  the other end - there are 5 miles in 8
  km!

But why does it work?

  Fibonacci numbers have a property that
  the ratio of two consecutive numbers
  tends to the Golden ratio as
  numbers get bigger and bigger. The
  Golden ratio is a number and it
  happens to be approximately 1.618.

  Coincidentally, there are 1.609
  kilometers in a mile, which is within
  0.5% of the Golden ratio.

They're a much easier way to evaluate the function

$f(x) = \dfrac{\varphi^x -(-\varphi)^x}{\sqrt{5}}$

by hand, where $\varphi$ is the golden ratio and $x$ is an integer :)


Probablity identities and $A$ given $B$ I'm stuck on a question in my homework:

  Assume that $\Pr(A) = 1/2$ and $\Pr(B\mid ¬A) = 3/5$. Determine $\Pr(A \cup B)$.

I'm assuming $\Pr(A)$ is equal to $\Pr(¬A)$ because $1-1/2 = 1/2$. I'm confused to why this doesn't mean $\Pr (B\mid ¬A) = \Pr(B\mid A)$ (according to my friend).  Assuming the previous isn't true, I got this far:

  $\Pr(¬A \cap B) =  3/5 \cdot 1/2$

Thanks for the help!

$Pr (B|\neg A)$ means "the probability of $B$ to happen given $A$ is false (or $A$ does not happen)." $B$ may be more likely to happen more if $A$ happens, $B$ may be less likely to happen if $A$ happens. Hence $Pr (B|\neg A)$ can have nothing to do with $Pr (B|A)$.

On the other hand, $Pr(A)$ does imply $Pr(\neg A)$, and $Pr (B|\neg A)$ does imply $Pr (\neg B|\neg A)$.

$Pr (A \cup B)$ means "the probability of $A$ or $B$ (or both) to happen". So you should find out the probability where both $A$ and $B$ do not happen (say $y$) by using $Pr(\neg A)$ and $Pr (\neg B|\neg A)$ found above, and then use $1-y$ to find out $Pr (A \cup B)$.


How do I explain 2 to the power of zero equals 1 to a child My daughter is stuck on the concept that $$2^0 = 1,$$ having the intuitive expectation that it be equal to zero. I have tried explaining it, but I guess not well enough.

How would you explain the concept to a child, other than the teachers "that is just the rule" approach?

I will give a different answer than the answer I gave in the other thread which tries to appeal to intuition.  I am sure your daughter has no problem accepting that $2\times 0 = 0$.  Intuitively this is because if you add $2$ to itself zero times, you get zero.  Or, to be concrete, if someone gives you two apples zero times, you have zero apples.

For repeatedly adding $2$, talking about collections of apples is a good model.  But for repeatedly multiplying by $2$, it isn't necessarily, since you can't multiply apples and apples (at least, not in a way that makes sense to a child).  But you can multiply apples by numbers; that is, you can start with $1$ apple, then double the number of apples you have to get $2$ apples, then double the number of apples you have to get $4$ apples, and so forth.  In general if you double your apples $n$ times, you have $2^n$ apples.

What happens if you double your apples zero times?  Well, that means you haven't started doubling them yet, so you still have $1$ apple.  If you want your notation to be consistent, then you should say $2^0 = 1$.

This is a subtly different argument from the argument I gave before.  It's intuitive what it means to add different amounts of apples, and it's intuitive what it means to have zero apples.  But the twos I am now working with aren't numbers of apples, they're just abstract numbers; in other words, they're unitless, so it's harder to get a grip on them.  What $2^n$ really represents above is an endomorphism of the free commutative monoid on an apple, which is much less concrete than an apple.

There is a way to gain intuition here which sort of involves units, but I don't know if you can convince your daughter that it makes sense.  One way to interpret $2^n$ is that it is the "size" of an $n$-cube of side length $2$ in dimension $n$.  For example, the length of a segment of side length $2$ is $2$, the area of a square of side length $2$ is $4$, and so forth.  One way to say this is that $2^n$ is the number of $n$-cubes of side length $1$ that fit into an $n$-cube of side length $2$.

To get a meaningful interpretation of the above when $n = 0$ we need to decide what $0$-dimensional objects are.  Well, if $2$-dimensional space is a plane and $1$-dimensional space is a line, then $0$-dimensional space must be... a point.  In particular, a $0$-cube, of any side length, is a point, and so exactly one $0$-cube of side length $1$ fits into a $0$-cube of side length $2$.  Hence $2^0 = 1$.

(I'm really curious what her response to this argument will be, actually.  Could you report back on this?)

easy

so for example in every exponetation/multiplication there is a hidden 1

so 2^2 is 2x2=

2^3=8

but 2^3=2x2x2x1

x1 is always there but invisible its still part of the value hidden in every single calculation, its just not written, adding x1 makes no difference

so 2^1=2(x1)

2^0=(x1)=1

so 0^1 is 0x1

0^2=0x0x1=0

but 0^0=1 and nothing

its ironic


Proof that a periodic function is bounded and uniformly continuous. I need to show that if $f:\mathbb{R}\to \mathbb{R}$ is continuous and $\forall x \in \mathbb R, f(x+1)=f(x)$, then:

$f$ is bounded,
$f$ is uniformly continuous,
there exists $c\in \mathbb{R}$ such that $f(c)=f(c+\pi)$.

Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function.

Now, let $c_0$ be a point where $f_{|[0,1]}$ has a minimum. Because of the periodicity, $c_0$ is a minimum for the whole of $f$. So $\forall x \in \mathbb R, f(x) \geq f(c_0)$. In particular $(x = c_0 + \pi)$, $g(c_0) \geq 0$.

The same argument with “maximum” instead of minimum gives a $c_1$ where $g(c_1) \leq0$.

The intermediate value theorem now gives a point $c$ on which $g$ vanishes, QED.

$f$ is bounded in the compact [0,1] and as $f(x)= f(x+1), x \in \mathbb{R}, f$  is bounded.Now note that $f$ is uniformly bounde in the compacts $[0,1/2],[1/2,1]$. Then given $\epsilon>0$ there are $\delta_1, \delta_2>0$ such that
\begin{equation}
|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [0,1]
\end{equation}
\begin{equation}
|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [1/2,3/2]
\end{equation}
As $f(x)= f(x+1), x \in \mathbb{R}, f$ take $\delta=\min\{\delta_1,\delta_2$. Hence, $f$ is uniformily continuous. After I do the rest.


For given $k,N\in\mathbb{N}$, how to compute $\sum_{i=0}^Ni^k$? For given $k$ and $N$, $k,N\in\mathbb{N}$, how to compute $\sum_{i=0}^Ni^k$?

We have: $\sum^N_{i=0}i=\frac{N(N+1)}2$

Also according to what I found in the Internet we have $\sum^N_{i=0}i^2=\frac{n(n+1)(2n+1)}{6}$ (I can prove this formula by induction, but I couldn't obtain it myself)

But how to generalize this for any $k$?

We have
\begin{eqnarray*}
\sum_{i=1}^{n} 1 =n \\
\sum_{i=1}^{n} i =\frac{n(n+1)}{2} \\
\end{eqnarray*}
The following is a telescoping sum
\begin{eqnarray*}
\sum_{i=1}^{n} [(i+1)^3-i^3] =(n+1)^3-1. \\
\end{eqnarray*}
It is now a bit of a grind but with these results you will get to
\begin{eqnarray*}
\sum_{i=1}^{n} i^2 =\frac{n(n+1)(2n+1)}{6}. \\
\end{eqnarray*}
A similar trick will allow you to calculate higher powers, but the calculation gets longer each time.

You are probably best to do as dxiv suggests & use the reference to Faulhaber's formula.

Here is an induction formula from which all sum can be derived:

Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$

then,
We know the following Binomial formula

$$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$
where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$.
Which implies that,

$$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$

But $$\sum_{k=1}^{n} (k+1)^p = \sum_{k=2}^{n+1} k^p = S_{n+1}(p) -1 = S_n(p) +(n+1)^p -1$$

Hence finally we get the formula :

$$\color{red}{(n+1)^p -1  =\sum_{i=0}^{p-1}\binom{p}{i} S_n(i)} $$

  From this it is possible to compute the sum for any $p\ge 1 $ in $ \mathbb N $.


You should not expand the power : as you said, you want to find $k$ such that
$$(k-10)^3=1$$
which is equivalent to
$$k-10=1$$
so
$$k=11$$

Your first insight is exactly the idea, though the algebra would be easier without the expansion you do:
$$f(-2) = 0 \implies -1 + (k - 10)^3 = 0 \implies (k - 10)^3 = 1$$
If your problem only asks for real solutions of $k$, then it is clear that $k = 11$ is the only solution. However, if complex numbers are allowed, you will need to consider the case where $k - 10$ is a complex third root of unity, i.e. $k - 10 = \frac{-1}{ 2} + \frac{\sqrt 3}{ 2} i$ or $k - 10 = \frac{-1}{ 2} - \frac{\sqrt 3}{ 2} i$.


Do we have $(gf)^*\mathcal F=f^*(g^*\mathcal F)$? Let $X\xrightarrow f Y\xrightarrow g Z$ be morphisms of schemes and $\mathcal F$ an $\mathcal O_Z$-module. Do we have $(gf)^*\mathcal F=f^*(g^*\mathcal F)$?

Yes since for a morphism of schemes $h:X \rightarrow Y$, $h^*$ is the left adjoint to $h_*$.

Thus we have

$$Hom((f \circ g)^* \mathscr F, \mathscr G) = Hom(\mathscr F, (f \circ g)_* \mathscr G) = Hom(\mathscr F, (f_* \circ g_*) \mathscr G) = Hom(f^*\mathscr F, g_* \mathscr G) =
 Hom((g^* \circ f^*)\mathscr F, \mathscr G)$$ for all $\mathscr F$ and $\mathscr G$ modules on the respective schemes.

And thus by yoneda lemma $(f \circ g)^* \mathscr F = (g^* \circ f^*)\mathscr F$

$(gf)^*\mathcal F=f^*(g^*\mathcal F)$

I can find it in Vakil's book.


Why does this set a compact? Consider the following set in $\mathbb{R}^n$:

$$S = \{ x \ :\ \|x\| = 1 \}$$

Why is this set a compact? I don't know about closed, but it doesn't look bounded to me.

HINT: $S$ is the boundary of a certain ball.

$S$ is representing a boundary and it is an inverse image continuous function $||x||$
Compact Set is a subset of a topological space if for every open cover of there exists a finite subcover of.


Prove or disprove that $A^n=B^n$ implies $A=B$ Given two square matrices $A,B$ such that $A^n=B^n$ for some $n>2$. Is it necessarily true that $A=B$?

It is obvious that if $A^n=B^n=0$ then $A\neq B$ is not necessarily true. What if $A,B\in GL(n,\mathbb{R})$?

Scratching my head trying to come up with counter-examples.

\begin{eqnarray*}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}^2 =\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}^2.
\end{eqnarray*}

What is $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}^2$ ?


geometry question on straight lines Three points X,Y,Z are on a straight line such that XY = 10 and XZ = 3. How can I find the product of all possible values of YZ?

There are only two possibilities, Z-X-Y or X-Z-Y.Then doing the necessary calculations we get YZ either 13 or 7 so answer is 91.

Ths collinearity is X-Z-Y now XY is 10 and XZ is  3 so YZ has to be 7.So proxuct has to be 7. And Z-X-Y so 13 so product =13.7=91


Rudin assumes $(x^a)^b=x^{ab}$(for real $a$ and $b$) without proof? I am currently self studying Baby Rudin and I'm having some problems with his proof of Theorem 3.20(a) on page 58. I have read all previous chapters and I can't find any mention of real exponents besides its definition in the exercise section of Chapter 1 (but nothing about order/this identity).
He defines $x^a$ for rational $a$ and $x>1$ then this is used to extend it to all real $a$ by $x^a = \sup \{x^t : t \leq a,~ t \in Q\}$.

Am I missing something or does he expect the reader to fill this huge gap in the proof?

Prefatory note on parts for Theorem 3.20: We shall now compute the limits of some sequences which occur frequently. The proofs will all be based on the following remark: If $0\leq x_n\leq s_n$ for $n\geq N$, where $N$ is some fixed number, and if $s_n\to 0$, then $x_n\to 0$.

Theorem 3.20 (a): If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.

Proof. Take $n>(1/\varepsilon)^{1/p}$. (Note that the archimedean property of the real number system is used here.)

First show that it holds for $a, b \in \mathbb N$.
Then show it holds for $a,b \in \mathbb Q$ (this is easy: since $x^{p/q} = x^p \cdot (x)^{1/q}$, and $p, q \in \mathbb N$)

Finally for reals $a, b$ you have that $x^a = \sup\{x^t, t\le a, t \in \mathbb Q\}$; hence you get

\begin{eqnarray*}(x^{a})^b &=& \sup\{\sup\{x^t, t\le a, t \in \mathbb Q\}^s, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{\sup\{(x^t)^s, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{\sup\{x^{ts}, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em]
&=& \sup\{x^{ts}, t\le a, s\le b, s \in \mathbb Q, t \in \mathbb Q\} \\[0.5em]
&=&\sup\{x^{k}, k\le ab, k \in \mathbb Q\} \\[0.5em]
&=& x^{ab}
\end{eqnarray*}

Well, when we consider integers...

Note that
$$
(x^a)^1 = x^a = x^{a \cdot 1}
$$

Use induction:

IF $(x^a)^{n} = x^{an}$, then also $(x^a)^{n+1} = x^{a(n+1)}$ , as
$$
(x^a)^{n + 1} = (x^a)^n x^a = x^{an} x^a = x^{a(n+1)}
$$

It is true for $n=1$, and by induction it is true for any positive integer.


Eigenvalues of a tridiagonal matrix I am currently studying this type of Jacobi matrix for all n

Example of the type of matrix $n=4$ $$\mathbf{X}=\left[\begin{array}{*{20}{c}}
{0}&{1}&{0}&{0}\\
{1}&{0}&{1}&{0}\\
{0}&{1}& {0} &{1}\\
{0}&{0}&{1}&{0}
\end{array}\right]$$

I am struggling to prove that all eigenvalues for all n will be; $ \lambda \leq |2|$

If all you need is to show that the spectral radius $\rho(X)$ is bounded above by 2, there are many ways to do it:

As pointed out in a user's comment to your question, this follows directly from  Geršgorin disc theorem.
The result also follows immediately from the inequality $\rho(X)\le\|X\|$ if you take the induced 1-norm (i.e. the maximum absolute column sum norm $\|X\|=\max_{j}\sum_i|x_{ij}|$).
Another approach is to add 1 to the top right and bottom left corners of $X$, and call the resulting matrix $Y$. Since $X$ and $Y$ are irreducible nonnegative matrices and $X$ is entrywise bounded above by $Y$, we get $\rho(X)\le\rho(Y)$. However, as $\frac12Y$ is a stochastic matrix, we have $\rho(\frac12 Y)=1$.
Alternatively, in (3) above, as $Y$ is also a circulant matrix, you can express its eigenvalues explicitly: they are $\{2\operatorname{Re}(\omega^k): k=0,1,2,\ldots,n-1\}$, where $\omega$ is the $n$-th root of unity. Hence $\rho(Y)=2$.

A computational way:

$$\det (tI-X)=\begin{vmatrix}t&-1&0&0\\
-1&t&-1&0\\
0&-1&t&-1\\
0&0&-1&t\end{vmatrix}=t\begin{vmatrix}t&-1&0\\
-1&t&-1\\
0&-1&t\end{vmatrix}+\begin{vmatrix}-1&0&0\\
-1&t&-1\\
0&-1&t\end{vmatrix}=$$

$$=t\left(t^3-2t\right)+\left(-t^2+1\right)=t^4-3t^2+1$$

Using the roots formula, we get that the eigenvalues are:

$$\left|\pm\sqrt{\frac{3\pm\sqrt5}2}\right|\le 2$$


How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Please help me to find a closed form for this integral:
$$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$
I suspect it might exist because there are similar integrals having closed forms:
$$\begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta(5)-\pi^2\zeta(3)\tag2\\
\int_0^1\frac{\ln^2(1+x)\ln x}x\mathrm dx&=\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)\tag3\\
\int_0^1\frac{\ln^3(1+x)\ln x}{x^2}\mathrm dx&=\frac34\zeta(3)-\frac{63}4\zeta(3)\ln2+\frac{23\pi^4}{120}\\&-\frac34\ln^42-2\ln^32+\frac{3\pi^2}4\ln^22-18\operatorname{Li}_4\!\left(\tfrac12\right).\tag4\end{align}$$
Thanks!

Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
I&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\
&=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\
&=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 (1-x)}{x}-\frac{2\ln^3x\ln(1-x)}{x}+\frac{\ln^4x}{x}\right]\ dx\quad\Rightarrow\quad\color{red}{x\mapsto\frac1x}\\
&=-\frac32\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\left.\frac3{10}\ln^5x\right|_{\large\frac12}^1\\
&=-\frac32\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\frac3{10}\ln^52.
\end{align}
Applying IBP again to evaluate the red integral by setting $u=\ln^2(1-x)$ and $dv=\dfrac{\ln^2 x}{x}\ dx$ yields
\begin{align}
\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}&=\frac13\ln^52+\frac23\color{blue}{\int_{\large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}.
\end{align}

  For the simplicity, let
  $$
\color{blue}{\mathbf{H}_{m}^{(k)}(x)}=\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^m}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=\sum_{n=1}^\infty H_{n}x^n,
$$
  Introduce a generating function for the generalized harmonic numbers for $|x|<1$
  $$
\color{blue}{\mathbf{H}^{(k)}(x)}=\sum_{n=1}^\infty H_{n}^{(k)}x^n=\frac{\operatorname{Li}_k(x)}{1-x}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=-\frac{\ln(1-x)}{1-x}
$$
  and the following identity
  $$
H_{n+1}^{(k)}-H_{n}^{(k)}=\frac1{(n+1)^k}\qquad\Rightarrow\qquad H_{n+1}-H_{n}=\frac1{n+1}
$$

Let us integrating the indefinite form of the blue integral.
\begin{align}
\color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&-\int\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\int x^n\ln^3x\ dx\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\int x^n\ dx\right]\\
=&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\frac{x^{n+1}}{n+1}\right]\\
=&-\sum_{n=1}^\infty H_n\left[\frac{x^{n+1}\ln^3x}{n+1}-\frac{3x^{n+1}\ln^2x}{(n+1)^2}+\frac{6x^{n+1}\ln x}{(n+1)^3}-\frac{6x^{n+1}}{(n+1)^4}\right]\\
=&-\ln^3x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{n+1}+\ln^3x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}+3\ln^2x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^2}\\&-3\ln^2x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}-6\ln x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^3}+6\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\\&+6\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^4}-6\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^5}\\
=&\ -\sum_{n=1}^\infty\left[\frac{H_nx^{n}\ln^3x}{n}-\frac{x^{n}\ln^3x}{n^2}-\frac{3H_nx^{n}\ln^2x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}\right.\\& \left.\ +\frac{6H_nx^{n}\ln x}{n^3}-\frac{6x^{n}\ln x}{n^4}-\frac{6H_nx^{n}}{n^4}+\frac{6x^{n}}{n^5}\right]\\
=&\ -\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\color{blue}{\mathbf{H}_{2}(x)}\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\color{blue}{\mathbf{H}_{3}(x)}\ln x+6\operatorname{Li}_4(x)\ln x+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x).
\end{align}
Therefore
\begin{align}
\color{blue}{\int_{\Large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}
=&\ 6\color{blue}{\mathbf{H}_{4}(1)}-6\operatorname{Li}_5(1)-\left[\color{blue}{\mathbf{H}_{1}\left(\frac12\right)}\ln^32-\operatorname{Li}_2\left(\frac12\right)\ln^32\right.\\&\left.\ +3\color{blue}{\mathbf{H}_{2}\left(\frac12\right)}\ln^22-3\operatorname{Li}_3\left(\frac12\right)\ln^22+6\color{blue}{\mathbf{H}_{3}\left(\frac12\right)}\ln 2\right.\\&\ -6\operatorname{Li}_4(x)\ln 2+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)\bigg]\\
=&\ 12\zeta(5)-\pi^2\zeta(3)+\frac{3}8\zeta(3)\ln^22-\frac{\pi^4}{120}\ln2-\frac{1}
{4}\ln^52\\&\ -6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}+6\operatorname{Li}_4\left(\frac12\right)\ln 2+6\operatorname{Li}_5\left(\frac12\right).
\end{align}
Using the similar approach as calculating the blue integral, then
\begin{align}
\int\frac{\ln^3x\ln (1-x)}{x}\ dx&=-\int\sum_{n=1}^\infty \frac{x^{n-1}}{n}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\int x^{n-1}\ln^3x\ dx\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\int x^{n-1}\ dx\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\frac{x^{n}}{n}\right]\\
&=-\sum_{n=1}^\infty \frac{1}{n}\left[\frac{x^{n}\ln^3x}{n}-\frac{3x^{n}\ln^2x}{n^2}+\frac{6x^{n}\ln x}{n^3}-\frac{6x^{n}}{n^4}\right]\\
&=\sum_{n=1}^\infty \left[-\frac{x^{n}\ln^3x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}-\frac{6x^{n}\ln x}{n^4}+\frac{6x^{n}}{n^5}\right]\\
&=6\operatorname{Li}_5(x)-6\operatorname{Li}_4(x)\ln x+3\operatorname{Li}_3(x)\ln^2x-\operatorname{Li}_2(x)\ln^3x.
\end{align}
Hence
$$
\int_{\large\frac{1}{2}}^1\frac{\ln^3x\ln (1-x)}{x}\ dx=\frac{\pi^2}{6}\ln^32-\frac{21}{8}\zeta(3)\ln^22-6\operatorname{Li}_4\left(\frac{1}{2}\right)\ln2-6\operatorname{Li}_5\left(\frac{1}{2}\right)+6\zeta(5).
$$
Combining altogether, we have

  \begin{align}
I=&\ \frac{\pi^4}{120}\ln2-\frac{33}4\zeta(3)\ln^22+\frac{\pi^2}2\ln^32-\frac{11}{20}\ln^52+6\zeta(5)+\pi^2\zeta(3)\\
&\ +6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}-18\operatorname{Li}_4\left(\frac12\right)\ln2-24\operatorname{Li}_5\left(\frac12\right).
\end{align}

Continuing my answer in: A sum containing harmonic numbers $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, we have
\begin{align}
\color{blue}{\mathbf{H}_{3}\left(x\right)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag1
\end{align}
Dividing $(1)$ by $x$ and then integrating yields
$$\small\begin{align}
\color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\ dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx+\frac{\pi^4}{60}\ln x\\
=&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}\right].\tag2
\end{align}$$
Evaluating the red integral using the same technique as the previous one yields
\begin{align}
\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}&=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}
Evaluating the purple integral yields
\begin{align}
\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ dx\right]\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\
&=\color{blue}{\mathbf{H}_{3}(x)}\ln x-\color{blue}{\mathbf{H}_{4}(x)}.
\end{align}
Evaluating the green integral using IBP by setting  $u=\ln x$ and $dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx$ yields
\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ dx\\
&=\operatorname{Li}_4(x)\ln x-\operatorname{Li}_5(x).
\end{align}
Evaluating the orange integral using IBP by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields
\begin{align}
\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}&=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}.
\end{align}
Applying IBP again to evaluate the maroon integral by setting $u=\operatorname{Li}_2(1-x)$ and
$$
dv=\dfrac{\ln^2 x}{1-x}\ dx\quad\Rightarrow\quad
v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x,
$$
we have
$$\small{\begin{align}
\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\
&-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}}$$

We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.

  \begin{align}
\int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ dx\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ dx\right]\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n+1}^{(k)}x^{n+1}\ln x}{n+1}-\frac{x^{n+1}\ln x}{(n+1)^{k+1}}-\frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+\frac{x^{n+1}}{(n+1)^{k+2}}\right]\\
&=\sum_{n=1}^\infty\left[\frac{H_{n}^{(k)}x^{n}\ln x}{n}-\frac{x^{n}\ln x}{n^{k+1}}-\frac{H_{n}^{(k)}x^{n}}{n^2}+\frac{x^{n}}{n^{k+2}}\right]\\
&=\color{blue}{\mathbf{H}_{1}^{(k)}(x)}\ln x-\operatorname{Li}_{k+1}(x)\ln x-\color{blue}{\mathbf{H}_{2}^{(k)}(x)}+\operatorname{Li}_{k+2}(x).
\end{align}

Dividing generating function of $\color{blue}{\mathbf{H}^{(k)}(x)}$ by $x$ and then integrating yields

  \begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx\\
&=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx.
\end{align}

Repeating the process above yields

  \begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2}
&=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx,
\end{align}

where it is easy to show by using IBP that

  \begin{align}
\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ dx\\
&=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x
\end{align}

and

  $$
\int\frac{\operatorname{Li}_3(x)}{1-x}\ dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ dx=-\frac12\operatorname{Li}_2^2(1-x)-\operatorname{Li}_3(1-x)\ln x.
$$

Now, all unknown terms have been obtained. Putting altogether to $(2)$, we have
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C.\tag3
\end{align}}$$
The next step is finding the constant of integration. Setting $x=1$ to $(3)$ yields
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(1)}
&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\
3\zeta(5)+\zeta(2)\zeta(3)&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac{19}{30}\operatorname{Li}_5(1)+\frac{3}{5}\operatorname{Li}_3(1)+C\\
C&=\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5).
\end{align}}$$
Thus
$$\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\tag4
\end{align}}$$
and setting $x=\frac12$ to $(4)$ yields
\begin{align}
\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \frac{\ln^52}{40}-\frac{\pi^2}{36}\ln^32+\frac{\zeta(3)}{2}\ln^22-\frac{\pi^2}{12}\zeta(3)\\&+\frac{\zeta(5)}{32}-\frac{\pi^4}{720}\ln2+\operatorname{Li}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5
\end{align}

Finally, we obtain

  \begin{align}
\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx=&\ \color{blue}{\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)},
\end{align}

which again matches @Cleo's answer.

References :

$[1]\ $ Harmonic number

$[2]\ $ Polylogarithm

Related problems and techniques: (I), (II). Here is a different form of solution

  $$ I = -3\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^k{ n\brack k}k(k-1) }{(n+1)^3n!} ,$$

where $ {n \brack k} $ is the Stirling numbers of the first kind.


why the vector's location doesn't matter? I'm wondering that why people doesn't care about vector's location.
but, when we add the vectors, we move two or one vector to unite the vector's start point. and this mean location is important.
but, when we learn very first of vector, we learn the vector doesn't matter where it is.
in this point, I'm so confused because I can't find the reason that why vector's location doesn't matter.
so i want to ask you - why the vector's location is not important?

thank you

What you are probably getting at are 'free vectors'. Their description in school math is usually something like $\vec{x} = (a,b)$ - that means, go right $a$ units and up $b$ units. Ok, but starting from where?? It doesn't matter. That is the meaning of 'freedom' in this sense. You can 'freely' pick the point on the plane from which you apply the vector $\vec{x}$, for instance the origin.

We also have vectors called 'bound vectors' for which location does matter.

The confusion is maybe due to some possible double meaning in the notation. Consider for instance:

$V=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $

In essence this double meaning only occurs when picking coordinates to represent a vector. Clearly a 3-tuple of numbers can be considered as a point in 3 dimensional space. However, it can also be thought of as a direction, i.e. line of length $\sqrt{2} $ starting at the origin. Generally speaking if we only consider(or care about) the data of length and direction, the point from which the direction comes from is of little importance, hence there should be an equivalence of vectors by translation. In fact, where i am from, it was back in the day not so uncommon to introduce vectors as just this data. I.e. a length and a direction with equivalence relation given by translation, and then afterwards introduce coordinate systems to describe the operations precisely. Furthermore, translation invariance is great for physics and geometry since directions, e.g. tangent directions and forces acting on objects in space have meaning not as a direction from the origin, but really at certain points.

I guess one could be more specific when one wants a concrete base point of the vector by including it in the notation (one does so for instance when one considers “vector fields“ in physics, describing how forces act and varies at points, or for instance in differential geometry, when one considers vector spaces attached to certain points of a space, giving a model for the directions and “speed” of curves passing through that point). However, in a standard vector space, the connection between the nice algebraic operations in a vector space,  and the arguably even nicer geometric interpretation of vectors would be lost if we do not require translation invariance.

For instance, in the usual interpretation of adding the vectors $v_{x}+w_{x+v}$(where $_{x} $ means the starting point of the vector ) could be defined algebraically and geometrically. However, we do NOT have commutativitity, and we cannot even make sense of$w_{x+v}+v_{x}$. Even worse, we cannot even rescale the $v_{x}$ vector by a number b and adding the other vector to it, since the endpoint of $bv_{x} $ would not match up with w_{x+v}.

I hope this was of some help


Correct proof of supremum property? Let $u$ be an upper bound of non-empty set $A$ in $\mathbb{R}$.
Prove that $u$ is the supremum of $A$
if and only if for all $\epsilon > 0$ there is an $a \in A$ such that
$u-\epsilon < a$.

Lemma

If $s = \sup(A)$, then
 $$ \exists \, a \in A, \, \,  \forall \epsilon > 0 \, \, \text{such that} \, \, |s - a| < \epsilon$$

Proof:

Suppose $\nexists \, a \in A $ such that $\forall \, \epsilon > 0$, $|\sup(A) - a| < \epsilon$. Then $\forall \, a \in A, \, \exists \, \epsilon_0> 0$ such that $|\sup(A) - a| > \epsilon_0$. This implies that $\sup(A) > a + \epsilon_0$. Thus, $\exists\, s' \not\in A$ such that $\sup(A) > s' > a + \epsilon_1$ if we take $\epsilon_1 = \frac{\epsilon_0}{2}$. But we cannot have $s' < \sup(A) : s' \not\in A$ so we have arrived at a contradiction and thus the lemma is proved.

Proof:

$(i \implies ii)$ if $u  = \sup(A)$, then, by the lemma, $\forall \, \epsilon > 0$, \, $\exists \, a \in A$ such that $|u-a|< \epsilon$. This implies $u-\epsilon < a$.

$(ii \implies i)$ If $\forall \, \epsilon > 0, \, \exists \, a \in A$ such that $u - \epsilon < a$, then $u-a < \epsilon$ and by the lemma, $u =\sup(A)$.

Is this a complete and correct proof?

You can't base the proof of the statement on an obviously false lemma (at least with an incorrect formulation).

The proof is much simpler than that.

Assume that $u$ is an upper bound for $A$ with the property that, for every $\varepsilon>0$, there exists $a\in A$ with $u-\varepsilon<a$.

By definition, the supremum is the least upper bound. So if $u$ is not the supremum, there is an upper bound $v$ of $A$ with $v<u$.

Now take $\varepsilon=(u-v)/2$. By assumption there exists $a\in A$ with $u-\varepsilon<a$; this becomes
$$
u-\frac{u-v}{2}<a
$$
that is,
$$
\frac{u+v}{2}<a
$$
which is a contradiction, because from $a\le u$ and $a\le v$ we obtain
$$
a\le\frac{u+v}{2}
$$

For the converse, suppose $u=\sup(A)$ and let $\varepsilon>0$. Then $u-\varepsilon<u$, so $u-\varepsilon$ is not an upper bound for $A$, which means there exists $a\in A$ with $u-\varepsilon<a$.

By supremum we mean the least upper bound. So if $u = \sup A$ then $u$ is an upper bound for $A$ and any number less than $u$ is not an upper bound for $A$. Now if $\epsilon > 0$ then $u' = u - \epsilon < u$ and hence $u'$ is not an upper bound for $A$ and therefore it is exceeded by some member of $A$. So there is a member $a \in A$ such that $a > u' = u - \epsilon$. This proves one part of the question.

For the converse let's assume that $u$ is an upper bound of $A$ such that if $\epsilon > 0$ then there is an $a\in A$ such that $u - \epsilon < a$. Then I prove that any number less than $u$ is not an upper bound for $A$. This is almost obvious. Let $u'$ be a number less than $u$ i.e. $u' < u$ then $\epsilon = u - u' > 0$. By what we have assumed it follows that there is an $a \in A$ with $u - \epsilon < a$ i.e. $u' < a$. This clearly shows that $u'$ is not an upper bound for $A$. So we have shown that $u$ is an upper bound for $A$ but any number $u' < u$ is not an upper bound for $A$. So $u$ is the least upper bound for $A$ i.e. $u = \sup A$.

Another point which I want to emphasize here is that unless one has a thorough understanding of the concepts in real analysis it is not advisable to use so many symbols like $\exists, \forall, \Rightarrow$ together with $\epsilon$'s and $\delta$'s in writing proofs. It is better to write proofs in natural language (say English) with bare minimum uses of symbols. This helps a lot in understanding the concepts. Otherwise it looks like a silly game of symbol shunting without any rhyme or reason.

You can see the problem of symbol shunting in your statement $$\exists a \in A, \forall \epsilon > 0\text{ such that }|s - a| < \epsilon$$ This statement means that there is an $a \in A$ such that for all $\epsilon > 0$ we have $|s - a| < \epsilon$. This tells us that there is at least one particular $a \in A$ such that the result $|s - a| < \epsilon$ holds for all $\epsilon > 0$. This is not correct. Instead what you perhaps wanted to say was this:

If $s = \sup A$ then for every $\epsilon > 0$ there exists an $a \in A$ such that $|s - a| < \epsilon$.

Here you want to emphasize that for each $\epsilon$ the corresponding $a \in A$ may (or may not) be different. Thus one $a$ does not need to work for all $\epsilon$'s. The way to write this is $$\forall \epsilon > 0, \exists a \in A\text{ such that }|s - a| < \epsilon$$ This amply demonstrates the problem of symbol shunting. Unless one is expert (which one can become with some patience and practice) in such symbol shunting game, one is better off writing statements in natural language.


If $f: A \to B$ is injective and $B$ is countable, then so is $A$. I would appreciate any hint or idea to prove this, thank you.

HINT

Create an enumeration of $A$ from your enumeration of $B$.

Any subset of a countable set has a least element.