In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula. So by easy means, I derived

$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$

But then I checked for the actual formula it is this;

$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2 + b^2}+a}{2}}± i(\sqrt{\frac{\sqrt{a^2 + b^2}-a}{2}})$

So how do you simplify that to this?
Btw here is the derivation;

  Let $\sqrt{a+ib} = x+iy$

  $a+ib=x^2 - y^2 + i2xy$

  We know, $(x^2-y^2)^2 + (2xy)^2 = x^2 + y^2$

  or, $x^2 + y^2 = a^2 + b^2$

  Hence we get $x = \sqrt{\frac{a+a^2+b^2}{2}}$ and
  $y=\frac{b}{\sqrt{2}\sqrt{a^2+a+b^2}}$

  Therefore;

  $\sqrt{a+ib} =
> \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$

You made a mistake in your derivation.  You should have $(x^2+y^2)^\color{red}2=a^2+b^2$.

This follows from $a=x^2-y^2$ and $b=2xy$ (or from known properties of complex modulus).

Thus, $a=x^2-\dfrac {b^2}{4x^2}$;

solving this quadratic equation in $x^2$ yields $x^2=\dfrac{a+\sqrt{a^2+b^2}}2$ as the correct answer.

One of the approache is use Euler form if Z is a complex numbers such that $ Z\, = \,r.e^{ix}$ then
Square root of Z will be $Z_{1}= \, \sqrt{r} . e^{iy} $
Where angle y will be x/2 and (x + 2π)/2.


Apparent Paradox in the Idea of Random Numbers This question is a bit less than rigorous, but it's only because I don't know how to formulate it rigorously.

Suppose there was some machine, or function, or whatever that could output a random positive whole number. Let's say that it has done its job and I have a number n.

Here's what seems like a paradox to me: if n really was randomly generated, I shouldn't be surprised by its value, no matter how big or small. After all, there is no number that I should have expected more than any other.

On the other hand, if I asked the question of whether I expected the number to be bigger or smaller than the one that I got, the answer would always be bigger. In fact, it was infinitely more likely that I would have gotten a number bigger than what I got. If I got a digit with one billion digits, there are still infinitely more numbers above n than there are below it. No matter the number, it is infinitely improbable that it could have been that small.

Is it just that the idea of picking randomly from an infinite set doesn't make any sense? Or is this a problem that one also runs in to when trying to pick a random element of a finite set?

Picking a number at random from an infinite set makes perfect sense. It just can't be done uniformly over all points. You would have to specify the probability of (in your case, the natural numbers) choosing each and every natural number. That is, for every $n\ge 1$ you must specify a probability $p_n\ge 0$. The only requirement is that $\sum_np_n=1$. Now, your paradox arises from the hidden supposition that the distribution is uniform, i.e., that $p_n=p_m$ for all $n,m$. This is of course inconsistent, so no such probability distribution exists.

Now that you know that, you must revise your question and see if the paradox survives. So, if you are now given a machine that generates a random natural number according to some distribution (which may or may not be known to you), are you surprised at the outcome? if you see the number $1005$, do you expect the next number to be larger or smaller?

I'm very rusty, but I'd say if you want the probability that a number is lower than your current pick, you'd use the following formula, where A = the maximum number in the set and B is the number you picked:

P(lower) = 1 - (A-(B-1))/A

So if you were picking from 1 to 10 and picked 9, the probability of picking a lower number would be 1 - (10-(9-1))/10, or 0.8.

How this works when A is infinite is a bit beyond me, but intuitively I'd agree with you: dividing infinity by infinity is 1 (except it isn't, but it feels right!), and then 1-1 is zero, so P(lower) will be 0.


Is there any formula for the infinite surd $\sqrt{k+\sqrt{k^2+\sqrt{k^3+\sqrt{k^4+...}}}}$? Is there any related studies about the topic?

Let
$$n=\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$
then
$$n^2=k+\sqrt{k^2+\sqrt{k^3+\cdots}}$$
By subtracting $k$ from both sides and then multiplying out $\sqrt{k}$ we get
$$n^2 - k=\sqrt{k}\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$
But now we got the same form as in the beginning, thus we can insert $n$ to get
$$n^2 - k=\sqrt{k}\cdot n$$
with the solutions
$$n_{1,2}=\frac{\sqrt{k}}{2} \pm \sqrt{\frac{k^2}{2}+k}$$

Its strange that there are two solutions, so please check my solution for errors. But the main idea in these kind of questions is always to get back to the first form.


Precision in $f : A \to B$ notation Here I have a nice function (also known as $f(x)=\sin (\pi x/2)+1$) that is continuous on the interval $[0,4]$:

￼

According to the definitions I've found, $f:A\rightarrow B$ means that $\forall n\in A:f(n)\in B$. Therefore, in describing this function’s domain and range, the most precise thing to do would be to say $f:[1,4]\rightarrow [0,2]$. However, from what I understand, this notation is not the most precise, and the following would also be true:

$f:[1,4]\rightarrow [-5,5]$
$f:[1,4]\rightarrow\Bbb{R}$
$f:[1,2]\rightarrow\Bbb{R}$

Basically, $A$ must include some—but not all of—$f$’s domain but must not include any number for which $f$ is not defined while $B$ can include any numbers so long as it includes the range of $f$ on $A$—which makes sense when looking at the definition of $f:A\rightarrow B$.

All of this research I’ve been doing (I’m only a junior in high school who just started AP Calculus BC) was originally to determine a concise and symbolic way to express that a function is continuous on a certain interval, but now my question has evolved to include something else: how can one clearly and symbolically define the full domain and range of a function without any ambiguity and in one fell swoop?

Update:

It has come to my attention that codomain and range are actually different. From what I understand at this point, codomain is a kind of restriction on the range. Here's an example:

Let $f:x\mapsto\sin\left(\dfrac{\pi x}{2}\right) +1$, as shown in red:

If one wishes to restrict the domain, simply graph $f:[0,4]\rightarrow\Bbb R$ (as shown in blue):

Want to restrict the range, too? Go ahead and meddle with the codomain. Here's $f:[0,4]\rightarrow [0.25,1.75]$ in green:

As of right now I'm confused as to whether or not you have to restrict the domain to exclude images of x that would be excluded anyways by restricting the codomain

The Range of $f:A\to B$ is sometimes denoted $f(A)$. $f(A)$ is defined as $\{f(a) \ | \  a\in A\}$. A well-defined function (i.e. a function) satisfies $f(A)\subseteq B$.

Requiring a function be defined as $f: A\to f(A)$ is too constricting. While this defines in a sense the "best" restriction of any $f:A\to B$, sometimes the set $f(A)$ is unknowable or tedious to describe.

Take for example the function

$$f:\mathbb N\to \mathbb N, \ \ \ \ f(n) = \text{the $n^{th}$ largest prime number}$$

We don't have a formulaic way to describe $f(\mathbb N)$. $\mathbb N$ is simply a good choice for a codomain, rather than say $\mathbb Z$ or $[0,\infty)$ because the others are unncessarily large or complicated.

One notation that I have seen is

$f([0, 4])
= [0, 2]
$.


Calculus optimization problem Help I'm stuck on how to proceed with this problem.

Consider the function $f(x)=9-x^2$ for $f(x) \leq 0$. The shaded region is an isosceles triangle formed by joining the points $(0,0)$, $(x,f(x))$ and $(-x,f(x))$. What is the area of the largest triangle that satisfies the stated conditions (Look at picture)?

My teacher gave us a hint and I tried to do it as $A=1/2(3)(9-x^2)$. But I just have my doubts that I'm doing it wrong and I've looked everywhere for help and the only one who has it was Chegg, and I don't have access to that.

Here's how the figure should look like

Let the base of the triangle as $2x$ and the height as $f(x)$
$$A(x) = xf(x)$$
Set the derivative of $A(x)$ equal to zero
$$
\frac{d}{dx}A(x) = 0
$$
You'll found the $x$ points.

Use symmetry. Find the biggest triangle possible in the first quadrant and we know that the area of the full triangle is double it. The base of our triangle is given by $x$ and the height is given by $f(x)$ so we want to maximize $\frac{xf(x)}{2}=area$.

$xf(x)=x(9-x^2)=9x-x^3$ take derivative to get $f’(x)=-3x^2+9$

and $x=\sqrt{3}$ makes it it zero which is the value that makes the area maximized.

By symmetry, the other point is at -$\sqrt{3}$


Prove two subsequent primes cannot be written as a product of two primes Suppose we have two subsequent primes, say $p$ and $p'$. Prove their sum cannot be written as a product of two primes, say $p_1$ and $p_2$.

I wanted to proof by contradiction. I started by thinking about parity of the sum. Suppose $p=2$, then $p'=3$. But this sum cannot be written as a product of two primes $p_1$ and $p_2$. So we know that $p>2$; this implies that both $p$ and $p'$ are odd, so $p+p'$ is even. This means that either $p_1$ or $p_2$ must equal $2$. This results in:
\begin{equation} p+p'=2p_1.\end{equation}

Now, how can I finish the proof?

For $p=2$ then it's immediate. For odd primes $p,p'$ you would have $p_1p_2=even$ namely $p_1=2$ and $p_2=\frac{p+p'}{2}$. But then $p_2$ is a prime that lies between two consecutive primes $p,p'$. Contradiction.

The lowest pair to check is $2$ and $3$. That sum can not be represented as the product of two primes. Therefore $p > 2$. Because of that $p$ must be $odd$. So the sum must be $even$. To get an $even$ product, one of the two factors must be $even$.
So we got two facts:

  $p > 2$, therefore $p$ is $odd$

and

  $p_1 \times p_2$ = $x$, whereas $x$ is $even$

That means either $p_1$ or $p_2$ must be $even$. That contradicts with our first conclusion that $p > 2$ and $p$ must be $odd$.

That means that we have proven this problem by contradiction.


How to find image, preimage, domain and range of function? Given:

(a) $f = \{(x, f(x))\mid x ∈ \mathbb R \setminus \{0\} , f(x) = 1/x^2\}$

(b) $f = \{((x_1, x_2),\, x_1−2 x_2)\mid x_1 > 0, x_2 > 0\}$

I need to find image, preimage, domain and range of these two and I don't know where to start.

I'll do (a), and answer your four questions in order:
Image
The image of $f$ is $\{f(x) \mid x \in \mathbb{R}\setminus\{0\}\}$. Note that for any $y > 0$, we have $y = f\left(\frac{1}{\sqrt{y}}\right)$, so $y$ is in the image of $f$. Notice also that for every $x \in \mathbb{R}\setminus\{0\}$, we have $f(x) = \frac{1}{x^2} > 0$, so there's nothing else in the image. Thus, the image of $f$ is exactly $\{y \in \mathbb{R} \mid y > 0\}$.
Preimage
This one's rather the odd one out: one doesn't have "a preimage of $f$", you have, for each subset $A \subseteq \mathop{\mathrm{Range}}(f)$, a preimage of $A$ under $f$, which is defined as $f^{-1}(A) = \{x \in \mathop{\mathrm{Domain}}(f) \mid f(x) \in A\}$. For your function, and any $A \subseteq \mathop{\mathrm{Range}}(f)$, we have $f^{-1}(A) = \{x \in \mathbb{R}\setminus\{0\} \mid \frac{1}{x^2} \in A\} = \{\frac{1}{\sqrt{a}},\frac{-1}{\sqrt{a}} \mid a \in A\}$.
Domain
This one's easy: it's explicitly given in the question. It's $\mathbb{R}\setminus\{0\}$. [Note that if it isn't given in the question, it's generally impossible to determine. This is the case, in particular, for (b). If you want to know why, ask away, and I'll have a rant].
Range
Unless you're using some variant definitions that I'm not familiar with, "range" and "image" are synonyms.


$\sum_{n=1}^{\infty} {2^n}{\sin(\frac{1}{3^nx})}$ is uniformly continuous on the Interval $(1,\infty)$? Is $\sum_{n=1}^{\infty} {2^n}{\sin(\frac{1}{3^nx})}$ uniformly continuous on the Interval $(1,\infty)$ ?

I have tried:

$$M_{n} = \max|{2^n}{\sin(\frac{1}{3^nx})}| < \max|{2^n}{\frac{1}{3^nx}}|  = \max (\frac{2}{3})^n \frac{1}{x}  < (\frac{2}{3})^n$$

So this should be uniformly convergent.

I don't know the correct answer.
Have I missed something here? This should be uniformly convergent correct?

You have proved that the series is uniformly convergent. This proves that the sum is continous. To prove that it is uniformly continuous on $[0,\infty)$ observe that $|\sum\limits_{n=1}^{\infty} 2^{n} \sin (\frac 1 {3^{n}x})| <\sum\limits_{n=1}^{\infty} 2^{n}  \frac 1 {3^{n}x} <\epsilon$ if  $x >\frac 2 {\epsilon}$. If a function is continuous on $[0,\infty)$ and vanishes at $\infty$ then it is uniformly continuous. To prove this let $f$ be the sum of the series and  $\epsilon >0$ and choose $M$ such that $|f(x)| <\epsilon /2$ for $x >M$.(Note that the series actually converges uniformly on $[1,\infty)$ there is no problem near $1$).  Since $f$ is continuous on the compact interval $[1 ,M+1]$ there exits $\delta \in (0,1)$ such that  $x,y \in [1  ,M+1],|x-y| <\delta$ implies $|f(x)-f(y)| <\epsilon$. You can now verify that $x,y \in (1  ,\infty],|x-y| <\delta$ implies $|f(x)-f(y)| <\epsilon$.

Yes, we have $|{2^n}{\sin(\frac{1}{3^nx})}| \le (\frac{2}{3})^n$ for all $x > 1$ and all $n$. Since $ \sum (\frac{2}{3})^n$ is convergent, the Weierstraß test shows that the series is uniformly convergent in $(1, \infty).$


How prove this matrix have a positive eigenvalue let
$$A=\begin{bmatrix}
a_{1}&a_{2}&a_{3}\\
a_{4}&a_{5}&a_{6}\\
a_{7}&a_{8}&a_{9}
\end{bmatrix}$$
where $a_{i}>0$,

show that the matrix $A$ At least one
  positive eigenvalue。

I think this problem is very interesting,and I can't prove it,Thank you

It is said this problem can be proved by Brouwer Fixed Point Theorem： http://www.math.hmc.edu/funfacts/ffiles/20002.7.shtml
and it's very nice, who can see it? Thank you.

and I think this problem can be solved using linear-algebra methods.Thank you

Everything you need is here: http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem

In a nutshell, the Perron-Frobenius theorem states that such a positive eigenvalue exists, and more specifically that there is only one eigenvalue with maximal modulus, which is real>0.

The trace of the matrix Tr$({\bf A})$ is the sum of the eigenvalues, and also the sum of the diagonals under any transformation. Since the sum of the diagonals is greater than zero, at least one eigenvalue must be greater than zero.


Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$ where the tangent plane is parallel to the plane $3x - y + 3z = 1$. Find the points on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$, where the tangent plane is parallel to the plane $3x - y + 3z = 1$.

I'm not sure how to go about solving this. I'd appreciate some help.

Let the contact point be $(X,Y,Z)$, now the tangent plane is

  $$Xx+2Yy+3Zz=1$$

Comparing coefficients,
$$X:2Y:3Z=3:-1:3$$

That is $$\frac{X}{3}=\frac{2Y}{-1}=\frac{3Z}{3}=k$$

Now substitute $(X,Y,Z)=\left( 3k,-\dfrac{k}{2},k \right)$ into $x^{2}+2y^{2}+3z^{2}=1$

We have
\begin{align*}
  (3k)^{2}+2\left( -\frac{k}{2} \right)^{2}+3(k)^{2} &= 1 \\
  \frac{25}{2}k^{2} &= 1 \\
  k &= \pm \frac{\sqrt{2}}{5} \\
  (X,Y,Z) &=
  \left(
    \pm \frac{3\sqrt{2}}{5},
    \mp \frac{1}{5\sqrt{2}},
    \pm \frac{\sqrt{2}}{5} \right)
\end{align*}

Consider the function $f(x,y,z) = x^2+2y^2+3z^2-1$ then the $E$=ellipsoid is given by $f^{-1}(0)$ i.e a level surface. Let $p \in E$ then $\nabla f(p)$ is a normal vector for the tangent plane. For the tangent plane to be parallel to $3x-y+3z=1$, you need $p \in E$ such that $\nabla f(p) = \lambda \langle 3,-1,3\rangle$.


Number of equations needed to define a rectangle? I've been wondering whether there is some sort of fundamental rule (such as n equations are needed to solve for n variables) underlying the seeming need for the definitions of different shapes.

For example, to define a right angled triangle given points in a plane, we need only the right angle - its characteristic - hence we can write an equation using Pythagoras' theorem which defines it:

  $$AB^2 + AC^2 = BC^2$$

However, what about the definition of a rectangle? What fundamental properties does a rectangle possess? I tried the idea of right angles for each of the vertices, but it seems that you can also define a square in just 3 equations. For example:

  $$AC=BD$$
  $$AB=CD$$
  $$AB^2 + AC^2 = BC^2$$

As well as:

  $$AD=BC$$
  $$AD^2 + BA^2 = BD^2$$
  $$CD^2 + CB^2 = BD^2$$

What fundamental truth requires three properties to define a rectangle in terms of points in a plane? Is it related to solving a problem in n variables? Can this rule, if any, be extended to shapes in general in arbitrary dimensions?

While not exactly what the OP wants, I would suggest the coordinate approach. Then the "triangle equation" becomes a circle equation in the form $x^2+y^2=R^2$ which completely defines a circle centered at the point $(0,0)$.

Because of the corners, there's no such nice equation for a square. But we can make some "square-like" shapes in a similar way using the equation:

$$x^{2n}+y^{2n}=R^{2n}, \qquad n>1$$

Here's an illustration for a few $n$, starting with a circle:

"But none of these are squares!" - anyone would say, and be right.

To define a square we can't just use a single equation. We'd have to deal with the absolute value function.

For example, here's a square with the diagonal $D$, rotated by $\pi/4$:

$$|x|+|y|=D$$

We can easily rotate this one right back, and expand it so it fits with the others:

  $$|x+y|+|x-y|=a$$

Where $a$ is now the length of the side.

For a rectangle (the edited question) we just need to scale one of the coordinates. For a general rectangle with the sides $a$ and $b$ we have:

  $$\left|\frac{x}{a}+\frac{y}{b} \right|+\left|\frac{x}{a}-\frac{y}{b} \right|=1$$

To get rid of the absolute value we can try squaring twice:

$$\left(\frac{x}{a}+\frac{y}{b} \right)^2+\left(\frac{x}{a}-\frac{y}{b} \right)^2+2\left|\frac{x^2}{a^2}-\frac{y^2}{b^2} \right|=1$$

$$\left(\left(\frac{x}{a}+\frac{y}{b} \right)^2+\left(\frac{x}{a}-\frac{y}{b} \right)^2-1\right)^2=4\left(\frac{x^2}{a^2}-\frac{y^2}{b^2} \right)^2$$

Unfortunately, as it often happens with squaring, we get extra solutions (the dashed lines), which don't lie on the original rectangle:

The number of equations needed to define a rectangle must equal the number of unknowns needed to define a rectangle. Three points can uniquely define a rectangle, and each point consists of two coordinates, so there are six unknowns, and therefore six equations.


Number theory primes and congruences: $p^{32} -1$ is divisible by $16320$ I had problem in the following problem any help or hint will highly be appreciated.

For $p>17$, $p^{32} -1$ is divisible by $16320$

Hint: we have $16320=2^6\cdot 3\cdot 5\cdot 17$, and already $p^{32}\equiv 1 \bmod k$ for each $k\in \{2^6,3,5,17\}$. For example, $p^2\equiv 1\bmod 3$, because $\gcd(p,3)=1$.

NOTE.  It would be absurd that this property is valid for all prime number greater than $17$; indeed $29^{32}-1=(29^{16}-1)(29^{16}+1)$. Calculating we have
$29^{16}=250246473680347348787 521$ and it is verify that neither $29^{16}-1$ nor $29^{16}+1$ is divisible by $17$ hence not divisible by a multiple of $17$ thus not divisible by 16320.

NOTE (bis).The first note was a mistake induced by both, certain idea about the anarchy of primes and the calculator. It is so clear that the property of divisibility by $17$ is true……. The "hidden" $16 = 17-1$, which could be applied to any prime p with the exponent appropriate  $2(p-1)$..... But this usually happens. See for example my answer to "Can a pre-calculus student prove this?" and compare it with the other ones. Here also the answer is immediate but qualified people did not see.

Therefore the question must be understand as looking for a prime satisfying the enunciated property. I find below the prime $p=16319$

The number $p=16320-1=16319$ is prime and $p^{32}=16320M+1$ for some $M$. Hence you have $p^{32}-1$ is a multiple of $16320$.

Note anyway that for all even k you have the same result.


How to import a sequence of files into a cell in matlab? I'm working on class assignment. I've to construct a hidden markov model for a give sequence of discrete data. The data is a 2-dimensional vector separated by comma in which each line represents x,y coordinates. I've been given 500 files each for 3 classes.

The name file looks like this "class.txt" so if the file belongs to class1 and it is 2nd example and has a sequence length of 42 then the file name look like this "class1_2_42.txt"

My question how to import data from those files into a cell or matrix of size (500X2).

I tried following this code from matlab help

For example, to read files named file1.txt through file20.txt with importdata:

numfiles = 20;`
mydata = cell(1, numfiles);

for k = 1:numfiles
  myfilename = sprintf('file%d.txt', k);
  mydata{k} = importdata(myfilename);
end

But I can't work it out because of the name of the given files. Please help

For reference I added the images of how my files and data look like

The code is self explicable.

res = dir('*.txt'); %to read all files with *.txt
nfiles = length(res);
mydata = cell(1, nfiles);
for k = 1:nfiles
   mydata{k} = importdata(res(k).name);
end


$\sqrt{x}$$\sqrt{x}$ = $x$ but $\sqrt{x^2}$ = $|x|$. Why? $\sqrt{x}$$\sqrt{x}$ = $x$ but $\sqrt{x^2}$ = $|x|$. Why is this?

I'm just learning algebra again after many years and I can't seem to figure out why this is. I'm sure this is trivial but if someone could explain it it would help me a lot. Thanks!

By definition $\sqrt x$ is the unique non-negative real number $y$ such that $y^2=x$, so $\sqrt x\cdot\sqrt x=x$ is true by definition.

Now apply the definition to $\sqrt{x^2}$: $\sqrt{x^2}$ is the unique non-negative real number $y$ such that $y^2=x^2$. If $x=0$, the only real number whose square is $x^2$ is $0$, so of course $\sqrt{x^2}=0=|0|$. If $x\ne 0$, there are always two real numbers $y$ such that $y^2=x^2$: one of them is $x$, and the other is $-x$. Exactly one of these two is positive. Since we don’t know whether $x$ is positive or not, we don’t know which of them is positive, but we know that whichever it is, it’s $|x|$. Therefore $|x|$ is the unique positive real number such that $|x|^2=x^2$, and by definition $\sqrt{x^2}=|x|$.

As an example, suppose that $x=-3$. Then $x^2=9$, and the two real numbers whose squares are $9$ are $-3$ (i.e., $x$) and $3$ (i.e., $-x$). The non-negative one is $3=-(-3)=|-3|$. Had we started with $x=3$, $x^2$ would still have been $9$, and we’d still have wanted the positive one of $x$ and $-x$, but this time that would be $x$, not $-x$. It’s still true, however, that $|x|=|3|=3$, the one that we want.

I think your source of confusion here comes from thinking of $\sqrt{x}$ as a number rather than a function. Often times, we write something as $\sqrt{x}$ to denote the squareroot(s) of the number $x$. However, in many other situations, we also write $\sqrt{x}$ to talk about the function which sends the number $x$ to its square root.

In particular, in your example, the first instance $\sqrt{x}\sqrt{x} = x$, is most likely talking about the number $\sqrt{x}$. Indeed, by definition of the number $\sqrt{x}$, we have that equation.

In the second instance, $\sqrt{x^2} = |x|$, it is more appropriate to think about $\sqrt{x}$ as a function, which is composed with the function $x^2$. In that case, $\sqrt{x}$ must be strictly positive, since there are no real numbers whose square roots are negative!


Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?

few thoughts on the first one:

$$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

This is just a Riemann sum, and thus its limit is $$\int_{0}^1 \ln(\sin(x)) dx$$

This is an improper integral though, so the RS approach might not be best, but I think it is a convergent improper integral, since $\int_0^1 \ln(x)dx $ is convergent..

Maybe someone can take over....

As $x\rightarrow 0$ , $\sin(x) \rightarrow x$

So $ \sin(\frac{1}{n}) \rightarrow \frac{1}{n}$ as $n \rightarrow \infty$

The product $ [ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] \rightarrow \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot...\frac{n}{n} =
\frac{n!}{n^n}$

As $n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n  $

So $\frac{n!}{n^n} = \sqrt{2\pi n} \cdot (\frac{1}{e})^n $

$\lim_{ n \rightarrow \infty}$ of $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] ^ \frac{1}{n} \rightarrow \frac{1}{e}$

Since $\sin(\frac{m}{n})$ is always positive  if $0< m \le n$ and less than $\frac{m}{n}$ .
 The limit will be less than (1/e)


Assume that $\alpha, \beta, \gamma \in [0,\pi/2]$, and $\sin\alpha+\sin\gamma=\sin\beta$, $\cos\beta+\cos\gamma=\cos\alpha$. Find $\alpha-\beta$. Assume that $\{\alpha, \beta, \gamma\} \subset \left[0,\frac{\pi}{2}\right]$, $\sin\alpha+\sin\gamma=\sin\beta$ and $\cos\beta+\cos\gamma=\cos\alpha$.
Try to find a value of $\alpha-\beta$.

Actually I have gotten that $\alpha+2\gamma+\beta=\pi$ and $\sin2\alpha+\sin2\beta=\sin(\alpha+\beta)$, $\cos2\alpha+\cos2\beta=\cos(\alpha+\beta)$
But I can't get further more.

we have $$\sin(\gamma)=\sin(\beta)-\sin(\alpha)$$ and $$\cos(\gamma)=\cos(\alpha)-\cos(\beta)$$
squaring and adding both we get
$$-1=-2(\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta))$$ therefore we obtain
$$\cos(\alpha-\beta)=\frac{1}{2}$$

If you have $$\sin2\alpha+\sin2\beta=\sin(\alpha+\beta)   \to 2\sin(\alpha+\beta)\sin(\alpha-\beta)=\sin(\alpha+\beta)$$and $$\cos2\alpha+\cos2\beta=\cos(\alpha+\beta)\to \cos(\alpha+\beta)\cos(\alpha-\beta)=\cos(\alpha+\beta) $$divide them
$$\frac{2\sin(\alpha+\beta)\sin(\alpha-\beta)}{2\cos(\alpha+\beta)\cos(\alpha-\beta)}=\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\\to \tan(\alpha-\beta)=1 \to \alpha-\beta=\frac{\pi}{4}$$


Determinant of $2 \times 2$ block matrix whose diagonal blocks are zero $\Bbb A$ is an $n\times n$ matrix, and $\Bbb B$ is an $m × m$ matrix.$\space$What is the determinant of matrix $\Bbb C$?
\begin{equation*}
\mathbb{C}=
\begin{pmatrix}
\begin{array}{@{}c|c@{}}
  \begin{matrix}
  0
  \end{matrix}
  & \mathbb{A} \\
\hline
  \mathbb{B} &
  \begin{matrix}
 0
  \end{matrix}
\end{array}
\end{pmatrix}
\end{equation*}
I thought it could be just
($\det\Bbb A\cdot\det\Bbb B$)
or ($-\det\Bbb A\cdot\det\Bbb B$),
but I'm not sure, as it seems too easy.

This determinant can be computed by the Laplace expansion theorem (the generalized form). Fix the first $1, 2, \ldots, n$ rows, and let columns range over $(j_1, j_2, \ldots, j_n) \in \{1, 2, \ldots, n + m\}$. Since the square minor is non-zero only if $(j_1, j_2, \ldots, j_n) = (m + 1, m + 2, \ldots, m + n)$,  it follows that
\begin{align*}
\det C = & C\begin{pmatrix}1 & 2 & \cdots & n \\
m + 1 & m + 2 & \cdots & m + n\end{pmatrix}(-1)^{1 + \cdots + n + m + 1 + \cdots + m + n}C\begin{pmatrix}n + 1 & n + 2 & \cdots & n + m \\
1 & 2 & \cdots & m \end{pmatrix} \\
= & (-1)^{n(n + 1) + mn}\det(A)\det(B).
\end{align*}

As Robert Israel noted,
$${\rm C} = \underbrace{\begin{bmatrix} {\rm O} & {\rm I}_n \\ {\rm I}_m & {\rm O} \end{bmatrix}}_{=: \rm P} \begin{bmatrix} {\rm B} & {\rm O} \\ {\rm O} & {\rm A} \end{bmatrix}$$
Let $\sigma := \det ({\rm P})$. Hence,
$$ \det({\rm C}) = \det\begin{bmatrix} {\rm O} & {\rm I}_n \\ {\rm I}_m & {\rm O} \end{bmatrix} \cdot \det\begin{bmatrix} {\rm B} & {\rm O} \\ {\rm O} & {\rm A} \end{bmatrix} = \sigma \cdot \det({\rm A}) \cdot \det({\rm B})$$
Note that $\rm P$ is an $(m+n) \times (m+n)$ permutation matrix. It is also circulant. Hence, its $k$-th eigenvalue is
$$\lambda_k = \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) k \right)$$
where $k \in \{0,1,\dots,m+n-1\}$. Since the determinant is the product of the eigenvalues,
$$\sigma = \prod_{k=0}^{m+n-1} \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) k \right) = \exp \left( -i 2\pi \left( \frac{m}{m+n} \right) \sum_{k=0}^{m+n-1} k \right)$$
Since $\displaystyle\sum_{k=0}^{m+n-1} k = \frac{(m+n)(m+n-1)}{2}$,
$$\sigma = \exp \left( -i \pi m (m+n-1) \right) = (-1)^{m (m+n-1)} = \begin{cases} +1 & \text{ if } m (m+n-1) \text{ is even}\\ -1 & \text{ if } m (m+n-1) \text{ is odd}\end{cases}$$
Note that $m (m+n-1)$ is odd only when both $m$ and $n$ are odd. Thus,
$$\sigma = \begin{cases} -1 & \text{ if } m \text{ and } n \text{ are odd}\\ +1 & \text{ otherwise}\end{cases}$$

Robert M. Gray, Toeplitz and Circulant Matrices: a Review.

linear-algebra matrices block-matrices circulant-matrices permutation-matrices determinant eigenvalues-eigenvectors permutations


What is the reasoning behind the definition of relative entropy? I do not understand the notion of relative entropy.

Relative Entropy. $D_{KL}(P||Q) = \sum_{i}^{}P(i)\log \frac{P(i)}{Q(i)}$.

I try to get some intuition why it looks the way it looks. I see that it works: if
I take $Q=P$ then $D_{KL}(P||Q)=0$, so the distance between identical distributions is 0.

I tried to find some intuition in wikipedia: KL divergence can be interpreted as the expected extra message-length per datum that must be communicated if a code that is optimal for a given (wrong) distribution Q is used, compared to using a code based on the true distribution P.

Very confusion description, and has no clue why it's actually $P(i)\frac{P(i)}{Q(i)}$.

I would appreciate if someone could give the reasoning about the definition of relative entropy.

In information theory, relative entropy $D(P\|Q)$ is the number of extra bits required per letter on average to encode a source with a distribution $Q=(q_1,\dots,q_n)$ when the true underlying distribution is $P=(p_1,\dots,p_n)$. $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ---- ~~~~(*)$

The optimal expected compressed length subject to unique decodability (which is equivalent to the Kraft inequality $\sum 2^{-l_i}\le 1$) is  given by $$l_i^*=\log \frac{1}{q_i}.$$

Hence the optimal expected code length (per letter) when coding with $Q$ is
\begin{eqnarray}
\sum p_i\log \frac{1}{q_i}
&=&\sum p_i\log \frac{1}{q_i}\\
&=&\sum p_i\log \frac{p_i}{p_i\cdot q_i}\\
&=& \sum p_i\log \frac{1}{p_i}+\sum p_i\log \frac{p_i}{q_i}\\
&=& H(P)+D(P\|Q).
\end{eqnarray}
Hence the statement $(*)$. It has got other interpretations in probability theory and statistics though.

As you can see in this video. Cross entropy: H(p,q) can be interpreted as average message length that is sent to receiver based on predicted q distribution. In information theory, to compress given massage in most effective way, we look at its probability distribution. So basically, entropy: H(p) of the given probability distribution is our ultimate limit. For example, if I was sending 4 bits of data on average to receiver about the wheater of Istanbul, and also if I know the entropy of my message words is actually 3.2, it means that I can actually can compress my message more.
Thus, the difference between average message lenght that was sent and actual entropy of the message is equal to KL Divergence.
Edit: The person who gave me a down vote, please can you explain why?


Dimension of irreducible affine variety is same as any open subset Let $X$ be an irreducible affine variety. Let $U \subset X$ be a nonempty open subset. Show that dim $U=$ dim $X$.

Since $U \subset X$, dim $U \leq$ dim $X$ is immediate. I also know that the result is not true if $X$ is any irreducible topological space, so somehow the properties of an affine variety have to come in. I have tried assuming $U=X$ \ $V(f_1,...,f_k)$ but I don't know how to continue on.

Any help is appreciated!

Let $X=\operatorname{Spec} A$, and pick some nonzero $f\in A$ with $\operatorname{Spec} A_f = D(f) \subset U$.  Then we can lift chains of closed sets to see that $\dim D(f) \leq \dim U
\leq \dim X$.

But as shown here, we have $\dim D(f) = \dim A_f = \dim A = \dim X$.  We can also see this quite quickly by noting that $A$ and $A_f$ have the same fraction fields, and therefore the same transcendence degree over $k$.

Note that this is false for general rings $A$, even if $A$ is assumed to be a noetherian domain.  In fact, any DVR is a counterexample.

You should note that any nonempty open subset of an irreducible variety (or topological space) is dense : https://en.wikipedia.org/wiki/Hyperconnected_space.
Then you can use the definition of the dimension to conclude.


partial derivatives and implicit differentiation My question is what is $F_x$.

When you find $y'$ from the implicit function $$F(x,y)= x^3+y^3-6xy=0$$

you can obtain $y'$ by $$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}$$

The text book says $F_x$ is $$3x^2-6y$$.
And $F_y$ is $$3y^2-6x$$ and obtain $$\frac{dy}{dx}=-\frac{3x^2-6y}{3y^2-6x}$$

Well, isn't $y$ a function of $x$?

Like F(x,y)=F(x,f(x)) and should it not go like $$ F_x=3x^2-6y-6x\frac{\partial y}{\partial x}$$ and not $$F_x=3x^2-6y$$?

although it would directly lead to the answer $\frac{dy}{dx}$.

What I am asking is why in implicit differentiation it is ignoring $y$ is a function of $x$ and treat $y$ as some constant?

It’s important to distinguish the function $F:\mathbb R^2\to\mathbb R$ defined as $F: (x,y)\mapsto x^2+y^2-6xy$ from the equation $F(x,y)=0$. For this function, $x$ and $y$ are independent variables—they can take any values whatsoever—so it makes perfect sense to compute its partial derivatives with respect to each variable by treating the other variable as a constant. The graph of this function is some surface in $\mathbb R^3$ and at points at which it’s differentiable (everywhere for this function), this surface has a tangent plane with normal vector $(F_x,F_y,-1)$.

The equation $F(x,y)=0$, describes a level curve of the function $F$, specifically the intersection of its graph with the $x$-$y$ plane. The tangent line to this curve at a point is just the intersection of the tangent plane to the graph of $F$ with the $x$-$y$ plane. The projection of the tangent plane’s normal onto the $x$-$y$ plane, namely $(F_x,F_y)=\nabla F$ is orthogonal to this tangent line. The slope of this vector is $m_1 = F_y/F_x$, and using the relation $m_1m_2=-1$ for the slopes of perpendicular lines, we see that the slope of the tangent line is $m_2=-F_x/F_y$.

To put it another way, when you compute $dy/dx$ via partial derivatives of $F$, you are viewing the curve implicitly defined by the equation $F(x,y)=0$ as a horizontal slice through the 3-D graph of $F$ and then finding a tangent line to this curve by taking the same slice through the tangent plane to the graph of $F$ at that point. By adding an extra dimension and stepping out of the $x$-$y$ plane in this way, you can temporarily ignore the fact that $x$ and $y$ are not independent for points on the level curve. This dependency doesn’t disappear completely, though. It still plays a role in computing the coordinates of a point on the level curve, which you need in order to compute the partial derivatives of $F$ there.

An important thing to note about the above approach is that nowhere does it require that there exists some function $f:\mathbb R\to\mathbb R$ such that $F(x,y)=0$ implies that $f(x)=y$. Indeed, there may not even be such a single function. E.g., $x^2+y^2=1$ can’t be the graph of any function since it fails the vertical line test. (It might still be possible to find such a function in the neighborhood of a point, though—that’s what the implicit function theorem is about.) When you perform implicit differentiation, you start off by assuming that there is such a function and then differentiate both sides of the equation $F(x,y)=0$ taking that into account.

These two approaches can be reconciled via the chain rule. Suppose that $y=f(x)$. To help keep things straight, we introduce the function $\phi:\mathbb R\to\mathbb R^2$ with $\phi:x\mapsto(x,f(x))$, and also $g=F\circ\phi$. Then $$g' = \nabla F\cdot\phi' = (F_x,F_y)\cdot(1,f') = F_x+F_yf'.$$ This expression takes into account the assumed dependence of $y$ on $x$, but it nevertheless involves partial derivatives of $F$. Since we have $F(x,y)=g(x)$, implicit differentiation is equivalent to differentiating both sides of the equation $g(x)=0$ and then solving for $f'$, but this gives $F_x+F_yf'=0$, from which $f'=-F_x/F_y$ as before.

If we assume that $y=y(x)$ is given, then we get
$$3x^2+3y^2y'-6y-6xy'=0$$


If $\gcd(m,n)=1$ and $q|mn$, then $\exists d,e$ such that $q=de$, $d|m$, $e|n$, $\gcd(d,e)=1$ and $\gcd(\frac{m}{d},\frac{n}{e})=1$. Let $m,n\in\mathbb{N}$ with $\gcd(m,n)=1$ and $q\in\mathbb{N}$ such that $q|mn$. Then there exists $d,e\in\mathbb{N}$ such that $q=de$, $d|m$, $e|n$, $\gcd(d,e)=1$ and $\gcd(\frac{m}{d},\frac{n}{e})=1$.

I am trying to prove the above claim but can only go as far as this:

As $q\in\mathbb{N}$, then there exists $d,e\in\mathbb{N}$ such that $q=de$. As $q|mn$, then there exists $x\in\mathbb{N}$ such that $mn=qx$. But $q=de$, so $mn=x(de)$. Since $mn=d(ex)=e(dx)$, then $d|mn$ and $e|mn$.

I even don't know if I'm going about it the right way. Some help would be greatly appreciated thank you.

Let $d=\gcd(q,m),\, e=\gcd(q,n)$. Because $\gcd(m,n)=1$, we get $de=\gcd(q,mn)$; and because $q\mid mn$, we get $\gcd(q,mn)=q$. Therefore $de=q$. Also $d\mid m,\, e\mid n,\, \gcd(d,e)=1$, as wanted.

$\gcd\left(\frac{m}{d},\frac{n}{e}\right)=1$ is unnecessary to consider: $\gcd(m,n)=1$ already implies it.

Hint:
Let
\begin{align*}
d &= \gcd(m,q) \\
e &= \gcd(n,q).
\end{align*}
Then:

To prove $de = q$, first show that $\gcd(m,q) \cdot \gcd(n,q) = \gcd(mn,q)$, since $m,n$ relatively prime. Then argue that $\gcd(mn,q) = q$.
$d | m$ and $e | n$ are immediate.
To prove $\gcd(d,e) = 1$, and $\gcd(\frac{m}{e}, \frac{n}{d}) = 1$, you can prove more generally that if $a$ divides $A$ and $b$ divides $B$ then $\gcd(a,b)$ divides $\gcd(A,B)$. Then take $A = m$ and $B = n$.


Show that $f(x) = x^p -x -1 \in \Bbb{F}_p[x]$ is irreducible over $\Bbb{F}_p$ for every $p$. Let $p$ be a prime.
a) Show that $f$ has no roots in $\Bbb{F}_p$.

Let $F^*$ be the multiplicative group of $\Bbb{F}_p$. Then, by lagrange's thoerem for all nonzero $\alpha \in \Bbb{F}_p$, $\alpha^{p-1} = 1 \implies \alpha^p=\alpha \implies \alpha^p-\alpha=0$. Of course $0^p=0$, so this is true for all elements of $F$ and not just the nonzero ones. But then $\alpha^p - \alpha - 1 = -1$ for all $\alpha \in \Bbb{F}_p$ and so it must have no roots in $\Bbb{F}_p$. I could have also done this using the Frobenius automorphism, right?

b) Let $\alpha$ be a root of $f$ (in some algebraic closure of $\Bbb{F}_p$). Show that $\alpha + s$ is also a root for all $s \in \Bbb{F}_p$.

Let $\alpha^p - \alpha -1 =0$. Let $E$ be an algebraic closure of $\Bbb{F}_p$. Since $E$ has characteristic $p$, $(\alpha + s)^p = \alpha^p + s^p$. So we have,
$$(\alpha + s)^p - (\alpha + s) -1 = \alpha^p + s^p - \alpha -s -1 = s^p - s = 0.$$

c) Conclude that $f$ is irreducible over $\Bbb{F}_p$, for every $p$.

By b) and the fact that $\Bbb{F}_p$ has $p$ distinct elements, we know that the roots of $f$ are $\alpha, \alpha+1, ... , \alpha + p-1$. So if $K$ is a splitting field, we have $$x^p - x -1 = (x-\alpha)(x-(\alpha+1))...(x-(\alpha+p-1)).$$
Now let's assume that $f$ is reducible over $\Bbb{F}_p$. Then $f=gh$ for some $g$ and $h$ with degrees less than that of $f$. So $g$ and $h$ must be of the form $(x-(\alpha+s_1))...(x-(\alpha+s_k))$ where k is less than n. Let's say that g has degree 2, because the other cases are similar.
So $g =(x-(\alpha + s_i))(x-(\alpha + s_j))$ and the constant term for $g$ is,
$$(\alpha + s_i)(\alpha + s_j) = \alpha^2 + s_is_j\alpha + s_is_j.$$
Since $\Bbb{F}_p$ is a field, if $\alpha s_is_j$ is an element of $\Bbb{F}_p$ , then so is $((\frac{1}{s_is_j})(\alpha s_is_j) = \alpha$, a contradiction.
We can show by induction that if we multiply $(x-(\alpha+s_1))...(x-(\alpha+s_k)$, we get a term that looks like $s_1s_2...s_k\alpha$. So this is also true for $k>2$.
Do you think that my answer is correct?
Thank you in advance

I like your idea very much, but I didn't understand how you managed to ignore the $\alpha^2$ term from your constant term.

I have written up an answer to this question with a similar idea, where I look at the next to highest degree term of $g(x)$ (=the term of degree $k-1$). In your degree two example, this would be the linear term. Its coefficient is
$2\alpha+(s_i+s_j).$ As $s_i,s_j$ are in the prime field, and $2$ is invertible, we can, as in your argument, conclude that $\alpha\in\mathbb{F}_p$, which is a contradiction.

By studying the lowest degree term, you get a lot of clutter from powers of $\alpha$. The degree $k-1$ term is IMHO easier to manage.

Another approach:

Suppose $x^p-x -1 = gh$ for $g, h \in \mathbb F_p[x]$ with degrees $\geq 1$. Then taking a (formal) derivative, we have $$-1 = g'h + gh'$$ since $px^{p-1} = 0$ in $\mathbb F_p[x]$. Note that $g'$ and $h'$ cannot both be $0$, otherwise, $-1 = 0$. This implies the degree of the right hand side is at least $1$ (verify!). But the degree of $-1$ is $0$. Contradiction.


Function definition notation When I've a function $\text{f}$ and it can be depended on $x$ (where $x$ will be a real number) or other real number(s) (constant(s)), can I define the function as follows?:

$$\exists\space\text{f}\left(x\right)\mapsto\left\{\text{f}\left(x\right)\in\mathbb{R}:\forall\space x\right\}$$

Or is there a more common or good notation?

The "signature" of the function (a programming term, I believe, which might not be standard amongst mathematicians; it basically means "domain and codomain" in this context) is written $f: S \to \mathbb{R}$, whatever $S$ is. Note that this does not involve the input to the function, so (for instance) it's really quite weird to write something like $f: x \to \mathbb{R} \ (x \in \mathbb{R})$; in this instance, $x$ is the input to $f$ and hence shouldn't appear in the function's signature. Notice also that $\mapsto$ tells us what happens to a particular input to the function, while $\to$ tells us what the domain and codomain of the function are.

For example, we may write the factorial function in the following ways (by no means an exhaustive list!):

$f: \mathbb{N} \to \mathbb{R}$ given by $n \mapsto n!$
$f: \mathbb{N} \to \mathbb{R}$ given by $x \mapsto x!$
$f: \{ 1, 2, 3, \dots \} \to \{ x: x \in \mathbb{R} \}$ given by sending $i$ to $i \times (i-1) \times \dots \times 2 \times 1$.
$G: \mathbb{N} \to \mathbb{R}$ given by $r \mapsto r!$
$n \mapsto n!$ (this expression, without specifying the "signature" of the function, will suffice if the domain and codomain are clear from context)

You could write $f:X\times Y \to \mathbb{R}$ where $$f(x,y)=\begin{cases} g(x) : y=0 \\ c : y\neq 0, \end{cases}$$ where $y\in \{0,1\}$ is a variable that determines whether $x$ is relevant, and $g:X\to \mathbb{R}$. We don't have functions that have optional arguments in math.


Patterns in pi in "Contact" In Carl Sagan's novel Contact, the main character (Ellie Arroway) is told by an alien that certain megastructures in the universe were created by an unknown advanced intelligence that left messages embedded inside transcendental numbers. To check this, Arroway writes a program that computes the digits of $\pi$ in several bases, and eventually finds that the base 11 representation of $\pi$ contains a sequence of ones and zeros that, when properly aligned on a page, produce a circular pattern. She takes this as an indication that there is a higher intelligence that imbues meaning in the universe and yaddayaddayadda.

I always thought that Sagan was pulling a fast one on us. Given that transcendental numbers (or, for that matter, irrational numbers) are infinite, non-repeating sequences of digits, they contain any possible sequence of numbers, including the one Arroway found. It's hard for me to infer anything philosophical/spiritual from the fact that you can find this sequence if you look hard enough (to make a somewhat facile comparison, if I look hard enough in my sock drawer, I will find both socks of any given pair, but you can't take this as evidence for a higher intelligence in the universe). But then, Sagan did know one or two things about math, so maybe I'm missing something here. Are there any circumstances in which finding a particular sequence in a certain position of $\pi$ would make mathematicians go "wow!"?

Given that transcendental numbers (or, for that matter, irrational numbers) are infinite, non-repeating sequences of digits, they contain any possible sequence of numbers, including the one Arroway found.

This is not necessarily true. A number whose digits contain, with equal frequency, all possible sequences of the same length, is a normal number. There are trancendental numbers that are known to be normal, and others that are known to be not normal, generally because they have been specifically constructed as such. However, whether pi is normal has been long suspected, and appears to be the case when you look at the digits already computed, but it's never been proven one way or the other.

So, in that sense, it is indeed possible for there to be some kind of message encoded in pi that we would recognise as being something other than pure chance. It's also one of the best places to put such a message, since pi is (believed to be) a universal constant that does not depend on local physics or units of measurement. However, there is a risk that other sentient races would use a different circle constant (such as some movements on Earth that claim that tau, $\tau=2\pi$, is a more natural choice) and any race that goes down that route would never see such a message (and if the message is actually hidden in tau, we won't see it either).

Actually, the aliens (call them $A$), were in the process of making $\pi$ rational starting at the Feymann point,

$$\pi = 3.1415926\dots9999998\dots$$

but they were rudely interrupted by the approach of the Death Star.

Seriously, regarding the whole question of a message being inserted in the digits of $\pi$ by an alien intelligence, this is covered by the post,

"Could pi have a different value in a different universe?"

as the question is essentially,

"Can pi have a different value at different times (pre-$A$ and post-$A$) in the same universe?

The excellent answer to the first also applies to the second.


A cylinder is circumscribed about a sphere. If their volumes are denoted by $C$ and $S$, find $C$ as a function of $S$ Here is the problem.

A cylinder is circumscribed about a sphere. If their volumes are denoted by $C$ and $S$, find $C$ as a function of $S$

My (Amended) Attempt:
[Based on the correction suggested by herbSteinberg]

Let $r$ be the radius of the sphere.

Then the height of the cylinder is $2r$, and the radius of the base  is $r$. So the volume $C$ of the cylinder is given by
$$
C = \pi r^2 (2r) = 2 \pi r^3. \tag{1}
$$

And, the volume $S$ of the sphere is given by
$$
S = \frac43 \pi r^3. \tag{2}
$$

From (2), we obtain
$$
r^3 = \frac{3}{4 \pi} S = \frac{ 3S }{4 \pi},
$$
and hence
$$
r = \sqrt[3]{ \frac{3S}{4 \pi} }. \tag{3}
$$

Finally, putting the value of $r$ from (3) into (1), we get
$$
C = 2 \pi \left( \sqrt[3]{ \frac{3S}{4 \pi} } \right)^3 = 2 \pi \left( \frac{3S}{4 \pi} \right) = \frac32 S.
$$

Is my solution correct in each and every detail? Or, are there any errors of approach or answer?

As noted in the comments, the radius of the cylinder is just $r$. Otherwise your work appears to be correct, but you've made things harder on yourself than necessary. Note that $r^3$ appears in both formulas, so once you have solved for $r^3 = \frac{3S}{4\pi}$, you can immediately use this expression in place of $r^3$ in the formula for $C$, going straight to the last equality.
In other words:
$$S = \frac{4}{3}\pi r^3 \Rightarrow r^3 = \frac{3S}{4\pi}$$
$$C = \pi r^2 \cdot (2r) = 2\pi r^3 = 2\pi\cdot\frac{3S}{4\pi}=\frac{3S}{2} $$

The radius of the sphere should be equal to the radius of the cylinder face. Though you are correct about the height of the cylinder being twice the radius of the sphere.


In a topological space $X, G$ is open iff $G\cap\overline{A}\subseteq \overline{G\cap A}$ In a topological space $X$, $G$ is open iff $G\cap\overline{A}\subseteq \overline{G\cap A}$, where
  $A$ is an arbitrary set in $X$.

To prove this I started from definition of an open set, I could not complete it.
I even tried to prove it in metric space using def I could not complete it

Take $A=X-G$. Then $G\cap A=\varnothing$, so $\overline{G\cap A}=\varnothing$. Therefore, you must have $G\cap\overline{A}=\varnothing$. This means that $\overline{X-G} = \overline{A}\subseteq X-G$, hence $\overline{X-G}=X-G$. That is, the complement of $G$ is closed.

Conversely, suppose that $G$ is open, and let $A$ be arbitrary. Let $x\in G\cap \overline{A}$. To show that $x\in \overline{G\cap A}$, we just need to show that any open set that contains $x$ will intersect $G\cap A$ nontrivially. Let $\mathcal{O}$ be an open set containing $x$. Then $\mathcal{O}\cap G$ is an open set containing $x$ (since $x\in G$), so $\mathcal{O}\cap G\cap A\neq\varnothing$ (since $x\in\overline{A}$). Thus, $\mathcal{O}\cap (G\cap A)\neq\varnothing$. This proves that $x\in\overline{G\cap A}$, giving the desired inclusion.

Assume that $G \cap \overline{A} \subset \overline{G\cap A} $ then
\begin{equation}\label{1}
\begin{aligned}
G \cap \overline{A} \subset \overline{G\cap A} & \Longleftrightarrow X \backslash (\overline{G\cap A}) \subset X \backslash ( G \cap \overline{A}) \Longleftrightarrow X \backslash int(G \cap A) \subset X \backslash ( G \cap \overline{A})
\end{aligned}
\end{equation}
Take $A = X$ then
\begin{equation}
X \backslash int(G) = X \backslash int(G \cap X) \subset X \backslash G \cap \overline{X} =  X \backslash G ,
\end{equation}
 i. e., $G \subset int(G)$ and therefore $G$ is open.


Finding coefficients of a quadratic with roots having certain intervals If the roots of the quadratic equation $$(4p−p^2  −5)x^2  −(2p−1)x+3p=0$$ lie on either side of unity, then the number of integral values of $p$ is?

Okay so I'm having a hard time what the question means by both sides of unity. Does it mean one root is on $ x>1 $ and other in $ x<1$? After that, how do I adjust the coefficent so that one root is greater tha one?

First notice that leading coefficient $a=-(p^2-4p+4+1) = -(p-2)^2-1$ is negative.
If the roots of quadratic function $$f(x)=(4p−p^2  −5)x^2  −(2p−1)x+3p$$ are $x_1$ and $x_2$ (say $x_1>x_2$) then we have $$x_2<1<x_1\implies 1-x_2>0 \;\;{\rm and}\;\; 1-x_1<0$$
So if we write write $f$ in factor form:$$f(x)=a(x-x_1)(x-x_2)$$
then we see that $$f(1) = a(1-x_1)(1-x_2)>0$$
So you need to solve $f(1)=-(p-4)(p-1)>0$ ...

If the leading coefficient is positive, then the condition $f(1)\lt 0$ suffices. If it is negative, then we want $f(1)\gt 0$. These conditions can be condensed into $$(4p-p^2-5) f(1)\lt 0 $$ Can you finish?


Does the word "spectrum" in linear algebra have different meanings? I'm reading several papers that refer to the spectrum as the set of all possible eigenvalues of a matrix, i.e., counting multiplicity, so that a list such as $\sigma = (\alpha_1, ... \alpha_n, 0, 0, 0,...,0)$ can be the spectrum of some matrix.

But I had learned in my linear algebra courses that the spectrum is the set of distinct eigenvalues.

Who is right?

Does the spectrum have different meanings?

This difference is not a triviality, for the paper that I am trying to work on - for example, adding more zeroes to the list changes the problem; adding more zeroes can increase the solvability of the problem.

So, I just want to make sure.

Thanks,

Spectrum is a subject in functional analysis rather than linear algebra.
Its concept in functional analysis is more extended than eigenvalue.
Eigenvalue is a special case of elements of spectrum. You can see functional analysis Rudin.


Special functions and diff eq's ...... They're are all these methods of dealing with linear second order diff eq's:

generating function;
recurrence relation;
Rodrigues differential form;
Schlafi integral form;
associated form;
second form;
shifted form;
series form;

you can use on differential equations with special names:

Airy;
Bessel;
Chebyshev;
Gauss hypergeometric;
Hermite;
Jacobi;
Laguerre;
Legendre

There are at least 100 ideas in these two little lists, and apparently one can start from anything in the first list and derive any other quantity in that list directly from it, using equations in the second list as examples - how in the world does one make sense of all this? Where should one start? What am I missing? I can't even begin with all this there's so much going on :(

Is there not some standard way to begin with something & derive everything from it in an obvious way, & a way to remember all the equations - or a good reasong why you shouldn't care about remembering their names? :(

I would recommend starting by learning how to solve sets of $n$ linear ordinary differential equations (and the "discrete time" analogue: difference equations) with constant coefficients of the form

$$\dot{x}(t)=Ax(t)+f(t),$$

where $x:\mathbb{R}\to\mathbb{R}^n$, $A\in\mathbb{R}^{n\times n}$ and $f:\mathbb{R}\to\mathbb{R}^n$. Then, I'd move on to the case with time-varying coefficients

$$\dot{x}(t)=A(t)x(t)+f(t),$$

where $A:\mathbb{R}\to\mathbb{R}^{n\times n}$. The former is covered in lectures $10-15$ of here while a good resource for the latter are chapter 1-3 of this book (however, it is not free -- maybe someone can suggest one that is?).

But be warned, there is no silver bullet, the solutions to the general cases will only take you so far--they depend integrals involving $f(t)$ and $A(t)$ and how you (or whether you can at all) evaluate these analytically very much depend on $f(t)$ and $A(t)$ specifically.

This said, one can always avoid the difficulties of trying to solve them analytically by obtaining instead approximations to the solutions (nowadays, very often, to an arbitrarily precision) by numerically integrating the ODEs on a computer. But I'm starting to go off on a tangent...

By the way, welcome to MSE.

EDIT: I should mention that you can always re-write a linear $n^{th}$ order differential equation as a set of $n$ first order differential equations as above (but not vice-versa). However, in my experience, it is not much easier to solve the more general set of $n$ differential equations than an $n^{th}$ differential equation and, I think, you'll get more insight doing the general case.

Also, not much changes from considering second order to $n^{th}$ order ODEs when it comes to solving them, so, similarly, I recommend you try directly the general case $n^{th}$ order case (in fact, I think it'll be more understandable). However, these are just my opinions, I'm sure other people think differently.


Prove $A$ is diagonalizable while $A^{2}=kA$ Let $A$ be an $n$-order real matrix, with:
  $$A^{2}=kA$$
  Prove that $A$ is diagonalizable, i.e., there exists an invertible matrix $P$ such that  $P^{−1}AP$ is a diagonal matrix.

My thoughts: I assume that $$A=P^{-1}\mathrm{diag}(a_{1},...,a_{n})P,$$ so $$A^{2}=P^{-1}\mathrm{diag}(a_{1}^{2},...,a_{n}^{2})P=kP^{-1}\mathrm{diag}(a_{1},...,a_{n})P$$ which leads to $$\mathrm{diag}(a_{1}^{2},...,a_{n}^{2})=\mathrm{diag}(ka_{1},...,ka_{n})$$
so we have a series of equations:$$a_{1}^2=ka_{1},...$$
OK, I stopped here and want some help because the conclusion form my assumption seems rather weird. Anyone have better idea with it?

This is false; take $k=0$ and $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Assuming $k\ne0$, reduce $A$ to Jordan form; let $\lambda E+N$ be a Jordan block; then
$\lambda^2 E+2\lambda N +N^2=kE+kN$, whence it follows that $N^2=0$, $(2\lambda-k)N=0$ and $\lambda^2=k\lambda$. If $N\ne0$, then $\lambda=k/2$, and hence $k^2/4=k^2/2\implies k=0$, a contradiction. Thus, $N=0$; i.e., there are no notrivial Jordan blocks and the matrix is diagonalizable.

If $A$ is invertible than it can be only a multiple if identity.

$$ A^2 = k A
$$

$$
A = A^2 A^{-1} = k A A^{-1} = k I
$$


How to show that an analytic function can be written as $g(z)^k$? Let $f$ be an analytic function with $f(z_0) = 0$ and $f \neq 0$. Let $k = gr(f, z_0) \geq 1$. Show that there exists an analytic function $g$ such that locally, $g(z)^k = f(z)$.

I know that if $f(z) \neq 0$ for all $z$, then i could let $h$ be the antiderivative of $f'/f$, so that $f(z)e^{-h(z)}$ is a constant, and finally write $f$ as $f(z) = ce^{h(z)}$, and then lastly, i could let $g(z) = c^{1/k}e^{h(z)/k}$

But this technique doesn't work in this case, since f can't be written as $e^{h(z)}$.

How should i go on about this?

Thank you.

I think i managed to solve it.

Since $f$ is analytic, it can be written as

$$f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n$$

And because $gr(f, z_0) = k$ we have that

$$f(z) = \sum_{n = k}^{\infty} a_n (z - z_0)^n$$

$$ = \sum_{n = 0}^{\infty} a_{n+ k} (z - z_0)^{n+k}$$

$$ = (z - z_0)^{k} \sum_{n = 0}^{\infty} a_{n+ k} (z - z_0)^{n}$$

$$ = (z - z_0)^{k} \sum_{n = 0}^{\infty} b_{n} (z - z_0)^{n}$$

where $b_n = a_{n+k}$. Let us name this new function as $r(z) = \sum_{n = 0}^{\infty} b_{n} (z - z_0)^{n}$. Now, since $b_0 = a_k \neq 0$, we have that $r(z_0) \neq 0$.

$r$ is clearly analytic, and since it is not $0$, then $\frac{r'}{r}$ is analytic. Let $h$ be it's antiderivative.

Now, notice that

$$(r(z)e^{-h(z)})' = r'(z)e^{-h(z)} - r(z)e^{-h(z)}h'(z) =r'(z)e^{-h(z)} - r(z)e^{-h(z)}\frac{r'(z)}{r(z)} = 0$$

Therefore, $r(z)e^{-h(z)} = c$, and we conclude that $r(z) = ce^{h(z)}$.

Replacing this in the expression of $f$ we conclude that

$f(z) = (z - z_0)^kce^{h(z)}$. Now writing $g(z) = (z - z_0)c^{1/k}e^{\frac{h(z)}{k}}$, we can see that $g(z)^k = f(z)$


Conditionally convergent - limit of series' Let $\sum_{n=1}^\infty a_n$ be conditionally convergent.
Let $k_n:= \max(a_n,0),l_n:=-\min(a_n,0)$ for $n\in \mathbb{N}$ and show that
$\sum_{n=1}^\infty k_n =\infty $ and $\sum_{n=1}^\infty l_n=\infty$.

I found this exercise on some german math forum and the corresponding post was closed for the lack of answers.
Now I tried myself to it and I can visualize that the limits of those series' is $\infty$ because, since it's not absolutely convergent it is $>0$ in case of $k_n$ and also in case of $l_n$ because $\min(a_n,0)<0$ and the $(-)$ would make it positive again.
But how can I prove that mathematically? Or is the argumentation that I gave enough?

First of all we have
$$\sum k_n+\sum l_n\stackrel{(1)}=\sum(k_n+l_n)\stackrel{(2)}=\sum |a_n|\stackrel{(3)}=\infty $$
where $(1)$ is valid because all $k_n,l_n$ are nonnegative, $(2)$ is valid because $k_n+l_n=|a_n|$ by definition of $k_n,l_n$, and $(3)$ is valid by assumption. Therefore, at least one of the series on the left diverges to $\infty$. If the other were finite (i.e., either $\sum k_n=K<\infty$ or $\sum l_n=L<\infty$), we'd have
$$\sum a_n=\sum k_n-\sum l_n=\begin{cases}K-\infty=-\infty\text{ or}\\\infty-L=+\infty\end{cases} $$
contrary to assumption.

If $\sum a_n$ is conditionally convergent, then $\sum a_n$ converges, but $\sum |a_n|=\infty$ or, in other words

$$\lim_{n\to\infty}\sum_{m=0}^na_m=\infty$$

Let $P=\{n\in\Bbb N: a_n\ge 0\}$.

$$k_n=\begin{cases}a_n, & n\in P\\0, & n\notin P\end{cases}$$

This gives
$$\sum_{n=0}^\infty k_n=\sum_{n\in P} |a_n|=\infty.$$

The last equality holds because $\lim_{n\to\infty}|a_n|=\infty$ implies that any subsequence of partial sums also approaches infinity.

The argument for $\sum l_n$ is similar.


$f(x) = \begin{cases} \frac{xy^2}{x^2+y^4}, (x,y) \ne0 \\0, (x,y) = 0 \end{cases}$ Determine if the function is continuous at the origin. Let $f: \mathbb{R^2} \to \mathbb{R}$, $$f(x) = \begin{cases} \frac{xy^2}{x^2+y^4}, (x,y) \ne0 \\0, (x,y) = 0 \end{cases}$$ Determine if the function is continuous at the origin.

Now when $(x,y) \ne 0$, letting $x = y^2$ I have that $\frac{y^4}{y^4+y^4} = \frac{1}{2}$. Similarly when $y=x$ I get $\frac{x^3}{x^2+x^4} = \frac{x}{1+x^2} \to 0$, hence the function is not continuous at the origin. Now I tried using epsilon-delta for this, but couldn't get it to work. What's would be wrong with my approach?
We have that $|f(x)-f(0)| < \varepsilon$ whenever $\|x-0\| = \|x\| = |x| < \delta$.
So let $\varepsilon >0$ and $\delta= \frac{\varepsilon}{y^2}$ $$|f(x)-f(0)| = |f(x)| = \frac{|x|y^2}{|x^2+y^4|} \leqslant |x|y^2 < \delta y^2 =\varepsilon.$$
But this is not true since just above we showed that the function is not continuous at the origin. I guess the problem is $\frac{|x|y^2}{|x^2+y^4|} \leqslant |x|y^2$ which isn't probably true?

The statement "$f$ is continuous at $(0,0)$" has the definition: "for every $\varepsilon>0$, there exists a $\delta>0$ such that for every $(x,y)\in\Bbb R^2$, if $(x,y)$ is within $\delta$ of $(0,0)$, then $f(x,y)$ is within $\varepsilon$ of $f(0,0)$".
The first quantified variable is $\varepsilon$, which must be an arbitrary constant (not depending on any other variables).
The next quantified variable is $\delta$, which can be chosen in terms of $\varepsilon$ (since that variable has already been quantified), but must be independent of $x$ and $y$ (since those variables have not been quantified yet when we get to $\delta$). This is the source of the error in the proof—you set $\delta= \varepsilon/y$ which is not allowed.
(And the next quantified variables are $x$ and $y$, which are allowed to be constrained by $\delta$ since that variable has already been quantified.)


For n≥3, n set of boys and girls has a stable matching (true or false) For some n ≥ 3 there exists a set of n boys, n girls, and preference lists for every boy
and girl such that every possible boy-girl matching is stable. If true, give a proof. If false, give a refutation.

This is obviously false as at n=3 I can find a unstable matching. But then I need to prove it for n≥3, no stable matching will appear.

I tried doing it like this: Assume I keep adding a stable pair of couples starting at 3 pairs, no matter how many pairs I add the matching remains unstable as an unstable pair already exist in the group. But what if I added unstable couples, then I cannot promise at some n the matching might suddenly become stable.

So is there a way to prove it?

You're asking about the Stable Marriage Problem.
Not only does there exist a stable setting, but you can also find it in ($n^2$) time!
First you fix a sex that "proposes", say men. Then every man, one after another, proposes to his first-choice woman. They accept provisionally. If, at a later point a woman, who has already accepted a proposal, gets an offer from a higher-preferred man, then she boots her first choice out. The rejected enters the loop again, then asking for his next-highest choice.
Now you only need to prove that the algorithm terminates, that everyone's married then and that the constellation is stable.


Solving $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$ How do you solve $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$?

Working
 in modulo $p$  we have $x^5 = 0 \pmod{p}$ and $x = 0$ which is the only solution in modulo $p$ since $a^{p} \equiv a \mod{p}$ for any prime $p$ and integer $a$. Taking this back we get $p^2y^5 = 0$ which implies $y=0$. Hence the only solution is $(x,y) = (0,0)$.

Edit/attempt 2:

Since $x = 0 \mod{p}$ we can let $x = pm$ where $m<x$ then we get  $p^2 (m^5 p^3 + m^2 p y^3 + y^5) = 0 \implies m^5 p^3 + m^2 p y^3 + y^5$ since $p$ can't be zero. But now $y^5 = 0 \mod{p}$ ; suppose $y = pn$ where $n < y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$

Your second attempt looks good (the moment you note explicitly that $(m,n)$ is also a solution), and leads to a contradiction in very much the same way as the standard proof that $x^2-2y^2=0$ has no non-zero integer solution (leading to $\sqrt2$ being irrational).

Above equation shown below:

$x^5+px^2y^3+p^2y^5 = 0$  -------$(1)$

Equation $(1)$ has numerical solution given below:

$(x, y, p)=(-6, 3, -8)$


If $a<b$, $c<d$ then $a+c<b+d$? This is how I managed to prove it:

I know $a \lt b$ and $c \lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c \gt 0$ and because of this $a+c \lt b+d$

Did I prove it right?

your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$

I would like to give a sweet and simple proof:

$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!


Find real constants $c$ and $k$ such that $y=cx^k$ passes through point $(a, b)$ with slope $m$ In the Cartesian plane, can a power function of the form $y=cx^k$ (where $c>0$ and $k>1$, not necessarily an integer) be found such that its graph passes through any arbitrary point $(a, b)$ with arbitrary slope $m$? [Assume that $a>0$, $b>0$, and $m>(b/a)$.] If so, how can $c$ and $k$ be calculated in terms of $a$, $b$, and $m$?

Note that this technique could be used to generate a type of spline curve that is tangent to each side of an obtuse angle, such as the one formed by the intersection of the lines $y=y_0$ and $y=mx$ (where $y_0>0$ and $m>0$).

The equations you get are

$$b=ca^k$$
$$m=cka^{k^-1}$$
dividing the first equation by the second yields
$$\frac{b}{m}=\frac{a}{k}$$
which can be solved for $k$, and you can then solve for c.


Can someone show me how to solve functions that include a set of ordered pairs? Im really stuck on this. Im unsure of how to go about solving these when given a set of

Given

$f= \{(-2,4),(0,-2),(1,2),(2,3)\}$ and $g(x)= 3x-5$

Find

A)  $f(-2)$

B) $(f+g)(2)$

C) $f(g(x))$

D) $f(1)-g(1-k)$

In each case I’m not sure how to calculate the $f$ values.

The part I think you are missing is that $f$ is a function which is defined only at the values -2, 0, 1, and 2, and nowhere else. $f(-2) = 4$, which is the answer to part (A); $f(2) = 3$, from which you can easily solve part (B); $f(1) = 2$, from which you can solve part (D).

This leaves only part (C), to determine $f(g(x))$.  $f$ is defined only at -2, 0, 1, and 2, so the composite function $f(g(x))$ is defined only at values of $x$ where $g(x)$ is one of -2, 0, 1, or 2.  If $x$ is such that $g(x)$ comes out to something else, say 37, then that is an invalid argument for $f$, and the entire expression $f(g(x))$ is undefined.  So what you need to do is make a list of the few values of $x$ that make $g$ yield a good argument for $f$, and then you can describe the behavior of the resulting $f(g(x))$ function in a way similar to the way that $f$ itself was described.

In principle, all functions can be described as a list (possible a very large infinite list) of ordered pairs in this way. In some contexts in mathematics we take an ordinary function like $g(x) = 3x -5$ and define it as the infinite set $$\{\ldots, (-2, -11), (-1, -8), (0, -5), (1, -2), \left(1\frac13, -1\right), \\
(2, 1), (\pi, 3\pi-5), \ldots, (57.89, 168.67), \ldots \}.$$  The function $f$ here is no different in principle; it just has a much smaller domain.


Is any property of the number TREE($3$) known? The number TREE($3$) is an insane large number occuring in Ramsey theory.

For the usual large numbers constructed with Knut's up-arrow-notation, Conway chain-notation and Bowers array-notation, it is usually easy to determine the last few digits of the constructed number because mostly power towers with many identical entries come into play.

  Is that also the case for TREE($3$) ? I have read that no upper bound is known, but it might be that the number is known to be a huge power tower. So, can we determine the last digits of TREE($3$) ? Or do we know nothing, not even whether TREE($3$) is even or odd ?

Yes, I have used the last binary (NOT decimal) digits of the tree steps and proven that it is odd and the last 3 binary digits are 011.


Compact inclusion in $L^p$ Is it true that there is a compact inclusion from $L^p$ to $L^q$ whith $q<p$?

What is the counterexample if what I said is wrong?

Thank you.

It is well known that $L^p \subseteq L^q$ with $1 \leq q < p < \infty$ on a measure space $(X,\Sigma,m)$ if and only if $m(X) < \infty$.

Compactness, however, doesn't have to hold.

Example: Let $X = [0,1]$ with the usual Lebesgue measure. Let $f_n(x) = \sin(2\pi n x)\in L^\infty(X) \subset L^p(X)$ for all $1\leq p$. So in particular it is a bounded sequence in $L^p(X)$.

Now, observe that if $n \neq m$
$$ \int_0^1 |f_n(x) - f_m(x)|^2 \mathrm{d}x = \int_0^1 f_n^2(x) + f_m^2(x) + f_n(x) f_m(x) \mathrm{d}x = 1 $$
and hence the sequence cannot have any Cauchy subsequence in $L^2$. (Similarly you can show that there exists no Cauchy subsequence in any $L^p$ using Holder's inequality and various interpolation inequalities.) Hence you have no compactness.

In fact, we have the following:

Theorem If a measure space $(X,\Sigma,m)$ is such for some $1 \leq q < p < \infty$, the inclusion $L^p(X)\subset L^q(X)$ holds and is compact, then every non-negligible subset of $X$ contains an atom.

The proof basically goes the same as the above example. By a theorem of Sierpinski's a non-atomic measure space can be evenly divided into two halves. So if $X$ contains a non-atomic subset $Y$ with positive measure, we can associate to every finite binary string $a_1a_2\cdots a_n$ a subset $Y_{a_1a_2\cdots a_n} \subset Y$ such that

For any two distinct strings of the same length, $Y_{a_1a_2\cdots a_n} \cap Y_{b_1b_2\cdots b_n} = \emptyset$
For any string of length $n$, $m(Y_{a_1 a_2\cdots a_n}) = \frac1{2^n} m(Y)$
For any string of length $n$, $Y_{a_1a_2\cdots a_n} \subsetneq Y_{a_1a_2\cdots a_{n-1}}$.

Defining $$f_n = \sum_{\text{strings of length } n} (2a_n - 1) \mathbf{1}_{Y_{a_1a_2\cdots a_n}}$$
we see that $f_n$ cannot admit any Cauchy subsequence, much in the same way as the example above.

On the other hand, if $X$ is made up only of atoms, then the inclusion can indeed be compact. For example, if $X$ is a finite set and $m$ is the counting measure on $X$. Then since the unit ball in $L^p(X,m)$ now is compact, automatically the inclusion $L^p(X) \subset L^q(X)$ is compact.

Note that, $L^{2}([0, 1])\subset L^{1}([0,1])$(using holders inequality); but take $f:(0, 1]\rightarrow \mathbb R$ such that $f(x)=x^{-\frac{1}{2}}$, $f\in L^{1}$ but $f\notin L^{2}$.


What does Arg {Inf I(d)} means? I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x)\ = \ \exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
with $$ W = \{d\ |\ d(0)=1,\ d(\pm \infty)=0\}$$
where I(d) is a functional defined as an integral
$$
I(d) = \frac{1}{2}\int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.

Can anyone tell me what does that mean?

It returns the argument that makes the functional $I\equiv I(d)$ infimum such that the function $d$ belongs to the space of kinematically acceptable functions $W$, i.e., it satisfies the prescribed boundary conditions. It just explicitly emphasises that we are after the function (or argument) that makes the functional infimum and not the infimum value itself.


Prove that for all real numbers $a$ and $b$, $|a| ≤ b$ iff $-b ≤ a ≤ b$ The problem is from exercise 12 of chapter 3.5 of Velleman's How To Prove It.
For the (---->) direction, I realise we are supposed to break the proof into two cases.
So, suppose $$|a|≤ b $$ Case 1:   $a=|a| $ which implies $a≤b.\;$ Now suppose $-b>a,\;$which implies $-a>b,\;$ but we know that $a≤b,\;$ so $-b>a\;$ is a contradiction. Therefore, $-b≤a.\;$Since we now know that $a≤b\;$ and $-b≤a,\;$we can conclude that $-b ≤ a ≤ b.$
Case 2: $-a=|a|,\;$which implies  $-a≤b,\;$which is equivalent to $-b≤a.$
This is where I'm stuck at. From the assumption of case 2, we have deduced that $-b≤a. \;$ But I don't know how to prove that $a≤b\;$ to reach the stated goal of $-b ≤ a ≤ b \:$ for this case. Assuming $a ≤ b \;$ is false (that is, assuming $a>b$) and trying to reach a contradiction doesn't seem to work.
I'm sure the solution is simple but I've spent 2 hours on this with no luck. ;
Thank you for your time.
EDIT: I should have clarified that since the text is about proof strategies, the author intended the reader to use proof by cases in this section of the book to get a better grasp of that particular strategy. Even so, thank you all so much for all the different ways of approaching the proof that you suggested! I've learned new things today!
For case 2, I ended up using the fact that $a<0\,$ to prove that $a≤b. \,$ That is, since $a<0\,$ iff $-a>0,\,$ and since $b≥-a,\,$ then $b>0.\,$ Thus we have $a<0<b,\,$ which implies $a≤b.\,$
Therefore, we can conclude that $-b ≤ a ≤ b.$

You can also use this maximum properties $\quad\begin{cases}\max(x,y)+z=\max(x+z,y+z)\\\max(u,v)\le w\iff (u\le w)\text{ and }(v\le w)\end{cases}$
$|a|=\max(a,-a)\le b\iff \max(a,-a)-b\le 0\iff \max(a-b,-a-b)\le 0 \iff\begin{cases}a-b\le 0\\-a-b\le 0\end{cases}\iff \begin{cases}a\le b\\-b\le a\end{cases}\iff -b\le a\le b$
This way of proceeding somehow 'hides' the multiple cases discussion.

I think the following reasoning is shorter.
For $b\geq0$ it's obvious by the definition of the absolute value:
$|a|$ it's a distance between the point, which is corresponded to $a$ on the number axis, and the origin, which gives immediately: $$|a|\leq b\Leftrightarrow -b\leq a\leq b$$
But for $b<0$ the statement $$|a|\leq b\Leftrightarrow -b\leq a\leq b$$ is also obvious and we are done!


Arrangement of points in a circle From the 2015 Moscow Mathematical Olympiad:

  The numbers $1$ to $1000$ are arranged on a circle such that each number divides the sum of its two neighbors. Suppose that the number $k$ has two odd neighbors. What's the parity of $k$ ?

I haven't been able to answer the question just yet.

Here are a few things I noticed. The constraint "$x$ divides its two neighbors $a$ and $b$" gets tighter when $x$ grows. For example, if $x=1000$, we must have $1000=a+b$. If $x=999$, we have either $999=a+b$ or $1998=a+b$ (in that case $(a,b)\in \{(1000,998),(998,1000)\}$).

Nevertheless, I can't derive anything useful from $k| a+b$ with $a,b$ being odd numbers.

diagram

the numbers from 1 to 1000 have equal number of even(E) and odd(O) numbers.
Let us assume an arrangement where there is at least one group of an odd number with two odd neighbors.This is arrangement A.
So we have 3 odds one after another. Let all evens in A have two odd neighbors. But then if we group two numbers in the anticlockwise direction we end up with an odd number for every even number, along with some odds extra.Thus it is not possible.
Again let in A, we have some evens with one odd & one even neighbor instead of all evens with odd neighbors. But since even doesn't divide even+odd, so this arrangement too is futile.
Thus, for our given arrangement A, we should have an even number with 2 even neighbors, like: ......,E1,E2,E3,..... In such case,E3 must be followed by E4,
for otherwise we have ...,E1,E2,E3,O,.... but E3 can't divide E2+O. Again, for E4 we must have a consequent E5, and for E5 we must have E6 ;thus this makes all numbers on the circle even,which is impossible.
*Note that we have exhausted all variations of arrangement for A, each time getting a contradiction. Thus arrangement A is impossible.
Therefore, a number with two odd neighbors must be even.*

Just write out the numbers is their order from $1$ to $1000$ on the circle and the condition is satisfied. The answer to the question is now even. You will have $1000$ between $1$ and $999$ and every other number $n$ will have $n+1$ and $n-1$ whose sum is $2n$.


evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
 \int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$

However the result seems to be $ \pi/\sqrt3 $ only.
Why is there this $ 2\ln2 $?

Detail:

$$ \begin{align*}
t  \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$

I think you made a simplification error. We have
$$\begin{align*}
\frac{d}{dt}\arcsin\left(\frac{2t}{1+t^2}\right) &= \frac{1}{\sqrt{1 - \frac{4t^2}{(1+t^2)^2}}}\left(\frac{2t}{1+t^2}\right)'\\
&= \frac{(1+t^2)}{\sqrt{(1+t^2)^2-4t^2}}\left(\frac{2(1+t^2)-4t^2}{(1+t^2)^2}\right)\\
&= \frac{(1+t^2)}{\sqrt{(1-t^2)^2}}\left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\\
&= \frac{2(1-t^2)}{(1+t^2)\sqrt{(t^2-1)^2}}
=\frac{2(1-t^2)}{(1+t^2)|t^2-1|}.
\end{align*}$$
You then cancelled to get
$$-\frac{2}{1+t^2}.$$
However, that cancellation is only valid if $t^2-1\geq 0$, i.e., if $|t|\geq 1$. Yet your integral covers a region that includes places where you get $t^2\lt 1$, so that the cancellation is not valid over the entire interval. Try doing it by splitting the integral as an integral over $[0,1]$ and over $[1,\sqrt{3}]$, being careful with the signs.

part of integral solution is $\ln(1+t^2)$. When you insert integral bounds you get $\ln(1+(\sqrt{3})^2)-\ln(1+(0)^2)$$=\ln(4)-ln(1)$$=2\ln(2)$


$f(x) = 0$ when $x$ is $0$, and $1$ otherwise I've been trying to create a function that will return $0$ when $x$ is $0$, and for any other $x$ value it should return $1$. I've searched for a pre-existing function online too and wasn't able to find one.

Do you know of any function that can do this?

You've already defined your function (assuming you've also chosen its domain).

One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.

Typical notation for a function created by the sort of description you give is a definition by cases:

$$ f(x) := \begin{cases} 0 & x = 0 \\ 1 & x \neq 0 \end{cases} $$

For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.

The complementary function

$$ g(x) := \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} $$

which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.

The Kronecker delta. A few different notations exist depending on the setting; e.g. $\delta_x$, $\delta[x]$, or $\delta_{x,0}$.
The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.
x == 0 computes this function in C and C++, and many other programming languages allow similar.

Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that

$$ g(x) = \mathop{\mathrm{sinc}}(\pi x) $$

holds whenever $x$ is an integer.

(if you're unfamiliar with it, $\mathop{\mathrm{sinc}}(z)$ is the continuous extension of $\sin(z) / z$)

Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all real, imaginary, and complex numbers.
$f(x) = \left \lceil \frac{1}{2\Gamma \left ( \left | x \right | \right )} \right \rceil$
*Note the absolute value inside of Gamma, as it's easy to go unnoticed


Is there an easier way of solving this problem? Take a look at this figure:

What's the length of the longest 1-dimensional rod which can pass through this corridor horizontally? I can solve this problem mathematically using first and second derivative tests, but is there an easier way, similar to what feynman does in many of his lectures? One approach might be to take advantage of the symmetry of the situation.

The longest such rod will make an angle of pi/4 with the original direction when it (almost) touches the inner corner. I need a valid logical explaination for that.

You can get an intuitive explanation by considering the extreme cases. Bring the rod into the corner parallel to the first corridor, and it can be infinitely long. Start to turn it around the corner, and it must be shorter and shorter to fit between the two outer walls (the lower one and the right-most one). When you have passed the 45 degree point, symmetry shows that this same logic repeats: the rod can now be longer and longer until it can be infinitely long when parallel with the second corridor. The shortest moment was obviously at the 45 degree point.

If the corridors weren't equal in width, it'd quickly get too complicated to intuitively figure it out this way.

Mathematically, I think I would parameterize a straight line (something like $s=\left(a\cos(v),a\sin(v)\right)$, with some proper x- and y-interval restrictions) and equate it with the parameterized corridor area and with the corner point. Solving those two equations simultaneously should give you the length $a$ which you can then minimize. But I would advice you to ask this on Mathematics SE if you seek more in-depth mathematical answers.

This is a "common sense" argument, which could be turned into rigorous math using calculus if you want to do that:

The angles $\dfrac \pi 4 - \alpha$ and $\dfrac \pi 4 + \alpha$ give the same length, for any angle $\alpha$. This is obvious from the symmetry of the situation - just draw a picture.

It follows that angle $\dfrac \pi 4$ corresponds to either a minimum or a maximum value for the length of the rod, and common sense says it can't be a maximum - therefore it is the minimum length.


Linear transformation and vectors on a certain type of 3 dimensional vector space over the rational number field Let $V$ be a $3$ dimensional vector space over $\mathbb Q$ and $T$ be a linear transform on $V$ such that for some $\vec x , \vec y , \vec z \in V$ with $\vec x \ne \vec 0$ , $T(\vec x)=\vec y , T(\vec y )=\vec z , T(\vec z)=\vec x + \vec y$ , then is it true that $\vec x , \vec y , \vec z$ are linearly independent  or equivalently saying that  $\vec x , \vec y , \vec z$  generates $V$ ( since $V$ is of dimension $3$  ) ?

The answer is YES.

Let us show first that $x$ and $y$ are linearly independent. If not, $y$
must be a multiple of $x$, $y=qx$ for some $q\in{\mathbb Q}$. We deduce
$z=T(y)=T(qx)=qT(x)=q(qx)=q^2x$, and $T(z)=x+y$ becomes $q^3x=(1+q)x$
whence $q^3=(1+q)$. But the polynomial $X^3-(1+X)$ has no rational roots,
so this is impossible.

Next, we show that $x,y$ and $z$ are linearly independent. If not, $z$
must be a linear combination of $x$ and $y$, $z=ax+by$ with $a,b\in{\mathbb Q}$.
Then $T(z)=x+y$ becomes $(ab)x+(a+b^2)y=x+y$, so $ab=1$ and $a+b^2=1$. We deduce
$b(a+b^2)=b$, or $1+b^3=b$. But the polynomial $X-(1+X^3)$ has no rational roots,
so this is impossible.

The conditions on the existence of $ \vec x$, $\vec y$ and $\vec z$ tells a lot more than tells the eye since not any $T$ satisfies this identities. In fact the problem can be reduced to the exisitence of a transformation $T$ that has the property $\det(T^3-T-T^0)=0$. Indeed in this case there exists $\vec x$ such that $(T^3-T-T^0)(\vec x)=\vec 0$ and we have $T^3(\vec x)=T(\vec z)=T(\vec x)+\vec x=\vec y+\vec x$. That such transformations exist can be seen from the companionmatrix of the polynomial $x^3-x-1$:$$T=\left(\begin{array}{rrr}%
0&0&1\\%
1&0&1\\%
0&1&0\\%
\end{array}\right)%
\text{so that for arbitary } \vec x=\left(\begin{array}{r}%
a\\%
b\\%
c\\%
\end{array}\right) \text{ we have }%
\vec y = \left(\begin{array}{c}%
c\\%
a+c\\%
b\\%
\end{array}\right)%
\text{ and } \vec z = \left(\begin{array}{c}%
b\\%
b+c\\%
a+c\\%
\end{array}\right)%
\text{ with } T(\vec z)=\left(\begin{array}{c}%
a+c\\%
a+b+c\\%
b+c\\%
\end{array}\right)%
=\left(\begin{array}{r}%
a\\%
b\\%
c\\%
\end{array}\right) +  \left(\begin{array}{c}%
c\\%
a+c\\%
b\\%
\end{array}\right)%
$$ and $\vec x, \vec y, \vec z$ linearly indepentent for $\vec x\neq \vec 0$. It is clear that any transformation similar to $T$ also meet this requirements. Remains the question if there are transformations $T$ such that $T^3-T-T^0$ has rank $<0$ instead of rank $0$ as it is the case here. Whith your permission I would open this here as a new question.


Difference of order statistics in a sample of uniform random variables Let $U_{1}, \, ... \, ,U_{n}$ be a random sample of uniform random variables $U_i \sim \mathrm{Uniform}(0,1)$. Let $U_{(1)}, \, ... \, , U_{(n)}$ be the order statistics of the sample. The goal is to prove that:

$$
W = U_{(s)}-U_{(r)} \sim \textrm{Beta}(s-r, \, n - s + r +1) \qquad 1 \leq r < s \leq n
$$

My "proof"

The joint pdf of $U_{(r)}, U_{(s)}$ is:

$$
f_{U_{(r)}, U_{(s)}} (u, v) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} u^{r-1} (v-u)^{s-1-r} (1 -v)^{n-s} \cdot \textbf{1}_{\{u < v\}}
$$

Consider the transformation:

$$
\begin{Bmatrix}
W = U_{(s)} - U{(r)}  & U_{(r)} = Z \\
Z = U_{(r)} & U_{(s)} = W + Z \\
\end{Bmatrix}
$$

The absolute value of the Jacobian determinant is 1. Therefore, the joint pdf of the transformation is:

$$
f_{W,Z}(w,z) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} z^{r-1} w^{s-1-r} (1 - w  - z)^{n-s}
$$

$$
f_W(w) = \int_{S} f_{W,Z}(w,z) \, \textrm{d}z
$$

For fixed $w$, we have  $S = \{z  \, : \, 0 \leq z \leq 1 - w\}$. Thus:

$$
f_W(w) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} w^{s-1-r} \int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z
$$

The problem is that I do not know how to evaluate the integral, but Wolfram Alpha does:

$$
\int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z= \frac{\Gamma(r)\Gamma(n-s+1)}{\Gamma(n+r-s+1)} (1- w)^{n+r-s}
$$

Now it is really easy to figure out the distribution of $W$. Indeed, the only thing that has to be done is to rewrite the factorials using gamma functions.

Initially I was going to ask how to evaluate the integral without the help of Wolfram Alpha, but then I realized that maybe there is a "better" proof that avoids it. I have tried to find an alternative proof by using multiple transformations which involved $W$ and another transformation, but it was useless.

Edit

I have corrected some mistakes I think I made:

1) The joint pdf I wrote was incorrect.

2) I wrote that I the pdf of $W$ could be found as a convolution, but $U_{(r)}, U_{(s)}$ are clearly dependent. What I actually did was an implicit change of variables.

One does not need to evaluate the integral. The change of variables $u=(1-w)v$ yields that $f_W(w)$ is a constant factor $C$ times $w^{s-1-r}$ times $(1-w)^{n+r-s}$. This also yields the value of the constant factor $C$ since $f_W$ must integrate to $1$.

(Or, after the change of variables, one can recognize the integral over $v$ from $0$ to $1$ as a Beta.)

Consider this alternative proof:
One can generate $n$ uniform random variables on a line by the following equivalent way:
1. Generating $n+1$ uniform random variables on a circle.
2. Choosing a random point as a references point ('$0$' point).

Therefore $U(s)−U(r)$ has an identical distribution to $U(s-1)−U(r-1)$, and to $U(s-r)−U(0)=U(s-r)-0 = U(s-r)$.
We already know $U(s-r) \sim \text{Beta}(s−r,n−(s+r)+1)$


Direct proof that the wedge product preserves integral cohomology classes? Let $H^k(M,\mathbb R)$ be the De Rham cohomology of a manifold $M$.

There is a canonical map $H^k(M;\mathbb Z) \to H^k(M;\mathbb R)$ from the integral cohomology to the cohomology with coefficients in $\mathbb R$, which is isomorphic to the De Rham cohomology. As a previous question already revealed, the images of this map are precisely the classes of differential $k$-forms $[\omega]$ that yield integers when integrated over a $k$-cycle $\sigma$,

$$ \int_{\sigma} \omega \in \mathbb{Z}  \quad\text{ whenever } d\sigma = 0$$

Let us call them "integral forms".

Motivated by the cup product on cohomology, my question/request is the following:

  Give a direct proof that the wedge product $[\omega\wedge\eta]\in H^{k+l}(M,\mathbb R)$ of two integral forms $\omega\in \Omega^k(M)$ and $\eta\in \Omega^l(M)$ is again an integral form.

This should be true because the cup product is mapped to the wedge product, but the point of the exercise is to prove this statement directly, without constructing the singular cohomology $H^k(M,\mathbb Z)$ or homology first.

Maybe I also have to make sure that the condition of being an integral form is something that can be "checked effectively" without singular homology; this might be subject to a new question.

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I'm not sure if this is an answer to the question, since it does refer to $H_{\ast}(M, \ZZ)$, but I think it sheds some interesting light on why the problem is hard.

We begin with a flawed proof attempt. Write $\delta$ for the diagonal map $M \to M \times M$, and $\pi_1$ and $\pi_2$ for the projections from $M \times M$ onto its first and second factor. Let $\rho$ be an integer $k+\ell$-cycle and let $\alpha$ and $\beta$ be a $k$-form and an $\ell$-form on $M$. Then $\int_{\rho} \alpha \wedge \beta = \int_{\delta(\sigma)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$. Suppose $\rho$ were homologous in $M \times M$ to $\sum \sigma_i \times \tau_i$, for various cycles $\sigma_i$ and $\tau_i$ in $H_{\ast}(M)$, with $\dim \sigma_i + \dim \tau_i=k+\ell$. Then we would have
$$\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta) =\sum_{(\dim \sigma_i, \dim \tau_i) = (k, \ell)} \int_{\sigma_i} \alpha \int_{\tau_i} \beta.$$
This would prove the result. (The integrals over terms where $\dim \sigma_i \neq k$ would drop out. If $\dim \sigma_i<k$, then $\pi_1^{\ast}(\alpha)|_{\sigma_i \times \tau_i}=0$ so $\int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)=0$, and likewise if $\dim \tau_i < \ell$.)

Unfortunately, this need not be true. Saying that $\delta(\rho)$ should always be homologous to such a class is saying that $\delta_{\ast} H_m(M)$ should lie in the image of $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$. But the map $\bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ)$ need not be surjective -- the cokernel is $\bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ))$ by the Kunneth theorem. There is no reason that $\delta_{\ast}$ needs to land in the image of $\bigoplus H_i \otimes H_j$; I give an example where it doesn't in this Mathoverflow thread,

So we have to work harder. The good news is that the full statement of Kunneth's theorem includes the statement that
$$0 \to \bigoplus_{i+j=m} H_i(M, \ZZ) \otimes H_j(M, \ZZ) \to H_{m}(M \times M, \ZZ) \to \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to 0$$
is (noncanonically) split. Choose such a splitting
$$\iota: \bigoplus_{i+j=m-1} \mathrm{Tor}_1(H_i(M, \ZZ), H_j(M, \ZZ)) \to H_m(M, \ZZ).$$
Then we can write $\delta(\rho) = \sum \sigma_i \times \tau_i + \iota(\sum \phi_j)$ for some $\phi_j$ in Tor groups. But the good news is that Tor groups are torsion, so an integer multiple of $\phi_j$ is homologous to zero. This means that $\int_{\iota(\phi_j)} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ must be zero. So we still have $\int_{\rho} \alpha \wedge \beta = \sum \int_{\sigma_i \times \tau_i} \pi_1^{\ast}(\alpha) \wedge \pi_2^{\ast}(\beta)$ and now the proof concludes as before. $\square$.

What I find interesting about this argument is two things:

(1) We never had to build cup product in $H^{\ast}(M, \ZZ)$, or prove its compatability with de Rham cohomology, we just tried to shove cycles around naively.

(2) Proving Kunneth with the full statement that the sequence is split is not hard. But I have only ever seen it done in a very algebraic way, not by explicitly writing a cycle on $X \times Y$ as $\sum \sigma_i \times \tau_j + \phi$ for some torsion $\phi$. Moreover, there is no canonical choice of splitting. This may explain why no one seems to be able to prove this in a natural way.

I mostly use cohomology as a blackbox so I may be overlooking technical difficulties but here's how I'd try to prove your statement.

First I would note that being integral is a local property: the class $[\omega]$ of a closed form is integral iff its restrictions $[\omega|_{U_i}]$ to a covering $M = \bigcup U_i$ is integral. One way is obvious: if $\sigma \subset U_i$, then $\int_\sigma \omega|_{U_i} = \int_\sigma \omega$. For the other, we can chose a (finite) triangulation $\sigma = \bigcup \sigma_\alpha$ such that each $\sigma_\alpha$ is included in one of the $U_i$ so that we can write $\int_\sigma \omega = \sum_\alpha \int_{\sigma_{\alpha}} \omega|_{U_{i(\alpha)}}$.

The point is that you can check integrality by using the triangulation you like.

Then I would write the cup product as $H^i_{dR}(X) \otimes H^j_{dR}(X) \to H^{i+j}_{dR}(X\times X) \to H^{i+j}_{dR}(X)$ the external product followed by restriction to the diagonal. And I'd check that each map sends integral integral classes to integral classes using triangulations.


What does "norms less than $\delta$" refers to in the context of Riemann sum? What does "norms less than $\delta$" refers to in the context of Riemann sum?

Given this problem "Prove if $\int_{a}^{b} f(x)$ exists, then for every $\epsilon>0$, there is a $\delta >0 $ such that $|\sigma_1 -\sigma_2|<\epsilon$,  if $\sigma_1$ and $\sigma_2$ are Riemann sums of $f$ over partitions P$_1$ and P$_2$ of $[a,b] $ with norms less than $\delta$."

I am trying to understand this problem that I posted Prove that if $\int_{a}^{b} f(x)$ exists, $\delta >0 $ such that $|\sigma_1 -\sigma_2|<\epsilon$ ), I did receive one answer but I am not understanding the how and the why of his argumentation leads to the conclusion that "the norms is less than $\delta$" (sub-question "norms" why plural?). What should be understood from by "the norms is less than $\delta$"?

The plural concerns the two subdivisions or partitions $P_1$ and $P_2$.

if $P=(x_0,x_1,x_2,...,x_n) $ is a subdivision of $[a,b] $ then what you call norm of $P $ is

$$\|P\|=\max_{0\le i\le n-1}\{x_{i+1}-x_i\} $$

Before this there should be a definition of the "norm of a partition". The norms being referred to here are the norm of $P_1$ and the norm of $P_2$; both are less than $\delta $.


Closure set in $\mathbb{R}$ Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function.

Let $$A=\left\{x\in\mathbb{R}, f(x)> 0\right\}$$

Can you please help me to find the closure set of $A$?

Thanks in advance.

for every set $A$ it is true that $A\subset \overline{A}$.

So you have $\{x: f(x)>0 \} \subset \overline{A}$. now, what about the points $x$ where there is equality, $f(x)= 0$ ?

Assume $A\ne \emptyset$ ,try to show that $\overline{A} = \{x: f(x)\ge 0\}$ you can do that using the complement of the set, showing $\{x : f(x)<0\}$ is open using the continuity of $f$.


Conceptual proof that $p\choose k$ ($1 < k < p$) is divisible by $p$ when $p$ is prime? (I.e., no equations). If $n,k\in \mathbf{N}$, then one defines $n\choose k$ to be the number of ways to choose $k$ elements from a set of size $n$. One can then show (by a combinatorial argument) that $${n\choose k} = \frac{n!}{k!(n-k)!}$$
whenever $n\choose k$ is defined. In particular, since this expression is an integer for all $n$ and $k$, when $n=p$ is a prime, $p\mid{p\choose k}$ when $1< k<p$.

I've wondered sometimes if there is also a proof of this fact which does not use the formula above; i.e., maybe one can see it by using the definition of $n\choose k$? What do you think?

You can group your subsets of size $k$ in families of size $p$ when they differ only by a "translation".

That is, if one of your subsets contains $a_1, \dots, a_k$, the other contains exactly the elements $a_i+d$, but regarded modulo $p$ to be again in the set $\{1,2,\dots, p\}$.

p(choose)k= p! / (k! (p-k)!) = r

Since k and p-k are both less than p, and p is prime, there can be no factors of p in k! (p-k)! since this is a product of numbers strictly less than p, and none of those (other than 1) can divide p.

   Hence, since p! = r (k! (p-k)!) has a factor of p, r must have a factor of p.


Does a sequence of vectors whose norms converge to zero necessarily converge to the zero vector? Let $V$ be an $N$-dimensional normed vector space, and let $(v_n)$ be a sequence of vectors such that $\lim (\left \| v_n \right \|)=0$. Given some basis $\{e^1,e^2,..., e^N \}$, we can write each $v_n$ as $a_1^{(n)}e^1+a_2^{(n)}e^2 +\dots  + a_N^{(n)}e^N$.

Is it necessarily true that for each fixed $k \in \{1, 2, \dots, N\}$, $\lim_{n \to \infty} a_k^{(n)}=0$ ? If so, how can we prove this using only the axioms for a generic norm?

Here is a proof that uses no theorem on normed linear spaces: Let $x_n=(a_1^{(n)},a_2^{(n)},\cdots,a_N^{(n)})$ and $r_n$ be its (usual) norm in $\mathbb R^{N}$. If $\{r_n\}$ is unbounded there is a subsequence $\{r_n'\}$ that tends to $\infty$. Now $\frac {v_n'} {r_n'} \to 0$ and we get unit vectors $(b_1^{(n')},b_2^{(n')},\cdots,b_N^{(n')})$ (obtained by normalizing $(a_1^{(n')},a_2^{(n')},\cdots,a_N^{(n')})$) such that $\sum b_i^{(n')}e_i \to 0$. But any sequence of unit vectors in $\mathbb R^{N}$ has  a subsequence which tends to  unit vector. We end up with $\sum c_i e_i=0$ where $(c_1,c_2,\cdots,c_n)$ is a unit vector contradicting linear independence of $e_i$'s. Hence $\{r_n\}$ is bounded. Now extract a convergent subsequence and take the limit in the equation $v_n=\sum a_1^{(n)}e_i$ to see that the limit of the subsquence must be $(0,0,\cdots,0)$. Now you can apply this argument to subsequences of $(a_1^{(n)},a_2^{(n)},\cdots,a_N^{(n)})$ to conclude that every subsequence of this has  a further subsequence converging to the zero vector. Hence $a_i^{(n)} \to 0$ for each $i$.

PS This approach can be used prove many of the basic theorem about finite dimensional normed linear spaces from scratch. Unfortunately it is not seen in text books. By this approach you can prove that finite dimensional subspaces are closed, that linear maps on finite dimensional spaces are continuous and all norms on finite dimensional spaces are equivalent.

Going off this question, is it possible for there to be an infinite sequence whose L2 norm goes to zero as N goes to infinity, but for every t in it's duration, the signal at t does NOT tend to zero?