Is there a closed form expression for the first zero of the first Bessel function? $j_{1,1}$ denotes the first zero of the first Bessel function of the first kind.  (That's a lot of firsts!)  It's approximately equal to $3.83$.  My question is, is there any closed form expression for its value?  Even a infinite series or infinite product that yields it would be good.

I ask because this value is used in physics, in the context of diffraction of light through a circular aperture, and students often make the mistake of thinking that the number just pops out of nowhere.

I like Claude's approach and approximation.  However for variety (and possibly interest) here is another approach which i pinched from Watson who pinched from Rayleigh who pinched from Euler.  And while Claude's approximation converges from above, here we'll approach from below.  First we can write the Bessel function as an infinite product with the form
$${{J}_{1}}\left( z \right)=\frac{1}{2}z\prod\limits_{n=0}^{\infty }{\left( 1-\frac{{{z}^{2}}}{j_{1,n}^{2}} \right)}$$
Taking the logarithmic derivative, we have then
$$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2}}\frac{1}{1-{{z}^{2}}/j_{1,n}^{2}}}$$
Converting this into a series we find
$$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2}}\sum\limits_{m=0}^{\infty }{\frac{{{z}^{2m}}}{j_{1,n}^{2m}}}}$$
And so we recover the definition of Rayleigh’s function
$$\frac{1}{{{J}_{1}}\left( z \right)}{{J}_{1}}'\left( z \right)=\frac{1}{z}-2z\sum\limits_{m=0}^{\infty }{{{z}^{2m}}{{\sigma }_{2m+2}}}$$
where

$${{\sigma }_{2m+2}}=\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{2m+2}}}$$
Using well known recurrence relationships we can simplify this to
$${{J}_{2}}\left( z \right)=2{{J}_{1}}\left( z \right)\sum\limits_{m=0}^{\infty }{{{z}^{2m+1}}{{\sigma }_{2m+2}}}$$
Substituting the series representation for the Bessel functions we have
$$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{z}^{2k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}}=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{2}^{2n}}n!\left( n+1 \right)!}}\sum\limits_{m=0}^{\infty }{{{\sigma }_{2m+2}}{{z}^{2m+2n}}}$$
And reversing the summation, i.e. $\sum\limits_{k=0}^{\infty }{\sum\limits_{j=0}^{\infty }{{{a}_{k,j}}}}=\sum\limits_{j=0}^{\infty }{\sum\limits_{k=0}^{j}{{{a}_{k,j-k}}}}$ one finds
$$\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{z}^{2k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}}=\sum\limits_{m=0}^{\infty }{{{z}^{2m}}}\sum\limits_{n=0}^{m}{\frac{{{\left( -1 \right)}^{n}}}{{{2}^{2n}}n!\left( n+1 \right)!}{{\sigma }_{2m-2n+2}}}$$
Equating coefficients
$$\sum\limits_{n=0}^{k}{\frac{{{\left( -1 \right)}^{n}}{{\sigma }_{2k-2n+2}}}{{{2}^{2n}}n!\left( n+1 \right)!}}=\frac{{{\left( -1 \right)}^{k}}}{{{2}^{2k+2}}k!\left( k+2 \right)!}$$
It’s very easy to create the sigma numbers in mathematica (and even by hand I suppose), so explicitly we have
$${{\sigma }_{2}}=\frac{1}{8}, {{\sigma }_{4}}=\frac{1}{192}, {{\sigma }_{6}}=\frac{1}{3072}, {{\sigma }_{8}}=\frac{1}{46080},…,{{\sigma }_{20}}=\frac{777013}{361634098839552000}$$
Now since $0<{{j}_{1,1}}<{{j}_{1,2}}<{{j}_{1,3}}...$ then ${{\sigma }_{M}}=\sum\limits_{n=0}^{\infty }{\frac{1}{j_{1,n}^{M}}}>\frac{1}{j_{1,1}^{M}}$and so $j_{1,1}^{{}}>{{\left( {{\sigma }_{M}} \right)}^{-1/M}}$.  You can bound this the other way as well (see Watson) but what we have is enough to get good approximations. So for a somewhat reasonable example (about 2 decimal places - when rounded)

$${{j}_{1,1}}\simeq {{2}^{5/4}}{{3}^{1/4}}{{5}^{1/8}}$$

a more unreasonable example, that yields about 6 places,
$${{j}_{1,1}}\simeq 2{{\left( \frac{1}{777013} \right)}^{1/20}}{{2}^{7/10}}{{3}^{7/20}}{{5}^{3/20}}{{7}^{1/20}}{{11}^{1/20}}$$

The series expansion is:

$$J_{1}\left(x\right)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2^{2n+1} n! (n+1)!}$$
$$=\frac{x}{2} - \frac{x^{3}}{16} + \frac{x^{5}}{384} - \frac{x^{7}}{18432} + …$$

The denominators are Sloan Sequence A002474.

Sloan Sequence A115369 is the first zero.


Proving that there is some $x$ such that $\gcd(a + bx, c) = 1$ Let $\gcd(a,b) = 1$ and $c>0$. Pove that there is an integer x such that $\gcd(a + bx, c) = 1$.
This question is from Section 2.3 (Covers the Chinese Remainder Theorem) Problem 15 of An Introduction to the Theory of Numbers by Niven and Zuckerman.
I didn't use the Chinese Remainder Theorem in my attempt, but I suspect that it should, since $\gcd(a,b) = 1$. In fact I didn't even use that $\gcd(a,b)=1$, either.
Attempt: Instead of finding $x$, we find what $x$ cannot be. That is, $y$ such that $\gcd(a+by,c) \neq 1$.
Reduce $c$ into its prime factors, so that $c = c_{1}^{i_{1}}c_{2}^{i_{2}}c_{3}^{i_{3}}...c_{n}^{i_{n}}$. Then we can solve, for each $1 \leq m \leq n$,
$a + by \equiv _{c_{m}}0 \Leftrightarrow by \equiv _{c_{m}} -a$
The point is that if $a + by$ and $c$ have a GCD other than $1$, then one prime factor must be in common.
$by \equiv _{c_{m}} -a$ has solutions when $gcd(b,c_{m})$ divides $-a$ and has no solution otherwise. Suppose there are no solutions for all $m$. Then we are done, for there are no restrictions on $x$. Suppose for some or all $m$, there are solutions. Then $by \equiv _{c_{m}} -a$, has solutions $y \equiv _{c_{m}} -(\frac{a}{g})y_{0}+t(\frac{c_{m}}{g})$. Where $g = \gcd(b,c_{m})$, $t = 0,1,..,g - 1$, and $y_{0}$ is a solution to $(\frac{b}{g})y \equiv _{c_{m}}1$. But the set of all these solutions cannot equal $\mathbb{Z}$, because there are infinite primes and finite $c_{m}$. So there must be some integer $x$ such that $gcd(a + bx, c) = 1$.
$\blacksquare$
I don't like this proof, because it's nonconstructive. Is there a way to use the CRT or some other basic number theorem?

Let $c=c_1c_2$ where every prime divisor of $c_1$ is a prime divisor of $a$ and $c_2$ is coprime with $a$.
Then $a+bc_2$ and $c$ are coprime.

This is almost certainly not what the problem in Niven and Zuckerman is looking for, but if you have Dirichlet's Theorem at your disposal, then from the assumption $\gcd(a,b)=1$ you can conclude there are infinitely many prime numbers of the form $a+bx$, at which point you can just take one that's greater than $c$.


7 friends are going to the cinema. They will be sitting in a row with 7 seats. What is the probability that John and Mary don't sit together? To watch a movie, John, Mary and 5 friends will sit randomly in a row with 7 seats. What is the probability John and Mary won't sit together?
$$(\mathbf A)\ \frac{2\times5!}{7!}\qquad(\mathbf B)\ \frac{5!}{7!}\qquad(\mathbf C)\ \frac27\qquad(\mathbf D)\ \frac57$$

I did:
$$1-\left(6\cdot 2\cdot\left(\frac{2}{7}\cdot\frac{1}{6}\right)\right) = \frac{3}{7}$$
But my book states the solution is D). I tried not multiplying by 2 and I get D), however I don't know exactly why the 2 is wrong.
You can make 2 permutations with Mary(M) and John(J), MJ and JM.
Then if you imagine the 2 of them as a block of 2 seats they can sit in $^6C_1=6$ places.
Why doesn't my book count those 2 permutations of JM and MJ?

If you seat John first, he sits on the end with probability $\frac 27$ then Mary has $\frac 56$ chance not to sit next to him, or he sits in the middle with probability $\frac 57$ and Mary has $\frac 46$ chance not to sit next to him.
$$\frac 27 \cdot \frac 56+\frac 57 \cdot \frac46=\frac {30}{42}=\frac 57$$

In the rest of your computation you are not considering order, so you shouldn't for JM either.


Question regarding the proof of chain rule using the $\lim_{h \rightarrow 0}\frac{ f(x+h) - f(x)}{h}$ definition of derivative Using the following definition of derivative:

$$\frac{dy}{dx}=\lim_{h \rightarrow 0}\frac{ f(x+h) - f(x)}{h}$$

I am trying to prove the chain rule, that is:

$$\frac{df(g(x))}{dx} =\color\red{\frac{df(g(x))}{dg(x)}}  \times\frac{dg(x)}{dx}$$

However, I don't understand where we get the definition of derivative in terms of two functions (colored in red). So that:

$$\frac{df(g(x))}{dg(x)} = \frac{df(g(x))}{d(x)} \times\frac{dx}{dg(x)}$$

$$= \lim_{h \rightarrow 0}\frac{ f(g(x+h)) - f(g(x))}{h} \times \lim_{h \rightarrow 0} \frac{h }{g(x+h) - g(x)}=\lim_{h \rightarrow 0} \frac{ f(g(x+h)) - f(g(x))}{g(x+h) - g(x)}  $$

From which the relation is easy to see. How is this definition/equality derived?¨

$-----------------------$

EDIT:

Here is an attempt but there may still be problems. First note that the term (again in red) is defined as first taking the derivative and then evaluating it at $g(x)$. So we replace every x in $f'(x)$ with $g(x)$. However this should the same as replacing every occurrence of x with $g(x)$ right in the beginning (Admittedly an argument with less hand-waving would be preferable).

As such the red term is equivalent to:

$$\lim_{h \rightarrow 0}\frac{ f(g(x)+h) - f(g(x))}{h}$$

Now substitute $h = g(x)-g(x+h)$

$$\lim_{g(x)-g(x+h) \rightarrow 0}\frac{ f(g(x+h) - f(g(x))}{g(x+h)-g(x)}$$

Which is equivalent to:

$$\lim_{h \rightarrow 0}\frac{ f(g(x+h) - f(g(x))}{g(x+h)-g(x)}$$

$\require{cancel}$

Perhaps look at it this way:

$$\frac{df(g(x))}{dx} =\color\red{\frac{df(g(x))}{dg(x)}}  \times\frac{dg(x)}{dx}$$

How we got the quantity in red is because notice how the denominator $\color{red}{dg(x)}$ cancels with the numerator in the next term. So we have that:

$$\frac{df(g(x))}{dx} =\color\red{\frac{df(g(x))}{\cancel{dg(x)}}}  \times\frac{\cancel{dg(x)}}{dx}$$

Basically, we have the following:

$$\frac{df(g(x))}{dx} =\frac{df(g(x))}{dx}  \times 1$$

$$1 =\frac{dg(x)}{dg(x)} $$

Hence we get:

$$\frac{df(g(x))}{dx} =\color\red{\frac{df(g(x))}{dg(x)}}  \times\frac{dg(x)}{dx}$$

Simply by moving around the denominator. We "get" this because it's required to cancel out terms, which you see further in your chain rule proof. I don't think it's as important that you understand "where it came from", rather then "why it's here and how it's applied"


If all mappings $f: A\to B$ are many-to-one, does there exist surjective $g: A\to B$? Suppose sets $A$ and $B$ [edit: for $B\ne \emptyset$] are such that all mappings $f: A\to B$ are many-to-one (i.e. not injective). Can we prove that there must exist a surjective $g: A\to B$? Ideally, I am hoping to be able to construct $g$.

Using the axiom of choice, you can show that $|A| \leq |B|$ or $|B| \leq |A|$. (Each of the two sets are in bijection to a certain ordinal, say $\alpha$ and $\beta$, because they can be well-ordered. Then $\alpha \leq \beta$ or $\beta \leq \alpha$). Hence, there is an injection of $A$ into $B$ or of $B$ into $A$.

Your assumption excludes the first case. Hence, there is an injection $f : B \to A$. Now any left inverse of $f$ will be a surjection $g : A \to B$.

I think the key point here is that there is no mapping from A to B that is injective, without knowing anything else about sets A and B. In other words there is no way we can construct an injective mapping from A to B. If they are finite, the only way that this can be true is only if |B|<|A| so no matter how we try there will always exist an element b in B such that: $ a_1 a_2\in A, f(a_1)=b=f(a_2) $ assuming f is a total function. Or if B is finite and A infinite. In any case if it is impossible for an injection to exist, set A must have a higher cardinality than set B; more elements. For if A has fewer elements than B I don't see a way that a injection cannot exist.

I think this come down to the more general 'axiom of choice' and its equivalent form: for sets A, B either $|A|\leq|B|\: or\: |B|\leq|A|$ as stated by Asaf.
If $|A|\leq|B|$ then must exist an injective function from A to B, so this is not the case and by the theorem must hold that $|B|\leq|A|$ and exists a function $f:B\rightarrow A$ that is injective and must exist a function $g:A\rightarrow B$ that is surjective.

If we do not accept the axiom of choice (or Zorn's lemma) how can you choose the elements for the construction of your surjection? If I have understood the axiom of choice, if you say something like $g(a)$='the first element b from B that ...' is invoking the axiom of choice. Where '...' you put what you want, like 'is bigger than g(a-1)' or 'is the first b such that there exist no a in A: g(a)=b' or 'find a random element in A, check if it has been given a pair from A, if not return it else find another'. All these are implying the axiom of choice.

So I we know nothing for A and B, e.g. if the can be ordered or any other information, and you cannot use the axiom of choice then I would say you cannot prove that such a surjection exists.


Limit evaluation: very tough question, cannot use L'hopitals rule I found a very tough limits question online. The question asks you to evaluate the limit $$\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x}$$ without using L'Hôpitals rule.

I tried to treat the top as a radical expression with the $e^x-9$ grouped and the other in root form to try to attempt rationalization. It did not work because you still get $\frac 0 0$.

I tried a trick of double rationalization but that did not work, got back to the starting.
Second attempt I tried to let $x=z-4$, a substitution, but it still did not lead to something that could remove a zero from the numerator.

Then I tried to break this up into three fractions, by dividing $x$ into each term in the numerator, and I basically got $+\infty$, then can't do $e^x/x$ and then $-\infty$.

So I have exhausted all the algebraic tricks I can think of.

Anybody out there think they they can crack this one?
Hope someone can.

Sincerely,

Palu

Let $f(x) = (x+4)^\frac{3}{2}+e^{_{x}}-9$. Your limit can be written as
$$\lim_{x \to 0}\frac{f(x)- f(0)}{x - 0}$$
Which is the definition of $f'(0)$. Thus the answer is ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$.


Square and triangles in a simple geometry problem Given $ABCD$ is a square with side "$a$"  and  both $\triangle ABE$ and $\triangle BCF$ are equilateral triangles, how can I find the area of $\triangle BEF$ ?

I know that $BE=BF=a$ ... but I would need $EF$ for the perimeter formula for the area .

The situation is represented here :

Thanks a lot!

Hints:

$$\angle EBF=90^\circ\implies \;\Delta BEF\;\;\text{is an isosceles straight-angle triangle}\implies$$

$$EF=\frac2{\sqrt2}a=\sqrt2\,a$$

Isn't it easier to divide the area of the square $BEDF$ in two?


Can two figures have the same area, perimeter, and same number of segments have different shape? I want to make an algorithm grouping all the details having the same shape.
each detail is defined by its surface, and a list of contour lines.

First I believed that having the same perimeter length and same surface would be enough, but I saw on that link that it is wrong hypothesis.

If I take as additional condition that the two shapes have the same number of segments, would it be enough? Or else how can I check that?

The problem is for each detail, they can get rotation, or symmetry.

Edit :

Thanks for your answers, I finally found a way to solve the problem (answer below)

In the spirit of the no-words answer to the linked question:

A famous problem was posed in 1966, "Can you hear the shape of a drum?" That is, can you find two different planar domains whose spectrum is the same? It was known that any two such domains must have the same area and the same perimeter. The question was answered negatively by Gordon, Webb and Wolpert in the early 1990s. Here is their example:

Notice that the number of sides is also the same, so this answer's the OP's question with even more conditions imposed.


Easier ways to prove $\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$ Prove that

$$\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$$

One way to do this is use the idea in the proof of Sophomore's dream. We have
$$x^{-x}=\exp(-x\log x)=\sum_{n=0}^\infty\frac{(-1)^nx^n\log^n x}{n!}$$
Therefore, using the change of variable $x=\exp(-t/(n+1))$ we have

$$\begin{aligned}\int_0^1 \frac{\log^2 x}{x^x}dx=&\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1x^n\log^{n+2}xdx\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}}{n!}\int_0^\infty t^{n+2}e^{-t}dt\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}(n+2)!}{n!}\\
=&\sum_{n=0}^\infty(n+1)^{-(n+2)}(n+2)\\
=&\sum_{n=1}^\infty n^{-(n+1)}(n+1)\\
=&\sum_{n=1}^\infty n^{-n}+n^{-(n+1)}\\
<&2\sum_{n=1}^\infty n^{-n}\\
=&\int_0^1\frac{2}{x^x}dx
\end{aligned}$$

Hence the result follows.

I am curious if there are any other methods to prove this, especially I am interested in easier approaches.

P.S. This was a bonus problem in an assignment from a multivariable calculus class.

Note that $f(x)=x^x$ is convex in $[0,1]$ with its minimum happening at $1/e$. Now since $2^e>e$ ($e>1$) it holds that $f(1/e)>1/2$.
Therefore
\begin{align}
\int_0^1 \frac{\log^2 x-2}{x^x}dx<2\int_0^1 (\log^2 x-2)dx=0
\end{align}

Using the inequality $\log(x)<x-1$ for all $x>0$ we have
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx}
\end{align}
Also we use the inequality $e^{x} \ge 1+x$ for all $x>0$ to express the denumeantor
$$x^x=\exp(x\ln(x))\ge 1+x\ln(x).$$

Therefore,
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx} <\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\ln(x)}}dx}.
\end{align}
Now, using the inequality $\log(x) > \frac{x-1}{x}$ for all $x > 0$, we egt
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx}  < \int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{x^x }}dx} &<\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\ln(x)}}dx}
\\
&<\int_0^1 {\frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\cdot \frac{x-1}{x}}}dx}
\\
&<\int_0^1 {e\cdot x\cdot \frac{{\left( {x - 1} \right)^2  - 2}}{{1+x\cdot \frac{x-1}{x}}}dx} \,\,\,\,\,\,\,\,\,\,\,\,(\text{since $0<x<1$})
\\
&<\int_0^1 {e\cdot \left({\left( {x - 1} \right)^2  - 2}\right)dx} = -\frac{5}{3} e <0
\end{align}


Big O notation and Proofs If $f \in O(h) $, does that mean there will always be some $c$, such that $f(n) \leq ch(n) $ is always true for all $n \geq 1$? That is, assuming $f$ and $h$ are always $>0$.

$f=O(h)$ means that there exists a constant $c$ such that $|f(n)|\leq c|h(n)|$ holds for all sufficiently large $n$.

No. The big O notation means that, for sufficiently large values of $x$, $f(x)$ and $g(x)$ behaves similar up to a constant, that is:

$$ f=O(h) \Longleftrightarrow \exists M\in\mathbb R^* : \lim_{x\to\infty} f(x) = M \lim_{x\to\infty} g(x) \Longleftrightarrow\\ \Longleftrightarrow M\in\mathbb R^* \mbox{ and } x_0\in R^+: f(x) \leq Mg(x) \; \forall x\geq x_0 . $$


Is there a constant known to be algebraic but unknown whether it is a surd? Is there a mathematical constant known to be algebraic irrational, but which is unknown to be a surd (root of an integer polnyomial of degree $2$) or not ?
Is there a mathemetical constant known to be algebraic, but unknown to be rational or not ?

Irrationality proofs or transcendental proofs can be extremely difficult, so I wonder whether verifying a constant to be a surd or not, can be extremely difficult as well (even if it is known that it is algebraic and irrational).
I have not much hope concerning the second question because a number known to be algebraic can probably be proven relatively easy to be rational or irrational, but maybe I am wrong.

What do you mean by "mathematical constant?" Do you mean something like $\pi$ or $e$? Or do you mean something without a name like 17 or $\sqrt{2}$?

An algebraic number is a real number that is a root of a nonzero polynomial equation with integer coefficients. I would say $\phi$, the golden ratio, is a mathematical constant that is algebraic. It is irrational because it is equal to $\frac{1+\sqrt{5}}{2}$.

If I recall correctly, a surd is an nth root of a number (not just a square root, but any root), and so is a root of some polynomial. Therefore, it is irrational. So it seems to me your first question is asking, "Is there something that is an apple that is not known whether it is an apple?" Maybe I'm not really understanding what it is you're asking.

Regarding the 2nd question, we know the only numbers with rational square roots are perfect squares (in fact, the roots are integers). Similarly, the only numbers with rational cube roots are perfect cubes (again, the roots are integers). So it would seem any algebraic number (any root of a polynomial) will be known to be irrational, unless it is an integer.


Crisis in my understanding of probability If I were to roll a die, what would be the probability of getting $2$? Certainly it would be $\dfrac 16$ (because there are $6$ numbers and sample space contains 6 numbers)
 But I think we can look at it another way.
We need $2$ right? and from $1-6$ there is only one number which is $2$ i.e. $2$. If I were to roll there would essentially be two types of numbers ie. one which is $2$ and one which is not ie $1,3,4,5,6$. So there are two basic outcomes.
P(E) = number $2$÷total outcomes ie $2$.
P(E)= $\dfrac 12$
So why is the probability $\dfrac 16$(when it is clear that the outcome we desire, can either happen or not happen.) Shouldn't the probability be $\dfrac 12$ instead, why not?

The probability is $\frac{1}{6}$ because there are $6$ elements in the sample space that are all equally likely.

If we take a sample space of $2$ and not $2$, then there are $2$ elements in the sample space, but they are not equally likely.

The classic example is that you don't have a $50\%$ chance of winning the lottery because the two options are "win" and "don't win".

P(E) = number 22÷total outcomes

The problem in the construction you have created is that the selection part (choose a side out of 6, the set {1,2,3,4,5,6}) doesn't match the total outcomes part (choose from the set {2, not 2}).


Consider a function $f \colon U \to X$, if $f$ is the identity when acting on $X$, then $df_x$ is also the identity for all $x \in X$? While doing an independent project, I ran into this question. Let $X$ be a $k$-dimensional compact manifold in $\mathbb{R}^n$ and let $U \subset \mathbb{R}^n$ be an open set such that $X \subset U$. As in the title, consider a smooth function $f \colon U \to X$ such that $f(x) = x$ for all $x \in X$. Then, is it true that $df_x$ is the identity matrix?

My first guess is that it is not always true. For this, recall what a lefschetz map is. The latter is a function such that on its fixed points $x$ (i.e. $x \in U$ such that $f(x) = x$) $df_x$ does not have $+1$ as an eiganvalue, in other words we do not have $df_x(v) = v$ for all $v$ in $T_x(X)$. So, if the above were true then we would not have lefschetz maps. Is this reasoning correct? Or this depends on the nature of $f$ or $X$?

Any help or advice would be greatly appreciated.

Let $v\in T_{x}U$ and take a curve $c: I \longrightarrow U$ such that $c(0) = x$ and $\dot{c}(0) = v \in T_{x} U \subset T_{x}X$. Like $c(t) \in U$ then $(f \circ c)(t) = c(t)$ and from here
\begin{equation}
df_{x}(v) = df_{x}(\dot{c}(0)) = \left. \frac{d}{dt}(\ f \circ c\ )(t) \right|_{t=0} =  \left. \frac{d}{dt}(\ c(t)\ ) \right|_{t=0} = \dot{c}(0) = v
\end{equation}
Therefore $df_{x}$ is the identity linear transformation.


If $f'(x)>0$ on $E$ , where $m(E)>0,$ then $m(f(E))>0$ Let $f:\mathbb R\to \mathbb R,$ and suppose $f$ is differentiable at every point of a measurable $E\subset \mathbb R,$ with $f'>0$ on $E$.
Suppose also that $m(E)>0$ (where $m$ is Lebesgue measure).
Prove that $m(f(E))>0$.

My proof:
Since $f$ is differentiable then it's continuous and hence it preserves both compact sets and intervals.
Now since $m(E)>0$ we can find compact interval inside it (is this true or not?)
If this is true so the proof is completed.
I know that if $E$ is measurable then $E$ is either Borel set or a set of measure zero.
So here $E$ is Borel, but still not necessarily to be an interval.

I will prove two points : first, in response to a comment of zhw, we check that $f(E)$ is measurable, and then that its measure is $>0$.

We recall that $m^*$ denotes here Lebesgue outer measure, defined for every subset $S \subset \mathbb{R}$. For $I$ an open interval of $\mathbb{R}$, we note $l(I) = \sup I - \inf I$. Then we define $$m^*(S) = \inf \left \{ \sum \limits_{k=1}^{+\infty} l(I_k),\ (I_k)_{k \ge 1} \textrm{ is a sequence of intervals with } S \subset \bigcup \limits_{k \in \mathbb{N}^*} I_k \right \}$$

For both results, the following lemma will be needed.

  Lemma : Given $f : \mathbb{R} \rightarrow \mathbb{R}$ differentiable at any point of a set $S$, assuming that there exists $C \ge 0$ such that $\forall x \in S,\ |f'(x)| \le C$, then $m^* \big( f(S) \big) \le C \cdot m^*(S)$.

Proof : Let $\varepsilon > 0$. We define an increasing sequence of sets $(S_n)_{n \ge 1}$ by : $$S_n = \left \{ x \in S,\ \forall t \in S,\ |t-x| \le \frac{1}{n} \Rightarrow |f(t)-f(x)| \le (C+\varepsilon)|t-x| \right \}$$

Using the hypothesis $|f'|\le C$, we get that $S = \bigcup \limits_{n=1}^{\infty} S_n$. For every $n \ge 1$, we can take a sequence of open intervals $(I_{n,k})_{k \ge 1}$ covering $S_n$ and such that
$\sum \limits_{k=1}^{+\infty} m^*(I_{n,k}) \le m^*(S_n)+\varepsilon.$

Without loss of generality, we can assume that for every $n$ and $k$, $m^*(I_{n,k}) \le \frac{1}{n}$.

Then, for $n \in \mathbb{N}^*$, for all $k \in \mathbb{N}^*$, for $x,y \in S_n \cap I_{n,k}$, we have $|y-x|\le \frac{1}{n}$ and $x,y \in S_n$, so we can write $|f(y)-f(x)| \le (C+\varepsilon)|y-x|\le (C+\varepsilon)\cdot m^*(I_{n,k})$.
Thus, for $n \ge 1$, $$m^*(f(S_n)) \le \sum \limits_{k=1}^{+\infty} m^*\big(f(S_n \cap I_{n,k}) \big) \le \sum \limits_{k=1}^{+\infty} (C+\varepsilon)\cdot m^*(I_{n,k}) \le (C+\varepsilon)\cdot (m^*(S_n)+\varepsilon)$$

Letting $n \to +\infty$ and then $\varepsilon \to 0^+$, we get $m^* \big( f(S) \big) \le C \cdot m^* (S)$.

Now we prove that $f(E)$ is measurable. Classically (see Problem about $G_{\delta}$ and $F_{\delta}$ sets),

  Claim 1 : There exists a subset $H \subset E$ which is $F_{\delta}$ (i.e. a countable union of closed sets) such that $N=E \backslash H$ is a null set.

Write $H = \bigcup \limits_{k=1}^{+\infty} F_k$ where the $F_k$ are closed. For $k \ge 1$, for all $M>0$, $[-M,M] \cap F_k$ is compact and $f$ is continuous (because it is differentiable) on this set, so $f([-M,M]\cap F_k)$ is closed (it is a compact set).
So for all $k$, $f(F_k) = \bigcup \limits_{M \in \mathbb{N}^*} f\big([-M,M]\cap F_k\big)$ is a Borel set, so $f(H)$ is measurable.

Now we prove that $f(N)$ is a null set. For $k \in \mathbb{N}^*$, we denote $N_k = \{ x \in N,\ f'(x)<k \}$. $N_k \subset N$ so $N_k$ is a null set, so $m^*(N_k)=0$ for $k \ge 1$. Plus, we can use the previous lemma on $N_k$, because $0 \le f' \le k$ on $N_k$, so $m^* \big(f(N_k) \big)\le 0$.
Thus $f(N_k)$ is a null set. As $f(E) = f(H) \cup f(N)$, we can conclude that $$f(E) \textrm{ is measurable}.$$

Now back to the original problem : we have some measurable set $E$ with positive measure, $f$ differentiable on $E$, $f'>0$ on E. We suppose that $m \big( f(E) \big)=0$.

For $x \in E$, $f'(x)>0$ so $\frac{f(x)-f(y)}{x-y}>0$ for all $y \in E \backslash \{x \}$ in some neighborhood of $x$. Thus
\begin{align*}
E & = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x > q \ \textrm{ and }\ \forall y \in ]q,x[,\ \frac{f(x)-f(y)}{x-y}>0 \right \}\\
& = \bigcup \limits_{q \in \mathbb{Q}} \left \{ x \in E\ | \ \ x > q \ \textrm{ and }\ \forall y \in ]q,x[,\ f(x)>f(y) \right \}
\end{align*}

because $\mathbb{Q}$ is dense. Moreover, $E$ has positive measure and $\mathbb{Q}$ is countable. Hence, there exists $q_0 \in \mathbb{Q}$ such that $B = \left \{ x \in E\ | \ \ x > q_0 \ \textrm{ and }\ \forall y \in ]q_0,x[,\ f(x)>f(y) \right \}$ has positive measure.

Plus, for $(x,y) \in B^2$ with $x<y$, we have $q_0 < x <y$, so $f(y)>f(x)$. Hence $f_{|B}$ is increasing.

Finally it is a well-known fact (see Can we have an uncountable number of isolated points) that $B$ has countably many isolated points, and thus we have a measurable subset $A \subset B$ such that $m(A)=m(B)>0$ and $A$ has no isolated points. Note that we also have $f_{|A}$ increasing, $f'>0$ on A, and $m \big( f(A) \big)=0$.

Now we just need a stronger version of our lemma :

  Lemma (bis) : Given $A \subset \mathbb{R}$ with no isolated points, and $f : A \rightarrow \mathbb{R}$, we say that $f$ is differentiable at $x \in A$ whenever $\lim \limits_{t \to x^{\neq}} \frac{f(x)-f(t)}{x-t}$ exists, and we note $f'(x)$ the limit. Assuming that $f$ is differentiable over $A$, and that there exists $C \ge 0$ such that $|f'| \le C$, we have $$m^* \big( f(A) \big) \le C \cdot m^*(A)$$

Proof : the proof is exactly the same as the one we gave for the first lemma.

Finally, we denote $g = f_{|A}$. As $g$ is stricly increasing, $g^{-1}$ is well defined. Moreover, as $f'>0$ on $A$, it is classical (see Inverse functions and differentiation) to show that $g^{-1}$ is differentiable in the sense of the Lemma bis on $g(A)$. As $g(A)$ is a null set, we can use our lemma (as we did with the set $N$ - see above) to conclude that $g^{-1}\big( g(A) \big)$ is a null set, so $A$ is a null set, which is absurd.

$$\textrm{Hence we have }\ m \big (f(E) \big) > 0.$$

From $f'>0$, there exists $c>0$ such that $f'>c$. Then $$m(f(E))=\int_{E} f' dm>\int_{E} c dm=c\int_{E} 1 dm=cm(E)>0.$$


How to prove sin(sin(sin...(sin(1))) converges to 0 without using continuity? Using continuity I was able to show the sequence $x_0 = 1$, $x_{n+1} = sin(x_n)$ converges to 0, but I was wondering if there was a way to prove it using only properties and theorems related to sequences and series, without using continuity.

So far, I know the sequence is monotonically decreasing and bounded below by 0, so it must converge to its infimum. From here I'm not exactly sure how to show 0 is the infimum of this set of numbers.

Alternatively, I could check convergence to $0$ by comparison, but no sequences come to mind that are greater than the given sequence for all $n \geq N$ and converge to $0$.

I've already seen the answers at the following:

Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$

Prove that $\sin(\sin...(\sin(x))..)$ converges asymptotically to zero

We know that for all $x \in (0,1]$, we have
$$
0 < \sin x < x
$$
From there, you can show that $x_0,x_1,\dots$ is a strictly monotonically decreasing sequence.  Is this enough? That is, can we forgo continuity? No. As a counterexample, consider the seuqence $x_k = f(x_{k-1}); x_0 =1$ with
$$
f(x) =
\begin{cases}
\sin(x) + 0.1 & x^*<0\leq 1\\
\sin(x) & 0 < x \leq x^*
\end{cases}
$$
Where $x^*$ is the positive solution to $= x^* = \sin(x^*) + 0.1$.
We note that $x_k$ is a strictly monotonically decreasing sequence converging to $x^*$ (and therefore bounded below by $0$), and we even have $f(x)<x$ for every point in $(0,1]$.  However, because of the lack of continuity, this is not enough to "force" $\{x_k\}$ to converge to $0$.


Determining hexes intersected by a line between two hexes on a hex grid Given two hexes within a hex grid of tiles, is it possible to determine all the hexes which are intersected by a line draw from the centre of first hex to the centre of the second if you are only given the row and column number of each?

For example, if you have some hexes h1 and h2 in the grid of tiles and you know h1 is located at (r1, c1) and h2 is located at (r2, c2) can all the hexes that will be intersected by a line draw from the centre of h1 to the centre of h2 be determine without any further information?

The reference image on this question may be helpful in visualizing the hex grid and the (x, y) coordinate system employed may make a possible solution easier.

Let $s$ be the slope of the line. Let us consider, by simplicity, only the case where $s$ is positive (the solution for the case with $s$ negative is similar and symmetric). Thus, we have two hexes with coordinates $r1, c1$ and $r2, c2$, with $r1\leq r2$ and $c1\leq c2$. Also, let us define $r$ and $c$ as the Cartesian coordinates (with standard horizontal and vertical lines) shown in the Figure posted in the previous question. Note that, according to this coordinate system, the vertices of all hexes in the plane have integer coordinates on the x-axis, whereas they can have coordinates of the form $i$, $i+1/3$, or $i-1/3$ (with $i$ integer) on the y-axis. Also assume that "crossing" a hex includes the possibility that the line overlaps with one of the hex's sides or even that the line crosses a vertex.

In the row where $y=r1$, the hex with center in $c1+2$ is crossed by the line if $s\leq\frac{2/3}{2}=1/3$. The hex with center in $c1+4$ is crossed if $s\leq\frac{2/3}{4}=1/6$. Continuing in this way, we get that any hex with center in $c1+2k$ is crossed if $s\leq\frac{2/3}{2k}=\frac{1}{3k}$.

In the upper row, where $y=r1+1$, the hex with center in $c1+1$ is crossed if $\frac {1/3}{1}<s$. The hex with center in $c1+3$ is crossed if $\frac{1/3}{3}<s\leq \frac{4/3}{2}$. However, for the next hex the formula changes because a line with decreasing slope crosses the hex at its upper vertex instead than at the left upper one. Thus, the hex with center in $c1+5$ is crossed if $\frac{1/3}{5}\leq s\leq \frac{5/3}{5}$, the hex with center in $c1+7$ is crossed if $\frac{1/3}{7}\leq s \leq \frac{5/3}{7}$, and so on. Continuing in this way we get that  any hex with center in $c1+(2k-1)$ is crossed if:

$\frac{1/3}{(2k-1)} \leq s \leq \frac{4/3}{(2k-2)}$, when the slope $s>1/3$ (note that for $k=1$ it gives $1/3<s<\infty$);
$\frac{1/3}{(2k-1)} \leq s \leq \frac{5/3}{(2k-1)}$, when the slope $s\leq1/3$.

In the row $r1+2$, similar considerations allow to obtain that  any hex with center in $c1+2k$ is crossed if:

$\frac{5/3}{(2k+1)}\leq s\leq \frac{7/3}{(2k-1)}$, when the slope $s>1/3$;
$\frac{4/3}{(2k)}\leq s\leq \frac{8/3}{2k}$, when the slope $s\leq 1/3$.

Generalizing, we have that, in the row $r1+j$ with $j$ odd,  any hex with center in $c1+(2k-1)$ is crossed if:

$\frac{(j-1/3)}{2k}\leq s\leq \frac{(j+1/3)}{(2k-2)}$, when the slope $s>1/3$;
$\frac{(j-2/3)}{(2k-1)}\leq s\leq \frac{(j+2/3)}{(2k-1)}$, when the slope $s\leq1/3$.

On the other hand, when $j$ is even, any hex with center in $c1+2k$ is crossed if:

$\frac{(j-1/3)}{(2k+1)}\leq s\leq \frac{(j+1/3)}{(2k-1)}$, when the slope $s>1/3$;
$\frac{(j-2/3)}{(2k)}\leq s\leq \frac{(j+2/3)}{2k}$, when the slope $s\leq1/3$.

Of course the number of intersected hexes can be determined from the information you mention, since it forms a complete description of the problem. What further information might there be to affect that count?

That doesn't mean that I know a simple formula to do that computation, though. I'm merely pointing out that you could certainly come up with a stupid algorithm which moves from one hex to the next and keeps track of the count. If you multiply row numbers by $3$, then all corners of the hexes would even be on integer coordinates, in which case the algorithm could operate on integers only as well, avoiding numeric issues.


Determine homogeneous transformation matrix for reflection about the line $y = mx + b$, or specifically $y = 2x - 6$ Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$,  or specifically $ y = 2x - 6$.

I use  $mx - y +b =0$:  $\text{slope} = m$,  $\tan(\theta)= m$
intersection with the axes:
$x =0$  is $y = -b$   and  $y =0$ is  $x = \dfrac{b}{m}$
My question is, what  can I do next?

Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $\vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $\vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:

$$\vec{r} \mapsto (\vec{r}-\vec{r}_0)-2\hat{n}( \hat{n}\cdot(\vec{r}-\vec{r}_0))+\vec{r}_0$$
I normalized the normal, $\hat{n}=\frac{\vec{n}}{|n|}=\frac{(m,-1)}{\sqrt{m^2+1}}$.

This can be simplified to:
$$\vec{r}-2\hat{n}( \hat{n}\cdot\vec{r})-2\hat{n}( \hat{n}\cdot\vec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).

Observe that because $\vec{r}_0$ is on the line, it satisfies the original equation $\vec{n}\cdot\vec{r}_0=-b$, so our choice of the origin point is irrelevant.

Now, you can simply write the matrix out:

$$\begin{bmatrix}x'\\y'\\1\end{bmatrix}=\begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \\ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \\0 & 0& 1\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix}$$

In our specific case:
$$\begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \\ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \\0 & 0& 1\end{bmatrix}$$

You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$\begin{bmatrix}1-2a^2 & -2ab & -2ac \\ -2ab & 1-2b^2 & -2bc \\0 & 0& 1\end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.

Note that any matrix $\mathbf A\cdot \vec{0}=\vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $\vec 0 \to (4 \frac{4}{5},-2\frac{2}{5})$.  You might want to do something about that.


characteristic curves for second-order equations Reading about characteristic curves for second-order equations, in particular semi-linear equations of second order with two independent variables:

$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{y,y}=f(x,y,u,u_{x},u_{y})$     $   $  $  $    $(1)$

My book reference, define characteristic curve to $(1)$, as plane curves along which the PDE can be written in a form containing only total derivatives of $u_{x}$ and $u_{y}$.

I do not understand this definition (only total derivatives of $u_{x}$ and $u_{y}$??? ), I do not know how to see it this, for example, in equation

$xu_{xx}+2xu_{xy}+xu_{yy}=u_{x}+u_{y}$.

I read some questions about characteristic curves here, but did not help.

Can anyone help me?
Thank you.

They just mean that the PDE can factored into something of the form $\prod_i\left(a_i\partial_t^{\alpha_i} +b_i\partial_x^{\beta_i}+c_i \right)$, where $\beta_i,\alpha_i\in\mathbb{N}$. So for example, the heat equation is not of this form, but the wave equation is. And once it's in this form, you just solve this nested sequence of characteristic problems.

If you want an example, see how Evans derives the solution for the easiest form of wave equation.

Since you haven't accepted, here are all the details. The example PDE you posted can be factored as $$(\partial_x+\partial_y)(x\partial_x+x\partial_y-1)u=0$$
Solve it by letting $v(x,y):=(x\partial_x+x\partial_y-1)u$. Then we have $v_x+v_y=0$ Solve this for $v$, and then solve $v=(x\partial_x+x\partial_y-1)u$ for u.

Another definition:
Let $$\sum_{i,j=1}^na_{ij}(x)U_{x_ix_j}+\sum_{i,j=1}^nb_i(x)U_{x_i}+c(x)U=f(x)$$
$$x=(x_1\dots x_n)$$

A smooth surface $S:\Phi(x)=0$ with the  domain of $\Phi$ being a subset of the domain of the coefficients in the equation is called a characteristic if $\forall x \in S$
$$\sum_{i,j=1}^na_{ij}(x){\partial\Phi\over\partial x_i}(x){\partial\Phi\over\partial x_j}(x)=0$$
For n=2 we talk of characteristic curves.

Under a twice smooth bijective change of variables $y=y(x)$, the characteristic of the transformed equation is $\Phi((x(y))=0$, where x(y) is the inverse change.


Prove that $gcd(\mid G:H\mid,\mid G:K\mid )=1\Rightarrow G=HK$ If $G$ be a group and $H\le G,$ $K\le G$ such that $\mid G:H \mid=m$ and $\mid G:K \mid=n$ then  we know the following two relsults from group theory.

$(1)$ $lcm(m,n) \le \mid G:H\cap K \mid \le mn$

$(2)$ $\mid G:H \cap K \mid \le mn.$ The equality of $(2)$ holds iff
$G=HK.$

I know the proof of these two results. Note that $gcd(m,n)=1 \Rightarrow \mid G:H \cap K \mid =mn(by (1)) \Rightarrow G=HK(by (2)).$ But I want to prove $gcd(m,n)=1 \Rightarrow G=HK$ without using $(2)$.
Any hint is accepted.

I have seen the solutions to similar question.But it's assuming that $G$ is finite.But what happen if $G$ is infinite

Remark :
Notice that $[G : HK] \ \mid \ [G : K] $ and $[G : HK] \ \mid \ [G : H]$ .
[It is only division of cardinality of sets, not groups.]

So we can conclude that $[G : HK] \ \mid m$ and $[G : HK] \ \mid n$ ;
by assumption of primitivity of $m$ and $n$  ($\gcd(m,n)=1$);
we can conclude that $[G : HK] \ \mid 1$ ;
equivalently  $[G : HK] \ \mid 1$ ;
so we are done.


A general formula for simple pendulum I am a student and hope you can explain in simple terms the formula for the pendulum:

wikipedia says that the period of a simple pendulum for angles of small amplitude is
$$T \approx 2\pi \sqrt\frac{L}{g}  \qquad 0\lt\theta\lt 12^{\circ}  ,$$

because the sine of an angle can be approximated to the angle
$$ {d^2\theta\over dt^2}+{g\over \ell} \theta=0\qquad $$
in the general formula
$${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0\\$$

wikipedia gives some solutions to that DE, and further to that I learned in another question that the exact general solution is given by the so-called Jacobi elliptic function

Those articles are too difficult for me, can you tell me what is the best/simplest formula to find the period of a pendulum? If you choose any angle between 45° and 90° and make a concrete example, that would be great

There is no simple solution as the period cannot be expressed with the ordinary functions.

The true expression is

$$T=4\sqrt{\frac Lg}K\left(\sin(\theta/2)\right)$$

where $K$ denotes a special function kwnon as the "complete elliptic integral of the first kind", defined as

$$K(k)=\int_0^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}.$$

(If you never heard of integrals, this will be meaningless to you.)

This function is tabulated or can be computed numerically with specific algorithms. Below, a plot of the function:

The starting value is $\pi/2$, showing that for small angles, the period is indeed $2\pi\sqrt{L/g}$. And given the "flatness" of the curve, the validity domain of the approximation isn't too bad.

Notice that for angles reaching $180°$, the period becomes infinite. Indeed an upside down pendulum will remain in equilibrium forever (at least in theory).

After you get to $\dfrac{d^2\theta}{dt^2}+\dfrac g{\ell} \theta=0$, which is derived at Wikipedia:pendulum in the boxes at the end of the paragraph, this becomes a fairly simple differential equation in $\theta$, with a solution of $\theta=\cos(\sqrt{\dfrac g{\ell}t})$, because $\dfrac{d^2\theta}{dt^2}=-\dfrac g{\ell}\cos(\sqrt{\dfrac g{\ell}t})$.

$\sqrt{\dfrac g{\ell}}$ is $2\pi$ times the frequency of the pendulum swing, and as a pendulum is a simple harmonic oscillator, we can say that $T=\dfrac1f=2\pi\sqrt{\dfrac{\ell}{g}}$.

The full solution is more complicated, involving the arithmetic-geometric mean. The Wikipedia link is here.


function over independent random variables this is my first post here in math.stackexchange. I have this problem and really no good idea:

$ X_1 ,...,X_n :\Omega \rightarrow \mathbb{R} $ are independent random variables, $m \in \{1,...,n-1\}$ and $f: \mathbb{R}^m \rightarrow \mathbb{R}, g: \mathbb{R}^{(n-m)} \rightarrow \mathbb{R}$. Show that $f(X_1,...,X_m)$ and $g(X_{m+1},...,X_n)$ are independent.

for understanding: f and g are meant as random variables themselves, defined as following: $f(X_1,...,X_m)(\omega)=f((X_1(\omega),...,X_m(\omega)))$

Thank you for your help,

EDIT: corrected it, thanks:)

For Borel sets $E$ and $F$ in $\mathbb R^{m}$ and $\mathbb R^{n-m}$ respectively consider the property $$P((X_1,X_2,...,X_m)^{-1}(E) \cap (X_{m+1},X_{m+2},...,X_n)^{-1}(F))=P((X_1,X_2,...,X_m)^{-1}(E) P(X_{m+1},X_{m+2},...,X_n)^{-1}(F)$$

First take $F$ to be a measurable rectanagle $F_{m+1}\times F_{m+2}\times ... \times F_n$ where $F_i$'s are Borel sets in $\mathbb R$. Verify that the collection of all Borel sets $E$ in $\mathbb R^{m}$ for which above equation holds is a monotone class containing finite unions of measurable rectangles. [This is where independence is used].  Use monotone class theorem to conclude that the equation holds for all Borel sets $E$ in $\mathbb R^{m}$. Now repeat the argument for $F$: consider the collection of all $F$ for which the equation holds and apply monotone class theorem again. We have proved that $P(U^{-1}(E)\cap V^{-1}(F)=P(U^{-1}(A))P(V^{-1}(B))$ for all Borel sets  $A$ and $B$ in $\mathbb R$ where $U=f(X_1,X_2,...,x_m)$ and $V=g(X_{M+1},X_{m+1},..,X_n)$. [I have used the fact $f^{-1}(A)$ and $g^{-1}(B)$ are Borel sets in $\mathbb R^{m}$ and $\mathbb R^{n-m}$ respectively].


Finding integer solutions to $n^2 = 2d^2 - 2d + 1$ I'm trying to find the smallest integer solution to

$$n^2 = 2d^2 - 2d + 1$$

Additional constraints: $$d > 10^{12}, n > 0$$

I wrote a computer program to bruteforce it, but that is too slow.

This page says that I can consider the equation modulo some prime to find whether it has solutions, but I already know that it has a solution, I just don't know what it is.

It's $$n^2=d^2+(d-1)^2$$ and we got Pythagorean triples because we can assume that $d-1>0$,

otherwise it's or obvious or the same.

Since $gcd(d,d-1)=1$, we have two cases.

$d=2mn$ and $d-1=m^2-n^2$, where $gcd(m,n)=1$, $m>n$ and $mn$ is even;
$d=m^2-n^2$ and $d-1=2mn$.

We got two Pell's equations:
$$(m+n)^2=1+2m^2$$ and $$(m-n)^2=1+2n^2.$$
They are $y^2=1+2x^2$ with a minimal solution $(2,3)$ and from here we can get all solutions.

Found a solution.

Valid values of d have a ratio of approx 5.82842712474619 between them. (Is this ratio significant in some way? Damned if I know.)

In other words, if you know that 1235216565974041 is a valid value, then you can guess that 7199369738058942 is a valid value, which is only 2 away from 7199369738058940, which is correct.

This makes the problem bruteforceable, and solves my problem.


Hilbert transform pair We know the Hilbert transform is defined as:
$$
y(t)=\mathcal{H}(x(t))=\operatorname{P.V.}\ \left\{x(t)*\frac{1}{\pi t}\right\}
$$
In the frequency domain this is equivalent to:
$$
Y(f)=X(f)\times (-j\operatorname{sgn}(f))
$$
Using this twice, we get :
$$
Z(f)=Y(f)(-j\operatorname{sgn}(f))=X(f)(-j\operatorname{sgn}(f))^2=-X(f)\operatorname{sgn}^2(f)
$$
Many textbooks say
$$\operatorname{sgn}^2(f)=1
$$
So $$Z(f)=-X(f)
$$
and thus
$$\hat{\hat{x}}(t)=-x(t)$$
But in my mind,$$
\operatorname{sgn}(f)|_{f=0}=0$$ and if$$X(0)\neq0,$$
$$Z(0)\neq-X(0)$$
So there is a single point where $$Z(f)\neq X(f).$$
Does this matter? Or Hilbert transform is defined in almost everywhere sense?

Any time you speak of integral operators, equality is almost always (hah) meant in the sense of almost everywhere. This makes sense since integrals don't see points; so as long as two functions only differ on some "small" set (of measure zero), we should expect to get the same output regardless.

As a note for future posts: sometimes in-line math is a good thing. Not every line of LaTeX code needs to be in an equation environment.


If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$? Problem :

  If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the
  value of $\cos(\theta -\frac{\pi}{4})$?

My approach :

Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$

$\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2} - (\pi \sin\theta) \} $

$\Rightarrow \pi \cos\theta = \frac{\pi}{2} - (\pi \sin\theta)$

$\Rightarrow \frac{1}{2} =\frac{1}{\sqrt{2}}[\sin\frac{\pi}{4} \cos\theta + \cos\frac{\pi}{4} \sin\theta] $

$\Rightarrow \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4} + \theta)$
$\Rightarrow \frac{\pi}{4} = \frac{\pi}{4} + \theta$

$\Rightarrow \theta = 0$

$\therefore \cos(\theta - \frac{\pi}{4})$ = $\frac{1}{\sqrt{2}}$ But this is wrong answer.. please suggest where I am wrong... thanks.

I would use

$$\frac{\sin{(\pi \cos{\theta})}}{\cos{(\pi \cos{\theta})}} = \frac{\cos{(\pi \sin{\theta})}}{\sin{(\pi \sin{\theta})}}$$

from which I get

$$\cos{[\pi (\cos{\theta}+\sin{\theta})]}=0$$

or, in one case,

$$\pi (\cos{\theta}+\sin{\theta}) = \frac{\pi}{2}$$

or,

$$\sqrt{2} \pi \cos{\left ( \theta-\frac{\pi}{4}\right )} = \frac{\pi}{2}$$

You can take it from here.

$$\tan (\pi \cos x)=\cot(\pi \sin x)$$

or,

$$\tan (\pi \cos x)=\tan (\frac{\pi}{2} - \pi \sin x)$$ [∵, tan(π/2+Ф)=cotФ]

or, πcosx=π/2-πsinx

or, π(sinx+cosx)=π/2

or, sinx+cosx=1/2

or, (sinx+cosx)1/√2=1/2√2

or, cosx(1/√2)+sinx(1/√2)=1/2√2

or, cosxcosπ/4+sinxsinπ/4=1/2√2

or, cos(x-π/4)=1/2√2

∴, a)1/2√2 is the right answer


Integral in $\mathbb R^3$ and $\Gamma$-function How one can show this equality?
$$
\iiint_V x^{a-1}y^{b-1}z^{c-1}\,dxdydz = \dfrac{\Gamma(a)\Gamma(b)\Gamma(c)}{\Gamma(a+b+c+1)},
$$
where $V$ is simplex $x\geqslant0, y\geqslant0, z\geqslant0, x+y+z\leqslant1 $.

Only thing I thought is changing variable in some way but it didn't help me.

That's actually quite a nice one. If you know Quantum Field Theory you might have heard of Feynman parameters? They come in handy here (beware rigour, I'm a physicist :D) So, how can you prove the above? First note

$$\frac{1}{AB}=\int_{0}^1dxdy \delta(x+y-1)\frac{1}{(xA+yB)^2}.$$ This can be proven straightforwardly. Moreover, it can be generalized

$$\frac{1}{A_1\cdots A_n}=\int_0^1 dx_1 \cdots dx_n \delta\left(\sum_i x_i -1\right)\frac{(n-1)!}{(x_1A_1+\ldots+x_nA_n)^n}$$ (Proof by induction.)

Now, you can differentiate wrt to the $A_i$ to find

$$\frac{1}{A_1^{m_1} \cdots A_n^{m_n}}=\int_0^1 dx_1 \cdots dx_n \delta\left(\sum_i x_i-1\right)\frac{\prod_i x_i^{m_i-1}}{\left(\sum_i x_i A_i\right)^{\sum m_i}}\cdot \frac{\Gamma(m_1+\ldots+m_n)}{\Gamma(m_1)\Gamma(m_2) \cdots \Gamma(m_n)}.$$

Set all $A_i$ to one and you essentially have your statement. This is also stated in the QFT book by Peskin/Schroeder on p. 190.

You remember beta-function? $B(z,w)=\displaystyle \int_0^1 t^{z-1}(1-t)^{w-1} dt$, you are going to use it and make some substitutions. You take the integral one at a time while making the substitution and see what happen. Am sure you will get it.


What does c represent in this differential equation? If I have this equation:

$p'(t)-p(t)\alpha =0$

I can say that $p$ is a function that represents the size of a population at time t.  The rate at which the population grows is constant.  The solution will show that the size of the population is proportional to the initial size.

If I have this equation:

$p'(t)-p(t)f(t) =0$

I can say the rate at which the population grows is determined by $f$.  The size of the population is still proportional to the initial size.

But if I have:

$p'(t)-p(t)\alpha = h(t)$

It's difficult to determine from the solution what role the initial size of the population, $p(0)$, has.  The solution is:

$p(t)=\bigl(\int e^{-\alpha t}h(t)\ dt + c\bigr)\ e^{\alpha t}$

So my question is this: if I'm interpreting these differential equations as growth functions, what does $c$ represent in the last equation?  In the previous equations, $c=p(0)$.

The equation
$$p'(t)=f(t)\ p(t)$$
says: The rate $p'$ by which the  population $P$ grows is proportional to the current size $p$ of $P$; but the proportionality factor $f$ valid at time $t$ depends on time. If, e.g., the growth rate depends on the seasons, the function $t\mapsto f(t)$ would be a certain periodic function with period 365 days. At any rate, the size of $P$ at some later time $t$ will be proportional to the initial size $p(0)$.

The equation
$$p'(t)=\alpha p(t)+h(t)$$
with constant $\alpha\in{\mathbb R}$ says: The population $P$ would rise (or decline) at the constant rate $\alpha$, and therefore would be given by $p(t)=p(0)\ e^{\alpha t}$, if it were not for an additional extraneous influx (or decrease) $h(t)$ dependent only on time $t$, but not on the current size $p(t)$ of the population. In this case it may very well be that for large $t>0$ the initial value $p(0)$ has almost no effect on the actual value $p(t)$. This is the case, e.g., if $\alpha<0$, and $h$ is some periodic function with period $T$. Then for large $t$ there will be a certain "stable" periodic behavior. This "limiting" behavior depends only on $h$, but not on the initial value $p(0)$.

Concerning your last question: When you write the solution in the "more correct" form
$$p(t)=\int_0^t e^{\alpha(t-\tau)}\ h(\tau)\ d\tau + p(0)e^{\alpha t}$$
then you can verify by inspection that $p(0)$ plays no rôle for large $t$ when $\alpha<0$.

In your first equation, the solution is $p(t)=ce^{at}$.  As you say, you choose $c$ to be $p(0)$ if that is the initial condition you are given.  If you are given that $p(2)=10000$, that there were $10000$ individuals at year $2$, you would find $c=10000e^{-2a}$  Similarly, $c$ in the last equation is a constant of integration which you set to match some initial condition you are given.


How can I differentiate a Poisson distribution to find the maximum that way? I want to find the maximum to a Poisson distribution using calculus. How can I do this? I want to do this algebraically.

Hint: The ratio of two consecutive Poisson probabilities is $\frac{p_{n+1}}{p_n}=\frac{e^{-\mu}\mu^{n+1}/(n+1)!}{e^{-\mu}\mu^{n}/n!}=\frac{\mu}{n+1}$. For what $n$ is this ratio greater/less than one?

The Poisson distribution is only defined for $k\in\mathbb{N}$ and therefore can't be differentiated.


Brouwer fixed point theorem If $f:\mathbb R^n \to\mathbb R^n$ is continuous and such that for each $x$ in the domain $\|f(x)-x\|\leq1$ holds, how can I prove that then $f$ must be onto?

As a hint, I have been told, for an $a\in\mathbb R^n$ look for a fixed point of the function

$$g_a(x):=x+a-f(x+a)$$

2nd hint: Suppose that the value $a$ is not attained.

What can you say about fixed points of $g_a$ ?

And what does Brouwer give you when you look at the map of the closed unit ball under $g_a$?


Definite Integral involving logarithm and tangent function Show that
$$\mathcal{I}:=\int_0^{\frac{\pi}2} \log |1-a^2\tan^2\theta| d\theta= \pi\log\sqrt{a^2+1}.$$

I tried to use the substitution $\tan\theta=z$, to get that
$$\mathcal{I}:=\int_0^{\infty} \frac{\log|1-a^2z^2|}{z^2+1}dz$$
This integral is quite similar to this one:
Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
However note that the sign inside the logarithm term is different and it seems none of the idea is applicable to this setting. May be there is a way to use the above link result to prove my integral. But I dont know.
Edit: Okay I think I have figured out one way to do it via contour integration. We can take a contour that looks like a large semicircle on the upper half plane so that it encloses only the pole at $z=i$. The contour should also have two holes around $z=\pm \frac1a$ to avoid the singularity coming from $\log$. Then Residue Calculus gives the desired result.
It would be nice to see a proof without residue calculus as well.

Note that we'll need to take the Cauchy Principal Value. Observe that by DUTIS,
\begin{align*}
    I(a) &= \int_0^{\frac{\pi}{2}} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
    I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\partial}{\partial a} \ln\left|1-a^2\tan^2\theta\right|\,\mathrm{d}\theta\\
    &= \int_0^{\frac{\pi}{2}} \frac{2a \tan^2\theta}{a^2\tan^2\theta-1}\,\mathrm{d}\theta\\
    &= \frac{2}{a} \int_0^{\frac{\pi}{2}} 1 + \frac{1}{a^2 \tan^2\theta - 1} \,\mathrm{d}\theta\\\\
    &= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \int_0^{\cot^{-1} a - \varepsilon} 1 + \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta + \int_{\cot^{-1} a + \varepsilon}^{\frac{\pi}{2}} 1 + \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta \right ).\\
\end{align*}
so it is sufficient to evaluate the integral of $\dfrac{1}{a^2 \tan^2\theta - 1}$. We have
\begin{align*}
    \int \frac{1}{a^2 \tan^2\theta - 1} \, \mathrm{d}\theta &= \int \frac{1}{(u^2 + 1)(a^2 u^2 - 1)} \, \mathrm{d}u && \text{using $u = \tan\theta$}\\
    &= -\frac{1}{a^2 + 1} \int \frac{1}{u^2 + 1} \, \mathrm{d}u - \frac{a^2}{2(a^2 + 1)} \int \frac{1}{au+1} \, \mathrm{d}u + \frac{a^2}{2(a^2+1)} \int\frac{1}{au-1} \, \mathrm{d}u\\
    &= -\frac{\tan^{-1}(u)}{a^2 + 1} - \frac{a \ln |au + 1|}{2(a^2+1)} + \frac{a \ln |au - 1|}{2(a^2+1)} + C\\
    &= \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} + C.
\end{align*}
Applying this to our original equation, we have
\begin{align*}
I'(a) &= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \left [\theta - \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} \right ]_0^{\cot^{-1} a - \varepsilon} + \left [\theta - \frac{a \ln \left | \frac{a \tan\theta - 1}{a \tan\theta + 1} \right | - 2\theta}{2(a^2 + 1)} \right ]_{\cot^{-1} a + \varepsilon}^{\frac{\pi}{2}}\right )\\
&= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \frac{2a^2 \cot^{-1} a - 2a^2 \varepsilon - a \ln \left | \frac{-(a^2+1)\tan\varepsilon}{(1-a^2)\tan\varepsilon+2a}\right |}{2(a^2+1)} + \frac{a^2 \pi}{2(a^2+1)} - \frac{2a^2 \cot^{-1} a + 2a^2 \varepsilon - a \ln \left | \frac{(a^2+1)\tan\varepsilon}{(a^2-1)\tan\varepsilon+2a}\right |}{2(a^2 + 1)} \right )\\
&= \frac{2}{a} \left ( \lim_{\varepsilon\to 0} \frac{-4a^2 \varepsilon - a \ln \left | \frac{(1-a^2)\tan\varepsilon+2a}{(a^2-1)\tan\varepsilon+2a}\right |}{2(a^2+1)} + \frac{a^2 \pi}{2(a^2+1)}\right )\\
&= \frac{a \pi}{a^2 + 1}\\
\implies I(a) &= \int \frac{a\pi}{a^2+1} \, \mathrm{d}a\\
&= \pi \ln\sqrt{a^2 + 1} + C.
\end{align*}
Finally, observe $I(0) = 0$, so $C = 0$. Hence, $I(a) = \pi \ln\sqrt{a^2 + 1}$.

$$
I=\int_0^{\pi/2}\ln|1-a^2\tan^2\theta|d\theta
$$

\begin{align}
I' & =\int_0^{\pi/2}\frac{2a\tan^2\theta}{a^2\tan^2\theta-1}d\theta
\\[5mm] & =\frac{2}{a}\int_0^{\pi/2}\frac{a^2\tan^2\theta}{a^2\tan^2\theta-1}d\theta
\\[5mm] & =\frac 2a\int_0^{\pi/2}1+\frac{1}{a^2\tan^2\theta-1}d\theta
\end{align}
can you do something with this? I fear it may diverge


Who invented special framed links as a way to specify four-manifolds? It is possible to specify (smooth, oriented, compact) four-manifolds with special framed links:

$$I \times \mathbb{R}P^3$$
Such a link is divided in two sublinks, one ordinary link representing the 2-handles (and their attaching curves) and an unlinked, unknotted link representing the 1-handles.
This notation is used e.g. in Eugénia César de Sá, A link calculus for 4-manifolds (1979), and in more modern works like the book by Gompf & Stipsicz (chapter 5.4), where it is attributed to Selman Akbulut (in works from 1977 to 1979), who in turn thanks Robion Kirby and cites his influential A calculus for framed links in $S^3$ (back then yet to appear), which doesn't talk about 1-handles in 4-manifolds though.
What intrigues me is that de Sá and Akbulut don't cite each other, so I'm somehow suspecting that maybe the notion was developed by someone else (Kirby? Rourke?) and de Sá and Akbulut used it without particularly citing it because everyone knew where it came from?.
Or is the situation yet different and Akbulut "only" developed special framed links and de Sá developed "only" the $\Gamma$-moves, which tell are the equivalence transformations of special framed links?

Here is my answer, hope it clarifies the question.

A Link Calculus for 4 Manifolds was taken from part of my PhD Thesis in Warwick, U.K, with the title Automorphisms of 3 Manifolds and Representations of 4 Manifolds, 1977, which, obviously, was written before.

This article is from the proceedings  of a Conference in Low Dimensional Topology that took place in Sussex 1977.

I knew the Kirby Calculus for 3 Manifolds, both from preprints that were circulating and from Kirby because I attended several seminars given by him.
I also knew the representation of a 1 handle as as circle with a dot from the same source. And that's all.

Kirby is referred in both my thesis and in this article.
There was a lot of "common knowledge" for orientable manifolds at the time.

All the other is original from my thesis (but I was only an unknown PhD student...)

The name Gamma moves that appear in the framed links for non-orientable manifolds and you refer to, comes from my name. A short name for Eugénia  is, in my country, Gena, so I put Gamma because it's the greek letter for G.

Eugénia  César de Sá

Can try to scan the 4 Manifold part of my thesis ( have to find the originals ☺) and send you by mail, if you want to. Or photopcopy them and post them to you


How to calculate $\lim\limits_{x\to5} 2\cdot \frac {x+2} {x-5}$ without L'Hopital? How to calculate  $\lim\limits_{x\to5} 2\cdot \frac {x+2} {x-5}$ without L'Hopital?

I'm making a lot of progress in learning how to calculate limits, but some still seem to be impossible for me. I don't want a concrete solution but if someone could provide a hint or a trick to progress in general it would be really helpful. I'm having a hard time here.

Direct substitution gives

$$\lim_{x\to5}2\cdot\frac{x+2}{x-5}=2\cdot\frac70\to\pm\infty$$

Can you check this reasoning?

A different approach..
$$
\frac{x+2}{x-5} = \frac{x-5 + 7}{x-5} = 1 + \frac{7}{x-5}
$$
So
$$
\lim_{x\to 5}\frac{x+2}{x-5} =\lim_{x\to 5} 1 + \frac{7}{x-5} = 1 +\lim_{x\to 5} \frac{7}{x-5}
$$
(if this makes more sense as we only varying the denominator)


Smallest no of balls in the box? A box contains white and black balls. When two balls are drawn without replacement suppose the probability that both are white is  1 /3.

a) Find smallest number of balls in the box ?
b) How small can the total number of balls be if black balls are even in  number  ?

I considered no of white balls as 'n' and no of black balls as 'm'.

So for first two balls to be white it would be
{n/(m+n)}*{(n-1)/(m+n-1)} = 1/3 . But how to solve further ? we have only one equation

Can anyone please explain in detail.

Thanks in advance!

There being 1 equation simply means that it has an infinite number of real solutions. But we don't want any real solution. The solutions need to be positive integers, and the problem asks for the smallest one, so you need to experiment.

Simplify the notation a bit by making $M$ the total number of balls. Your equation is $\frac{n}{M}\frac{n-1}{M-1} = \frac{1}{3}$, or $3n(n-1) = M(M - 1)$. Since everything is in integers in this last equation, you must have 3 divides either $M$ or $M-1$. What is the smallest value of $M$ such that 3 divides $M$ or $M-1$? And does that value of $M$ allow you to find an integer $n$ satisfying $3n(n-1) = M(M - 1)$?

there is another solution to this problem:

Let a and b denote the number of white and black balls in the box. We can thus define  the probability of the event as.

$$P(W_2|W_1)P(W_1) = ((a-1)/(a+b-1))\cdot (a/(a+b))$$
As $a/(a+b) > (a-1)/(a+b-1)$
Since $\min(x,y)\le GM(x,y)\le\max(x,y)$
we can say that  $((a-1)/(a+b-1))^2 < 1/3 < (a/(a+b))^2$

Solving these inequalities we can say that

$$b / (\sqrt3-1) < a <  (b-1+\sqrt3) / (\sqrt3-1)$$
if we compute $b = 1$ we find an interval of $1.366 < a < 2.366$ so we can say that $a = 2$ and the minimum number of balls is $3$.

You can compute as many $b$ as you wish...


Prove: the density operator of a pure state has exactly one non-zero eigenvalue equal to unity What is the proper way of proving : the density operator $\hat{\rho}$ of a pure state has exactly one non-zero eigenvalue and it is unity, i.e,

the density matrix takes the form (after diagonalizing):
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 \\
\end{bmatrix}}
\end{equation}

For mixed state: $\hat{\rho}=\sum \limits_{i}P_{i}|\psi_{i}\rangle\langle\psi_{i}|$

For any state: $Tr(\hat{\rho})=\sum\limits_{i}P_{i}=1$

For pure state: $\hat{\rho}=|\psi\rangle\langle\psi|$

$|\psi\rangle$ is the statevector of the system

$P_{i}$ is the probability to be in the state $|\psi_{i}\rangle$, which are the eigenvalues of the density operator.

Since $\langle \psi|\psi\rangle=1$, we have
$$
(|\psi\rangle\langle\psi|)^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|
$$
Then $|\psi\rangle\langle\psi|$ is a projection, and its eigenvalues are $0$ and $1$. Since we know that the trace of $|\psi\rangle\langle\psi|$ is one we conclude that, counting multiplicities, the eigenvalues of $|\psi\rangle\langle\psi|$ are $1,0,\ldots,0$. So its diagonal form is
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 \\
\end{bmatrix}}
\end{equation}

Do you not know how to find the eigenvalues of a matrix by solving the "characteristic equation" of the matrix?   The characteristic equation of any diagonal matrix is just the product of linear terms, each the number on the diagonal minus x so the eigenvalues [b]are[/b] the numbers on the diagonal.


Derivative of a series I am having trouble with a homework (so no full solutions please) question. Having trouble starting, so if someone could point me in the right direction, I would really appreciate it.

f(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ... Find the smallest value of n such that f$^{(n)}$ = f(x).

I am unsure how to handle this since it appears to be a series.

For power series like this, it turns out that you can differentiate term-by-term; the formal justification takes a bit of work - but for this series, the radius of convergence is infinite, so differentiating termwise will work.

To get you started, note that

\begin{align*}
f'(x) &= \frac{d}{dx} x - \frac{d}{dx} \frac{x^3}{3!} + \frac{d}{dx} \frac{x^5}{5!} - \dots \\
&= 1 - \frac{3 x^2}{3!} + \frac{5x^4}{5!} - \dots \\
&= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
\end{align*}


Set of functions from $SS = \{ A_{1}, A_{2}, A_{3},...\}$ to $\{ \mathbf{T}, \mathbf{F} \} $ is countable? Let $SS=\{ A_1,A_2,A_3,\ldots\}$, and let $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ .Is the set  V countable?  Justify your answer.
My instinct is to say that $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ is uncountable, and to prove this using a diagonalization argument, i.e. create a table of values of the functions $v_{1}, v_{2},...$ for natural numbers $n_1, n_2,\ldots$ and define a function $v_{m} : SS \to \{ \mathbf{T}, \mathbf{F} \}$ where it takes all values in the diagonals of the table $T, F$ and flips them, and then show that $v_{m}$ cannot appear anywhere in the list. Is my guess correct, and if so would this be a reasonable approach to the problem?

Your approach is correct. Note that $V$ is bijective to the power set of $SS$ by mapping $v \in V$ to $v^{-1}(\mathbf{T}) \in 2^{SS}$, the subset of $SS$ that maps to $\mathbf{T}$.

Yes, this is (essentially) the countably infinite case of Cantor's Theorem.


Why the expectation of distance to center of disk is $r/3$ and not $r/2$? A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $\frac{r}{3}$, whereas I found $\frac{r}{2}$ as follow

We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$\mathbb E[Y]=\int_0^r (r-x)f_X(x)dx=\frac{1}{r}\int_0^r (r-x)dx=\frac{r}{2}.$$

Maybe there is a subtlety than I don't see ?

$$\frac{\int_0^r(r-x)x.dx}{\int_0^r x.dx}$$

This gives $r/3$.

The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2\pi x.dx$ ... but then the $2\pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.

Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.


Let $f$ and $g$ be two group homomorphisms from $G$ to $G'$. Is $H$ a subgroup of $G$? Can someone please verify this?

  Let $f$ and $g$ be two group homomorphisms from $G$ to $G'$. Let $H \subset G$ be the subset $\{x \in G: f(x)=g(x)\}$. Is $H$ a subgroup of $G$?

Let $e$ and $e'$ denote the identity elements of $G$ and $G'$, respectively. Using the fact that a homomorphism maps the identity to the identity, we have

\begin{eqnarray}
f(e) = g(e) = e'
\end{eqnarray}

Therefore, $e \in H$. Now, let $x \in H$. Then,

\begin{eqnarray}
f(x) &=& g(x) \\
\end{eqnarray}
Since we proved that $e \in H$,
\begin{eqnarray}
f(x \cdot x^{-1}) &=& g(x \cdot x^{-1}) \\
f(x)*f(x^{-1})&=&g(x)*g(x^{-1}) \\
f(x)*f(x^{-1}) &=& f(x)*g(x^{-1}) \\
f(x^{-1})&=&g(x^{-1})
\end{eqnarray}
This implies that $x^{-1} \in H$. Now, let $x$ and $y$ be elements in $H$. Then,

\begin{eqnarray}
f(x \cdot y) &=& f(x)*f(y)\\
&=& g(x)*g(y)\\
&=& g(x \cdot y)
\end{eqnarray}

So, $x \cdot y \in H$.

Since $H$ is closed under the operation $\cdot$, contains the identity and inverse elements, it is a subgroup of G. The associativity of $\cdot$ in $H$ carries over from the associativity of $\cdot$ in $G$.

I'd like to point out a more abstract answer, which only works for abelian groups (a similar argument as in the comment is described beneath):

Let's consider the set $Hom(G, G')$ consisting of all the group homomorphisms from $G$ to $G'$ of abelian groups. This becomes a group if we define $(f*g)(x) := f(x) * g(x)$ for all $f, g \in Hom(G, G')$ and all $x \in G$.
(Maybe you already learned that. If not, then proving that $Hom(G, G')$ is a group is as difficult/easy as your prove that $H$ is a subgroup of $G$)

Now we have $H = \{x \in G | f(x) = g(x)\} = \{x \in G \mid (f*g^{-1})(x) = 1)\} = ker(f*g^{-1})$, were $ker(h)$ means the kernel of $h \in Hom(G, G')$. A kernel of a group homomorphism is always a (even normal) subgroup.

Another proof follows the idea of k.stm. It works for all groups:

Let $f \times g: G \to G' \times G’$ be the diagonal map (which is again a group homomorphism) given by $x \mapsto (f(x), g(x))$. $G' \times G'$ is a group by componentwise multiplication. Now the proof consists of the following parts:

$1.$ Proving that the diagonal $\Delta = \{(y, y) \mid y \in G'\}$ is a subgroup of $G' \times G'$

$2.$ Proving the equality $(f \times g)^{-1}(\Delta) = H$

$3.$ Proving or knowing that the preimage of a subgroup under a group homomorphism is a subgroup.


Find $k$ such that $\lim_{n\to\infty}\tan n/n^k=0$. Find $k\in\mathbb{R}$ such that $$\lim_{n\to\infty}\frac{\tan n}{n^k}=0.$$

Does such a $k$ exist? And is it possible to find all such $k$?

Edit: It is $\lim_{n\to\infty}\tan n/n^k=0$, the limit of the sequence, instead of $\lim_{n\to\infty}\tan x/x^k=0$. I was wrong. I'm really sorry for the trouble I caused...

I expect you really mean $$\lim_{n\to\infty}\frac{\tan n}{n^k}=0.$$
Here the limit is over integers, not reals. The difficulty here is that
an integer $n$ is close to an odd multiple of $\pi/2$. If that is so
then $|n-m\pi/2|$ is small and then $|\tan n|\approx|n-m\pi/2|^{-1}$.
If we had a theorem stating that $|n-m\pi/2|>Cn^{-r}$ then for $k>r$
your limit would be zero. This leads to the "irrationality measure of $\pi$". Theorems of this nature are known but difficult. I am not up to speed on recent developments, but you might want to start with
Zudilin's survey.

An easy proof of this is given by an intuitive reasoning using the definition of a limit. We write:

$$\lim_{x\rightarrow\infty}f(x)=L \text{ iff } \forall\epsilon>0, \exists N \text{ s.t. } \forall x\geq N, |L-f(x)|<\epsilon$$

In this case:

$$L=0, f(x)=\frac{\tan(x)}{x^k}$$

And let's have $\epsilon=1$, an easy number. Now, all we have to do is prove that there is no $N$ s.t. for all $x\geq N$,

$$\left|-\frac{\tan(x)}{x^k}\right|<1$$

Remember, $1$ is our $\epsilon$.

Proof by contradiction. Suppose $N$ exists. Then, $N$ cannot be represented by $\pi/2+k_1\pi$, $k_1 \in \mathbb{N}$, because that our above statement is undefined and therefore not less than $1$. In fact, if our function in the absolute value sign (let's call it $A(x)$ for convenience), is undefined at any point $c$ larger than $N$, it isn't smaller than $1$.

Since $N$ cannot be represented by $\pi/2+k_1\pi$ for all $k_1$, it can be said that:

$$\frac{\pi}2+m\pi < N < \frac{\pi}{2}+(m+1)\pi, m\in\mathbb{N}$$

Let's have $v=\frac{\pi}{2}+(m+1)\pi$

Remember two things:

$v>N$
$A(x)$ is undefined at $v$, because $A(x)$ is undefined at all points $\pi/2+k_1\pi$ for all $k_1 \in \mathbb{N}$

This is a contradiction.

So, there does exist an $\epsilon$ such that no $N$ exists to fill the requirement, no matter what $k$ is.

This means that it is not the case that for all epsilon that there exists an $N$ to fufill the requirement.

Q.E.D

This is really rigorous, but that, in my opinion, is the best kind of proof. It's like responding to the YouTube comments with your pretentious a****** vocabulary. Can I swear on StackExchange? I should check. Bleeping it out anyway, I guess.

Have a nice day, whoever's reading this.

OH NO, THE STACK EXCHANGE POLICE ARE BREAKING DOWN THE DOOR! THEY'RE CO


simple vector problem Let V=(0,3,2) and A=(-1,0,1). Let P be the point on the line passing through A with direction vector (1,1,1) that is closest to V. Find AP, P, and the distance between V and the line.
This is a pretty basic question, but I haven't done any problem like these. Can someone give me a pointer?
For my try, I got an answer of (3,0,2) which is wrong.

The equation of line containing P is
$\textbf r=(-1,0,1)+\lambda (1,1,1) $
This gives the point P as $(\lambda-1,\lambda,\lambda+1)$
Now P is nearer to V when the direction ratio of vector joining V and P is perpendicular to direction ratio of the line.
This gives,
$(\lambda-1,\lambda-3,\lambda-1)\cdot(1,1,1)=0$
Using this you can find $\lambda$ and hence the point P.


Inverse of linear tranformation Let $T$ be a linear transformation from $\mathbb R^n$ to $\mathbb R^n$.Let ${x_{1},x_{2},......,x_{n}}$ be a basis of $\mathbb R^n$ suppose that $Tx_{i}\neq{0}$. Then we say that that $\mathrm{null}(T)=0$ and $\mathrm{rank}(T)=n$. Thus $T$ is invertible. This is true.

Well, this is not true for $n>1$.

Take e.g. $v\in\mathbb R^n\setminus\{0\}$ and define $T$ s.t. $$Tx_i=v\text{ for all }i\in\{1,\dots,n\}.$$.

Then $\mathrm{img}(T)=\mathrm{span}(v)\subsetneq\mathbb R^n$ and thus $\mathrm{rank}(T)=1<n$ and followingly $\mathrm{null}(T)=n-\mathrm{rank}(T)=n-1>0$.

consider T for ex. $ \begin{vmatrix}  A_1 &  A_2 &  A_3 \\ B_1 & B_2 &  B_3\\C_1 & C_2 &  C_3 \end{vmatrix}$, also consider the basis B $ \begin{vmatrix}  1 &  0 &  0 \\ 0 & 1 &  0\\0 & 0 &  1 \end{vmatrix}$, clearly $TB=T$, $Tx_i\ne0$ but column vectors of T can be completely in some 2d plane or 1d, for ex. they can lie completely in xy plane. So $TB$ is not a basis then. if $TB$ is a basis, then off course T is full rank. Let $TB=C$ so $T=CB^{-1}$ as $C,B^{-1}$ both are full rank then T is also full rank.


Uniform convergence of $\sqrt{x^2+n^2}^{-1}-n^{-1}$ Assume the following sum:
$$
\sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{x^2+n^2}}-\frac{1}{n}\right),
$$
where $x \in \mathbb{R}$. The problem is to determine whether the sum converges uniformly, i.e. whether there exists $s(x)$ such as $\sum f_n(x) \rightrightarrows s(x)$.
My approach is following: I show $f_n \rightarrow 0$ for every $x$, as
$$
f_n \sim \frac{1}{n^3}, \lim_{n \rightarrow \infty} \frac{f_n}{n^{-3}} = -\frac{x^2}{2}.
$$
Then I can show $f_n \rightrightarrows 0$, as
$$
|f_n(x)| \leq \frac{1}{n}, \forall x,
$$
thus the sequence $f_n$ converges uniformly to zero, and here I got stuck.
Is my approach correct? Could you help me find out on which interval the sum converges uniformly? Is the convergence absolute/local? What is the prescription of $s(x)$?

Thank you for any advice.

For fixed $x$,

$$f_n(x)=\frac{1}{n}((1+(\frac{x}{n})^2)^{-\frac{1}{2}})-\frac{1}{n}$$

$$=\frac{1}{n}(1-\frac{1}{2}\frac{x^2}{n^2}(1+\epsilon(n)))-\frac{1}{n}$$

$$=-\frac{x^2}{2n^3}(1+\epsilon(n))$$

thus the series $\sum f_n(x)$ converges.

Now let $a>0$.

for large enough $n$

$\forall x\in [-a,a]$

$|f_n(x)|\leq \frac{a^2}{2n^3}$

thus the series converges uniformly at

$ [-a,a]$ and also at $[a,b]$ but not on $\mathbb R.$


Bijective map from a set to a subset of reals? There is a concept that I have been thinking about quite a lot lately as I am currently self-studying point-set topology:
Say we have a bijective map from one interval, $[a,b]$, to another interval, $[c,d]$, both of which are in $\mathbb{R}$. Also set $c$ and $d$ so that $[c,d] \subseteq [a,b]$.
How can it be that function maps to a subset which is a proper subset of the map's preimage bijectively? I.e. How can the map be both one-to-one and onto when the image should contain "less" elements than the domain?
One example would be $f(x) := \frac{x}{1+x}: [0,10] \to [0, \frac{10}{11}]$
I'm hoping someone can show me why this isn't such a strange concept? Is there a theorem or result that explains this or provides some intuition?

The same applies to any infinite set. In fact "infinite set" can can be defined as a set that contains a proper subset with the same cardinality. More precisely these are known as Dedekind-infinite sets.
For example take naturals $\mathbb{N}=\{0,1,2,3,\ldots\}$ and its proper subset $\mathbb{N}_+=\{1,2,3,\ldots\}$ and note that there's a simple bijection between them $x\mapsto x+1$. You can even remove infinitely many elements from $\mathbb{N}$ and still end up with the same cardinality, e.g. for $2\mathbb{N}=\{0,2,4,6,\ldots\}$ we have a bijection $x\mapsto 2x$ even though there are infinitely many elements in $\mathbb{N}\backslash 2\mathbb{N}$.
And so "being a subset" and "being equinumerous" are loosely related concepts. At most we know that $|A|\leq |B|$ when $A\subseteq B$. But $A\subsetneq B$ doesn't imply $|A|<|B|$, unless $B$ is finite.

I'm hoping someone can show me why this isn't such a strange concept?

The idea may be strange to you. Infinities are weird. But most people simply accept that and move on. There's not really anything more to do about it. After some time you get used to it and it becomes a simple fact of mathematical reality.

Consider as an example there exists a bijection between $\mathbb{Z_+}$ and $\mathbb{Z}$.  While these are subsets of $\mathbb{R}$, the concept is the same.
Let $f: \mathbb{Z} \to \mathbb{Z_+}$ defined by
$f(i) = \{1$ if $i = 0$; $2i$ if $i$ is positive; $1-2i$ if $i$ is negative


Complicated Percentage Question In a certain town some people were affected by a ’flu’ epidemic. In the first
month $20$% of the population contracted the flu whilst $80$% were healthy.
In the following month $20$% of the sick people recovered and $20$% of the
healthy people contracted the disease.
What fraction of the population is healthy at the end of the second month?

Can someone help me tackle this problem as I am struggling to proceed.

$\frac15$  (20%) of $\frac45$ (80% of affected are still affected) plus $\frac45$ (healthy after first month) of $\frac15 $ (20% became sick):
$$\frac15\cdot \frac45+\frac45\cdot\frac15 = \frac8{25} = 32\%$$
 32% are affected;  68% are not.

$20\%$ of the $20\%$ of the population that was sick became healthy.

$80\%$ of the $80\%$ of the population that was healthy stayed healthy.

Multiply then add.  It's not that complicated.


Differentiating the term, $y=\sqrt{\frac{1-x^2}{1-x}}$ In my calculus book, the above question was given. Since the the term can be simplified, I simplified it to $$\sqrt{(1+x)}$$ which differentiates to $$0.5\sqrt{{\frac{1}{1+x}}}$$ But the answer in the book is $$0.5\sqrt{\frac{1-x}{1-x^2}}$$  I know, if we simplify it gives the same answer.  Then why didn't the author simplify the answer? Is there any special reason?

Hint What if $x=1$ would you do like $\frac{0\times 2}{0}$ and then cancel $0$ .There are indeterminate forms so its safe to keep it like that. Hint you can also make that function continuous by removing singularity at $x=1$ ie(removal of discontinuity at $1$).

Ted's comment is in fact the answer: you didn't differentiate $\sqrt{x}$ correctly. You made it $\frac{1}{2} \sqrt{x}$; it is in fact $\frac{1}{2 \sqrt{x}}$.

Although actually the answer in the book is also incorrect. The author should probably have simplified an expression of the form $\frac{1-x^2}{1-x}$, since the singularity at $x=1$ is removable.


Why Laplacian Matrix need normalization and how come the sqrt of Degree Matrix? Why Laplacian matrix needs normalization and how come the sqrt-power of degree matrix? The symmetric normalized Laplacian matrix is defined as
$$\ L = D^{1/2}AD^{-1/2}$$
where L is Laplacian matrix, A is adjacent matrix. Element $A_{ij}$ represents a measure of the similarity between data points with indices $i$ and $j$. D is diagonal matrix, defined as $\ D = \sum\limits_i A_{ij}$. In other words, it is the sum of similarity from node $j$ to all its neighbor node $i$.
I can't figure out some question about the formula:

Why we need to normalize Laplacian Matrix like that? $\ {1/2}$-power ?
Is there any special reason?
Is there any references?

Unnormalized Laplacian serve in the approximation of the minimization of RatioCut, while normalized Laplacian serve in the approximation of the minimization of NCut.
Basically, in the unnormalized case, you optimize an objective relative to the number of nodes in each cluster.
And, in the normalized case, you optimize the objective relative to the volume of each cluster.
The square root comes from this: $f^\top(I-D^{-1}A)f = g^\top(I-D^{-1/2}AD^{-1/2})g$ where $g=D^{1/2}f$.
So, if your optimization method relies on a symmetric PSD matrix, you use the symmetric normalized Laplacian, and then remove the bias by computing $D^{-1/2}g$ to recover $f$.

More insight about RatioCut and NCut.
Let $G=(V,E)$ be a graph with vertex set $V=\{v_1,\dots,v_n\}$ and edge set $E$, $w_{ij}$ denote the positive weight on the edge between $v_i$ and $v_j$. Let $\{C_i:1 \le i \le k\}$ be a disjoint partition of $V$. Define
$$
Cut(C_i:1\le i\le k)\triangleq\frac{1}{2}\sum_{c=1}^k \sum_{i \in C_c,j\in \bar C_c}A_{ij}
$$
$$
RatioCut(A_i:1\le i \le k)\triangleq\sum_{i=1}^k\frac{Cut(A_i,\bar A_i)}{|A_i|}
$$
$$
NCut(A_i:1\le i \le k)\triangleq\sum_{i=1}^k\frac{Cut(A_i,\bar A_i)}{vol(A_i)}
$$
RatioCut:
Let
$$\tag{1}\label{1}
f_i=\begin{cases}\sqrt{|\bar A|/|A|} & \text{if } v_i\in A\\\sqrt{|A|/|\bar A|} & \text{if } v_i \in \bar A\end{cases}
$$
\begin{align*}
f^\top L f & = \frac{1}{2}\sum_{i,j=1}^n w_{ij}(f_i-f_j)^2 \\
& = \frac{1}{2}\sum_{i\in A,j\in \bar A}w_{ij}\left(\sqrt{\frac{|\bar A|}{|A|}}+\sqrt{\frac{|A|}{|\bar A|}}\right) + \frac{1}{2}\sum_{i\in \bar A,j\in A}w_{ij}\left(-\sqrt{\frac{|\bar A|}{|A|}}-\sqrt{\frac{|A|}{|\bar A|}}\right) \\
&=Cut(A,\bar A)\left(\frac{|\bar A|}{|A|}+\frac{|A|}{|\bar A|}+2\right)\\
&=Cut(A,\bar A)\left(\frac{|A|+|\bar A|}{|A|}+\frac{|A|+|\bar A|}{|\bar A|}\right)\\
&=|V| RatioCut(A,\bar A)
\end{align*}
You can also see that $\sum_{i=1}^n f_i=0$, so $f\perp \mathbb 1$.
So minimizing RatioCut is equivalent to the following problem:
$$
\min_{A\subset V}f^\top L f\text{ subject to $f\perp\mathbb 1$, $f_i$ as defined in Eq.\eqref{1}},\|f\|^2=n
$$
NCut:
For this case, define
$$\tag{2}\label{2}
f_i=\begin{cases}
\sqrt{\frac{vol(\bar A)}{vol(A)}} &\text{if $v_i\in A$}\\
\sqrt{\frac{-vol(A)}{vol(\bar A)}} &\text{if $v_i\in \bar A$}\\
\end{cases}
$$
Similar to above, we have $Df\perp \mathbb 1$, $f^\top D f=vol(V)$ and $f^\top L f=vol(V)NCut(A,\bar A)$.
Minimizing NCut is equivalent to
$$
\min_{A\subset V} f^\top L f \text{ subject to $f$ as in Eq.\eqref{2}, $Df\perp \mathbb 1$ and $f^\top Df=vol(V)$}
$$
Substituting $g=D^{1/2}f$ we get
$$
\min_{A\subset V} g^\top D^{-1/2}LD^{-1/2} g \text{ subject to $g=D^{1/2}f$, $f$ as in Eq.\eqref{2}, $g\perp D^{1/2}\mathbb 1$ and $\|g\|^2=vol(V)$}
$$
Then observe that $D^{-1/2}LD^{-1/2}$ is the symmetric normalized Laplacian.

Note that relaxed version of NCut problem is that of minimization of generalized eigenproblem

$Lx = \lambda Dx$

consult the paper by Shi and Malik for a justification.

To solve this problem, i.e. to find eigenvector $x$ and eigenvalue $\lambda$, multiply both sides of this equation by $D^{-1/2}$:

$D^{-1/2}Lx=\lambda D^{1/2}x$

This equation can be rewritten as

$D^{-1/2}LD^{-1/2}D^{1/2}x=\lambda D^{1/2}x$

Setting $y=D^{1/2}x$ you obtain

$D^{-1/2}LD^{-1/2}y=\lambda y$

that is you reduced the generalized eigenproblem $Lx=\lambda Dx$ to the standard eigenproblem $\mathcal{L} y = \lambda y$, where $\mathcal{L} = D^{-1/2}LD^{-1/2}$ is the normalized Laplacian matrix.

Knowing eigenvector $y$ you can find the eigenvector $x=D^{-1/2}y$.


Is there a(n elementary) function whose derivative we cannot integrate? Say, for example, I take a reasonably-complicated function $f(x)=\tanh[\ln(x^x)]$, and differentiate it to get $$f'(x)=\frac{4x^{2x}
[1+\ln(x)]}{(x^{2x}+1)^2}.$$

Now, to integrate this, I imagine, would be very difficult and time-consuming.

My question is: does there exist a function $f$ whose derivative $f'$ we can't integrate (without differentiating $f$ in the first place), using substitution, parts and/or partial fractions (or other integration methods)?

My motivation for this is that it's very easy to differentiate even a ridiculously-complicated function (using the chain and/or product rule), but I've often wondered whether we could get back to the original function by integrating this derivative.

If I'm not articulating myself clearly enough, please ask me to explain further.

Thanks!

Edit

I know, from the Fundamental Theorem of Calculus, that such a function can be integrated, but, other than knowing the fact that this is the derivative of a suitable function, could it be impossible to reverse-engineer the problem, to get $f'$ back to $f$ (using only methods of integration)?

The answer is no.

While the methods you learn in calculus like partial fractions, integration by parts, etc. are going to be limited, there is a complete algorithm, called the Risch algorithm for integrating elementary functions. Complete means that for any elementary function, the algorithm will either find an elementary antiderivative, or prove that none exists (note that an antiderivative always exists, but it will not always be an elementary function, for instance, it's well known that $\int e^{-x^2}\, dx$ is not an elementary function).

Since the derivative of an elementary function is always an elementary function, the Risch algorithm applied to any derivative of an elementary function will always work. It may not produce exactly the same function back, since the antiderivative is only defined up to an additive constant (i.e., you might get your original function back plus $C$).

Note that the Risch algorithm itself is very complicated, and involves some deep algebra, and a ton of different technical cases. It's much more complicated than the tricks you learn in calculus.

Many computer algebra systems implement some subset of the Risch algorithm. For instance, Mathematica (or Wolfram Alpha) should be able to integrate just about any derivative of an elementary function that you throw at it.

If you're interested in a high level view of the Risch algorithm, I recommend Manuel Bronstein's Symbolic integration tutorial. If you want a more detailed view (with pseudocode to implement in a compute algebra system), I recommend his book, Symbolic Integration I: Transcendental Functions.

The fundamental question is: "Given a function, does its integral exist?" If the function F' is created by taking the derivative of F, then, obviously, the integral of F' MUST exist, as F has already been proven to exist.


Minkowski sum of a positive Lebesgue measure set and $\mathbb{Q}$. Let $A\subset \mathbb{R}$ be of positive Lebesgue measure, i.e. $\mu(A)>0$. Is it then true that $\mu(\mathbb{R}\setminus (A+\mathbb{Q})) = 0$?

I am quite sure that if $\mu(A)>0$, then $A-A$ contains a rational, as well as an irrational number, but I'm not sure if this actually helps.

Suppose to the contrary that the set $ \mathbb R\setminus (A+\mathbb Q)$ has positive measure. By the Lebesgue density theorem, it has a point of density, call it  $x$. Also pick a point of density of $A$, call it $a$. For sufficiently small $r>0$, we have
$$
\mu((x-2r,x+2r)\cap (A+\mathbb Q))<r
$$
and
$$
\mu((a-r,a+r)\cap A)> r
$$
Let $q$ be a rational number such that $|a+q-x|<r$. Then
$$((a-r,a+r)\cap A)+q \,\subseteq \,(x-2r,x+2r)\cap (A+q)  $$
but the set on the left has greater measure, a contradiction.

This proof is for measurable $ A $ :

It suffice to prove it for some measurable $ B \subset A $ and $ B\subseteq I $ where $I$ is an interval and $ m (B) > 0$:

There some measurable $ S$ with $ m (S) > 0$  and by Lebesgue density theorem there some $ x \in S \subseteq I $ and some $ 0<r<\frac{m(I)}{2},1>\epsilon > 0$ such that $ I_r=(x-r, x+r)$,  $ m (S \bigcap I_r)  > (1-\epsilon)m (I_r)$

It also obvious that for $ S_q =S + q $, $ I_{r, q}=I_r + q $,$ m(S_q)=m (S) , m(I_{r, q})=m (I_r) $, and $ (S_q\bigcap I_{r, q})  >(1-\epsilon)m (I_r)$ ,where $q \in \mathbb {Q} $

Now we can take subsequence  {$ I_{r, q} $}that is pairwise disjoint such that $\bigcup_q I_{r, q} \supseteq I $

Define $ B = S \bigcap I_r $ ,  $ B_q=B+q $, It follows $ m(\bigcup_q B_{q})  >(1-\epsilon)m (I)$

$ m(I-\bigcup_q B_{q}) = m (I) - m (  I\bigcap\bigcup_q B_{q}) $

$ m(I-\bigcup_q B_{q}) \le \epsilon m (I)+ 2r $

$A_q=A+q$

$ I- \bigcup_q A_q \subseteq I-\bigcup_q B_{q} $

Therefore $ m(I-\bigcup_q A_q) \le \epsilon m (I)+2r $

Letting $r,\epsilon \to 0$ the result follows


Kernel of ring homomorphism is an ideal I am asked to show that if f is a ring homomorphism from R to R' then kernel of f is an ideal of R.

According to definition of ideal : A non empty subset of R is an ideal for any two elements of ideal their substraction must be in that ideal and
the product of any element of R and an element of ideal must be in ideal,
I am not able to prove second condition. Please help

Let $f: R\rightarrow R'$ be a ring homomorphism. We assume that $R$ and $R'$ both have an identity. Since $0 \in\ker(f)=\{x\in R\mid f(x)=0\}, \ker(f)$ is non-empty.
Let $u,v \in \ker(f), r \in R$, then $f(ru)=f(r)f(u)=0,\ f(ur)=f(u)f(r)=0,\ f(u-v)=f(u)-f(v)=0$.

Let $\varphi: (R_1, +_1, \circ_1) \rightarrow (R_2, +_2, \circ_2)$ be a ring homomorphism.
Then the kernel of $\varphi$ is an ideal of $R_1$.
Proof:
By Kernel of Ring Homomorphism is Subring, $\ker(\varphi)$ is a subring of $R_1$.
Let $s \in \ker(\varphi)$, so $\varphi(s) = 0$ in $R_2$.
Suppose $x \in R_1$. Then:
\begin{align*}
\varphi(x \circ_1 s) & = \varphi(x) \circ_2 \varphi(s) & \text{Definition of Morphism Property} \\
& = \varphi(x) \circ_2 0_{R_2} & \text{as}\ s \in \ker(\varphi)  \\
& = 0_{R_2} & \text{Properties of}\ 0_{R_2}
\end{align*}
and similarly for $\varphi(s \circ_1 x)$.
The result follows.


How do I compute the rank of this matrix involving trigonometric functions? When is the rank of this matrix equal to $2$?
$\begin{bmatrix}-\sin u + \sin (u+v) & \sin(u+v) \\ \cos u - \cos(u+v) & -\cos(u+v)\\ 1 & 0 \end{bmatrix}$
I’ve computed the two $2 \times 2$ determinants, which give $\sin v$ and $-\cos(u+v)$. However, if I equal them to $0$, I obtain $v=\pi n$ for the first case and $v+u=\pi/2 +\pi n$ for the second one (with $n\in \mathbb{N}$). How should I conclude?

I think it is easier to compute the kernel of the matrix. So, let $v_1 = \begin{pmatrix} -\sin u + \sin(u+v) \\ \cos u - \cos(u+v) \\ 1 \end{pmatrix}$ and $v_2 = \begin{pmatrix} \sin(u+v) \\ -\cos(u+v) \\ 0 \end{pmatrix}$ and consider the equation
$$\lambda_1v_1 + \lambda_2v_2 = 0.$$
The matrix has maximal rank (rank $2$ in this case) if and only if this equation only has the trivial solution $\lambda_1 = \lambda_2 = 0$. From the third row we get
$$\lambda_1 + 0 = 0,$$
hence $\lambda_1 = 0$ and $\lambda_2v_2 = 0$. As $v_2$ is never $0$, this implies that $\lambda_2 = 0$ as well. Thus your matrix always has rank $2$.

Note that rank of this matrix is at least one (last row) and at most two (because its dimension is 3×2).
Now, you need to check when matrix rank is 1. If it is one, then first and second rows has to be linearly dependent on the last row, therefore $\sin(u+v) = -\cos(u+v) = 0$, but this can never happen as $\sin^2 x + \cos^2 x = 1$.
Therefore, this matrix has rank 2 for all values of $u$ and $v$


Clarifying meaning of "mod" I'm having trouble understanding exactly how $mod$ works. I know it's remainder, but I guess my problem is in how to write it out.

For example, $\frac{1}{3} = 0$ remainder $1$.

Would I write $1 = 0(mod 3)$, $1 = 1 (mod 3)$...?

Edit: So, the base goes in the (mod) brackets, the number being divided is on the left, and the remainder is on the right?

Though the question is already “answered”, it’s worth noting there are two uses of $\bmod$:

$a\bmod b$. This is a number, the remainder when dividing $a$ by $b$.
$a\equiv a'\pmod b$. This is a statement “$a$ is congruent to $a'$ modulo $b$”, that is, $a-a'$ is a multiple of $b$. Here, $a'$ can, but doesn’t need to be equal to $a\bmod b$. For example, $3\equiv 7\pmod2$. The number $b$ is called modulus in this case.

Sometimes, $a\pmod b$ is used. It means a residue class modulo $b$ which contains $a$, that is, a set of all $a'$ such that $a\equiv a'\pmod b$.

By the way, in the latter case, one can also write many congruences together, e. g. $28\equiv 8\equiv -12\equiv 578\pmod{10}$, like it’s done with $=$ or ordering relations.

You can also speak of a zero modulus: $a\equiv a'\pmod0$ means that $a-a'$ is a multiple of $0$—that is, $a = a'$. On the other hand, typically $a\bmod0$ is treated as undefined.

Based of its definition, Modulo operation is to find the "REMAINDER AFTER DIVISION"
Since it's target is the remainder, "You mod something" = "the remainder".

You mod something is between something which in turns produce that remainder. Obviously its division. Now who divided by who gets that remainder? Those in the mod would by definition be equivalent to what we can call the BASE.

How would you figure out for the rest? :D


Is there any trick to calculate $\int\ln(x+\sqrt{x^2-1})dx$ faster? $$\int\ln(x+\sqrt{x^2-1})dx$$
To evaluate the integral we can apply integral by parts (using $u=\ln(x+\sqrt{x^2-1})$ and $dv=dx$). but my question is: is there other approach to evaluate the integral faster? I think there might be something because we can write it as:
$$\ln\left(x+\sqrt{x^2-1}\times\dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right)=\ln\left(\cfrac1{x-\sqrt{x^2-1}}\right)=-\ln(x-\sqrt{x^2-1})$$
But I don't know how to use this.

Use $x=\cosh t$
$$\int \sinh t \ln(\cosh t + \sinh t) dt = \int t \sinh t dt = t\cosh t - \sinh t$$
then substitute back.

Here is the solution of your integral after you simplified it. I used Symbolab


derivative over nested integrals I have the following problem. First an example for two-variable functions

Let $A=A(t,\tau)$ and $B=B(t,\tau)$

If I want to compute the following

$\frac{\mathrm{d}}{\mathrm{d}t} \int_0^t \mathrm{d}\tau A(t,\tau) B(t,\tau)$ one can apply the following rule

\begin{equation}\frac{\mathrm{d}}{\mathrm{d}t} \int_0^t \mathrm{d}\tau A(t,\tau) B(t,\tau) =  \int_0^t \mathrm{d}\tau \Big( \frac{\partial}{\partial t} A \Big) B + \int_0^t \mathrm{d}\tau A \Big( \frac{\partial}{\partial t} B \Big) + \Big( A(t,\tau) B(t,\tau)\Big)\rvert_{\tau=t} \end{equation}

Now the problem: the functions depend on four time arguments, that is

$A=A(t,\tau,\tau_1,\tau_2)$ and $B=B(t,\tau,\tau_1,\tau_2)$

and I want to compute

\begin{equation}  \frac{\mathrm{d}}{\mathrm{d}t} \int_0^t \mathrm{d}\tau \int_\tau^t \mathrm{d}\tau_1 \int_\tau^{\tau_1} \mathrm{d}\tau_2~ A(t,\tau,\tau_1,\tau_2) B(t,\tau,\tau_1,\tau_2)     \end{equation}

What would be the rule for this case?

I would really appreciate your help!!!

Stefan

Think of it as

$$\frac{\mathrm{d}}{\mathrm{d}t} \int_0^t \mathrm{d}\tau F(t,\tau,\tau_1)$$

with

$$
F(t,\tau,\tau_1)=\int_\tau^t \mathrm{d}\tau_1 \int_\tau^{\tau_1} \mathrm{d}\tau_2~ A(t,\tau,\tau_1,\tau_2) B(t,\tau,\tau_1,\tau_2)
$$

and apply the same principles that you applied in the other case.

As a general principle to avoid mistakes, I would recommend to compute
$$ F(t+\delta t)-F(t)$$
and expand it to the first order in $\delta t$.

In your case, for $F(t)=\int_0^t f(t,\tau)\mathrm d\tau$ this gives
$$F(t+\delta t)-F(t)=\int_0^t \left[f(t+\delta t,\tau)-f(t,\tau)\right]\mathrm d\tau+\int_t^{t+\delta t}f(t+\delta t,\tau)\mathrm d\tau.$$
From this you deduce immediately that $F'(t)=\int_0^t\partial_tf(t,\tau)\mathrm d\tau+f(t,t)$.


Graph theory: adjacency vs incident Okay, so I think if 2 vertices are adjacent to each other, they are incident to each other....or do I have it wrong? Is this just different terminology. I thought I was totally clear on this for my class, but now I am doubting myself reading the book and looking at my notes. I just want to know if I have it correct, and if I don't could someone explain to me what the difference is between the two. I found several wiki's and different university definitions, but none ever said that the two are alike and I'm confused and would like some reassurance. Thanks in advance.

Usually one speaks of adjacent vertices, but of incident edges.

Two vertices are called adjacent if they are connected by an edge.
Two edges are called incident, if they share a vertex.

Also, a vertex and an edge are called incident, if the vertex is one of the two vertices the edge connects.

An edge "e" in a graph (Undirected or directed ) that is associated with the pair of vertices n and q is said to be incident on n and q, and n and q are said to be incident on e and to be adjacent vertices.