An Expected Value problem II

We saw the explicit expression for the expected distance from a random point inside a rectangle to an edge of the rectangle in the previous post. We deal with a similar problem here.

There are a lot of reference in the internet to find the expected distance between two random points in a rectangle (Stack Exchange post). Here we ask 'what is the expected distance between two random points in a right angled triangle?'.

Basically the question is conceptually simple and the expression for the expected value can be readily written as follows:

$\mathbb{E}_T(L)=\displaystyle\int\limits_0^a \int\limits_0^{mx}\int\limits_0^a\int\limits_0^{mu}\sqrt{(x-u)^2+(y-v)^2}\,dv\,du\,dy\,dx$

where the right angled triangle is formed by the points $(0, 0)$, $(a, 0)$, and $(a, b)$. Also, for simplicity we used $m=b/a$.

Now, the difficulty of this integral (and the one that makes it different from the rectangle case) is the variable limits of the integral. In case of a rectangle, all the limits would be constants which makes it a little easier.

I'm not going to go through all the details and trouble I went through to solve this. But it is a combination of both Mathematica and a lot of paper work. Finally, after a few days and hell-a-lot completely messed up ideas later, I ended up with the following explicit expression for the expected value.

$\mathbb{E}_T(L)=\displaystyle\frac{a^3+b^3+2 d^3}{15 d^2}+\frac{a^2 }{15 b}\left(\frac{b^3}{d^3}+1\right)\text{csch}^{-1}\left(\frac{a}{b}\right)+ \frac{b^2}{15 a}\left(\frac{a^3}{d^3}+1\right) \text{csch}^{-1}\left(\frac{b}{a}\right)$

where $d=\sqrt{a^2+b^2}$.

I also wrote a small python program in an Online IDE to verify the correctness of this result.

For example, for $a=2$ and $b=3$, the formula gives

$\mathbb{E}_T(L)=\displaystyle\frac{1}{195} \left(35+26 \sqrt{13}\right)+\frac{4}{45} \left(1+\frac{27}{13 \sqrt{13}}\right) \text{csch}^{-1}\left(\frac{2}{3}\right)+\frac{3}{10} \left(1+\frac{8}{13\sqrt{13}}\right) \text{csch}^{-1}\left(\frac{3}{2}\right) \approx 1.047156999$

whereas the simulation gives $1.047117963$ after $10^8$ trials (takes about 200 secs).

Now for completion sake we also give the expected length of two random points in a rectangle. Here we have,

$\mathbb{E}_R(L)=\displaystyle\frac{a^5+b^5-\left(a^4-3 a^2 b^2+b^4\right)d}{15 a^2 b^2}+\frac{a^2}{6 b}\text{csch}^{-1}\left(\frac{a}{b}\right)+\frac{b^2 }{6 a}\text{csch}^{-1}\left(\frac{b}{a}\right)$

One of the nice things of putting these two together is, now we can also get an expression for the length of two random points lying on opposite sides of diagonal $AC$ of rectangle $ABCD$. Denoting this expected length by $\mathbb{E}_D(L)$ (D denoting diagonal), we have this relation from which we can find the required value.

$\mathbb{E}_R(L)=\displaystyle\frac{1}{4}\mathbb{E}_T(L)+\frac{1}{4}\mathbb{E}_T(L)+\frac{1}{2}\mathbb{E}_D(L)$

I tried to get an expression but it doesn't factor nicely but so it better left out. Hope you enjoyed this.


Until then
Yours aye
Me

An Expected Value Problem

Hi All... Recently I thought of a problem and tried to solve it. To my surprise, I ended up with a closed form solution for the same which I would like to share with you by this post.

Consider a rectangle $ABCD$ with the longest side $AB=CD=a$ units and shorter side $BC=DC=b$ units. Now pick a random point in this rectangle and a draw a straight line from this point at a random angle until the line meets the edge of the rectangle. What will be the expected value of this line?

We can attack the problem head-on with multiple integrals. Let $L$ be the length of the line. Then we have,

$\mathbb{E}(L)=\displaystyle\frac{1}{ab}\frac{1}{2\pi}\int\limits_{0}^{a}\int\limits_{0}^{b}\int\frac{y}{\sin t}\,dt\,dy\,dx$

The limits of the innermost integral have been left out purposely. We have to decompose the innermost integral. Let's do it this way. Consider the point $(x,y)$. We'll call it $P$.

Now draw a perpendicular from this point to each of the four sides of the rectangle. Let the perpendicular meet the side $X$ at $P_X$. Also join this point to each of the four vertices of the rectangle. This splits the entire rectangle into eight regions. 

Consider the integral $I(a,b)$ (ignoring constant factors for time being) for the 'random lines' that end up in region $PAP_{AB}$

$I(a, b)=\displaystyle\int\limits_{0}^{a}\int\limits_{0}^{b}\int\limits_{\tan^{-1}(y/x)}^{\pi/2}\frac{y}{\sin t}\,dt\,dy\,dx$

But reflecting the 'random lines' about the $y$-axis, this integral also represents the $PP_{AB}B$, reflecting about the $x$-axis, this integral represents region $PP_{CD}D$, reflecting w.r.t both the axes it represents region $PCP_{CD}$. Solving this one integral covers four of the eight regions.

This is integral is pretty simple to solve with standard tables (or atmost with Mathematica). We get,

$I(a,b)=\displaystyle\frac{1}{6}\left(2b^3-a^3+d^3-3b^2d+3ab^2\ln{\frac{a+d}{b}}\right)$

where $d$ is the length of the diagonal of the rectangle.

Now for the other regions. We don't have to solve anything separately. Just interchange the values of $a$ and $b$. This amounts to rotating the rectangle by $90$ degrees and reasoning as before for the four other regions.

Note the nice thing that in going over the eight regions, we have made the full $2\pi$ radians possible for the 'random line'. So finally we have,

$\mathbb{E}(L)=\displaystyle\frac{4I(a,b)+4I(b,a)}{2\pi ab}$

Simplifying things we finally we end up with,

$\mathbb{E}(L)=\displaystyle\frac{a^3+b^3-d^3}{3\pi ab}+\frac{a}{\pi}\ln{\frac{b+d}{a}} + \frac{b}{\pi}\ln{\frac{a+d}{b}}$

(or)

$\mathbb{E}(L)=\displaystyle\frac{a^3+b^3-d^3}{3\pi ab} +\frac{a}{\pi}\text{csch}^{-1}\left(\frac{a}{b}\right)+ \frac{b}{\pi}\text{csch}^{-1}\left(\frac{b}{a}\right)$

The rectangle had a lot of symmetry that we were able to exploit. I'm trying to do the same for a given arbitrary triangle but it seems so very difficult with complicated integrals cropping at all places. I'll update if end up with something.

Until then,

Yours aye

Me

Division of Dirichlet Series with Python

We saw how to do Dirichlet Series multiplication in the previous post. This post is about dividing two Dirichlet series. Using pretty much the same notation as before, here we are concerned about the following.

$H(s)=\displaystyle\frac{F(s)}{G(s)}$

We can use the following Python code to get the Summatory function $H(n)$.

Note: We know that Dirichlet inverse exists if and only if $g(1)\ne0$. In fact, for most of the multiplicative functions $g(1)=1$. So I have made this assumption in the code. If it is not, changes have to made accordingly.

from timeit import default_timer

start = default_timer()

n = 1000000000
s = int(n ** 0.5)

Divs = [0] * (1 + s)

for k in range(1, 1 + s):
    temp = [1] if k == 1 else [1, k]
    stemp = int(k ** 0.5)
    for j in range(2, 1 + stemp):
        if k % j == 0:
            if j * j == k:
                temp += [j]
            else:
                temp += [j, k // j]
    Divs[k] = sorted(temp)


def DirDiv(Lowsn, Highsn, Lowsd, Highsd, n):
    s = int(n ** 0.5)
    Lows, Highs = [0] * (1 + s), [0] * (1 + s)
    Lows[1] = 1 // (Lowsd[1] - Lowsd[0])
    for k in range(2, 1 + s):
        for j in Divs[k][:-1]:
            Lows[k] -= (Lows[j] - Lows[j - 1]) * (Lowsd[k // j] - Lowsd[k // j - 1])
        Lows[k] += Lowsn[k] - Lowsn[k - 1]
        Lows[k] *= Lows[1]
        Lows[k] += Lows[k - 1]
    for k in range(s, 0, -1):
        tempn = n // k
        res = Highsn[k]
        u = int(tempn ** 0.5)
        v = tempn // (1 + u)
        for i in range(1, 1 + v):
            res -= (Lows[i] - Lows[i - 1]) * (Highsd[k * i] if k * i <= s else Lowsd[tempn // i])
        for i in range(2, 1 + u):
            res -= (Lowsd[i] - Lowsd[i - 1]) * (Highs[k * i] if k * i <= s else Lows[tempn // i])
        i = 1 + u
        res += Lows[tempn // i] * Lowsd[u]
        Highs[k] = res
    return Lows, Highs


Lowsn, Highsn = [0] + [1] * s, [0] + [1] * s
Lowsd, Highsd = [k for k in range(1 + s)], [0] + [n // k for k in range(1, 1 + s)]
Lows, Highs = DirDiv(Lowsn, Highsn, Lowsd, Highsd, n)
print(Highs[1])
print(default_timer() - start)

As in the multiplication case, here again the 'Lows' and 'Highs' mean the same. In the sample code given above, we have $F(s)=1$ and $G(s)=\zeta(s)$. This means $H(s)$ is the Dirichlet series of the Moebius function.

As a result, we get the values of the Merten's function as the output. Here again, the runtime of the algorithm is about $O(n^{\frac{3}{4}})$.

Until then
Yours Aye
Me

Multiplication of Dirichlet Series with Python

It is well known that polynomial multiplication can be much more efficiently using Fast Fourier Transform. I was searching for similar to use in case Dirichlet series. For example, if I have two functions $f(n)$ and $g(n)$, with Dirichlet series $F(s)$ and $G(s)$ respectively, what would be efficient way to get the summartory function of the Dirichlet convolution of the two functions?

Again, we could simply use the Dirichlet's Hyperbola method to get a very generic algorithm to solve this.

Let $f(n)$ and $g(n)$ be two arithmetic functions. Define $h(n)=(f*g)(n)$ as the Dirichlet convolution of $f$ and $g$. Also, let $F(n)$, $G(n)$ and $H(n)$ be the summatory functions for the respective functions.

Now using the formulas we saw in Dirichlet's Hyperbola method, I wrote the following Python code that does the job. The runtime of the following algorithm is about $O(n^{\frac{3}{4}})$

from timeit import default_timer

start = default_timer()

Lim = 1000000000
s = int(Lim ** 0.5)

Divs = [0] * (1 + s)

for k in range(1, 1 + s):
    temp = [1] if k == 1 else [1, k]
    stemp = int(k ** 0.5)
    for j in range(2, 1 + stemp):
        if k % j == 0:
            if j * j == k:
                temp += [j]
            else:
                temp += [j, k // j]
    Divs[k] = sorted(temp)


def DirMult(Losf, Highsf, Losg, Highsg, n):
    s = int(n ** 0.5)
    Highs = [0]
    Los = [0] * (1 + s)
    for k in range(1, 1 + s):
        temp = 0
        for j in Divs[k]:
            temp += (Losf[j] - Losf[j - 1]) * (Losg[k // j] - Losg[k // j - 1])
        Los[k] = Los[k - 1] + temp
        temp, tempn = 0, Lim // k
        u = int(tempn ** 0.5)
        v = tempn // (1 + u)
        mini, maxi = min(u, v), max(u, v)
        for j in range(1, 1 + mini):
            temp += (Losf[j] - Losf[j - 1]) * (Highsg[k * j] if k * j <= s else Losg[tempn // j]) + (Losg[j] - Losg[
                j - 1]) * (Highsf[k * j] if k * j <= s else Losf[tempn // j])
        for j in range(1 + mini, 1 + maxi):
            temp += (Losg[j] - Losg[j - 1]) * (Highsf[k * j] if k * j <= s else Losf[tempn // j])
        j = 1 + u
        temp -= Losg[u] * Losf[tempn // j]
        Highs += [temp]
    return Los, Highs


Losf, Highsf = [k for k in range(1 + s)], [0] + [Lim // k for k in range(1, 1 + s)]
Losg, Highsg = [k for k in range(1 + s)], [0] + [Lim // k for k in range(1, 1 + s)]

Lows, Highs = DirMult(Losf, Highsf, Losg, Highsg, Lim)

print(Highs[1])
print(default_timer() - start)

We'll see how to use this code. To use the function $\text{DirMult}$, we need four parameters. These parameters defined as below.

$\text{Losf}[k]=F(k)$, $\text{Highsf}[k]=F\left(\left\lfloor\frac{n}{k}\right\rfloor\right)$, $\text{Losg}[k]=G(k)$, and $\text{Highsg}[k]=G\left(\left\lfloor\frac{n}{k}\right\rfloor\right)$

We get two outputs and they too are interpreted the same way.

$\text{Los}[k]=H(k)$, $\text{Highs}[k]=H\left(\left\lfloor\frac{n}{k}\right\rfloor\right)$

In the example code above, we have $F(s)=G(s)=\zeta(s)$. Therefore, the output we get is the Divisor Summatory function.

Note that as the algorithm is made for an almost no-brainer approach. It doesn't take advantage of the properties of the functions that we are multiplying and hence the increased runtime.

I'll see you in the next post with Division of Dirichlet Series.


Until then
Yours Aye
Me

Divisor functions - Continued..

This post continues the discussion we had in my previous post.

I discussed only $k=0$ case before dismissing the others as uninteresting. They aren't after all. After some further work, I think I've figured out how the first DGF given above generalized to the other cases as well.

Following the same notations given before we have,

$\displaystyle\sum_{n=1}^\infty\frac{\sigma_k(m\cdot n)}{n^s}=\zeta(s)\zeta(s-k) \prod_{p_j|m}\left(1+(p_j^{k}+p_j^{2k}+p_j^{3k}+\cdots+p_j^{m_jk})(1-p_j^{-s})\right)$

Quite simple after all!! But I don't think the second DGF simplifies as nicely. That should be one for another post. But there is something else.

The first formula given in the previous post works even when $n$ is replaced by $n^2$. We are gonna make that replacement because of the closed form DGF we have for $\sigma_k(n^2)$.

Again following the same notations, this time we have,

$\displaystyle\sum_{n=1}^\infty\frac{\sigma_k(m\cdot n^2)}{n^s}=\frac{\zeta(s)\zeta(s-2k)\zeta(s-k)}{\zeta(2s-2k)} \prod_{p_j|m}\left(1+(p_j^{k}+p_j^{2k}+\cdots+p_j^{m_jk})\left(\frac{1-p_j^{-s}}{1+p_j^{-(s-k)}}\right)\right)$

As before, these DGFs can be used in combination with Dirichlet's Hyperbola method to find their summatory functions. Hope you find these interesting and worth your while. See ya soon.

Till then
Yours Aye,
Me

Divisor functions

Oh... I didn't realize I'vnt posted anything this year. Anyway, here's my first post of the year - my thoughts on the Divisor function.

The Divisor functions are well known and have been studied extensively in Mathematics. Their relation to the Partition function is one of best identities I've seen. I was playing around with them as part of my effort to solve a problem and came up with some identities that I can't find anywhere on the Web. So this post will be about them.

The Divisor function is known to be multiplicative, but not completely multiplicative. But I needed an some formulas where I wanted to find the value of the Divisor functions of product of two numbers that are not co-prime. After some extensive calculations, I ended with the following:

Let $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$ and $n=p_1^{n_1}p_2^{n_2}p_3^{n_3}\cdots$. Then,

$\displaystyle\frac{\sigma_k(m\cdot n)}{\sigma_k(n)}=\prod_{p_j|m}\left(1+\frac{p_j^k+p_j^{2k}+p_j^{3k}+\cdots+p_j^{m_jk}}{1+p_j^{-k}+p_j^{-2k}+p_j^{-3k}+\cdots+p_j^{-n_jk}}\right)$

We can easily check a couple of easy cases. For example, when $n=1$, all of the $n_j$'s become $0$ and the above formulae reduces to the definition of Divisor function. Similarly, when $m$ and $n$ are co-prime, for any prime that divides $m$, the exponent of that prime in $n$ will be $0$, again reducing to a form that shows that divisor functions are multiplicative.

I found the case $k=0$ especially very interesting. We'll see why in a short while. First of all, when $k=0$, the above formula reduces to,

$\displaystyle\frac{\sigma_0(m\cdot n)}{\sigma_0(n)}=\left(1+\frac{m_1}{n_1+1}\right)\left(1+\frac{m_2}{n_2+1}\right)\left(1+\frac{m_3}{n_3+1}\right)\cdots$

As $\sigma_0(n)=(1+n_1)(1+n_2)(1+n_3)\cdots$, we can write

$\sigma_0(m\cdot n)=\sigma_0(n)+m_1\sigma_{0,p_1}(n)+m_2\sigma_{0,p_2}(n)+\cdots+m_1m_2\sigma_{0,p_1p_2}(n)+\cdots+m_1m_2\cdots m_k\sigma_{0,p_1p_2\cdots p_k}(n)+\cdots$

where $\sigma_{0,B}(n)$ is (defined only for square-free $B$) a function that counts the number of divisors of $n$ that are not divisible by any of the prime factors of $B$. Note that we have eliminated  everything related to prime factorization of $n$.

Actually this is why I said the case $k=0$ is interesting. Had $k$ been greater than $0$, we would have those annoying prime factors of $n$ in the above expression. Now with what we have above we can make an interesting Dirichlet Generating function.

Given $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$, we have

$\displaystyle\sum_{n=1}^\infty\frac{\sigma_0(m\cdot n)}{n^s}=\zeta^2(s) \prod_{p_j|m}\left(1+m_j(1-p_j^{-s})\right)$

Even better, for a given $m$, if we define a function $f(n)$ as

$f(n)=\displaystyle\sum_{d|m}\sigma_0(d\cdot n)$

again we end up with a nice DGF. This we have,

$\displaystyle\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sigma_0(m)\zeta^2(s)\prod_{p_j|m}\left(1+\frac{m_j}{2}(1-p_j^{-s})\right)$

All of the above DGFs in combination with Dirichlet's Hyperbola method can be used to find their respective summatory function.

We can see how this generalizes in the next post. See ya soon.

Till then
Yours Aye,
Me

SquareFree numbers with exactly $k$ factors

A positive integer is squarefree if and only if it is not divisible by any perfect square other than 1. That makes them interesting. This post will be about finding the DGF squarefree numbersthat has exactly $k$ factors. We'll use symmetric polynomials to our advantage.

First, the Dirichlet Generating function of square numbers is quite straight forward. By the definition of square free numbers, every prime can be present atmost once in its prime factorization. Let $\xi_2(s)$ be the DGF of square free numbers. Then it is easy to see that,

$\xi_2(s)=\displaystyle\prod_p(1+p^{-s})$

Now, let's review about symmetric polynomials. We need two of the most fundamental symmetric polynomials. The Elementary symmetric polynomials (denoted by $e_k(x_1,x_2,\cdots)$) and the Power sum symmetric polynomials (denoted by $p_k(x_1,x_2,\cdots)$.

Consider these polynomials with finitely many variables. Then,

$e_1=x_1+x_2+x_3+\cdots$
$e_2=x_1x_2+x_1x_3+\cdots+x_2x_3+x_2x_4+\cdots$ and so on.

Similarly,

$p_k=x_1^k+x_2^k+\cdots$

Now notice something interesting. Let $x_k$ be the inverse of the $k$th prime raised to the power $s$. That is,

$x_1=2^{-s}$, $x_2=3^{-s}$, $x_3=5^{-s}$, and so on.

It is now very easy to see that $e_k =e_k(s)$ becomes the Dirichlet generating function of the square free numbers with exactly $k$ factors. It is also clear that the

$p_k =p_k(s)=\zeta_P(ks)$

where $\zeta_P(s)$ is the Prime Zeta function.

Therefore by the fundamental realtion connecting the Elementary symmetric polymials and the Power sum symmetric polynomials, we know

$e_k=\displaystyle\frac{1}{k}\sum_{j=1}^k(-1)^{j-1}e_{k-j}p_j$

This equation can now be directly translated as,

$e_k(s)=\displaystyle\frac{1}{k}\sum_{j=1}^k(-1)^{j-1}e_{k-j}(s)\zeta_P(js)$

which gives the sought after DGF in a recursive fashion. We'll use this DGF in combination with the Dirichlet Hyperbola method in the next post to find the number of squarefree numbers with exactly $k$ factors.


Until then
Yours Aye
Me