# Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

• $\displaystyle{\sum_{p}\frac{1}{p}}$   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

• $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}$, where $s$ is any complex number with $\text{Re}(s)>1$.

The outline of proof,   when $s$ is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html

# Bernoulli Numbers

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I have referred to them twice (here and here) so far. Also, the Euler-Maclaurin formula I discussed in second post, explains a lot about their occurrences (for example). Now I think it’s time to dive deeper and try to understand them.

In 1631, Johann Faulhaber   published Academia Algebra (it was a German text despite the Latin title). This text contains a generalisation of sums of powers, which in modern notations reads:

$\displaystyle{\sum_{m=1}^{n} m^{k-1} = \frac{1}{k}\left[n^k + \binom{k}{1} n^{k-1} \times \frac{1}{2} + \binom{k}{2}n^{k-2} \times \frac{1}{6} +\binom{k}{3}n^{k-3} \times 0+\binom{k}{4}n^{k-4} \times \frac{-1}{30}+ \ldots\right]}$

Observe that the expression on the right hand side in square brackets appears like binomial expansion, but there are some constant terms multiplied to them (which can also be 0). These constant terms were named “Bernoulli Numbers” by Abraham de Moivre, since they were intensively discussed by Jacob (Jacques) Bernoulli in Ars Conjectandi published in Basel in 1713, eight years after his death.

I will follow the notation from The book of numbers (published  in 1996). So we will denote $n^{th}$ Bernoulli number by $B^{n}$ where

$\displaystyle{B^0 = 1, B^1 = \frac{1}{2}, B^2 = \frac{1}{6}, B^3= B^{5} = \ldots = B^{odd} = 0, B^4 = B^8 = \frac{-1}{30}, B^6 = \frac{1}{42}, B^{10} = \frac{5}{66}, \ldots}$

This notation enables us to calculate sum of $k^{th}$ power of first $n$ natural numbers quickly. We can re-write above summation formula as:

$\displaystyle{\sum_{m=1}^n m^{k-1} = \frac{(n+B)^k - B^k}{k}}$

To illustrate, how to use this formula, let’s calculate sum of $5^{th}$ powers of first 1000 natural numbers:

$\displaystyle{\sum_{m=1}^{1000} m^{5} = \frac{(1000+B)^6 - B^6}{6}}$

$\displaystyle{ = \frac{1}{6}\left[1000^6 + 6B^1 1000^5 + 15B^2 1000^4 + 15 B^4 1000^2\right]}$

So, we have done binomial expansion of the right hand side and used the fact that $B^{odd>1} = 0$. Now we will replace corresponding values of Bernoulli Numbers to get:

$\displaystyle{\sum_{m=1}^{1000} m^{5} =\frac{1}{6}\left[10^{18} + 3\times 10^{15} + 2.5 \times 10^{12} - 0.5 \times 10^6\right]=\frac{1003002499995\times 10^5}{6}}$

$1^5 + 2^5 + \cdots + 1000^5 = 16716708333250000$

(This answer was cross-checked  using SageMath)

There are many ways to calculate the value of Binomial numbers, but the simplest one is to using the recursive definition:

$(B - 1)^k = B^k$ for k>1, gives value of $B^{k-1}$

There is another definition of Bernoulli  Numbers using power series:

$\displaystyle{\frac{z}{e^z-1} = \sum_{k=0}^{\infty} \frac{B^k z^k}{k!}}$

This gives slightly different sequence of Bernoulli numbers, since in this $B^{1}=\frac{-1}{2}$, and the recursive definition is

$(B+1)^{k} = B^{k}$ for k>1, gives value of $B^{k-1}$

This definition can be used to calculate val value of  $\tan(z)$,  since its infinite series expression has Bernoulli numbers in coefficients.

$\displaystyle{\tan(z)=\sum_{n=0}^{\infty}\frac{B^{2n}(-4)^n(1-4^n)}{2n!}z^{2n-1}}$

# Integration & Summation

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A few months ago I wrote a series of blog posts on “rigorous”  definitions of integration [Part 1, Part 2]. Last week I identified an interesting flaw in my “imagination” of integration in terms of “limiting summation” and it lead to an interesting investigation.

While defining integration as area under curve, we consider rectangles of equal width and let that width approach zero. Hence I used to imagine integration as summation of individual heights, since width approaches zero in limiting case. It was just like extending summation over integers to summation over real numbers.

My Thought Process..

But as per my above imagination, since width of line segment is zero,  I am considering rectangles of zero width. Then each rectangle is of zero area (I proved it recently). So the area under curve will be zero! Paradox!

I realized that, just like ancient greeks, I am using very bad imagination of limiting process!

The Insight

But, as it turns out, my imagination is NOT completely wrong.  I googled and stumbled upon this stack exchange post:

There is the answer by Jonathan to this question which captures my imagination:

The idea is that since $\int_0^n f(x)dx$ can be approximated by the Riemann sum, thus $\displaystyle{\sum_{i=0}^n f(i) = \int_{0}^n f(x)dx + \text{higher order corrections}}$

The generalization of above idea gives us the Euler–Maclaurin formula

$\displaystyle{\sum_{i=m+1}^n f(i) = \int^n_m f(x)\,dx + B_1 \left(f(n) - f(m)\right) + \sum_{k=1}^p\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)}(n) - f^{(2k - 1)}(m)\right) + R}$

where $m,n,p$ are natural numbers, $f (x)$ is a real valued continuous function, $B_k$ are the Bernoulli numbers and $R$ is an error term which is normally small for suitable values of $p$ (depends on $n, m, p$ and $f$).

Proof of above formula is by principle of mathematical induction. For more details, see this beautiful paper: Apostol, T. M.. (1999). An Elementary View of Euler’s Summation Formula. The American Mathematical Monthly, 106(5), 409–418. http://doi.org/10.2307/2589145

# [poem] Harmonic Noise

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———-
Harmonic Noise

One day I heard something musical,
It sounded to me mystical.
Since it was ordered,
A harmonic pattern was being followed.
The pattern was elegant,
It was inverse of each natural number element.
I made the number elements bigger hero,
The pattern approached symmetrical zero.
When I summed up the pattern without bound,
I got only noise without any musical sound.

-Gaurish
————
For more details about Harmonic series refer Wikipedia: https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Edit (20 May 2018): A video by Marc Chamberland discussing a puzzle involving harmonic series.

# Kempner Series & MMC

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In Madhava Mathematics Competition 2015 (held in January 2015), we were asked to prove the convergence of Kempner Series (first time proved in 1914). Recently I discovered the paper by A. J. Kempner (http://www.jstor.org/stable/2972074), so in this blog post I will state and prove that problem.

The basic idea behind proof is to divide whole series into chunks (finding symmetry) and then construct a converging geometric series which will act as upper bound of Kempner Seires.

A detailed study of this cartoon has been done by Ben Orlin (http://bit.ly/1KD4shF)

Theorem (Kempner Series, 1914). Harmonic Series,$\sum_{k=1}^{\infty} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} +\ldots$, converge, if the denominators do not include all natural numbers $1,2,3,\ldots$, but only those numbers which do not contain any figure 9.

Proof: Given Series is:
$K = \frac{1}{1}+\ldots +\frac{1}{8}+\frac{1}{10}+\ldots +\frac{1}{18}+\frac{1}{20} + \ldots + \frac{1}{28}+\frac{1}{30}+\ldots +\frac{1}{38}+\frac{1}{40}+\ldots +\frac{1}{48}+\frac{1}{50} + \ldots + \frac{1}{58}+\frac{1}{60}+\ldots+\frac{1}{68}+\frac{1}{70}+\ldots +\frac{1}{78}+\frac{1}{80} + \ldots + \frac{1}{88}+\frac{1}{100}+\ldots$
Now we can rewrite above series as:
$S = a_1 + a_2 + \ldots + a_n + \ldots$
where $a_n$ is the sum of all terms in $K$ of denominator $d$ with $10^{n-1} \leq d < 10^{n}$.
Observe that, each term of $K$ which forms part of $a_n$, is less than or equal to $1/10^{n-1}$.

Now count the number of terms of $K$ which are contained in $a_1$, in $a_2$, $\ldots$, in $a_n$. Clearly, $a_1$, consists of 8 terms, and $a_1 < 8 \cdot 1 < 9$. In $a_2$ there are, as is easily seen, less than $9^2$ terms of $K$, and $a_2 < (9^2/10)$. Altogether there are in $K$ less than $9^2 + 9$ terms with denominators under 100.

Assume now that we know the number of terms in $K$ which are contained in $a_n$ to be less than $9^n$, for $n = 1, 2, 3, \ldots, n$. Then, because each term of $K$ which is contained in $a_n$ is not greater than $1/10^{n-1}$, we have $a_n < (9^n/10^{n-1})$, and the total number of terms in $K$ with denominators under $10^n$ is less than $9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9$.

Now, let’s go for induction. For $n = 1$ and $n = 2$ we have verified all this, and we will now show that if it is true for $n$, then $a_{n+1} < (9^{n+1}/10^n)$. $a_{n+1}$ contains all terms in $K$ of denominator $d$, $10^n \leq d < 1^{n+1}$. This interval for $d$ can be broken up into the nine intervals,  $\alpha \cdot 10^n \leq d < (\alpha + 1)10^n$, $\alpha = 1,2, \ldots, 9$. The last interval does not contribute any term to $K$, the eight remaining intervals contribute each the same number of terms to $K$, and this is the same as the number of terms contributed by the whole interval $0 < d < 10^n$, that is, by assumption, less than $9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9$.

Therefore, $a_{n+1}$ contains less than $8(9^n + 9^{n-1} + 9^{n-2} +\ldots + 9^2 + 9) < 9^{n+1}$ terms of $K$, and each of these terms is less than or equal to $1/10$, we have $a_{n+1} < (9^{n+1}/10^n)$.

Hence, $S = a_1 + a_2 + a_3 + \ldots < 9 + \frac{9}{10} + \ldots + \frac{9^{n+1}}{10^n}+ \ldots = 90$
Thus, $S$ converges, and since, $K=S$, $K$ also converges.

——————————–
Note: There is nothing special about 9 here, the above method of proof holds unchanged if, instead of 9, any other figure $1, 2, \ldots, 8$ is excluded, but not for the figure $0$.

I have figured out that $\sqrt{1- \sqrt{1+ \sqrt{1- \sqrt{\ldots } }}}$ is a diverging series. But now I am struck with another problem: