Saturday, January 10, 2015




Wednesday, December 31, 2014

Divergence of sum of reciprocals of primes

We know that Harmonic series $\sum \frac{1}{n} $ is diverging. The proof is very simple by grouping the terms. Whereas for geometric sequence $\frac{1}{2^n}$ is converging. Our question is what happens when we have $\sum_{p \in prime} \frac{1}{p}$.
We will prove this by contradiction. Assuming that the sum of reciprocals of prime $S_p$ is a constant value. As $\frac{1}{p_i}$ are positive constants. That means for some $\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_{n-1}} < S_n -\frac{1}{2}$ and
$\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_n}+\frac{1}{p_{n}} \ge S_n - \frac{1}{2}$.
Thus $\frac{1}{p_{n+1}}+\frac{1}{p_{n+2}}+ ... < \frac{1}{2}$ and multiplying both sides by positive $x$ we get $\frac{x}{p_{n+1}}+\frac{x}{p_{n+2}}+ ... < \frac{x}{2}$
To prove this we first define a function $N(x)$ which counts the number of numbers whose prime factor is among the first $ k$ primes. For example if $k = 4$ the primes are $\{ 2,3,5,7 \}$. So the function $N(10) = 10, N(15) = 13, N(27) = 20$. ie for $10$ all the factors are from prime numbers $\{ 2,3,5,7 \}$. Let $k$ be the number of first k primes and $x$ be any number and we try to find $N(x)$. We see that any number $y \le x$ can be written as $y = {p_1}^{a_1}{p_2}^{a_2}...{p_t}^{a_t}$. We have to count all those $y$ whose factors are among the first $k$ primes. We realise that any number can be written as $uv$ where $w^2v$ or $u = w^2$ is a square number and so $v$ is a product of primes less than $k^{th}$ prime and has the form $p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}...p_k^{\alpha_k}$ where $\alpha_1,\alpha_2 ...\alpha_k \in \{0,1\}$. Thus $N(x) \le \sqrt{x}2^{k}$ and the numbers which are not divisible by first k primes are $x-N(x) \le \frac{x}{p_{k+1}}+\frac{x}{p_{k+2}}+\frac{x}{p_{k+3}}+....$.
So there are two equations as below
$\frac{x}{p_{n+1}}+\frac{x}{p_{n+2}}+ ... < \frac{x}{2}$
$x-N(x) \le \frac{x}{p_{k+1}}+\frac{x}{p_{k+2}}+\frac{x}{p_{k+3}}+....$.
Combining we have $x-N(x) < \frac{x}{2} \Rightarrow \frac{x}{2} < N(x)$ and holds for all values of $x$. Let $k = n $ then $N(x) \le 2^n\sqrt{x}$. Therefore we havce $\frac{x}{2} < N(x) \le 2^n\sqrt{x}$. When $n = 2^{2n+2}$ we get $\frac{1}{2}2^{2n+1} < N(x) \le 2^{2n+1}$. This contradiction proves that the series P diverges. ( A similar contradiction is obtained for $x$ equal to any value greater than $2^{2n+2}$).

Relation between Gamma Function and Zeta Function

We know that $\Gamma(x) = \int_0^\infty e^{-t}t^{x-1} dt$
Let $t = ru \implies dt = r du$  Which means
$\Gamma(x) = \int_0^\infty e^{-ru} {(ru)}^{x-1} r du$
$\Gamma (x) = \int_0^\infty r^x u^{x-1} e^{-ru} du$
$\frac{1}{r^x} \Gamma (x) = \int_0^\infty u^{x-1} e^{-ru} du$
Taking sum on both the sides we get
$\sum_{r=0}^\infty \frac{1}{r^x} \Gamma (x) = \sum_{r=0}^{\infty}\int_0^\infty r^x u^{x-1} e^{-ru} du$
$\Gamma(x) \sum_{r = 0}^{\infty} \frac{1}{r^x} = \int_0^{\infty}u^{x-1} \sum_{r=0}^{\infty}e^{-ru} du$
$\Gamma(x) \zeta(x) = \int_{0}^{\infty}u^{x-1}\frac{e^{-u}}{1-e^{-u}} du$
$\Gamma(x) \zeta(x) = \int_{0}^{\infty} \frac{u^{x-1}}{e^u-1}du$

Monday, December 29, 2014

Arithmetic Geometric Series

This is a famous inequality that one should know. The case for two numbers is really easy $(\sqrt{a}-\sqrt{b})^2 \ge 0 \Rightarrow a+b \ge 2\sqrt{a}\sqrt{b} \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}$
The case for 4 variables is also pretty straight forward $\dfrac{a+b+c+d}{4} =\dfrac{ \dfrac{a+b}{2} +\dfrac{c+d}{2}}{2} \ge \dfrac{\sqrt{ab}+\sqrt{cd}}{2} \ge \sqrt{ \sqrt{ab} \sqrt{cd}} \ge \sqrt[4]{abcd}$
The case for 3 variables is a little bit tricky but can be proved quite easily if we use use the 4 variable proof. Let $d = \sqrt[3]{abc}$. Then $\dfrac{a+b+c+d}{4} \ge \sqrt[4]{abcd} =\sqrt[4]{abc(abc)^{\frac{1}{3}}} = \sqrt[4]{(abc)^{\frac{4}{3}}} =(abc)^{\frac{1}{3}} =d$ and so we have $\dfrac{a+b+c+d}{4} \ge d \Rightarrow \dfrac{a+b+c}{4}+\dfrac{d}{4} =d \Rightarrow \dfrac{a+b+c}{4} = \dfrac{3d}{4} \Rightarrow \dfrac{a+b+c}{3} \ge d \Rightarrow \dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$

 

Wednesday, August 06, 2014

Tolstoy on fraction

A man is like fraction. The numerator corresponds to who he is and the denominator corresponds to what he honks of himself. He greater the numerator, smaller is the fraction.

Sunday, June 22, 2014

Goethe Quote

A man sees in the world what he carries in his heart.

Tuesday, May 20, 2014

Watching this song


Tuesday, March 25, 2014

Google hangout class




Things we did today for my Math studies students
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