## Calculating Square Root of 2

I had so much fun deriving that series which converges to pi the other day that I thought it’d be fun to occasionally whip up such series for other irrational numbers. Today I’ll do $\sqrt{2}$.

There are many different series that would do the job, the one I’ll derive is found by expanding $\sqrt{x}$ around $x=1$ as a power series.

Indeed, for $x$ satsfying $|x-1| < 1$ we have

$\begin{array}{ccl} \sqrt{x} & = & 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3-\frac{5}{128}(x-1)^4+ ... \\ & = & 1+ \frac{1}{2}\sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}(x-1)^{n+1}\end{array}$.

This series is guaranteed to converge for all $0, and to diverge for all $x<0$ and $x>2$.   The number 2 itself sits right on the boundary of the circle of convergence, so although we would like to say

$\sqrt{2} = 1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}$,

we first need to show that the series is convergent. Consider the sequence $a_n = \frac{(2n)!}{4^n(n!)^2(n+1)}$. This sequence is monotonically decreasing, and we will show that for all $n>0, a_n < \frac{1}{n}$. Firstly,

$a_1 = \frac{2!}{4\cdot 2} = \frac{1}{4} < 1$.

We now assume that $a_k < \frac{1}{k}$ for some integer $k$ and show that $a_{k+1} < \frac{1}{k+1}$. This is true, since

$\begin{array}{ccl}\frac{(2k+2)!}{4^{k+1}((k+1)!)^2(k+2)} & = & \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}\frac{(2k)!}{4^k(k!)^2(k+1)} \\ &=& \frac{(2k+1)(2k+2)}{4(k+1)(k+2)}a_k\\& < & \frac{(2k+1)(2k+2)}{4k(k+1)(k+2)}\\ & = & \frac{1}{4}(\frac{1}{k}+\frac{3}{k+2}) \\ & \leq & \frac{1}{k+1}\end{array}$,

where the last inequality holds for $k\geq1$. Thus $a_{k} < \frac{1}{k}$ implies $a_{k+1}<\frac{1}{k+1}$, and so by induction $a_n < \frac{1}{n}$ for all $n$.

To show convergence of our series, we note we have shown enough to invoke the alternating series test. So it is true that

$1 + \sum_{n=0}^\infty (-\frac{1}{4})^n\frac{(2n)!}{(n!)^2(n+1)}$

converges, and by continuity  it must equal $\sqrt{2}$. Truncating the series at n=10,000,000 gives

$\sqrt{2}\approx 1.4142135623775118$

which is correct up 11 decimal places.

## Happy Belated Pi Day

I didn’t create this blog until after Pi day, so unfortunately I’m a bit late. I celebrate Pi day by deriving the same series expression for pi every year. I’m not sure who to attribute this particular method to, but I definitely learned in from some old calculus book once upon a time. It’s as simple as calculating $\int_0^1\frac{1}{x^2+1}dx$ two different ways and equating the results. First up, we can make the substitution $x = \tan\theta$ and simplify yielding $\int_0^{\pi/4}d\theta = \frac{\pi}{4}$. On the other hand, if $|x|<1$ then $\sum_{n=0}^\infty(-1)^nx^{2n}$ converges (uniformly on all intervals $|x|\leq r < 1$) to $\frac{1}{x^2+1}$, and so may proceed with

$\int_0^1\frac{1}{1+x^2}dx = \sum_{n=0}^\infty(-1)^n\int_0^1x^{2n}dx$. Each $\int_0^1x^{2n}dx = \frac{1}{2n+1}$, and so putting it all together we’re left with

$\pi = 4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...)$

And there it is!

## Generating Function for Fibonacci Sequence

We seek closed form for the power series $\sum_{n=0}^\infty F_nz^n$, where $F_0 = 1$, $F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$. First we note that the ratio test yields

$\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_n}|z| = \varphi |z|$, where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. Thus the series converges on the region of the complex plane where $|z| < \frac{1}{\varphi}$. Now let the power series be denoted by $F(z)$, and note that

$F(z) - zF(z) - z^2F(z) = 1 + \sum_{n=0}^\infty (F_{n+2}-F_{n+1}-F_n)z^{n+2} = 1$

and so $F(z) = \frac{1}{1-z-z^2}$. It seems so unlikely, before you know better at least, that things like that can be done. Why should it be that the nth Fibonacci number is given by $\frac{1}{2\pi i}\oint\frac{1}{z^{n+1}(1-z-z^2)}dz$, the integral along some closed contour in the complex plane of a function that otherwise seems unrelated. That’s maths!