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<x<2, 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.

On Complex vs Real Differentiability

Once upon a time I was confused about what really made a holomorphic function different from a function on \mathbb{R}^2 which is differentiable in the real sense. I mean obviously I knew how they were different but I wanted to put it in a context that was more meaningful for me, and after some thinking I thought the following and was satisfied:

Let’s cover some definitions; it’s likely I’ll write these classic definitions in more detail in later posts. Let f:\mathbb{C}\rightarrow\mathbb{C} be a complex function and let z_0 be a complex number. f is called (complex) differentiable at z_0 if the limit

f'(z_0) = \lim_{z\rightarrow z_0}\frac{f(z) - f(z_0)}{z-z_0}

exists, where the limit is taken along whatever path you like. The function f is differentiable on a subset A\subset\mathbb{C} if it is differentiable for all z_0\in A. Now a complex number can be written as z = x+iy, where x,y\in\mathbb{R}, and a complex function can be written as f(z=x+iy) = u(x,y)+iv(x,y) where u and v are both functions from \mathbb{R}^2 to \mathbb{R}. As \mathbb{R}-vector spaces, \mathbb{C} and \mathbb{R}^2 are isomorphic and we can think of f as a function f:\mathbb{R}^2\rightarrow\mathbb{R}^2 given by

(x,y)\mapsto (u(x,y),v(x,y)).

Now we it can be said that a complex function is complex differentiable if and only if it is differentiable in the real sense (i.e. all partial derivatives of u and v exist and are continuous) and u and v satisfy the Cauchy-Riemann equations (\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} and \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}).

This observation is well known and does a good job of contrasting the two senses of differentiability, but I wanted to know the answer to “why is the derivative of the real version of the function a linear transformation (given by the Jacobian matrix) but the derivative of the complex version another complex function?” The answer to that is obvious in hindsight, but I was happy when I went from not knowing to knowing so I thought I better recount it here. The trick is, of course, that we construct a field isomorphism (which I’ll call \varphi) from \mathbb{C} into \mathrm{M}_2(\mathbb{R}) (the ring of 2 by 2 matrices with real entries) given by

\varphi(z) = \begin{pmatrix} \mathrm{Re}z & -\mathrm{Im}z \\ \mathrm{Im}z& \mathrm{Re}z \end{pmatrix}.

Now taking the derivative  of f gives us a function \mathrm{D}f:\mathbb{R}^2\rightarrow\mathrm{M}_2(\mathbb{R}), but the Cauchy-Riemann equations mean that if f is complex differentiable, then \mathrm{D}f(\mathbb{R}^2) \subset \varphi(\mathbb{C})\subset\mathrm{M}_2(\mathbb{R}). So in that case we can think of \mathrm{D}f as a function from \mathbb{R}^2\cong\mathbb{C} to \varphi(\mathbb{C})\cong\mathbb{C}, and so my question is resolved! The real version of the function is complex differentiable precisely when the Jacobian matrix can be identified with a complex function through the isomorphism \varphi.