## 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.

## 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 $\varphi$.