Dear Uncle Colin,

I have a binomial expansion of $(1+x)^\frac{1}{2}$ and need to approximate $\sqrt{5}$. Apparently you need to substitute in $x=\frac{1}{4}$, but I’d have thought $x=4$ was a more obvious choice. What gives?

-- Roots Are Dangerous If Understood Sloppily

That does seem to be the obvious substitution, but for one small difficulty that you’ll discover if you try it: it doesn’t work. $(1+x)^\frac{1}{2} \approx 1 + \frac 12 x - \frac 18 x^2 + \frac{1}{16}x^3 + …$, and if you put $x=4$ into that you get $1 + 2 - 2 + 4 …$. The terms continue to get bigger in magnitude and oscillate between positive and negative - and they never converge to a single value.

This is because of something called the radius of convergence ((The reasons behind this are, quite literally, complex)) - if the exponent $n$ of a binomial expansion $(a+bx)^n$ is anything except a positive integer, the expansion is only defined for $|x| < \frac{a}{b}$. For this example, $a=b=1$, and when $x=4$, it’s certainly not within the radius!

Instead, we need to find a value of $x$ that makes the bracket a simple square multiple of 5 - and picking $x=\frac14$ makes the bracket $\frac{5}{4}$, the square root of which is $\frac12\sqrt{5}$.

Putting this into the expansion gives $\frac12\sqrt{5} \approx 1 + \frac{1}{8}-\frac{1}{128}+\frac{1}{1024}$, or 1.11816 or so. Doubling that gives 2.23633, compared to a true value of 2.23607. Even after four terms, it’s pretty good!

It’s possible to get a closer answer by picking an even cleverer value of $x$ - for example, $x = -\frac{1}{81}$ would make the bracket $\frac{80}{81}$, the square root of which is $\frac{4}{9}\sqrt{5}$. (The closer the value of $x$ is to 0, the quicker the convergence.)

I hope that helps!

-- Uncle Colin

* Edited 2016-12-21 to fix LaTex, formatting and categories.