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

Hi RADIUS, and thanks for your message!

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.