Binomial party tricks

February 6, 2021. Sketchy hacker notes on the binomial approximation. The flashy payoff: party trick arithmetic for estimating roots in your head.


The binomial approximation is the result that, for any real $\alpha$, and $|x| \ll 1$,

\[(1 + x)^\alpha \approx 1 + \alpha x.\]

The usual proof involves calculus. Here, we present a sketchy shortcut and an elementary longcut, neither of which involves calculus, strictly speaking. We also derive the quadratic term, and end with a fun party trick for finding roots.

Sketchy shortcut

We begin with the shortcut. In an earlier post, I derived the following result for the exponential, and $|x| \ll 1$:

\[e^x \approx 1 + x.\]

Rather than go off and read the post, we can do even better and simply define the exponential by this property. If it’s true, then for any $r$, we can set $x = r/n$ for very large $n$ to get

\[e^r = (e^{r/n})^n \approx \left(1 + \frac{r}{n}\right)^n.\]

In the limit of infinite $n$, the expression should be exact. And indeed, this is the standard definition of $e^r$:

\[e^r = \lim_{n\to\infty} \left(1 + \frac{r}{n}\right)^n.\]

Let’s proceed with a proof of the binomial approximation. The natural logarithm is the inverse function, so that

\[x = \log e^x \approx \log(1 + x).\]

Recall that

\[x^n = (e^{\log x})^n = e^{n\log x} \quad \Longrightarrow \quad \log x^n = n \log x.\]

Thus, taking the logarithm $(1 + x)^\alpha$, we have

\[\log [(1+x)^\alpha] = \alpha \log (1+ x) \approx \alpha x,\]

and hence

\[(1+x)^\alpha \approx e^{\alpha x} \approx 1 + \alpha x.\]

This works since all the corrections are at higher order in $x$.

Elementary longcut

This is a bit high brow, and we can get to the same conclusion using simple algebra. First note that, from the binomial theorem,

\[(1 + x)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots x^n \approx 1 + nx\]

for $|x| \ll 1$, neglecting higher order terms which are much smaller. So the binomial approximation is true for whole numbers $n$. If we consider a fraction $q = m/n$, then $(1 + x)^q$ raised to the power $n$ should equal

\[(1 + x)^{qn} = (1 + x)^{m} \approx 1 + mx \tag{1}\label{m}\]

by the binomial theorem. Let’s assume

\[(1 + x)^{q} \approx 1 + \beta x,\]

with some higher order terms we can ignore. Raising to the power $n$, we can use the binomial approximation for $n$ to get

\[(1 + x)^{qn} \approx (1 + \beta x)^n \approx 1 + \beta n x.\]

Comparing to (\ref{m}), we find that $\beta = m/n$, and hence the binomial approximation is true for positive rationals. We can add negative powers using the geometric series:

\[\frac{1}{1 - x} = 1 + x + x^2 + \cdots \approx 1 + x,\]

and hence for a negative rational $q = -m/n$,

\[(1 + x)^q \approx (1 - x)^{m/n} \approx 1 - \frac{m}{n}x = 1 + qx,\]

as required. Finally, there is arbitrary real $\alpha$. This is actually trivial, in some sense. Unlike whole numbers (repeated multiplication), fractions (roots), or negative numbers (reciprocals), an irrational power has no obvious interpretation. The most reasonable thing to do is define it as a limit of rational powers that approximate it:

\[(1 + x)^r = \lim_{n \to \infty} (1 + x)^{q_n},\]

where $q_n$ is a sequence of rational numbers (e.g. the decimal expansion) approximating $r$. In this case, the binomial approximation gives

\[(1 + x)^r = \lim_{n \to \infty} (1 + x)^{q_n} \approx 1 + x \lim_{n \to \infty} q_n = 1 + rx,\]

and so the result holds for all real numbers.

Higher terms

It’s possible, if messy, to extend these methods to determine the next term in the approximation. We’ll do the longcut, and use big-O notation, with $O(x^3)$ in this context meaning “terms with powers of $x^3$ or higher”. The binomial theorem gives

\[(1 + x)^n = 1 + nx + \frac{n(n-1)}{2} x^2 + O(x^3), \tag{2} \label{second}\]

since the coefficient of the $x^2$ term is the number of ways of choosing $2$ items (the $x$ terms) from $n$ items (the factors in the power). For a rational $q = m/n$, we have

\[(1 + x)^{qn} = (1 + x)^m = 1 + mx + \frac{m(m-1)}{2} x^2 + O(x^3),\]

and if we assume

\[(1 + x)^{q} = 1 + qx + \gamma x^2 + O(x^3),\]

then the binomial theorem again gives

\[(1 + x)^{qn} = \left[1 + qx + \gamma x^2 + O(x^3)\right]^n = 1 + nqx + \left[n\gamma + \frac{n(n-1)}{2}q^2 \right]x^2 + O(x^3).\]

The coefficient of the linear term $nq = m$ matches, but the quadratic term requires more work. Comparing to (\ref{second}) and rearranging for $\gamma$, we have

\[\begin{align*} \gamma & = \frac{1}{n}\left[\frac{m(m-1)}{2}- \frac{n(n-1)}{2}q^2\right] =\frac{m(m-1)}{2n}- \frac{m^2(n-1)}{2n^2} =\frac{q(q - 1)}{2}. \end{align*}\]

Thus, we find that to second order,

\[(1 + x)^q = 1 + qx + \frac{q(q-1)}{2} x^2 + O(x^3)\]

The extension to real and negative powers is easy. The extension to higher terms in $x$ is not. They obey something called the binomial series,

\[(1 + x)^\alpha = \sum_{k = 0}^\infty \frac{\alpha(\alpha - 1)\cdots (\alpha-k +1)}{k!} x^k,\]

and I have no idea how to get this without calculus. (One can use “analytic continuation” but this feels too much like cheating to me, partly because it’s not clear why this continuation is unique.) Any tips appreciated!

Rooting out the answer

The applications are many and various, but the simplest thing we can try is quickly calculating powers $y^\alpha$. The general trick is to find a power near $y$ that is simpler to evaluate, factor out the simple answer, then use the binomial approximation. I think there are actually better ways to estimate positive powers, but the binomial approximation really shines in the estimation of roots. It can even be a good party trick, depending on the kind of parties you go to!

Suppose someone asks you to find the square root of $8$. You look for a nearby perfect square, in this case $9$, then factor eight into $9$ times one minus something small:

\[\sqrt{8} = \sqrt{9\left(1 - \frac{1}{9}\right)} = 3 \left(1 - \frac{1}{9}\right)^{1/2}.\]

We can take $\alpha = 1/2$ and $x = -1/9$ in the binomial approximation, and see how we go, noting that

\[\sqrt{1 - x} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 + O(x^3).\]

To first order, we get

\[3 \left(1 - \frac{1}{9}\right)^{1/2} \approx 3\left[1 - \frac{1}{2} \cdot \frac{1}{9}\right] = \frac{17}{6} \approx 2.83.\]

To second order,

\[3 \left(1 - \frac{1}{9}\right)^{1/2} \approx 3\left[1 - \frac{1}{2} \cdot \frac{1}{9} - \frac{1}{8} \cdot \frac{1}{9^2}\right] = \frac{611}{216} \approx 2.829.\]

The actual answer is $\sqrt{8} = 2.828$, so even the first term in the binomial approximation is very good! We’ll finish with a somewhat more involved example. Let’s approximate the fifth root of six, $6^{1/5}$. I only know one fifth power of the top of my head, $2^5 = 32$, and this happens to be near $6^2 = 36$. We can chain these observations together as follows:

\[\begin{align*} 6^{1/5} = 36^{1/10} = 32^{1/10}\left(1 + \frac{1}{9}\right)^{1/10} & =\sqrt{2}\left(1 + \frac{1}{9}\right)^{1/10} \approx \sqrt{2} \cdot \left(1 + \frac{1}{10\cdot 9}\right). \end{align*}\]

At this point, we could separately approximate $\sqrt{2}$, but I happen to know it’s about $1.414$, so I can divide by $90$ (or even just $100$ for a quick mental estimate), and add them together to get

\[\sqrt[5]{6} \approx 1.414 + \frac{1.414}{90} \approx 1.43.\]

Consulting a calculator, this is correct to two decimal places! With the power of the binomial approximation, you can do it in your head.

Written on February 6, 2021
Mathematics   Physics   Hacker