Binomial estimation

In a nutshell

When $t$ is a value between $0$ and $1$ , large powers of $t$ become so small that they can be effectively treated like $0$ . Using the binomial expansion formula, this allows you to make quick yet accurate approximations of functions or constants.

Example 1

In ascending powers of $t$ , work out the first four terms of the expansion of $\left(1-\dfrac{t}{5}\right)^{12}$ and use this to approximate the value of $(0.96)^{12}$ to $2$ decimal places.

Use the general terms of the binomial expansion formula to get:

$\begin{aligned}(1-\dfrac{t}{5})^{12} &=\binom{12}{0}\left(-\dfrac{t}{5}\right)^0 + \binom{12}{1}\left(-\dfrac{t}{5}\right) + \binom{12}{2}\left(-\dfrac{t}{5}\right)^2 + \binom{12}{3}\left(-\dfrac{t}{5}\right)^3 + \dots \\&= 1 - \dfrac{12}{5}t + \dfrac{66}{25}t^2 - \dfrac{44}{25}t^3 + \dots\end{aligned}$

The term inside the brackets is $0.96$ , so solve $\left(1 - \dfrac{t}{5}\right) = 0.96$ for $t$ :

$\begin{aligned}\left(1-\dfrac{t}{5}\right) &= 0.96 \\t &= \dfrac{1}{5}\end{aligned}$

Substitute $t = \dfrac{1}{5}$ into the first four terms of the expansion you computed:

$\begin{aligned}(1 - \dfrac{t}{5})^{12} &= 1 - \dfrac{12}{5}\times \dfrac{1}{5} + \dfrac{66}{25}\left(\dfrac{1}{5}\right)^2 - \dfrac{44}{25}\left(\dfrac{1}{5}\right)^3 \\&= 0.61152\end{aligned}$

So $(0.96)^{12}$ is equal to $0.61$ to $2$ decimal places. In fact, $(0.96)^{12} = 0.6127 \dots$ , so the approximation is valid to $2$ places.

Therefore, the approximation is given by $\underline{0.61}$ .

Example 2

Supposing that $t$ is sufficiently small that terms of degree $4$ or higher are negligible, show that $(1-t^2)(4-5t)^5 \approx 1024 - 6400t + 14976t^2 -13600t^3$ .

Use the binomial expansion formula to expand $(4-5t)^5$ up to the $t^3$ term (as higher terms can be ignored).

$\begin{aligned}(4-5t)^5 &\approx \binom{5}{0}4^5 + \binom{5}{1}4^4(-5t) + \binom{5}{2}4^3(-5t)^2 + \binom{5}{3}4^2(-5t)^3 \\&\approx 1024 - 6400t + 16000t^2 - 20000t^3\end{aligned}$ 

Multiply this by $(1-t^2)$ , and set the terms of degree $4$ or higher equal to $0$ :

$\begin{aligned}(1-t^2)(1024 - 6400t + 16000t^2 - 20000t^3) &= 1024 -6400t + 14976t^2 - 13600t^3 -16000t^4 + 20000t^5 \\&\approx 1024 - 6400t + 14976t^2 - 13600t^3\end{aligned}$ 

Therefore, $\underline{(1-t^2)(4-5t)^5 \approx 1024 - 6400t + 14976t^2 - 13600t^3}$ .