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