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Binomial expansion I

Binomial estimation

Binomial estimation

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Explainer Video

Tutor: Dylan

Summary

Binomial estimation

In a nutshell

When ttt is a value between 000​ and 111​, large powers of ttt become so small that they can be effectively treated like 000. Using the binomial expansion formula, this allows you to make quick yet accurate approximations of functions or constants.



Example 1

​In ascending powers of ttt​, work out the first four terms of the expansion of (1−t5)12\left(1-\dfrac{t}{5}\right)^{12}(1−5t​)12​and use this to approximate the value of (0.96)12(0.96)^{12}(0.96)12 to 222 decimal places.


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


(1−t5)12=(120)(−t5)0+(121)(−t5)+(122)(−t5)2+(123)(−t5)3+…=1−125t+6625t2−4425t3+…\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}(1−5t​)12​=(012​)(−5t​)0+(112​)(−5t​)+(212​)(−5t​)2+(312​)(−5t​)3+…=1−512​t+2566​t2−2544​t3+…​​​


The term inside the brackets is 0.960.960.96, so solve (1−t5)=0.96\left(1 - \dfrac{t}{5}\right) = 0.96(1−5t​)=0.96 for ttt:

​(1−t5)=0.96t=15\begin{aligned}\left(1-\dfrac{t}{5}\right) &= 0.96 \\t &= \dfrac{1}{5}\end{aligned}(1−5t​)t​=0.96=51​​​​


Substitute t=15t = \dfrac{1}{5}t=51​ into the first four terms of the expansion you computed:

(1−t5)12=1−125×15+6625(15)2−4425(15)3=0.61152\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}(1−5t​)12​=1−512​×51​+2566​(51​)2−2544​(51​)3=0.61152​​​


So (0.96)12(0.96)^{12}(0.96)12 is equal to 0.610.610.61 to 222 decimal places. In fact, (0.96)12=0.6127…(0.96)^{12} = 0.6127 \dots(0.96)12=0.6127…​, so the approximation is valid to 222 places.


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

​


Example 2

Supposing that ttt is sufficiently small that terms of degree 444​ or higher are negligible, show that (1−t2)(4−5t)5≈1024−6400t+14976t2−13600t3(1-t^2)(4-5t)^5 \approx 1024 - 6400t + 14976t^2 -13600t^3(1−t2)(4−5t)5≈1024−6400t+14976t2−13600t3.


Use the binomial expansion formula to expand (4−5t)5(4-5t)^5(4−5t)5 up to the t3t^3t3 term (as higher terms can be ignored).

(4−5t)5≈(50)45+(51)44(−5t)+(52)43(−5t)2+(53)42(−5t)3≈1024−6400t+16000t2−20000t3\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}(4−5t)5​≈(05​)45+(15​)44(−5t)+(25​)43(−5t)2+(35​)42(−5t)3≈1024−6400t+16000t2−20000t3​​​


Multiply this by (1−t2)(1-t^2)(1−t2), and set the terms of degree 444 or higher equal to 000:

(1−t2)(1024−6400t+16000t2−20000t3)=1024−6400t+14976t2−13600t3−16000t4+20000t5≈1024−6400t+14976t2−13600t3\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}(1−t2)(1024−6400t+16000t2−20000t3)​=1024−6400t+14976t2−13600t3−16000t4+20000t5≈1024−6400t+14976t2−13600t3​​​


Therefore, (1−t2)(4−5t)5≈1024−6400t+14976t2−13600t3‾\underline{(1-t^2)(4-5t)^5 \approx 1024 - 6400t + 14976t^2 - 13600t^3}(1−t2)(4−5t)5≈1024−6400t+14976t2−13600t3​.


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FAQs - Frequently Asked Questions

When can I use the binomial expansion formula to estimate?

When x is a value between 0 and 1, large powers of x become so small that they can be effectively treated like 0.

How do I use the binomial expansion to estimate?

When x is a value between 0 and 1, large powers of x become so small they can be effectively treated like 0.

How do I do binomial estimation?

Using the binomial expansion formula, this allows you to make quick yet accurate approximations of functions or constants.

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