More Binomial Expansions

Partial Fractions

We can find the binomial expansion of complicated functions by first decomposing them into partial fractions.

Example: Find the first three terms of the expansion of \dfrac{2x+1}{(x-1)(x+2)}.

\begin{aligned}\dfrac{2x+1}{(x-1)(x+2)}&=\dfrac{1}{x-1}+\dfrac{1}{x+2}\\[1.2em]&=(x-1)^{-1}+(x+2)^{-1}\end{aligned}

Now we can do binomial expansion on (x-1)^{-1} and (x+2)^{-1}

\begin{aligned}&\dfrac{2x+1}{(x-1)(x+2)}=(-1)^{-1}(1-x)^{-1}+2^{-1}\left(1+\dfrac{1}{2}x\right)^{-1}\\[1.2em]&=-\left(1-(-x)+\dfrac{-1\times(-2)}{1\times2}(-x)^{2}+...\right)\\[1.2em]&+\dfrac{1}{2}\left(1-\dfrac{1}{2}x+\dfrac{-1\times(-2)}{1\times2}\left(\dfrac{1}{2}x\right)^{2}+...\right)\\[1.2em]&=-\left(1+x+\dfrac{2}{2}x^{2}+...\right)\\[1.2em]&+\dfrac{1}{2}\left(1-\dfrac{1}{2}x+\dfrac{2}{2}\times\dfrac{1}{4}x^{2}+...\right)\\[1.2em]&=-(1+x+x^{2}+...)+\dfrac{1}{2}\left(1-\dfrac{1}{2}x+\dfrac{1}{4}x^{2}+...\right)\\[1.2em]&=-1-x-x^{2}+\dfrac{1}{2}-\dfrac{1}{4}x+\dfrac{1}{8}x^{2}+...\\[1.2em]&=-\dfrac{1}{2}-\dfrac{5}{4}x-\dfrac{7}{8}x^{2}+...\end{aligned}

Approximations from Binomial Expansions

By substituting in certain values for x, we can use the binomial expansion to approximate things.

Example: Use the binomial expansion of (1-3x)^{\frac{1}{4}} to four terms to find \sqrt[4]{0.97}

 

\begin{aligned}(1-3x)^{\frac{1}{4}}&=1-\left(\dfrac{1}{4}\times3x\right)+\left(\dfrac{\dfrac{1}{4}\times -\dfrac{3}{4}}{1\times2}\times(-3x)^{2}\right)\\[1.2em]&+\left(\dfrac{\dfrac{1}{4}\times -\dfrac{3}{4}\times-\dfrac{7}{4}}{1\times2\times3}\times(-3x)^{3}\right)+...\\[1.2em]&=1-\dfrac{3}{4}x+\left(\dfrac{-\dfrac{3}{16}}{2}\times9x^{2}\right)-\left(\dfrac{\dfrac{21}{64}}{6}\times27x^{3}\right)+...\\[1.2em]&=1-\dfrac{3}{4}x-\dfrac{3\times9}{16\times2}x^{2}-\dfrac{21\times27}{64\times6}x^{3}\\[1.2em]&=1-\dfrac{3}{4}x-\dfrac{27}{32}x^{2}-\dfrac{567}{384}x^{3}\\[1.2em]&=1-\dfrac{3}{4}x-\dfrac{27}{32}x^{2}-\dfrac{189}{128}x^{3}\end{aligned}

 

Now substitute x=\dfrac{1}{100}

 

\left(1-3\times\dfrac{1}{100}\right)^{\frac{1}{4}}=1-\left(\dfrac{3}{4}\times\dfrac{1}{100}\right)-\left(\dfrac{27}{32}\times\left(\dfrac{1}{100}\right)^{2}\right)-\left(\dfrac{189}{128}\times\left(\dfrac{1}{100}\right)^{3}\right)

 

(1-3\times0.01)^{\frac{1}{4}}=1-\dfrac{3}{400}-\left(\dfrac{27}{32}\times\dfrac{1}{10000}\right)-\left(\dfrac{189}{128}\times\dfrac{1}{1000000}\right)

 

(1-0.03)^{\frac{1}{4}}=1-\dfrac{3}{400}-\dfrac{27}{320000}-\dfrac{189}{128000000}

 

0.97^{\frac{1}{4}}=0.9916547734

 

So our estimate is \sqrt[4]{0.97}=0.9916547734, which is very close to the real value of 0.9924141173. Clearly, this approximation method is a very powerful tool.