Binomial Expansion – negative & fractional powers

1. Firstly, write the expression as $\left(1+2x\right)^{-2}$. The power  $n=-2$ is negative and so we must use the second formula. We can then find the expansion by setting $n=-2$ and replacing all $x$ with $2x$: $\begin{array}{l}&&\left(1+2x\right)^{-2}\\&=&1-2(2x)+\frac{-2(-3)}{1\times 2}(2x)^2+\frac{-2(-3)(-4)}{1\times 2\times 3}(2x)^3+…\\&=&1-4x+12x^2-32x^3+…\end{array}$
2. We find the range of validity by replacing $x$ with $2x$ in the expression $\vert x\vert <1$ to give $\vert 2x\vert <1$. It follows that the expansion is valid for $\vert x\vert <\frac{1}{2}$. We can also write this as $-\frac{1}{2}<x<\frac{1}{2}$.

Begin by writing the expression as $\left(4-3x\right)^{\frac{1}{2}}$. The power $n=\frac{1}{2}$ is fractional so we must use the second formula. However, this first requires us to remove a factor of 4:

$\left(4-3x\right)^{\frac{1}{2}}=\left(4\left(1-\frac{3x}{4}\right)\right)^{\frac{1}{2}}=4^{\frac{1}{2}}\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}$

Note that the expansion of $\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}$ is given by:

$\begin{array}{l}\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}&=&1+\frac{1}{2}\left(-\frac{3x}{4}\right)+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{1\times 2}\left(-\frac{3x}{4}\right)^2+…\\&=&1-\frac{3}{8}x-\frac{9}{128}x^2+…\end{array}$

Hence, multiplying by the factor of $4^{\frac{1}{2}}=2$ gives:

$\left(4-3x\right)^{\frac{1}{2}}=2\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}\approx 2-\frac{3}{4}x-\frac{9}{64}x^2$

This expansion is valid for $\vert -\frac{3x}{4}\vert <1$, that is $\vert x\vert <\frac{4}{3}$.

Finally, by setting $x=0.1$, we can find an approximation to $\sqrt{3.7}$:

$\left(3.7\right)^{\frac{1}{2}}\approx 2-\frac{3}{4}\times 0.1-\frac{9}{64}\times 0.1^2\approx 1.9246 $

to 4 decimal places. Note that we can do this since $-\frac{4}{3}<0.1<\frac{4}{3}$. The exact value of $\sqrt{3.7}=1.9235$ to 4 decimal places, which is a reasonable approximation.

1. Firstly, revise partial fractions by clicking here. Note that $\frac{A}{1-x}+\frac{B}{1+\frac{1}{2}x}=\frac{A\left(1+\frac{1}{2}x\right)+B(1-x)}{(1-x)(1+\frac{1}{2}x)}$. Hence, $A+B=3$ and $\frac{1}{2}A-B=5$. It follows that $A=\frac{16}{3}$ and $B=-\frac{7}{3}$ and so $\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}=\frac{16}{3(1-x)}-\frac{7}{3\left(1+\frac{1}{2}x\right)}$.
2. From part 1, we can write $\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}=\frac{16}{3}(1-x)^{-1}-\frac{7}{3}\left(1+\frac{1}{2}x\right)^{-1}$. Now, take binomial expansions for each term: $(1-x)^{-1}=1+x+x^2+…$ $\left(1+\frac{1}{2}x\right)^{-1}=1-\frac{1}{2}x+\frac{1}{4}x^2+…$ Hence, $\begin{array}{l}\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}&\approx&\frac{16}{3}(1+x+x^2)-\frac{7}{3}\left(1-\frac{1}{2}x+\frac{1}{4}x^2\right)\\&=&3+\frac{13}{2}x+\frac{19}{4}x^2\end{array}$ as given.
3. The first expansion is valid for $\vert -x\vert <1$ (or $-1<x<1$) while the second expansion is valid for $\vert \frac{1}{2}x\vert <1$ (or $-2<x<2$. In order for both of these inequalities to be satisfied, we must have $-1<x<1$.