Binomial series for any rational powerEdexcel International A Level Maths: Revision notes

For any rational number $n$, including negative and fractional values, $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$ The series is valid only for $|x|<1$. When $n$ is a non-negative integer the series stops and is valid for all $x$. For any other $n$ it goes on for ever and is an approximation that improves as more terms are used. For example $(1+x)^{-1}=1-x+x^2-x^3+\ldots$ and $(1+x)^{\frac12}=1+\frac12x-\frac18x^2+\frac1{16}x^3-\ldots$ for $|x|<1$.

First take out the constant: $(a+bx)^n=a^n\left(1+\frac bax\right)^n$. Then expand with $u=\frac bax$: $a^n\left[1+nu+\frac{n(n-1)}{2!}u^2+\ldots\right]$. It is valid for $\left|\frac bax\right|<1$, that is $|x|<\left|\frac ab\right|$. In the form $(ax+b)^n$ the condition is $|x|<\left|\frac ba\right|$. Example: $(2+x)^{-2}=\frac14\left(1+\frac x2\right)^{-2}=\frac14\left(1-x+\frac34x^2-\frac12x^3\ldots\right)=\frac14-\frac x4+\frac{3x^2}{16}-\frac{x^3}{8}+\ldots$, valid for $|x|<2$.

(1) $(1+4x)^{-\frac12}$: put $u=4x$. $1-\frac12(4x)+\frac{\left(-\frac12\right)\left(-\frac32\right)}{2}(4x)^2+\frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6}(4x)^3=1-2x+6x^2-20x^3$, valid for $|x|<\frac14$. (2) $\sqrt[3]{8+3x}=2\left(1+\frac{3x}{8}\right)^{\frac13}=2\left[1+\frac x8-\frac{x^2}{64}+\ldots\right]=2+\frac x4-\frac{x^2}{32}$, valid for $|x|<\frac83$. Put brackets round the whole term when substituting, particularly for negative terms or fractions.

To estimate a number, choose $x$ so that the function takes the required value and lies inside the range of validity. For $\sqrt[3]{8+3x}$ with $x=0.1$ we get $\sqrt[3]{8.3}\approx2+0.025-0.0003125=2.0247$. To estimate $\sqrt[3]{23}$ you would need $x=5$, which is outside $|x|<\frac83$, so the series cannot be used. Smaller $|x|$ gives better accuracy. In a question you must justify the choice of $x$ and give the answer to the stated accuracy.

For a rational function such as $f(x)=\frac{10+5x}{(1+2x)(2-x)}$ first write it in partial fractions: $\frac{3}{1+2x}+\frac{4}{2-x}$. Expand each term separately: $3(1+2x)^{-1}=3-6x+12x^2-\ldots$ and $\frac{4}{2-x}=2\left(1-\frac x2\right)^{-1}=2+x+\frac{x^2}{2}+\ldots$. Add the results: $5-5x+\frac{25}{2}x^2+\ldots$. The combined expansion is valid only where every part is valid: the first needs $|x|<\frac12$ and the second $|x|<2$, so the expansion is valid for $|x|<\frac12$. Choose the smaller range.

Questions may give a coefficient and ask for an unknown constant. For $(2+ax)^{-3}=\frac18\left(1+\frac{ax}{2}\right)^{-3}$ the $x^2$ term is $\frac18\times6\times\frac{a^2}{4}=\frac{3a^2}{16}$. If this is $\frac34$, then $a^2=4$ and $a=2$ (when $a>0$). Then the $x^3$ term is $\frac18\times(-10)\times1=-\frac54$, and the expansion is valid for $|x|<1$ since $\left|\frac{ax}{2}\right|=|x|$. Form an equation from the given coefficient, solve, then continue.