Standard Maclaurin seriesAQA A-Level Further Maths: Revision notes

A Maclaurin series expresses a function as a power series in $x$. You must recognise and use these standard results: $\mathrm{e}^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^r}{r!}+\dots$ $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots+(-1)^{r+1}\frac{x^r}{r}+\dots$ $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots+(-1)^r\frac{x^{2r+1}}{(2r+1)!}+\dots$ $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots+(-1)^r\frac{x^{2r}}{(2r)!}+\dots$ $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\dots$ $\sin x$ has only odd powers and $\cos x$ only even powers, because $\sin$ is an odd function and $\cos$ is even. Proof of these series is not required.

Each series only equals the function for certain $x$:

• $\mathrm{e}^x$, $\sin x$ and $\cos x$: valid for all real $x$.
• $\ln(1+x)$: valid for $-1<x\le1$.
• $(1+x)^n$: valid for $|x|<1$ when $n$ is not a non-negative integer. If $n$ is a positive integer the series stops after the $x^n$ term and is exact for all $x$. The endpoint $x=1$ is included for $\ln(1+x)$ (it gives $\ln2=1-\frac12+\frac13-\dots$) but not $x=-1$, where $\ln0$ is undefined.

Replace $x$ by a function $u$ to get new series, and replace the range too. If the standard series is valid for a range of $u$, solve that range for $x$.

• $\mathrm{e}^{3x}=1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}+\dots$ valid for all $x$.
• $\ln(1+2x)=2x-2x^2+\frac83x^3-\dots$ valid for $-1<2x\le1$, i.e. $-\frac12<x\le\frac12$.
• $\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$ valid for $-1\le x<1$.
• $(1+4x)^{-1/2}=1-2x+6x^2-20x^3+\dots$ valid for $|4x|<1$, i.e. $|x|<\frac14$. Subtracting $\ln(1-2x)$ from $\ln(1+2x)$ gives $\ln\frac{1+2x}{1-2x}=4x+\frac{16}{3}x^3+\dots$, in which the even terms cancel.

To expand a product, write each series to the required power and multiply, collecting like powers. Only keep terms up to the order asked for. Example: $\mathrm{e}^x\sin x=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\left(x-\frac{x^3}{6}\right)=x+x^2+\frac13x^3+\dots$ The range of validity of the product is the overlap of the ranges of the factors. For $\frac{\cos x}{\sqrt{1+4x}}$ the cosine series holds for all $x$, so the result holds for $|x|<\frac14$. You can integrate a series term by term to estimate an integral over a small interval: $\int_0^{0.2}\mathrm{e}^x\sin x\,\mathrm{d}x\approx\left[\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{12}\right]_0^{0.2}=0.0228$.

Substituting a small $x$ into a truncated series gives an approximation. The smaller $|x|$ is, the better the approximation, and more terms give more accuracy. Choose $x$ so that the expression becomes the required number: for $\frac{1}{\sqrt{1.04}}$ use $1+4x=1.04$, so $x=0.01$: $\frac{1}{\sqrt{1.04}}\approx1-0.02+0.0006-0.00002=0.98058.$ Check that your $x$ lies inside the range of validity. Using $x=0.1$ in $\mathrm{e}^{3x}\approx1+3x+\frac92x^2$ gives $\mathrm{e}^{0.3}\approx1.345$ (true value $1.350$).