Taylor series and limitsEdexcel A-Level Further Maths: Revision notes

The Taylor series of $f(x)$ about $x=a$ writes $f$ as a power series in $(x-a)$: $f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+\dots+\frac{(x-a)^r}{r!}f^{(r)}(a)+\dots$ The case $a=0$ is the Maclaurin series. Taylor series are useful when the function is easy to evaluate at a point $a$ other than $0$, and when you want an approximation close to $a$. The coefficient of $(x-a)^r$ is always $\frac{f^{(r)}(a)}{r!}$; a common error is to leave out the factorial.

Expand $\sin x$ in ascending powers of $\left(x-\frac{\pi}{6}\right)$ up to the term in $\left(x-\frac{\pi}{6}\right)^3$. The derivatives are $\cos x$, $-\sin x$, $-\cos x$. At $x=\frac{\pi}{6}$ the values are $f=\frac12$, $f'=\frac{\sqrt3}{2}$, $f''=-\frac12$, $f'''=-\frac{\sqrt3}{2}$. $\sin x\approx\frac12+\frac{\sqrt3}{2}\left(x-\frac{\pi}{6}\right)-\frac14\left(x-\frac{\pi}{6}\right)^2-\frac{\sqrt3}{12}\left(x-\frac{\pi}{6}\right)^3.$ Write exact values: keep $\sqrt3$ and $\pi$ rather than decimals. The coefficients are $\frac12,\frac{\sqrt3}{2},\frac{-1/2}{2},\frac{-\sqrt3/2}{6}$.

A truncated Taylor series gives a good approximation when $x$ is close to $a$, because $(x-a)^r$ shrinks rapidly. Example: $\ln x=\ln2+\frac12(x-2)-\frac18(x-2)^2+\frac1{24}(x-2)^3-\dots$. With $x=2.2$, $x-2=0.2$, so $\ln2.2\approx0.6931+0.1-0.005+0.00033=0.7885$ (true value $0.78846$). The first omitted term estimates the error. A series gets worse as $|x-a|$ gets larger.

Standard Maclaurin series (valid for the ranges stated in the formula booklet) include $\mathrm{e}^x=1+x+\frac{x^2}{2!}+\dots$, $\sin x=x-\frac{x^3}{3!}+\dots$, $\cos x=1-\frac{x^2}{2!}+\dots$ and $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$. To find a limit as $x\to0$ of a quotient that gives $\frac00$, write the top and bottom as series, cancel the lowest power of $x$, then let $x\to0$: every remaining term in $x$ vanishes.

(1) $\displaystyle\lim_{x\to0}\frac{x-\arctan x}{x^3}$: $x-\arctan x=\frac{x^3}{3}-\frac{x^5}{5}+\dots$, so the quotient is $\frac13-\frac{x^2}{5}+\dots\to\frac13$. (2) $\displaystyle\lim_{x\to0}\frac{\mathrm{e}^{2x^2}-1}{x^2}$: $\mathrm{e}^{2x^2}=1+2x^2+2x^4+\dots$, so the quotient is $2+2x^2+\dots\to2$. Replace $x$ by the inner expression ($-x$, $2x^2$, $x^2$) in the standard series, then check the resulting power.