Taylor seriesEdexcel International A Level Further Maths: Revision notes

A Maclaurin series expands $f(x)$ about $x=0$, so it is only a good approximation for small $|x|$. A Taylor series expands about any point $x=a$ at which $f$ and its derivatives exist, in ascending powers of $(x-a)$. Derivation. Suppose $f(x)=a_0+a_1(x-a)+a_2(x-a)^2+a_3(x-a)^3+\dots$ Putting $x=a$ gives $a_0=f(a)$. Differentiating and putting $x=a$ gives $a_1=f'(a)$, then $2a_2=f''(a)$, and in general $n!\,a_n=f^{(n)}(a)$. Hence $f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+\dots+\frac{(x-a)^n}{n!}f^{(n)}(a)+\dots$ Equivalently, writing $x=a+h$: $f(a+h)=f(a)+hf'(a)+\frac{h^2}{2!}f''(a)+\dots$ The Maclaurin series is the special case $a=0$.

Method: (1) differentiate $f$ as many times as required; (2) evaluate $f(a),f'(a),f''(a),\dots$; (3) substitute into the formula; (4) divide by the factorials and tidy up. Example (specification): $f(x)=\sin x$ about $x=\pi$. Here $f'=\cos x$, $f''=-\sin x$, $f'''=-\cos x$, so $f(\pi)=0$, $f'(\pi)=-1$, $f''(\pi)=0$, $f'''(\pi)=1$. Therefore $\sin x=-(x-\pi)+\frac{(x-\pi)^3}{6}-\dots$ Check: $\sin(\pi+h)=-\sin h=-h+\frac{h^3}{6}-\dots$, which agrees. Example: $f(x)=\ln x$ about $x=2$ gives $f(2)=\ln2$, $f'(2)=\frac12$, $f''(2)=-\frac14$, $f'''(2)=\frac14$, so $\ln x=\ln2+\frac12(x-2)-\frac18(x-2)^2+\frac{1}{24}(x-2)^3-\dots$

For functions such as $\tan x$ the derivatives grow in complexity, so use trigonometric identities and keep the working organised. Example: $f(x)=\tan x$ about $x=\frac{\pi}{4}$. At $x=\frac{\pi}{4}$, $\tan x=1$ and $\sec^2x=2$.

• $f'=\sec^2x$, so $f'\left(\frac{\pi}{4}\right)=2$.
• $f''=2\sec^2x\tan x$, so $f''\left(\frac{\pi}{4}\right)=4$.
• $f'''=4\sec^2x\tan^2x+2\sec^4x$, so $f'''\left(\frac{\pi}{4}\right)=8+8=16$. Hence $\tan x=1+2h+2h^2+\frac83h^3+\dots$ with $h=x-\frac{\pi}{4}$. A series can then be differentiated or integrated term by term to give related series, such as $\sec^2x=2+4h+8h^2+\dots$ (which also follows from $\sec^2x=1+\tan^2x$).

To estimate $f(x_0)$, choose the centre $a$ close to $x_0$ where $f(a)$ and its derivatives are exact and easy.

• $\sqrt{4.2}$: use $f(x)=\sqrt x$ about $x=4$, where $f(4)=2$. Then $\sqrt x=2+\frac14(x-4)-\frac{1}{64}(x-4)^2+\frac{1}{512}(x-4)^3-\dots$ gives $\sqrt{4.2}\approx2.0494$.
• $\ln2.1$: use $\ln x$ about $x=2$.
• $\sin3$: use $\sin x$ about $\pi$ with $x-\pi=3-\pi$, giving $\sin3\approx0.14112$. The smaller $|x-a|$ is, the better the approximation, and more terms improve accuracy. Limits can also be found by dividing the series by a power of $(x-a)$: $\frac{\sin x}{x-\pi}=-1+\frac{(x-\pi)^2}{6}-\dots\to-1$.