5.19 Maclaurin seriesIB Maths: Analysis and Approaches HL: Revision notes

A Maclaurin series writes a function as an infinite polynomial built from its derivatives at $x=0$: $f(x)=f(0)+xf'(0)+\frac{x^{2}}{2!}f''(0)+\frac{x^{3}}{3!}f'''(0)+\dots$ The general term is $\frac{f^{(n)}(0)}{n!}x^{n}$. Truncating after a few terms gives a polynomial approximation that is very good close to $x=0$ and usually gets worse further away.

The formula is in the formula booklet; what is tested is finding the derivatives accurately and using the series.

These are given in the formula booklet, but you must be fluent with them:

• $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\dots$ (all $x$)
• $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\dots$ (all $x$)
• $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\dots$ (all $x$)
• $\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\dots$ (valid for $-1<x\le1$), with no factorials
• $(1+x)^{p}=1+px+\frac{p(p-1)}{2!}x^{2}+\dots$, $p\in\mathbb{Q}$ (valid for $|x|<1$)

Example: $\sqrt{1+x}=1+\frac x2-\frac{x^{2}}{8}+\dots$

Replace $x$ in a standard series by an expression, and simplify every power carefully:

• $e^{2x}=1+2x+\frac{(2x)^{2}}{2!}+\frac{(2x)^{3}}{3!}+\dots=1+2x+2x^{2}+\frac43x^{3}+\dots$
• $e^{x^{2}}=1+x^{2}+\frac{x^{4}}{2!}+\dots$
• $\ln(1+3x)=3x-\frac92x^{2}+9x^{3}-\dots$, valid for $-\frac13<x\le\frac13$

The interval of validity changes too: substituting $3x$ for $x$ in $-1<x\le1$ gives $-\frac13<x\le\frac13$.

Products: multiply two series and collect terms up to the power you need. For $e^{x}\sin x$: $\left(1+x+\tfrac{x^{2}}{2}+\tfrac{x^{3}}{6}\right)\left(x-\tfrac{x^{3}}{6}\right)=x+x^{2}+\tfrac13x^{3}+\dots$ Differentiation: differentiate a series term by term. From $\ln(1+3x)=3x-\frac92x^{2}+9x^{3}-\dots$ we get $\frac{3}{1+3x}=3-9x+27x^{2}-\dots$

Integration: integrate term by term. This gives approximations to integrals that cannot be done exactly, e.g. $\int_0^{0.5}e^{-x^{2}}\,dx\approx\int_0^{0.5}\left(1-x^{2}+\frac{x^{4}}{2}\right)dx$.

If $y$ is defined by a differential equation and an initial value, you can find its series without solving the equation:

1. Use the equation to find $y'(0)$.
2. Differentiate the equation implicitly to get $y''$, then again for $y'''$.
3. Substitute $x=0$ and the known values at each stage.
4. Put the values into $y(0)+y'(0)x+\frac{y''(0)}{2!}x^{2}+\dots$

Example: $y'=x+y^{2}$, $y(0)=1$. Then $y''=1+2yy'$ and $y'''=2(y')^{2}+2yy''$, so $y'(0)=1$, $y''(0)=3$, $y'''(0)=8$ and $y\approx1+x+\frac32x^{2}+\frac43x^{3}$.

Approximations: substitute a small value of $x$, e.g. $\ln1.3\approx0.264$ from three terms of $\ln(1+3x)$ with $x=0.1$ (true value $0.2624$). Accuracy falls as $x$ moves away from 0, because the neglected terms are no longer small.

Limits: replace each function by its series, cancel the lowest power of $x$, then let $x\to0$: $\lim_{x\to0}\frac{e^{2x}-1-2x}{x^{2}}=\lim_{x\to0}\left(2+\tfrac43x+\dots\right)=2.$