Series Expansions

Taylor's series and Maclaurin's series expand a function that is infinitely differentiable as an infinite power series.

Theorem 7.5(a) — Taylor's Series. Let $f(x)$ be a function infinitely differentiable at $x=a$. Then $f(x)$ can be expanded, in an interval $(x-a,x+a)$, as

Theorem 7.5(b) — Maclaurin's Series. If $a=0$, the expansion becomes

Proof sketch (why the coefficients are $A_n=\dfrac{f^{(n)}(a)}{n!}$). Write the series in powers of $(x-a)$ as $f(x)=A_0+\sum_{n\ge1}A_n(x-a)^n$. Substituting $x=a$ isolates $A_0=f(a)$ (every other term carries a positive power of $(x-a)$ and vanishes). Differentiating once and substituting $x=a$ isolates $A_1=f'(a)$. Differentiating a second time and substituting $x=a$ isolates $2!A_2=f''(a)$, i.e. $A_2=\dfrac{f''(a)}{2!}$. Continuing this process — differentiate $k$ times, substitute $x=a$ — isolates $A_k=\dfrac{f^{(k)}(a)}{k!}$ in general, which is exactly the stated coefficient.

Worked illustrations from the textbook (the pattern behind Exercise 7.4):

• Maclaurin expansion of $\log(1+x)$: tabulate $f,f',f'',\ldots$ at $x=0$, substitute into the series, giving $\log(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$ for $-1<x\le1$.
• Maclaurin expansion of $\tan x$: the same tabulation (derivatives of $\tan x$ get progressively messier, but each is still evaluated at $0$) gives $\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots$ for $-\tfrac{\pi}{2}<x<\tfrac{\pi}{2}$.
• Taylor expansion of $\dfrac1x$ about $x=2$: tabulate the derivatives of $\dfrac1x$ at $x=2$ and substitute into powers of $(x-2)$, giving $\dfrac1x=\dfrac12-\dfrac{x-2}{4}+\dfrac{(x-2)^2}{8}-\dfrac{(x-2)^3}{16}+\cdots$. …