4.1: Taylor Series
- Page ID
- 8357
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One series you have encountered before is Taylor’s series,
\[f(x) = \sum_{n=0}^{\infty} f^{(n)}(a)\frac{(x-a)^n}{n!}, \label{eq:IV:taylor} \]
where \(f^{(n)}(x)\) is the \(n\)th derivative of \(f\). An example is the Taylor series of the cosine around \(x=0\) (i.e., \(a=0\)),
\[\begin{aligned} &&\qquad&\cos(0) &= 1,\nonumber\\ \cos'(x) &= -\sin(x),&&\cos'(0)&=0,\nonumber\\ \cos^{(2)}(x) &= -\cos(x),&&\cos^{(2)}(0)&=-1,\\ \cos^{(3)}(x) &= \sin(x),&&\cos^{(3)}(0)&=0,\nonumber\\ \cos^{(4)}(x) &= \cos(x),&&\cos^{(4)}(0)&=1.\end{aligned} \nonumber \]
Notice that after four steps we are back where we started. We have thus found (using \(m=2n\) in (\(\PageIndex{1}\))) )
\[\cos x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} x^{2m}, \nonumber \]
Show that \[\sin x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} x^{2m+1}. \nonumber \]
- Answer
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TBA