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Mathematics LibreTexts

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., \(a=0\)), \[\begin{aligned} &&\qquad&\cos(0) &= 1,\nonumber\\ \cos'(x) &= -\sin(x),&&\cos'(0)&=0,\nonumber\\ \cos^{(2)}(x) &= -\cos(x),&&\cos^{(2)}(0)&=-1,\\ \cos^{(3)}(x) &= \sin(x),&&\cos^{(3)}(0)&=0,\nonumber\\ \cos^{(4)}(x) &= \cos(x),&&\cos^{(4)}(0)&=1.\end{aligned}\] Notice that after four steps we are back where we started. We have thus found (using \(m=2n\) in (\(\PageIndex{1}\))) ) \[\cos x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} x^{2m},\]

    Exercise \(\PageIndex{1}\)

    Show that \[\sin x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} x^{2m+1}.\]

    Answer

    TBA