Maclaurin Series by Successive Differentiation
Maclaurin Series by Successive Differentiation
Maclaurin series
The Maclaurin series is the Taylor series of a sufficiently differentiable function about \(x=0\):
\[\mathrm{f}(x)=\mathrm{f}(0)+x\mathrm{f}'(0)+\frac{x^2}{2!}\mathrm{f}''(0)+\cdots+\frac{x^n}{n!}\mathrm{f}^{(n)}(0)+\cdots\]
Method
- Differentiate \(\mathrm{f}(x)\) successively to obtain \(\mathrm{f}'(x),\mathrm{f}''(x),\mathrm{f}'''(x),\ldots\).
- Evaluate the function and derivatives at \(x=0\).
- Substitute these values into the Maclaurin formula, retaining the terms required by the question and checking the interval of validity.
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Answer:\(\displaystyle f(x)=\sum_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}x^r\); \(\displaystyle (1+x)^n=\sum_{r=0}^{\infty}\binom{n}{r}x^r\) for \(|x|<1\); \(\displaystyle e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}\); \(\displaystyle \sin x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r+1}}{(2r+1)!}\); \(\displaystyle \cos x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r)!}\); \(\displaystyle \ln(1+x)=\sum_{r=1}^{\infty}\frac{(-1)^{r+1}x^r}{r}\) for \(-1<x\le1\). For \(\displaystyle \frac{\ln(1-2x)}{1-\frac{x}{3}}\), the combined expansion is valid for \(-\frac12\le x<\frac12\).
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