Differentiation and Integration of Power Series

Theory

Let

$$f(x)=\sum_{n=0}^{\infty} c_n (x-a)^n$$

be a power series with radius of convergence R>0. Then $f(x)$ may be differentiated or integrated term-by-term within its interval of convergence. The resulting series for $f'(x)$ and $\int f(x)\,dx$ have the same radius of convergence $R$ (though endpoints must be checked separately).

Common Maclaurin Series

1. Geometric Series

\frac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n, \qquad -1<x<1

2. Exponential Function

e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \qquad -\infty<x<\infty

3. Sine Function

\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \qquad -\infty<x<\infty

4. Cosine Function

\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}, \qquad -\infty<x<\infty

5. Natural Logarithm

\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}, \qquad -1<x\le1

6. Inverse Tangent

\tan^{-1}x = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}, \qquad -1<x<1

Exercises

Exercise 1

Find the Maclaurin series, radius, and interval of convergence for $f(x)=\tan^{-1}x$.

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Solution

Recall the geometric series

\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \qquad |x|<1.

Since

$$\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2},$$

integrate term-by-term:

$$\tan^{-1}x = \int \sum_{n=0}^{\infty} (-1)^n x^{2n}\,dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}.$$

Radius of convergence:

$$R=1.$$

Interval of convergence:

$$(-1,1).$$