Example 1: Find the Taylor series for f(x)=\cos x centered at the given value of a=\pi. [Assume that f has a power series expansion. Do not show that R_n(x)\to 0.] Also find the associated radius of convergence.
Solution: (1) f(x)=\cos x and f(\pi)=-1; (2) f'(x)=-\sin x and f'(\pi)=0; (3) f''(x)=-\cos x and f''(\pi)=1; (4) f'''(x)=\sin x and f'''(x)=0; and this pattern repeats indefinitely. Therefore the Taylor series at \pi is -1+\frac{1}{2!}(x-\pi)^2-\frac{1}{4!}(x-\pi)^4+\frac{1}{6!}(x-\pi)^6-\frac{1}{8!}(x-\pi)^8+\ldots=\sum_{n=0}^\infty(-1)^{n-1}\frac{1}{(2n)!}(x-\pi)^{2n}.
The binomial series theorem: if k is any real number and |x|<1, then (1+x)^k = \sum_{n=0}^\infty {k\choose n}x^n = 1+\frac{k}{1!}x+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3+\ldots.
Example 2: Use the binomial series to expand the function \frac{1}{(2+x)^3} as a power series. State the radius of convergence.
Hint: Use \frac{1}{(2+x)^3}=\frac{1}{8}(1+x/2)^{-3} and the binomial series theorem.
Important Maclaurin series:
- \frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+\ldots;
- e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots;
- \sin x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots;
- \cos x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\ldots;
- \tan^{-1}x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots;
- \ln(1+x)=\sum_{n=0}^\infty(-1)^{n-1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots.
Example 3: Use the important Maclaurin series to obtain the Maclaurin series for the given function (a) f(x) = x\cos(x^2/2); (b) f(x)=\frac{x}{4+x^2}.
Hint: (a) Use the Maclaurin series of \cos x; (b) Use the Maclaurin series of 1/(1-x).
Example 4: Use series to evaluate the limit \lim_{x\to 0}\frac{\sin x-x+x^3/6}{x^5}.
Hint: Use the Maclaurin series of \sin x.
Example 5: Find the sum of the series \sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{n!}.
Solution: Use the Maclaurin series of e^x=\sum_{n=0}^\infty\frac{x^n}{n!}. By replacing x with -x^4, we obtain \sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{n!}=e^{-x^4}.