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Recitation 25

Example 1: Find the Taylor series for f(x)=cos⁡xf(x)=\cos x centered at the given value of a=πa=\pi. [Assume that ff has a power series expansion. Do not show that Rn(x)→0R_n(x)\to 0.] Also find the associated radius of convergence.

Solution: (1) f(x)=cos⁡xf(x)=\cos x and f(π)=−1f(\pi)=-1; (2) f′(x)=−sin⁡xf'(x)=-\sin x and f′(π)=0f'(\pi)=0; (3) f′′(x)=−cos⁡xf''(x)=-\cos x and f′′(π)=1f''(\pi)=1; (4) f′′′(x)=sin⁡xf'''(x)=\sin x and f′′′(x)=0f'''(x)=0; and this pattern repeats indefinitely. Therefore the Taylor series at π\pi is −1+12!(x−π)2−14!(x−π)4+16!(x−π)6−18!(x−π)8+…=∑n=0∞(−1)n−11(2n)!(x−π)2n-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 kk is any real number and ∣x∣<1|x|<1, then (1+x)k=∑n=0∞(kn)xn=1+k1!x+k(k−1)2!x2+k(k−1)(k−2)3!x3+…(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 1(2+x)3\frac{1}{(2+x)^3} as a power series. State the radius of convergence.

Hint: Use 1(2+x)3=18(1+x/2)−3\frac{1}{(2+x)^3}=\frac{1}{8}(1+x/2)^{-3} and the binomial series theorem.

Important Maclaurin series:

  1. 11−x=∑n=0∞xn=1+x+x2+x3+…\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+\ldots;
  2. ex=∑n=0∞xnn!=1+x1!+x22!+x33!+…e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots;
  3. sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!=x−x33!+x55!−x77!+…\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;
  4. cos⁡x=∑n=0∞(−1)nx2n(2n)!=1−x22!+x44!−x66!+…\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;
  5. tan⁡−1x=∑n=0∞(−1)nx2n+12n+1=x−x33+x55−x77+…\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;
  6. ln⁡(1+x)=∑n=0∞(−1)n−1xnn=x−x22+x33−x44+…\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)=xcos⁡(x2/2)f(x) = x\cos(x^2/2); (b) f(x)=x4+x2f(x)=\frac{x}{4+x^2}.

Hint: (a) Use the Maclaurin series of cos⁡x\cos x; (b) Use the Maclaurin series of 1/(1−x)1/(1-x).

Example 4: Use series to evaluate the limit lim⁡x→0sin⁡x−x+x3/6x5\lim_{x\to 0}\frac{\sin x-x+x^3/6}{x^5}.

Hint: Use the Maclaurin series of sin⁡x\sin x.

Example 5: Find the sum of the series ∑n=0∞(−1)nx4nn!\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{n!}.

Solution: Use the Maclaurin series of ex=∑n=0∞xnn!e^x=\sum_{n=0}^\infty\frac{x^n}{n!}. By replacing xx with −x4-x^4, we obtain ∑n=0∞(−1)nx4nn!=e−x4\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{n!}=e^{-x^4}.