Exercises 8.8 (6)

\[f(x) = tan^{-1}(x); c = 0\]

Key Idea:

\[tan^{-1}(x) = \sum^\infty_{n = 0}(-1)^n\frac{x^{2n+1}}{2n+1}\]

Derivatives for the first few n’s. Find the trend.

\[\begin{align} f(x) &= tan^{-1}(x), &f(0) &= 0 \\ \\ f'(x) &= \frac{1}{x^2+1}, &f'(0) &= 1 \\ \\ f''(x) &= -\frac{2x}{(x^2+1)^2}, &f''(0) &= 0 \\ \\ f'''(x) &= -\frac{2(-3x^2+1)}{(x^2+1)^3}, &f'''(0) &= -2 \\ \\ f''''(x) &= \frac{24x(-x^2+1)}{(x^2+1)^4}, &f''''(0) &= 0 \\ \\ f'''''(x) &= \frac{24(5x^4-10x^2+1)}{(x^2+1)^5}, &f'''''(0) &= 24 \\ \\ f''''''(x) &= \frac{240x(-3x^4+10x^2-3)}{(x^2+1)^6}, &f'''''(0) &= 0 \\ \\ f'''''''(x) &= \frac{720(7x^6-35x^4+21x^2-1)}{(x^2+1)^7}, &f''''''(0) &= -720 \\ \\ \end{align}\]

Now substitute into Taylor’s formula:

\[ \begin{align} tan^{-1}(x) &= \sum^\infty_{n = 0}\frac{f^n(0)}{n!}x^n \\ \\ &= \frac{0}{0!}x^0 + \frac{1}{1!}x^1 + \frac{0}{2!}x^2 - \frac{2}{3!}x^3 + \frac{0}{4!}x^4 + \frac{24}{5!}x^5 + \frac{0}{6!}x^6 - \frac{720}{7!}x^7 + ...\\ \\ &= 0 + \frac{x^1}{1} + 0 - \frac{x^3}{3} + 0 + \frac{x^5}{5} + 0 - \frac{x^7}{7} + ... \\ \\ &= \frac{x^1}{1} - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} \\ \\ &= \sum_{n = 0}^{\infty}(-1)^n*\frac{x^{2n + 1}}{2n + 1} \end{align} \]