Using a geometric series and properties of a power series, a power series representation of $f ( x ) = \tan ^ { - 1 } x$ is

A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } x ^ { 2 n } } { 2 n + 1 }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 2 n + 1 }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 3 } } { 2 n + 1 }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 2 n + 1 }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 3 } } { 2 n + 3 }$

Question

Using a geometric series and properties of a power series, a power series representation of $f ( x ) = \tan ^ { - 1 } x$ is

A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } x ^ { 2 n } } { 2 n + 1 }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 2 n + 1 }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 3 } } { 2 n + 1 }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 2 n + 1 }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 3 } } { 2 n + 3 }$

Quizplus · Accepted Answer

The power series representation of $f(x) = \tan^{-1}x$ is derived from the integral of its derivative, which is $\frac{1}{1+x^2}$. The series for $\frac{1}{1+x^2}$ is a geometric series with a common ratio of $-x^2$, and integrating term by term gives the series for $\tan^{-1}x$, which is $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$.

Using a Geometric Series and Properties of a Power Series f(x)=tan⁡−1xf ( x ) = \tan ^ { - 1 } xf(x)=tan−1x

Using a Geometric Series and Properties of a Power Series $f ( x ) = \tan ^ { - 1 } x$