Let $f ( x ) = \ln \left( 1 + x ^ { 3 } \right)$. (a) The Maclaurin series for $\ln ( 1 + x )$ is $x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } + \cdots + ( - 1 ) ^ { n + 1 } \cdot \frac { x ^ { n } } { n } + \cdots$. Use the series to write the first four nonzero terms and the general term of the Maclaurin series for $f$. (b) The radius of convergence of the Maclaurin series for $f$ is 1 . Determine the interval of convergence. Show the work that leads to your answer. (c) Write the first four nonzero terms of the Maclaurin series for $f ^ { \prime } \left( t ^ { 2 } \right)$. If $g ( x ) = \int _ { 0 } ^ { x } f ^ { \prime } \left( t ^ { 2 } \right) d t$, use the first two nonzero terms of the Maclaurin series for $g$ to approximate $g ( 1 )$. (d) The Maclaurin series for $g$, evaluated at $x = 1$, is a convergent alternating series with individual terms that decrease in absolute value to 0 . Show that your approximation in part (c) must differ from $g ( 1 )$ by less than $\frac { 1 } { 5 }$.
Let $f ( x ) = \ln \left( 1 + x ^ { 3 } \right)$.
(a) The Maclaurin series for $\ln ( 1 + x )$ is $x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } + \cdots + ( - 1 ) ^ { n + 1 } \cdot \frac { x ^ { n } } { n } + \cdots$. Use the series to write the first four nonzero terms and the general term of the Maclaurin series for $f$.
(b) The radius of convergence of the Maclaurin series for $f$ is 1 . Determine the interval of convergence. Show the work that leads to your answer.
(c) Write the first four nonzero terms of the Maclaurin series for $f ^ { \prime } \left( t ^ { 2 } \right)$. If $g ( x ) = \int _ { 0 } ^ { x } f ^ { \prime } \left( t ^ { 2 } \right) d t$, use the first two nonzero terms of the Maclaurin series for $g$ to approximate $g ( 1 )$.
(d) The Maclaurin series for $g$, evaluated at $x = 1$, is a convergent alternating series with individual terms that decrease in absolute value to 0 . Show that your approximation in part (c) must differ from $g ( 1 )$ by less than $\frac { 1 } { 5 }$.