Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (a)/Solution 1

Since ${\frac {df}{dx}}=x\left({\frac {1}{1+3x^{3}}}\right)$, a Maclaurin series for ${\frac {df}{dx}}$ can be computed from the geometric series

${\frac {1}{1+y}}={\frac {1}{1-(-y)}}=\sum _{n=0}^{\infty }(-y)^{n}=\sum _{n=0}^{\infty }(-1)^{n}y^{n}.$

By substituting $y=3x^{3}$, we obtain:

${\frac {df}{dx}}=x\left({\frac {1}{1+3x^{3}}}\right)=x\sum _{n=0}^{\infty }(-1)^{n}(3x^{3})^{n}=\sum _{n=0}^{\infty }(-3)^{n}x^{3n+1}.$

To obtain a series for $f(x)$, we integrate the series for ${\frac {df}{dx}}$ term by term and apply the initial condition $f(0)=1$. Integrating the series for ${\frac {df}{dx}}$ implies that

$f(x)=\sum _{n=0}^{\infty }(-3)^{n}\int x^{3n+1}\,dx=\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}+C$

for some constant C determined by the initial condition. Since $\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}$ evaluates to 0 at $x=0$, we must have $C=1$ for $f(0)=1$. Therefore, a Maclaurin series for $f(x)$ is given by

${\color {blue}1+\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}=1+{\frac {x^{2}}{2}}-{\frac {3x^{5}}{5}}+{\frac {9x^{8}}{8}}+\cdots .}$