Science:Math Exam Resources/Courses/MATH103/April 2014/Question 07 (b)/Solution 1

By part (a), (with $t$ replacing $x$)

$e^{-t^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}{t}^{2n}.$

Integrating term-by-term yields

$\underset{0}{\overset{x}{\int }}{e}^{-t^2}\mathrm{d}t=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}\underset{0}{\overset{x}{\int }}{t}^{2n}\mathrm{d}t=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}\cdot \frac{1}{2n+1}{x}^{2n+1}.$

Since only odd powers appear in this series, $b_{12}=0$.

On the other hand, $b_{11}{x}^{11}=b_{11}{x}^{2n+1}$ for $n=5$. But the coefficient of $x^{2n+1}$ is $\frac{(-1{)}^n}{n!}\cdot \frac{1}{2n+1}$. For $n=5$, this yields $b_{11}=\frac{(-1{)}^5}{5!}\cdot \frac{1}{2(5)+1}=-\frac{1}{11\cdot 5!}$.