MATH101 April 2006
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Question 01 (j)

Find the first three nonzero terms in the power series representation in powers of x (i.e. the Maclaurin series) for $\int _{0}^{x}t^{2}e^{t^{2}}\,dt$

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Hint

The Maclaurin series for $e^{y}$ is
${\begin{aligned}e^{y}=1+y+{\frac {y^{2}}{2!}}+{\frac {y^{3}}{3!}}+..=\sum _{n=0}^{\infty }{\frac {y^{n}}{n!}}\end{aligned}}$

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Solution

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Starting with the Maclaurin series for $e^{y}$, we have
${\begin{aligned}e^{y}=1+y+{\frac {y^{2}}{2!}}+{\frac {y^{3}}{3!}}+..=\sum _{n=0}^{\infty }{\frac {y^{n}}{n!}}\end{aligned}}$
Plugging in t^{2}, we have
${\begin{aligned}e^{t^{2}}=1t^{2}+{\frac {t^{4}}{2!}}{\frac {t^{6}}{3!}}+..=\sum _{n=0}^{\infty }{\frac {(1)^{n}t^{2n}}{n!}}\end{aligned}}$
Multiplying by $t^{2}$, we have
${\begin{aligned}t^{2}e^{t^{2}}=t^{2}t^{4}+{\frac {t^{6}}{2!}}{\frac {t^{8}}{3!}}+..=\sum _{n=0}^{\infty }{\frac {(1)^{n}t^{2n+2}}{n!}}\end{aligned}}$
Now integrating gives
${\begin{aligned}\int t^{2}e^{t^{2}}&=\int t^{2}t^{4}+{\frac {t^{6}}{2!}}{\frac {t^{8}}{3!}}+..\,dt=\sum _{n=0}^{\infty }\int {\frac {(1)^{n}t^{2n+2}}{n!}}\,dt\\&=C+{\frac {t^{3}}{3}}{\frac {t^{5}}{5}}+{\frac {t^{7}}{7\cdot 2!}}{\frac {t^{9}}{9\cdot 3!}}+...=C+\sum _{n=0}^{\infty }{\frac {(1)^{n}t^{2n+3}}{(2n+3)n!}}\end{aligned}}$
Plugging in endpoints, we have
${\begin{aligned}\int _{0}^{x}t^{2}e^{t^{2}}&={\frac {x^{3}}{3}}{\frac {x^{5}}{5}}+{\frac {x^{7}}{7\cdot 2!}}{\frac {x^{9}}{9\cdot 3!}}+...=\sum _{n=0}^{\infty }{\frac {(1)^{n}x^{2n+3}}{(2n+3)n!}}\end{aligned}}$
and so the first three nonzero terms are
${\frac {x^{3}}{3}},{\frac {x^{5}}{5}},{\frac {x^{7}}{14}}$

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