MATH101 April 2007
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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 $\int _{0}^{x}t\cos(t^{3})\,dt$.

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Hint

The Maclaurin series for $\cos x$ is
$\cos x=1{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}...=\sum _{n=0}^{\infty }(1)^{n}{\frac {x^{2n}}{(2n)!}}$

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Solution

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We apply $t^{3}$ for 'x in the Maclaurin series for cos(x) to get
$\cos(t^{3})=1{\frac {t^{6}}{2!}}+{\frac {t^{12}}{4!}}...=\sum _{n=0}^{\infty }(1)^{n}{\frac {t^{6n}}{(2n)!}}$
Now we multiply by t to get
$t\cos(t^{3})=t{\frac {t^{7}}{2!}}+{\frac {t^{13}}{4!}}...=\sum _{n=0}^{\infty }(1)^{n}{\frac {t^{6n+1}}{(2n)!}}$
Now we integrate to get
$\int t\cos(t^{3})=C+{\frac {t^{2}}{2}}{\frac {t^{8}}{8\cdot 2!}}+{\frac {t^{14}}{14\cdot 4!}}...=\sum _{n=0}^{\infty }(1)^{n}{\frac {t^{6n+2}}{(6n+2)(2n)!}}$
Applying the endpoints gives
$\int _{0}^{x}t\cos(t^{3})={\frac {x^{2}}{2}}{\frac {x^{8}}{8\cdot 2!}}+{\frac {x^{14}}{14\cdot 4!}}...=\sum _{n=0}^{\infty }(1)^{n}{\frac {x^{6n+2}}{(6n+2)(2n)!}}$
and so the first three nonzero terms are
${\frac {x^{2}}{2}},{\frac {x^{8}}{8\cdot 2!}},{\frac {x^{14}}{14\cdot 4!}}$
completing the question.

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