The question is attached. I know the power series representation of cos(x) and therefore can get cos(6x^2)/x^2 but I don't know what to do about the -1.
Any help would be greatly appreciated!
Thanks in advance!
Start with the known Maclaurin series of $\displaystyle \cos x $
$\displaystyle \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + ... $
then $\displaystyle \cos 6x^{2} = 1 - \frac{(6x^{2})^{2}}{2!} + \frac{(6x^{2})^{4}}{4!} - \frac{(6x^{2})^{6}}{6!} + ... = 1 - \frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ... $
and $\displaystyle \cos 6x^{2} -1 = -\frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...$
finally $\displaystyle \frac{\cos 6x^{2}-1}{x^{2}} = -\frac{6^{2}x^{2}}{2!} + \frac{6^{4}x^{6}}{4!} - \frac{6^{6}x^{10}}{6!} + ...$
For a Macluarin series, the coefficent of the $\displaystyle x^{10} $ term is $\displaystyle \frac{f^{(10)}(0)}{10!} $
The coefficient of the $\displaystyle x^{10} $ term in the above Macluarin series is $\displaystyle -\frac{6^{6}}{6!} $
so $\displaystyle \frac{f^{(10)}(0)}{10!} = -\frac{6^{6}}{6!} $
EDIT: and $\displaystyle f^{(10)}(0) = - \frac{6^{6} 10!}{6!} = - 235146240$