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!

Printable View

- May 17th 2010, 11:22 AMs3aCompute the 10th derivative of ...
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! - May 17th 2010, 12:23 PMRandom Variable
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 $\displaystyle f^{(10)}(0) = - \frac{6^{6} 10!}{6!} = - 235146240$ (Surprised) - May 17th 2010, 12:48 PMs3a
How do you get from the before-last step to the last one? (From the one that says "so" to the one with the emoticon)

- May 17th 2010, 12:53 PMlilaziz1
the 10! came from the exponent on $\displaystyle \frac{6^{6}x^{10}}{6!} $. If you were to take the derivative of that 10 times, you would get $\displaystyle \frac{6^{6} 10!}{6!} $

- May 17th 2010, 01:00 PMRandom Variable
- May 17th 2010, 01:04 PMs3a
Oh that 10! was also tricky but I get it now! Thanks!