Compute the 10th derivative of ...

s3a

Nov 2008
624
5
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!
 

Attachments

May 2009
959
362
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)
 
Last edited:
  • Like
Reactions: s3a

s3a

Nov 2008
624
5
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)
 
Mar 2010
107
14
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 \(\displaystyle \frac{6^{6} 10!}{6!} \)
 
  • Like
Reactions: s3a
May 2009
959
362
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)
It should be \(\displaystyle f^{(10)} (0) = - \frac{6^{6} 10!}{6!} \)
 
  • Like
Reactions: s3a

s3a

Nov 2008
624
5
Oh that 10! was also tricky but I get it now! Thanks!