Help finding high order derivatives using Taylor series?

**Mattpd** · April 29th 2010, 06:49 PM

I need to find the 7th derivative $f^{(7)}(0)$ :

$f(x)=(1/14)x cos(2x)$

I know it has something to do with Taylor/Maclaurin series, but I don't know the process. Can someone fill me in?

**tonio** · April 29th 2010, 07:29 PM

Originally Posted by Mattpd

I need to find the 7th derivative $f^{(7)}(0)$ :

$f(x)=(1/14)x cos(2x)$

I know it has something to do with Taylor/Maclaurin series, but I don't know the process. Can someone fill me in?

If you know the MacClaurin series for $\cos 2x$ then you're done:

$\cos 2x =\sum^\infty_{k=0}(-1)^k\,\frac{(2x)^{2k}}{(2k)!}\Longrightarrow \frac{1}{14}x\cos 2x=$ $\frac{1}{14}\sum^\infty_{k=0}(-1)^k\,\frac{2^{2k}x^{2k+1}}{(2k)!}$ ...

Tonio

**Random Variable** · April 29th 2010, 07:30 PM

$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - ...$

$\cos 2x = 1 - \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} - \frac{(2x)^{6}}{6!} + \frac{(2x)^{8}}{8!} - ...$

$\frac{1}{14} x \cos 2x = \frac{1}{14} \Big(x - \frac{2^{2}x^{3}}{2!} + \frac{2^{4}x^{5}}{4!} - \frac{2^{6}x^{7}}{6!} - \frac{2^{8}x^{9}}{8!} - ...\Big)$

the coefficient of the $x^{7}$ term is $\frac{f^{(7)}(0)}{7!}$

so $\frac{f^{(7)}(0)}{7!} = - \frac{1}{14} \frac{2^{6}}{6!}$

which implies that $f^{(7)}(0) = -32$

**Mattpd** · April 29th 2010, 07:35 PM

I understand up until you state the coefficient of the $x^{7}$ term is $\frac{f^{(7)}(0)}{7!}$ . Can you more simply explain what this means and how you ended up with the term you did?

**Random Variable** · April 29th 2010, 09:00 PM

Originally Posted by Mattpd

I understand up until you state the coefficient of the $x^{7}$ term is $\frac{f^{(7)}(0)}{7!}$ . Can you more simply explain what this means and how you ended up with the term you did?

The Taylor series of a function $f(x)$ about $x=0$ is ${f(0)} + {f'(0)}x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + ...$

so for the $x^{7}$ term the coefficient is $\frac{f^{(7)}(0)}{7!}$

the coefficient of the $x^{7}$ term in our series is $-\frac{1}{14} \frac{2^{6}}{6!}$ , but this has to be equal to $\frac{f^{(7)}(0)}{7!}$

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