I need to find the 7th derivative $\displaystyle f^{(7)}(0)$:
$\displaystyle f(x)=(1/14)x cos(2x)$
I know it has something to do with Taylor/Maclaurin series, but I don't know the process. Can someone fill me in?
I need to find the 7th derivative $\displaystyle f^{(7)}(0)$:
$\displaystyle f(x)=(1/14)x cos(2x)$
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 $\displaystyle \cos 2x$ then you're done:
$\displaystyle \cos 2x =\sum^\infty_{k=0}(-1)^k\,\frac{(2x)^{2k}}{(2k)!}\Longrightarrow \frac{1}{14}x\cos 2x=$ $\displaystyle \frac{1}{14}\sum^\infty_{k=0}(-1)^k\,\frac{2^{2k}x^{2k+1}}{(2k)!}$ ...
Tonio
$\displaystyle \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - ... $
$\displaystyle \cos 2x = 1 - \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} - \frac{(2x)^{6}}{6!} + \frac{(2x)^{8}}{8!} - ... $
$\displaystyle \frac{1}{14} x \cos 2x = \frac{1}{14} \Big(x - \frac{2^{2}x^{3}}{2!} + \frac{2^{4}x^{5}}{4!} - \frac{2^{6}x^{7}}{6!} - \frac{2^{8}x^{9}}{8!} - ...\Big)$
the coefficient of the $\displaystyle x^{7} $ term is $\displaystyle \frac{f^{(7)}(0)}{7!} $
so $\displaystyle \frac{f^{(7)}(0)}{7!} = - \frac{1}{14} \frac{2^{6}}{6!} $
which implies that $\displaystyle f^{(7)}(0) = -32 $
I understand up until you state the coefficient of the $\displaystyle x^{7}$ term is $\displaystyle \frac{f^{(7)}(0)}{7!}$. Can you more simply explain what this means and how you ended up with the term you did?
The Taylor series of a function $\displaystyle f(x)$ about $\displaystyle x=0$ is $\displaystyle {f(0)} + {f'(0)}x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + ... $
so for the $\displaystyle x^{7} $ term the coefficient is $\displaystyle \frac{f^{(7)}(0)}{7!} $
the coefficient of the $\displaystyle x^{7} $ term in our series is $\displaystyle -\frac{1}{14} \frac{2^{6}}{6!} $, but this has to be equal to $\displaystyle \frac{f^{(7)}(0)}{7!} $