# Use taylor series to evaluate f^11 (0) for f(x) = x^3 cos(x^2) ?

Jun 10, 2018

(11!)/(4!) = 1663200

#### Explanation:

We know that the Taylor expansion if $\cos x$ around $x = 0$ is given by

cos x = 1-x^2/(2!)+x^4/(4!)-... implies

cos (x^2) = 1-x^4/(2!)+x^8/(4!)-...

This yields the Taylor expansion of the function ${x}^{3} \cos \left({x}^{2}\right)$ around $x = 0$ as

x^3cos( x^2) = x^3-x^7/(2!)+x^11/(4!)-...

Since the coefficient of ${x}^{n}$ in the Taylor expansion is given by

c_n = f^{(n)}/(n!)

we have

f^{(11)}(0)/(11!) = 1/(4!)

and thus

f^{(11)}(0) = (11!)/(4!) = 1663200