# How do you find f^8(0) where f(x)=cos(x^2)?

Mar 12, 2017

${f}^{\left(8\right)} \left(0\right) = 1680$

#### Explanation:

We know that the power series for a function is unique. It does not matter how we obtain the power series (it could from the Binomial Theorem, a Taylor Series, or a Maclaurin series).

As we are looking for ${f}^{\left(8\right)} \left(0\right)$ let us consider the Taylor Series for $f \left(x\right)$ about $x = 0$, ie its Maclaurin series.

By definition:

 f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...

The (well known) Maclaurin Series for $\cos \left(x\right)$ is given by:

 cosx = 1 - x^2/(2!) + x^4/(4!) - ... \ \ \ \ \ \ AA x in RR

And so it must be that the Maclaurin Series for $\cos \left({x}^{2}\right)$ is given by:

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

And if we equate the coefficients of ${x}^{8}$ from this derived series and the definition then we have:

 (f^((8))(0))/(8!)= 1/(4!)

 :. f^((8))(0) = (8!)/(4!)
 " " = (8*7*6*5*4!)/(4!)
$\text{ } = 8 \cdot 7 \cdot 6 \cdot 5$
$\text{ } = 1680$