Question

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

Answer 1

`f^((8))(0) = 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^((8))(0)` let us consider the Taylor Series for `f(x)` 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(x)` 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(x^2)` 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!)`
`" " = 8*7*6*5`
`" " = 1680`