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

##### 1 Answer

# 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

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

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

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

And if we equate the coefficients of

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

# :. f^((8))(0) = (8!)/(4!) #

# " " = (8*7*6*5*4!)/(4!) #

# " " = 8*7*6*5 #

# " " = 1680 #