Home > Calculus > Power Series Representations of Functions

How do you find the power series representation for the function #f(x)=cos(2x)# ?

Answer 1

Luke Alvarez

Since

#cosx=sum_{n=0}^infty (-1)^n{x^{2n}}/{(2n)!}#,

by replacing #x# by #2x#,

#f(x)=cos(2x)=sum_{n=0}^infty (-1)^n{(2x)^{2n}}/{(2n)!}#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Samantha Adkison

To find the power series representation for the function ( f(x) = \cos(2x) ), we can use the Maclaurin series expansion of ( \cos(x) ) and then substitute ( 2x ) for ( x ):

[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} ]

Substituting ( 2x ) for ( x ) gives:

[ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} ]

Simplify the expression:

[ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n}}{(2n)!} ]

[ \cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n x^{2n}}{(2n)!} ]

Therefore, the power series representation for ( f(x) = \cos(2x) ) is:

[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n x^{2n}}{(2n)!} ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.