Home > Calculus > Power Series Representations of Functions

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

Answer 1

Emilia Aguilar

Since

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

by replacing #x# by #x^2#,

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

#=sum_{n=0}^infty(-1)^n{x^{4n+2}}/{(2n+1)!}#

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

Answer 2

Isabella Ackerman

To find the power series representation for the function ( f(x) = \sin(x^2) ), we'll use the Taylor series expansion for ( \sin(x) ) and substitute ( x^2 ) for ( x ). The Taylor series expansion for ( \sin(x) ) is:

[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots ]

Substitute ( x^2 ) for ( x ) in the above series:

[ \sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \frac{(x^2)^7}{7!} + \ldots ]

[ = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \ldots ]

Thus, the power series representation for ( f(x) = \sin(x^2) ) is:

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

This series converges for all real numbers ( x ).

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.