Maclaurin series homework help

[syeh]

Homework Statement

the maclaurin series for f(x) is given by 1/2! - x²/4! + x⁴/6! - x⁶/8! + ... + (-1)ⁿx²ⁿ/(2n+2)! + ...

a) Let g'(x) = 1-x² * f(x)
Write the Maclaurin series for g'(x), showing the first three nonzero terms and the general term.

b) write g'(x) in terms of a familiar function without using series. then write f(c) in terms of the same familiar function.

c) given that g(0)=3 write g(x) in terms of a familiar function without series.

The Attempt at a Solution

the solution is:

a) 1 - x²/2! - x⁴/4! -...+ (-1)ⁿx²ⁿ/(2n)!

b) g'(x) = cosx, f(x) = { (1-cosx)/x² if x≠0
{ 1/2 if x=0

c) g(x) = sinx+3

i tried doing part A but could not figure out how to find the maclaurin series for g'(x)=1-x² * f(x)
first you have to find the g'(0), g''(0), g'''(0), etc., then continue to find the maclaurin series by multiplying the terms with xⁿ and dividing by n!
but i couldn't figure out how to do it. any help will be appreciated, thanks!

 

[micromass]

Let me do a different example:

The Taylor series of $e^x$ is $1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...$.

If we substitute in $-x$ instead of $x$, then we get the Taylor series of $e^{-x}$:

1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ...

Multiply by $x$ to find the Taylor series of $x*(e^{-x})$:

xe^{-x} = x(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ... ) = x - x^2 + \frac{x^3}{2}- \frac{x^4}{6} + ...

We can also find $(1+x)e^{-x}$:

(1+x)e^{-x} = e^{-x} + xe^{-x} = (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + ...) + (x - x^2 + \frac{x^3}{2}- \frac{x^4}{6} + ...) = (1 - \frac{x^2}{2} + \frac{x^3}{3} + ...)

Does that help you a bit?

 

[BiGyElLoWhAt]

just multiply g'(x) through, and you'll see you're mistake, and hopefully you'll see the general solution as well =]
g'(x)=(1-x^2)f(x)=\frac{1}{2!}(1-x^2) - \frac{x^2}{4!}(1-x^2)... = \frac{1-x^2}{2!} +\frac{-x^2+x^4}{4!}...