# Maclaurin series homework help

1. Apr 16, 2014

### syeh

1. The problem statement, all variables and given/known data

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.

3. 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 couldnt figure out how to do it. any help will be appreciated, thanks!

2. Apr 16, 2014

### micromass

Staff Emeritus
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 $(1+x)e^{-x}$:

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

$g'(x)=(1-x^2)f(x)=\frac{1}{2!}(1-x^2) - \frac{x^2}{4!}(1-x^2)... = \frac{1-x^2}{2!} +\frac{-x^2+x^4}{4!}...$