# use a Maclaurin to find a Maclaurin

• April 23rd 2009, 09:59 AM
diroga
use a Maclaurin to find a Maclaurin
Quote:

Use a known Maclaurin polynomial to find a Maclaurin polynomial of degree 8 for f(x) = xsin(2x)
I am not sure what "Use a known Maclaurin polynomial" means. I know that sin(2x) = 2sin(x)cos(x). I know how to find a nth degree Maclaurin polynomial, keep diriving the function n times, plug in 0 for x, and divide by the n! multiple the result to the corrisponding x^n.
• April 23rd 2009, 12:23 PM
diroga
i tried finding the second dirivative and that took a while. finding the 8th will take a lot of work. Swapping xsin(2x) with 2xsin(x)cos(x) doesnt make it any easier. So can i break it part like g = x, h = sin (2x), f = g * h. find the 8th deg maclaurin for g and h and multiply them together?
• April 23rd 2009, 01:04 PM
redsoxfan325
The polynomial for $\sin x$ is $x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}$

Plug in $2x$ for $x$ to get $2x-\frac{(2x)^3}{6}+\frac{(2x)^5}{120}-\frac{(2x)^7}{5040} = 2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}$

Multiply through by $x$ to get your final answer: $2x^2-\frac{4x^4}{3}+\frac{4x^6}{15}-\frac{8x^8}{315}$