maclaurin series calculus 2?
the maclaurin series for sin x is series from n=0 to infiinity is ( (-1)^n*(x^(2n+1)/(2n+1)!).
Find the maclaurin series for sin 3x.
find the maclaurin series for the integral from 0 to x of sin t^2 dt.
uuse the maclaurin series for sinx to approximate sin 1/2 with four decimal precision
Comments
sin 3x = sum from n=0 to infiinity is ( (-1)^n*((3x)^(2n+1)/(2n+1)!)
integral from 0 to x of sin t^2 dt
= integral from 0 to x of ( (-1)^n*((t^2)^(2n+1)/(2n+1)!) dt
= [(-1)^n/(4n+3)]t^(4n+3)/(2n+1)! from 0 to x
= [(-1)^n/(4n+3)]x^(4n+3)/(2n+1)!