14Chapter 2Fourier SeriesSolutions to Exercises 2.41.The even extension is the function that is identically 1. So the cosine Fourierseries is just the constant 1. The odd extension yields the function in Exercise 1,Section 2.3, withp= 1. So the sine series is4π∞±k=0sin((2k+1)πx)2k+1.This is also obtained by evaluating the integral in (4), which givesbn=2²10sin(nπx)dx=-2nπcosnπx³³³10=2nπ(1-(-1)n).9.We havebn²10x(1-x) sin(nπx)dx.To evaluate this integral, we will use integration by parts to derive the followingtwo formulas: fora±=0,²xsin(ax)dx=-xcos(ax)a+sin()a2+C,and²x2sin(ax)dx=2 cos()a3-x2cos()a+2xsin()a2+C.So²x(1-x) sin(ax)dx=-2 cos()a3-xcos()a+x2cos()a
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.