Question

Half range fourier sine series f(x)=x(-x) in the interval 0<=x<= prove that

Answer 1

Answer:

Step-by-step explanation:

For a) I have solved it by using:

bn=2π2∫π20cos(x)sin(2nx)dx

So,

bn=4π∫π20cos(x)sin(2nx)dx

I solved bn by using the formula for sinAcosB=12(sin(A+B)+sin(A−B).

So I got,

12∫π20sin(2nx+x)sin(2nx−x)dx

This gives:

sin(πn)+12n+1+1−sin(πn)2n−1

After simplifying, I got:

bn=4π−sin(πn)+2n4n2−1

This is where my problem is. I'm thinking that what I get from part a) should be similar to part b), but with my n=2m+1. From part b), it seems the answer should be:

nsin(πn2)4n2−1

Hope it helps you....