{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem Differential Eq HW Solutions Fall 2011 16

# Chem Differential Eq HW Solutions Fall 2011 16 - 16 Chapter...

This preview shows page 1. Sign up to view the full content.

16 Chapter 2 Fourier Series Solutions to Exercises 2.5 1. We have f ( x ) = 1 if 0 <x< 1 , - 1 if - 1 <x< 0; The Fourier series representation is f ( x ) = 4 π k =0 1 2 k + 1 sin(2 k + 1) πx. The mean square error (from (5)) is E N = 1 2 1 - 1 f 2 ( x ) dx - a 2 0 - 1 2 N n =1 ( a 2 n + b 2 n ) . In this case, a n = 0 for all n , b 2 k = 0, b 2 k +1 = 4 π (2 k +1) , and 1 2 1 - 1 f 2 ( x ) dx = 1 2 1 - 1 dx = 1 . So E 1 = 1 - 1 2 ( b 2 1 ) = 1 - 8 π 2 0 . 189 . Since b 2 = 0, it follows that E 2 = E 1 . Finally, E 1 = 1 - 1 2 ( b 2 1 + b 2 3 ) = 1 - 8 π 2 - 8 9 π 2 0 . 099 . 5. We have E N = 1 2 1 - 1 f 2 ( x ) dx - a 2 0 - 1 2 N n =1 ( a 2 n + b 2 n ) = 1 - 1 2 N n =1 b 2 n = 1 - 8 π 2 1 n odd N 1 n 2 . With the help of a calculator, we find that E 39 = . 01013 and E 41 = . 0096. So take N = 41. 9. We have
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online