{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem Differential Eq HW Solutions Fall 2011 7

# Chem Differential Eq HW Solutions Fall 2011 7 - g x By...

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

Section 2.2 Fourier Series 7 s n_,x_ : Pi 2 4 PiSum 1 2k 1 ^2Cos 2k 1 x , k,0,n partialsums Table s n,x , n,1,7 ; f x_ x 2PiFloor x Pi 2Pi g x_ Abs f x Plot g x , x, 3Pi, 3Pi Plot Evaluate g x ,partialsums , x, 2Pi,2Pi The function g(x) = | x | and its periodic extension Partial sums of the Fourier series. Since we are summing over the odd integers, when n = 7, we are actually summing the 15th partial sum. 9. Just some hints: (1) f is even, so all the b n ’s are zero. (2) a 0 = 1 π π 0 x 2 dx = π 2 3 . (3) Establish the identity x 2 cos( ax ) dx = 2 x cos( ax ) a 2 + ( - 2 + a 2 x 2 ) sin( ax ) a 3 + C ( a = 0) , using integration by parts. 13. You can compute directly as we did in Example 1, or you can use the result of Example 1 as follows. Rename the function in Example 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g ( x ). By comparing graphs, note that f ( x ) =-2 g ( x + π ). Now using the Fourier series of g ( x ) from Example, we get f ( x ) =-2 ∞ ² n =1 sin n ( π + x ) n = 2 ∞ ² n =1 (-1) n +1 n sin nx. 17. Setting x = π in the Fourier series expansion in Exercise 9 and using the fact that the Fourier series converges for all x to f ( x ), we obtain π 2 = f ( π ) = π 2 3 + 4 ∞ ² n =1 (-1) n n 2 cos nπ = π 2 3 + 4 ∞ ² n =1 1 n 2 , where we have used cos nπ = (-1) n . Simplifying, we ±nd π 2 6 = ∞ ² n =1 1 n 2 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online