Hw 1 - Math 445 - Solutions to selected problems from HW 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 445 - Solutions to selected problems from HW 1 Exercise 13, page 486. We need to find the Fourier series of the 2 -periodic function f ( x ) defined by f ( x ) = 1 if- 2 < x < 2 , and f ( x ) = 0 if- x - 2 or 2 x . Solution Note that f ( x ) is even ! a = 1 2 R - f ( x ) dx = 1 2 R 2- 2 1 dx = 1 2 = 1 2 . a n = 1 R- f ( x ) cos( nx ) dx = 2 R 2 cos( nx ) dx = 1 sin( n 2 ) n- sin(0) n = 2 n sin( n 2 ). b n = 0 for all n 1 because the function is even. Hence the Fourier series of f is: f ( x ) = 1 2 + 2 cos( x )- 2 3 cos(3 x ) + 2 5 cos(5 x )- 2 7 cos(7 x ) + ... Exercise 2, page 490. We need to find the Fourier series of the p = 2 L-periodic function f ( x ) defined by f ( x ) = 0 if- 2 < x < 0 and f ( x ) = 4 if 0 < x < 2. Here p = 4. Solution Since p = 4 we have L = 2. Then: a = 1 2 L R L- L f ( x ) dx = 1 4 R 2 4 dx = 1 4 8 = 2. a n = 1 L R L- L f ( x ) cos( nx L ) dx = 1 2 R 2 4 cos( nx 2 ) dx = 4 n (sin( n )- sin(0)) = 0. b n = 1 L R L- L f ( x ) sin( nx L ) dx = 1 2 R 2 4 sin( nx 2 ) dx = 4 n (- cos( n ) + cos(0)) = 4(1- (- 1) n ) n ....
View Full Document

This note was uploaded on 02/27/2008 for the course MATH 445 taught by Professor Friedlander during the Fall '07 term at USC.

Page1 / 4

Hw 1 - Math 445 - Solutions to selected problems from HW 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online