This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 18.03 Problem Set 6 Fall 2009 Solutions In the entire problem set, f will be a function defined for all real numbers t , taking usually real values (but complex values would work fine). Ill always assume that f is piecewise continuous in the sense explained in section 8.1 of the text; this will ensure that the integrals I write down make sense. Finally, f will always be assumed to be periodic of period 2 . There are three kinds of Fourier coefficients a n = 1 Z  f ( t )cos( nt ) dt ( n = 0 , 1 , 2 ,... ) . b n = 1 Z  f ( t )sin( nt ) dt ( n = 1 , 2 , 3 ,... ) . c n = 1 2 Z  f ( t ) e int dt ( n = 0 , 1 , 2 ... ) . (If it isnt clear what function f is meant, Ill write a n ( f ) and so on.) This leads to three kinds of Fourier series: a / 2 + X n =1 a n cos( nt ) (Cosine series of f ) X n =1 b n sin( nt ) (Sine series of f ) c + X n =1 ( c n e int + c n e int ) (Complex series of f ) 1. (5 points) a) Write a formula expressing c in terms of a . We have c = 1 2 Z  f ( t ) dt and a = 1 Z  f ( t ) dt, so c = a / 2 . b) For n = 1 , 2 , 3 ,... , write a formula expressing c n and c n in terms of a n and b n , and explain why your formula is correct. c) For n = 1 , 2 , 3 ,... , write a formula expressing a n and b n in terms of c n and c n , and explain why your formula is correct. Note that e int = cos( nt ) i sin( nt ) and e i ( n ) t = cos( nt ) + i sin( nt ) . Thus, for n = 1 , 2 , 3 ,... , we have c n = ( a n ib n ) / 2 and c n = ( a n + ib n ) / 2 . Adding and subtracting these equations, we get e int + e i ( n ) t = 2cos( nt ) and e int e i ( n ) t = 2 i sin( nt ) ....
View
Full
Document
 Spring '09
 unknown

Click to edit the document details