ps06sol

ps06sol - 18.03 Problem Set 6 Fall 2009 Solutions In the...

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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). I’ll 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 π . There are three kinds of “Fourier coefficients” a n = 1 π Z π- π f ( t )cos( nt ) dt ( n = 0 , 1 , 2 ,... ) . b n = 1 π Z π- π f ( t )sin( nt ) dt ( n = 1 , 2 , 3 ,... ) . c n = 1 2 π Z π- π f ( t ) e- int dt ( n = 0 , ± 1 , ± 2 ... ) . (If it isn’t clear what function f is meant, I’ll 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 π Z π- π f ( t ) dt and a = 1 π Z π- π 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 ) ....
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ps06sol - 18.03 Problem Set 6 Fall 2009 Solutions In the...

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