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# SolSec3.3 - 3 21 Problems and Solutions Section...

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3- 21 Problems and Solutions Section 3.3 (problems 3.22-3.28) 3.22 Derive equations (3.24). (3.25) and (3.26) and hence verify the equations for the Fourier coefficient given by equations (3.21), (3.22) and (3.23). Solution: For n m , integration yields: 0 0 2 2 2 2 2 2 2 T T T T T T n t m tdt n m t n m n m t n m n m T T n m n m T T n m n m = ( ) ( ) + ( ) + ( ) = ( ) ( ) + ( ) + ( ) = ( ) sin sin sin sin sin sin sin ω ω ω ω π π π ( ) [ ] ( ) + ( )( ) [ ] + ( ) = 2 2 2 0 n m n m n m sin π Since m and n are integers, the sine terms are 0, so this is equal to 0. Equation (3.24), for m = n : 0 2 0 1 2 1 4 2 2 8 2 2 2 8 4 2 T T T T T n tdt t n n t T T n T T T T n n T = ( ) = = [ ] = sin sin sin sin ω ω ω π π π π π Since n is an integer, the sine term is 0, so this is equal to T /2. So, 0 0 2 T T T n t m tdt m n T m n = = sin sin / ω ω Equation (3.25), for m n 0 0 2 2 2 2 2 2 2 T T T T T T n t m tdt n m t n m n m t m n m T T n m n m T T n m n m = ( ) ( ) + ( ) + ( ) = ( ) ( ) + ( ) + ( ) = ( ) cos cos sin sin sin sin sin ω ω ω ω η π π π π ( ) [ ] ( ) + ( )( ) [ ] + ( ) = 2 2 2 0 n m n m n m sin Since m and n are integers, the sine terms are 0, so this is equal to 0.

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