Paper Topic:

FOURIER SERIES & PART.DIF.EQ

Section 4 .2 .9

2

I . Find the Fourier series expansion :Section 4 .2 .11

9 . We have that

f (t , with f (t f (t 2L , defined over -L to L by

f (t -K /L (L t ) for (-L t 0 K /L (L-t ) for (0 t L

i . Find the fourier series expansion Odd function implies an 0 for all n . Moreover , it also tells us that we only need to integrate from 0 to L due to the symmetry (and just multiply the bn 's by 2

)Thus , the series expansion for f (t ) is ii . illustrate f (t ) graphically for -3L t 3L

- ?o /oo

No symmetry this time , because f (t /f (-t ? f (-t /f (t . so we 're looking for : Finding the b 's :ii . Illustrate graphically for -12 t 12 34

i . First we find the complex form of the Fourier-Series expansion of

f (t t^2 (- t

f (t 2 f (t

T 2 , so the complex form is just :Doing integration by parts for n 0 for n 0 , we have Thus ii 36 . Finding the complex fourier series for each set

a . f (t (- t 0 , t (0 t

f (t 2 f (t )b . f (t a sin (wt (0 t T /2 , 0 (T /2 t T , f (t T f (t , T 2 /w c . f (t 2 (- t 0 , 1 (0 t , f (t 2 f (t d . f (t sin...

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