Fourier Serileri 6.pdf

= 1 ⎡ 0 nπx 5 nπx ⎤0. cos dx + 45 ⎢∫−55∫ cos dx⎥⎣0 5 ⎦= 4 5 nπxcos5∫ dx0 54 1Sinn − Sin0 = 4 0 − 0 = 0 ⇒ an= 0π n π.n= [ π ] .[ ]a 0 = 1 55 1 5 1 5 4 5 4f ( x) dx = f(x)dx 4dx-5 5∫ = = dx = x0 5 0 5 0 5∫ ∫ ∫ Ι50= 4 5 .5 = 4 ⇒ a = 4 0b n = 1 55∫ f ( x)sin-5nπ 1 5 π 1 5 nπxx dx =5 5∫ f(x)sin n dx = 4sin dx0 5 5∫ 0 5= 4 5 nπxsin dx5∫ =0 54πn[ 1 − ( −1)]n⇒ b n =4πn[ 1 − ( −1)]nbulunur. Buna görefin Fourier serisia02∞nπxnπx+ ∑ ( ancos + bnsin )5 5n=1=42∞ ⎧⎪ nπx4+ ∑ ⎨0.cos+n=1 5 π⎩⎪n[ 1- (-1) ]nsin n π x ⎫⎪⎬5⎭⎪48