1) Multiplying :
f(x) = å¥ n = 1 { a n cos npx/ L + b n sin npx/ L}
By sin mpx/ L and integrating from -L to L we get:
ò L-L f(x) (cos mpx/ L) dx = A ò L-L (cos mpx/ L) dx +
å¥ n = 1 [ a n ò L-L cos mpx/ L)( cos npx/ L) dx +
b m ò L-L (cos mpx/ L)( sin npx/ L)dx ]
= b m L
Thus: b m = 1/L ò L-L f(x) (sin mpx/ L) dx for m = 1, 2, 3...
(2) We let x = h = 0, then: a o = 1, a n = 0,
b n = {1/ p (0) = 0 (n even)
{ 2/ n p (n odd)
Then for the general representation of the Fourier series we get:
f(x) = a o / 2 + å¥ n = 1 a n cos nx + b n sin nx
To obtain specific series result we substitute values, i.e.
a o / 2 + å¥ n = 1 a n cos nx + b n sin nx =
1/2 + å n = odd 2/ n p (sin nx} But sin nx = 0 at x = 0
Therefore: At jump point series = 1/2
Thus f(0) = 1 but Fourier series = 1/2
(3) The graph of the function is shown below:
a n = 1/L ò c+2Lc f(x) (cos npx/ L) dx = 1/5 ò 5-5 f(x) (cos npx/ 5) dx
a n = 1/5 {ò 0-5 f(0) (cos npx/ 5) dx + ò 50 f(3) (cos npx/ 5) dx}
= 3/5 {ò 50 (cos npx/ 5) dx} = 3/5 [5 /np sin npx/ 5 ]50 if n ≠ 0
If n = 0 then a n = a o = 3/5 ò 50 cos (0)px/ 5) dx = 3/5 ò 50 dx = 3
Further: b n = 1/L ò c+ 2Lc f(x) (sin npx/ L) dx = 1/5 ò 5-5 f(x) (sin npx/ 5) dx
b n = 1/5 {ò 0-5 f(0) (sin npx/ 5) dx + ò 50 f(3) (sin npx/ 5) dx}
b n = 3/5 ò 50 sin npx/ 5 dx = 3/5 [5 /np cos npx/ 5 ]50
b n = 3(1 - cos np )/ np
(c) The corresponding Fourier series is written:
a o / 2 + å¥ n = 1 (a n cos npx/ L + b n sin npx / L)
= 3/2 + å¥ n = 1 3(1 - cos np )/ np (sin npx/ 5) =
3/2 + 6/p (sin px/ 5 + 1/3 sin 3px/ 5 + 1/5 sin 5px/ 5 + .....)
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