The fourier series of the function f(x)
a(0) / 2 + remainder(n) = f(x) - Sn(x) = 1/PI Sn(x) = 1/PI 
(k=1..
) (a(k) cos kx + b(k) sin kx)
a(k) = 1/PI
Remainder of fourier
series. Sn(x) = sum of first n+1 terms at x.
f(x) cos kx dx
b(k) = 1/PI f(x) sin
kx dx f(x+t) Dn(t) dt
f(x+t) Dn(t) dt
Dn(x) = Dirichlet kernel = 1/2 + cos x + cos 2x + .. + cos nx = [ sin(n + 1/2)x ] / [ 2sin(x/2) ]Riemann's Theorem. If f(x) is continuous except for a finite # of finite jumps in every finite interval then:
lim(k->) 
f(t) cos kt dt = lim(k->)f(t) sin kt dt = 0
The fourier series of the function f(x) in an arbitrary interval.
A(0) / 2 + 1/PI (k=1..) [ A(k) cos (k(PI)x / m) + B(k) (sin k(PI)x /
m) ]
a(k) = 1/m
f(x) cos (k(PI)x / m) dxb(k) = 1/m
Parseval's Theorem.
If f(x) is continuous; f(-PI) = f(PI) then
f(x) sin (k(PI)x / m)
dx f^2(x) dx = a(0)^2 /
2 + (k=1..) (a(k)^2 + b(k)^2)
Fourier Integral of the function f(x)
f(x) = 
( a(y) cos yx + b(y) sin yx ) dy
a(y) = 1/PI
f(t) cos ty dtb(y) = 1/PI
f(x) =
1/PI f(t) sin ty dt dy f(t) cos (y(x-t)) dt
Special Cases of Fourier Integral
if f(x) = f(-x) then
f(x) = 2/PI
if f(-x) = -f(x) then
cos xy dy f(t) cos yt
dtf(x) = 2/PI
Fourier Transforms
sin xy dy sin yt
dt
Fourier Cosine Transform
g(x) = 
(2/PI)f(t) cos xt dt
Fourier Sine Transform
g(x) = (2/PI)f(t) sin xt dt
Identities of the Transforms
If f(-x) = f(x) then
Fourier Cosine Transform ( Fourier Cosine Transform (f(x)) ) = f(x)If f(-x) = -f(x) then
Fourier Sine Transform (Fourier Sine Transform (f(x)) ) = f(x)
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