math

The fourier series of the function f(x)

a(0) / 2 + (k=1..) (a(k) cos kx + b(k) sin kx)

a(k) = 1/PI  f(x) cos kx dx
b(k) = 1/PI  f(x) sin kx dx

 Remainder of fourier series. Sn(x) = sum of first n+1 terms at x.
remainder(n) = f(x) - Sn(x) = 1/PI  f(x+t) Dn(t) dt

Sn(x) = 1/PI  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 + (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) dx

b(k) = 1/m f(x) sin (k(PI)x / m) dx

 Parseval's Theorem. If f(x) is continuous; f(-PI) = f(PI) then

1/PI  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 dt

b(y) = 1/PI  f(t) sin ty dt

f(x) = 1/PI  dy f(t) cos (y(x-t)) dt

 Special Cases of Fourier Integral

if f(x) = f(-x) then

f(x) = 2/PI  cos xy dy f(t) cos yt dt

if f(-x) = -f(x) then

f(x) = 2/PI  sin xy dy sin yt dt

 Fourier Transforms

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)