Wikipedia:Reference desk/Archives/Mathematics/2007 June 3

= June 3 =

Laplace transform
In a Laplace transform, you take a function of time, t, and make it a function of a complex variable, s, which is defined as &sigma; + i&omega;. &omega; is the angular frequency, but what is &sigma;? --h2g2bob (talk) 20:48, 3 June 2007 (UTC)


 * If you have a function f(x) that is nonzero only for x>0, then the Laplace transform G(s) (where s = &sigma; + i&omega;) of that function equals the Fourier transform F(&omega;) of the product f(x) e-&sigma; x. In other words, you multiply the function by a decaying exponential function before you take its Fourier transform, and &sigma; is the decay constant of that exponential function. Multiplying by the exponential can make the Fourier integral convergent in cases where f itself does not have a Fourier transform. For example, f(x)=x2 does not have a Fourier transform, but it does have a Laplace transform (namely G(s) = 2/s3 according to Laplace transform), because the exponential factor makes the integral converge. --mglg(talk) 21:44, 3 June 2007 (UTC)


 * Many thanks --h2g2bob (talk) 10:48, 4 June 2007 (UTC)