ASKSAGE: Sage Q&A Forum - Latest question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 14 May 2018 04:04:41 -0500inverse_laplace of a fraction whose denominator has real roots (or complex)http://ask.sagemath.org/question/42331/inverse_laplace-of-a-fraction-whose-denominator-has-real-roots-or-complex/ Hi, this is the simplest and very important method of Laplace inversion . It can be applied to any Laplace transform
(LT), by starting with a Pade approximation of the LT, then partial fractions and inversion. For some reason, a program I had written last year stopped working
def Ruin(Fx, rho, x, u):#assumes rational survival function Fx, i.e. a hyperexponential density
var('s')
L_F=laplace(Fx,x,s)
m1=L_F(s=0) #some algebraic manipulations known as Pollaczek-Khinchine formula
fe=L_F/m1
Fe=factor((1-fe)/s)
L_rui=rho*Fe/(1-rho*fe)
C=ComplexField(53);
dec=Frac(C['s'])(L_rui).partial_fraction_decomposition();
n=len(dec)
par=[inverse_laplace(dec[1][i],s,u) for i in [0..n]];
rui=sum(par)
return rui, L_rui, fe
It's pretty easy to repair the program when roots are rational -- see question "partial_fraction_decomposition" with possibly "complex roots", again. For nonrational roots, I proposed there the following test case
var('s,u')
R.<s> = PolynomialRing(QQbar)
F = R.fraction_field()
L=3/4*(19*s^2 + 156*s + 284)/(19*s^3 + 174*s^2 + 422*s + 228)
whole,LL=L.partial_fraction_decomposition()
show(LL[0])
inverse_laplace(LL[0],s,u)
Thanks in advance :)florinMon, 14 May 2018 04:04:41 -0500http://ask.sagemath.org/question/42331/