Re: Solving g(x+1)=g(x)+x+1/x , x in R+

Pawel Gladki <gladki@xxxxxxxxxxxxxxxxxxxx> writes:

Hello,

alainverghote@xxxxxxxx wrote:

Are there known ways to solve
g(x+1)=g(x)+x+1/x , x in R+ ,
g a continuous function g:R+ -> R+

Yes, there are :) This is an equation of the type:

g(x+a) - bg(x) = f(x)

and the general solution is of the form:

g(x) = T(x)*b^{x/a} + g_0(x),

where T(x) = T(x + a) is an arbitrary periodic function of period a, and
g_0 is any particular solution of the nonhomogeneous equation. For more
details see:

http://eqworld.ipmnet.ru/en/solutions/fe/fe1108.pdf

and the references cited there.

.... and a particular solution in this case is g(x) = (x^2-x)/2 + Psi(x),
where Psi(x) = (d/dx) ln(Gamma(x)).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.