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

On 9 août, 19:10, Robert Israel <isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Pawel Gladki <gla...@xxxxxxxxxxxxxxxxxxxx> writes:
Hello,

alainvergh...@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 isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada- Masquer le texte des messages précédents -

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Bonjour,
Thanks for your quick and very clear solutions.
A very 'naîve' approach here could be ;
from f(x) = f(x -1) + (x -1) + 1/(x -1)
f(x-1)= f(x -2) + (x -2) + 1/(x- 2) ,

f(x)= sum( i +1/i, i = 1+ frac(x) to x -1) , (1)
this solution seems very near the case g(x) =(x^2-x)+Psi(x)
proposed by Robert .
We may add to (1) any continuous function
g(x mod.1/p) p any positive integer ,

Alain

.

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