Re: Solving g(x+1)=g(x)+x+1/x , x in R+
- From: Pawel Gladki <gladki@xxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 09 Aug 2007 10:45:43 -0600
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.
--
_______ ---+
) . )__) Pawel Gladki : "There is no nature at an instant."
( _/(_ \ Homepage: http://math.usask.ca/~gladki/ : A.N. Whitehead
\_/\____) pl.sci.matematyka FAQ: www.math.us.edu.pl/~pgladki/faq/
.
- Follow-Ups:
- Re: Solving g(x+1)=g(x)+x+1/x , x in R+
- From: Robert Israel
- Re: Solving g(x+1)=g(x)+x+1/x , x in R+
- References:
- Solving g(x+1)=g(x)+x+1/x , x in R+
- From: alainverghote@xxxxxxxx
- Solving g(x+1)=g(x)+x+1/x , x in R+
- Prev by Date: Re: how to list all of the real numbers
- Next by Date: Re: Solving g(x+1)=g(x)+x+1/x , x in R+
- Previous by thread: Solving g(x+1)=g(x)+x+1/x , x in R+
- Next by thread: Re: Solving g(x+1)=g(x)+x+1/x , x in R+
- Index(es):
Loading