User Portlet

Thanks Gianluca, that's what I mean. From the recursion p[t+1] = ( x[t] p[t] ) / ( 1 + b p[t] ) p[0] = p0 I expect p[1] = ( x[0] p0 ) / ( 1 + b p0 ) but using RSolve I find p[1] = ( x[0] p0 ) / ( x[0] + b p0 )