# Solve the following equation with RSolve and non-autonomous series?

Posted 10 months ago
1066 Views
|
5 Replies
|
3 Total Likes
|
 In this example, RSolve is not giving the result I'd expect (the parenthesis!). Am I missing something? In:= RSolve[{p[t + 1] == ( x[t] p[t] ) / ( 1 + b p[t] ), p == p0}, p[t], t] /. t -> 1 Out= {{p -> p0 / ( 1 + ( b p0 ) / x )}} I am using Mathematica 11.3 on Linux. Answer
5 Replies
Sort By:
Posted 10 months ago
 HiYou have 2 variables p[t] and x[t] then you need a 2 equations to solve.Regards,MI Answer
Posted 10 months ago
 The answer given by RSolve seems wrong to me, unless x=1: Clear[p1, p, p0]; eq = {p[t] == (x[t - 1] p[t - 1])/(1 + b p[t - 1]), p == p0}; sol = RSolveValue[eq, p, t]; p1 = p0; p1[t_] := (x[t - 1] p1[t - 1])/(1 + b p1[t - 1]); Simplify[Table[sol[t] == p1[t], {t, 0, 3}]] Answer
Posted 10 months ago
 It is difficult to respond in the absence of a clear indication as to what was the expected result. Answer
Posted 10 months ago
 I would expect the result to be like this: p1 = p0; p1[t_] := (x[t - 1] p1[t - 1])/(1 + b p1[t - 1]); Answer
Posted 10 months ago
 Thanks Gianluca, that's what I mean. From the recursion p[t+1] = ( x[t] p[t] ) / ( 1 + b p[t] ) p = p0 I expect p = ( x p0 ) / ( 1 + b p0 ) but using RSolve I find p = ( x p0 ) / ( x + b p0 ) Answer
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.