The problem involves finding all functions f: R->R that satisfy the functional equation f(x+y)=f(x)+f(y)+1 with the condition f(1)=0. Participants are exploring the nature of the solutions and the implications of continuity and differentiability on the function's form.
The discussion is active, with participants sharing insights and questioning assumptions about the nature of the function. Some suggest that there may be multiple solutions, while others are trying to derive a general formula for f(nx) and explore the implications of differentiability.
Participants note the challenge of proving the uniqueness of the function and the difficulty in conceptualizing the problem. There is an ongoing exploration of whether the function can take forms other than y=x-1, and the discussion includes attempts to manipulate the functional equation to derive further insights.
Dick said:True. But I was really hoping you'd say limit(q_i-1)=limit(q_i)-1=c-1. That would be the punch line. So f(c)=c-1. You've got all of the parts of this. Can you try and string them together in a logical order??
iceblits said:So:
a sequence of rational numbers q_i approaches any other number. That is, lim(q_i)=c By continuity, lim f(q_i)=f(c). By the formula: f(q_i)=q_i-1. So then lim (q_i-1)=lim(q_i)-1=c-1, so, since f(c)=lim(q_i-1), f(c)=c-1
Dick said:Change the first sentence to "Given any irrational number c there is a sequence of rational numbers q_i that approaches c." From there on, I like it.
iceblits said:yes! so does this complete the proof? thank-you so much for your help by the way..you've been great..hopefully I didn't take up too much of your time and/or frustrate you too much