Find F: R->R satisfying F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y)

  • Thread starter Thread starter iN10SE
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves finding a function F: R->R that satisfies the functional equations F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y). Participants are exploring the implications of these equations and discussing potential forms of the function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various forms of F, including polynomial and linear functions, and question whether F can take on other forms. There is an exploration of the implications of setting specific values for x and y, and the use of the hint regarding positive real numbers.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning each other's reasoning. Some have suggested alternative approaches and highlighted potential issues with the logic presented. There is no clear consensus, but several lines of reasoning are being explored.

Contextual Notes

There is a mention of the lack of continuity or differentiability of F, which complicates the discussion. Participants also note the possibility of discontinuous functions that satisfy the given conditions, indicating that additional constraints may be necessary for a definitive conclusion.

  • #31
I screwed up most of all by throwing a monkey wrench into peoples thinking by suggesting it might not be true. Mea culpa again.
 
Physics news on Phys.org
  • #32


intuitively, the solution is f(kx)=kf(x). i am able to prove it for all integers k.but for a rational number k i am facing problem.so i assumed k to be p/q and later that i can't proced.would you give me a small hint
 
  • #33
Given that f(x+ y)= f(x)+ f(y), you can prove, by induction on the positive integer n, that f(nx)= nf(x), x any real number. From that, for n a non-zero integer, f(n/n)= nf(1/n)= f(1)= 1. Thus, f(1/n)= 1/f(n). You can get the case for f(m/n) from that.

This is an entirely different problem from before, isn't it? Requiring that f be continuous at x= 0 rather than that f(xy)= f(x)f(y).
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
Replies
14
Views
3K
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K