homeschool girl
Junior Member
- Joined
- Feb 6, 2020
- Messages
- 123
The problem:
The function [imath]f : \mathbb{R} \to \mathbb{R}[/imath] satisfies [math]f(x) f(y) = f(x + y) + xy[/math]for all real numbers [imath]x[/imath] and [imath]y.[/imath] Find all possible functions [imath]f.[/imath]
Hint: The first strategy for attacking a functional equation is to substitute values. Combinations of 0s and 1s are usually helpful.
What I've done so far:
I started by trying [imath]x=0, y=1[/imath] and got
[imath]f(0) f(1) = f(1)[/imath]
which implies that [imath]f(0)=1[/imath] and/or [imath]f(1)=0[/imath]
if [imath]f(1) = 0[/imath], then i think you could write [imath]f(n)[/imath] as
[imath]f(n-1)f(1)=f(n-1+1)+n-1,[/imath]
[imath]f(n)=-n+1.[/imath]
I'm not sure if that's correct though, and I'm stuck on how to find the other possible functions.
The function [imath]f : \mathbb{R} \to \mathbb{R}[/imath] satisfies [math]f(x) f(y) = f(x + y) + xy[/math]for all real numbers [imath]x[/imath] and [imath]y.[/imath] Find all possible functions [imath]f.[/imath]
Hint: The first strategy for attacking a functional equation is to substitute values. Combinations of 0s and 1s are usually helpful.
What I've done so far:
I started by trying [imath]x=0, y=1[/imath] and got
[imath]f(0) f(1) = f(1)[/imath]
which implies that [imath]f(0)=1[/imath] and/or [imath]f(1)=0[/imath]
if [imath]f(1) = 0[/imath], then i think you could write [imath]f(n)[/imath] as
[imath]f(n-1)f(1)=f(n-1+1)+n-1,[/imath]
[imath]f(n)=-n+1.[/imath]
I'm not sure if that's correct though, and I'm stuck on how to find the other possible functions.
