# Problem 41

• Nov 18th 2007, 04:59 PM
ThePerfectHacker
Problem 41
1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.
• Nov 18th 2007, 05:26 PM
angel.white
Quote:

Originally Posted by ThePerfectHacker
1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.

$\displaystyle f(x)=c^{x}$ for some constant c.

then
$\displaystyle f(x+y)=c^{x+y}$

and
$\displaystyle f(x)f(y)=c^{x}c^{y}=c^{x+y}$
• Nov 18th 2007, 05:33 PM
ThePerfectHacker
Quote:

Originally Posted by angel.white
$\displaystyle f(x)=c^{x}$ for some constant c.

then
$\displaystyle f(x+y)=c^{x+y}$

and
$\displaystyle f(x)f(y)=c^{x}c^{y}=c^{x+y}$

How do you know there are no other solutions? :eek:
• Nov 18th 2007, 05:50 PM
angel.white
Quote:

Originally Posted by ThePerfectHacker
How do you know there are no other solutions? :eek:

I suppose It's not.

I'll add another function while I'm at it, though :)

f(x)=0

then f(x+y)=0

and f(x)f(y)=0*0 = 0

:)
• Nov 20th 2007, 05:21 AM
math sucks
Let $\displaystyle n$ be a positive integer. Then
$\displaystyle f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)= f\left(\frac{1}{n}\right)^n$
So
$\displaystyle f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}$
Now let $\displaystyle \frac{m}{n}$ be a positive rational number. Then
$\displaystyle f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldot s+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f (1)^{\frac{m}{n}}$
Now let $\displaystyle x$ be a positive real number. By continuity,
$\displaystyle f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x$
where the last result is established by our definition of $\displaystyle f$ over the rational numbers.
Assume $\displaystyle f(1)\neq0$, then it is trivial that $\displaystyle f(0)=1$. Thus, for any positive real number $\displaystyle y$,
$\displaystyle f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}$
Therefore, $\displaystyle f(x)=f(1)^x$ for all $\displaystyle x$ or assume $\displaystyle f(1)=0$, then it is trivial that $\displaystyle f(x)=0$ for all $\displaystyle x$
• Nov 20th 2007, 11:22 AM
math sucks
Quote:

Originally Posted by topsquark
The notation "f(x)" does not imply that it is f times x, as you appear to be using.

-Dan

? I don't understand. How did you think that?
Maybe this will clarify,
$\displaystyle f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1 }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$
The second equality is straight from the problem.
• Nov 21st 2007, 08:42 AM
topsquark
Quote:

Originally Posted by math sucks
? I don't understand. How did you think that?
Maybe this will clarify,
$\displaystyle f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1 }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$
The second equality is straight from the problem.

Sorry. Now I see what you are doing. My bad!

-Dan
• May 26th 2009, 08:39 PM
Rebesques
[cheater] What of discontinuous f? (Angry)

[/cheater]