Problem 41

ThePerfectHacker · #1 $Old$ November 18th, 2007, 06:59 PM

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

angel.white · #2 $Old$ November 18th, 2007, 07:26 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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

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

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

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

ThePerfectHacker · #3 $Old$ November 18th, 2007, 07:33 PM

Quote:

Originally Posted by angel.white $View Post$

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

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

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

How do you know there are no other solutions?

angel.white · #4 $Old$ November 18th, 2007, 07:50 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

How do you know there are no other solutions?

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

math sucks · #5 $Old$ November 20th, 2007, 07:21 AM

Let $n$ be a positive integer. Then
$f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^n$
So
$f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}$
Now let $\frac{m}{n}$ be a positive rational number. Then
$f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f(1)^{\frac{m}{n}}$
Now let $x$ be a positive real number. By continuity,
$f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x$
where the last result is established by our definition of $f$ over the rational numbers.
Assume $f(1)\neq0$ , then it is trivial that $f(0)=1$ . Thus, for any positive real number $y$ ,
$f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}$
Therefore, $f(x)=f(1)^x$ for all $x$ or assume $f(1)=0$ , then it is trivial that $f(x)=0$ for all $x$

math sucks · #6 $Old$ November 20th, 2007, 01:22 PM

Quote:

Originally Posted by topsquark $View Post$

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,
$f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldots f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$
The second equality is straight from the problem.

topsquark · #7 $Old$ November 21st, 2007, 10:42 AM

Quote:

Originally Posted by math sucks $View Post$

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

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

-Dan

Rebesques · #8 $Old$ May 26th, 2009, 10:39 PM

[cheater] What of discontinuous f?

[/cheater]