# Functional equation

• October 18th 2008, 04:23 AM
alexmahone
Functional equation
Find all functions of $R\to R$ with the property
$f(xf(x) + f(y)) = (f(x))^2 + y$
• October 18th 2008, 05:33 AM
mr fantastic
Quote:

Originally Posted by alexmahone
Find all functions of $R\to R$ with the property
$f(xf(x) + f(y)) = (f(x))^2 + y$

I might be barking up the wrong tree here but can't you apply $f^{-1}$ to both sides and get $x f(x) + f(y) = x f(x) + f^{-1}(y)$ .....?

In which case the problem boils down to finding all functions such that $f(y) = f^{-1}(y)$, that is, functions which are their own inverse ......

Such functions are called involutions.
• October 18th 2008, 07:09 AM
alexmahone
Quote:

Originally Posted by mr fantastic
$x f(x) + f(y) = x f(x) + f^{-1}(y)$

No, the RHS of this equation is incorrect.
We do not know that $f^{-1}(a+b)=f^{-1}(a)+f^{-1}(b)$
• October 18th 2008, 01:22 PM
Showcase_22
$
f(xf(x) + f(y)) = (f(x))^2 + y
$

$f(xf(x)+f(y))-(f(x))^2=y$

$(\sqrt{f(xf(x)+f(y)}+f(x))(\sqrt{f(xf(x)+f(y))}-f(x)=y$

Does this help in any way? I was thinking that somehow I could get the $f(xf(x)+f(y))$ to cancel out but I couldn't get it to work.

The RHS can be multiplied by $\frac{\sqrt{f(xf(x)+f(y))}-f(x)}{\sqrt{f(xf(x)+f(y)}-f(x)}$ to give:

$\sqrt{f(xf(x)+f(y)}+f(x)=\frac{y(\sqrt{f(xf(x)+f(y ))}-f(x))}{f(xf(x)+f(y))-(f(x))^2}$

but I don't think that's helping.

It's an interesting question. I hope someone here can solve it!

EDIT: Perhaps you could try case analysis. When x=0:

[MATh]f(f(y))=(f(0))^2+y[/tex]

f(0) has an unknown value so it might not help. However, we have successfully eliminated a term which is beneficial!
• October 18th 2008, 02:28 PM
mr fantastic
Quote:

Originally Posted by alexmahone
No, the RHS of this equation is incorrect.
We do not know that $f^{-1}(a+b)=f^{-1}(a)+f^{-1}(b)$

Hmmm ..... you're right.

I know f(x) = x works (an involution, by the way). I haven't checked involutions of the form f(x) = (ax + b)/(cx + d), where there's a relationship between a, b, c, and d that I can't be bothered remembering .... might be worth a check.
• October 19th 2008, 03:53 PM
SimonM
I believe that in the case of f(0)=0 a function being an involution is a necessary but not sufficient criteria..
• October 20th 2008, 02:22 AM
NonCommAlg
Quote:

Originally Posted by alexmahone
Find all functions of $R\to R$ with the property
$f(xf(x) + f(y)) = (f(x))^2 + y$

1) $f$ is onto. (it's actually one-to-one as well, but we don't need that!)

Proof: let $z \in \mathbb{R}.$ choose $y=z-(f(0))^2.$ then: $f(f(y))=(f(0))^2+y=z. \ \ \Box$

2) $f(0)=0.$

Proof: by 1) there exists $x \in \mathbb{R}: \ f(x)=0.$ thus: $x=x+(f(x))^2=f(xf(x)+f(x))=f(0)=f(f(x))=(f(0))^2+x .$ thus: $f(0)=0. \ \ \Box$

3) $f(x)=\pm x.$

Proof: we have: $f(f(x))=(f(0))^2+x=x.$ thus: $x^2=(f(f(x)))^2=f(f(x)f(f(x))+f(0))=f(xf(x))=(f(x) )^2.$ thus: $f(x)=\pm x. \ \ \Box$ (Sleepy)