Property that f(x)*f(-x)=1

[Mr Davis 97]

Out of curiosity, I'm trying to find functions of a real variables such that $f(x)f(-x) = 1$. One obvious example is $f(x) = e^x$, and all other exponential functions. Are there any other examples? How would I go about generating them?

 

[BvU]

You already did. $f(0)=\pm 1$ and there is a differential equation

 

[mfb]

Freely choose f(x) for x>0 as long as no value is zero. Choose -1 or 1 for f(0). Fix f(x) for negative x according to your equation.

 

[hilbert2]

If $f(x)$ has to be continuous, the functions $f(x) = \exp (x^n )$, where $n$ is an odd integer, are one example of a solution.

Piecewise continuous examples include functions like

$f(x) = \left\{\begin{array}{c} 2^x, \hspace{30pt} |x|<5\\3^x, \hspace{30pt}|x|\geq 5\end{array}\right.$

 

[BvU]

Ugly, but correct. I was looking for too decent functions and thought to exploit $f'(x) f(-x) + f(x) f'(-x) = 0$ which is satisfied if $f'(x) = \pm f(x)\ \forall x\$ and also if $f'(x) = 0\ \forall x\$ . But compared to mfb, that's far too limited

 

[mfb]

If f(x) has to be continuous then we have the options f(0)=1 and f(x)>0 arbitrary but continuous for x>0, or f(0)=-1 and f(x)<0 arbitrary but continuous for x>0. In both cases we can again find the negative function values with the given constraint and it will always be continuous there as well.

If f(x) for x>0 is differentiable and the limit of the derivative for x->0 is finite then I would expect f(x) to be differentiable everywhere.

This works both for real and complex function values.

 

[Svein]

$f(x)=x^{2n}$, $f(x)=\lvert x \rvert$, $f(x)= \cos(\pi x)$,...

 

[BvU]

Huh ?

 

[mfb]

You might want to check that @Svein.

 

[Svein]

You might want to check that @Svein.

Sorry, did not read the specs closely enough (read it as $f(1)f(-1)=1$).

But then the constant function $f(x)=1$ is a trivial solution.

 

[jbriggs444]

But then the constant function $f(x)=1$ is a trivial solution.

Right. It is a special case of $f(x)=k^x$ which was mentioned in post #1 ("all exponential functions").

 

[hilbert2]

Even the function $f(x) = |x|^x$ is a solution if value $f(x) = 1$ is assumed at $x=0$ where it is undefined.

 

[Svein]

Right. It is a special case of $f(x)=k^x$ which was mentioned in post #1 ("all exponential functions").

Another trivial solution is $f(x)=-1$...