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

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

mfb
Mentor
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
Gold Member
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
Homework Helper
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
Mentor
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.

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

BvU
Homework Helper
Huh ?

mfb
Mentor
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.

Last edited:
jbriggs444
Homework Helper
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
Another trivial solution is $f(x)=-1$...