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

  • #1
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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?
 

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  • #2
BvU
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You already did. ##f(0)=\pm 1## and there is a differential equation
 
  • #3
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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.
 
  • #4
hilbert2
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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.##
 
  • #5
BvU
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Ugly, but correct. I was looking for too decent :rolleyes: 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
 
  • #6
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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.
 
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  • #7
Svein
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[itex]f(x)=x^{2n} [/itex], [itex] f(x)=\lvert x \rvert[/itex], [itex]f(x)= \cos(\pi x) [/itex],...
 
  • #8
BvU
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Huh ?
 
  • #10
Svein
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You might want to check that @Svein.
Sorry, did not read the specs closely enough (read it as [itex]f(1)f(-1)=1 [/itex]).

But then the constant function [itex]f(x)=1 [/itex] is a trivial solution.
 
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  • #11
jbriggs444
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But then the constant function [itex]f(x)=1 [/itex] is a trivial solution.
Right. It is a special case of ##f(x)=k^x## which was mentioned in post #1 ("all exponential functions").
 
  • #12
hilbert2
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Even the function ##f(x) = |x|^x## is a solution if value ##f(x) = 1## is assumed at ##x=0## where it is undefined.
 
  • #13
Svein
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Right. It is a special case of ##f(x)=k^x## which was mentioned in post #1 ("all exponential functions").
Another trivial solution is [itex]f(x)=-1 [/itex]...
 

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