How do i find a function such as #f(a)f(b)f(c) = f(sqrt(a^2+b^2+c^2))f^2(0)# ?

1 Answer
Apr 24, 2016

Answer:

#f(x) = p k^(x^2)# for constants #p in RR#, #k > 0#

Explanation:

I will assume that by #f^2(0)# you mean #(f(0))^2# rather than #f(f(0))# or #f^((2))(0)#

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f(x) is an even function

First note that if #f(0) = 0# then #(f(a))^3 = 0# for all #a#, hence #f(a) = 0# for all #a#. So one option for #f(x)# is the constant function #f(x) = 0#.

Otherwise, if we let #b = c = 0# then we find:

#f(a)f(0)f(0) = f(sqrt(a^2+0^2+0^2))f(0)f(0)#

and hence:

#f(a) = f(sqrt(a^2)) = f(abs(a))#

So we can deduce that #f(x)# is an even function.

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Any constant function is a solution

Suppose #f(x) = k# for all #x in RR#

Then:

#f(a)f(b)f(c) = k^3 = f(sqrt(a^2+b^2+c^2))f(0)f(0)#

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Are there any non-constant solutions?

Suppose #f(0) = 1# and #f(1) = k# for some constant #k > 0#

Notice that:

#f(sqrt(2)) = f(sqrt(1^2+1^2+0^0))f(0)f(0) = f(1)f(1)f(0) = k^2#

#f(sqrt(3)) = f(sqrt(1^2+1^2+1^2))f(0)f(0) = f(1)f(1)f(1) = k^3#

Observing this pattern, we can define

#f(x) = k^(x^2)#

To find:

#f(a)f(b)f(c) = k^(a^2)k^(b^2)k^(c^2)#

#= k^(a^2+b^2+c^2)#

#= k^((sqrt(a^2+b^2+c^2))^2)#

#= f(sqrt(a^2+b^2+c^2))#

#= f(sqrt(a^2+b^2+c^2))f(0)f(0)#

Note that if #p in RR# is any constant then:

#f(x) = p k^(x^2)#

will also be a solution.

The case #k=1# then covers the previously identified constant solution.