Find a non-trivial function f(x) that satisfies:
f(ab) = a*f(b) + b*f(a)
(trivial being obviously f(x) = 0 would work -- pick something else)
>>8092974
Looks like the product rule of differentiation...
>>8092974
Let a = b = 1/2, f(x) = 1. Your welcome.
>>8092982
Good one, I didn't think of that.
Now suppose a and b are real numbers not functions.
>>8092983
that does not work for all a and b
>>8092974
>f(ab) = a*f(b) + b*f(a)
f(a*0) = a*f(0) + 0*f(a)
f(0)=af(0) for any a so f(0)=0
f(x^2)=2xf(x)
f(1)=2f(1)
f(1)=0
f(-1^2)=0=-2f(-1)
f(-1)=0
f(-1*x)=-1*f(x) so f is odd
>>8092974
f(x) = C x log(x) / e^2
How about f(x) = x*ln(x)
>>8093033
this
>>8092974
Let [math]f: \mathcal C^2(\mathbb R) \to \mathcal C^1(\mathbb R)[/math] be the function which takes a function to its derivative.
Let g(x) = f(x)/x then
g(ab) = f(ab)/(ab) = (a*f(b) + b*f(a))/(ab) = f(b)/b + f(a)/a = g(a) + g(b)
Now let h(x) = g(e^x)
h(a+b) = g(e^(a+b)) = g(e^a * e^b) = g(e^a) + g(e^b) = h(a) + h(b)
which is Cauchy's Functional Equation with known solutions:
https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation
>>8092974
identity over the Zero ring works
