Find a non-trivial function f(x) that satisfies: f(ab) = a*f(b)

>>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