A Nice Functional Equation

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[michalchik]

Though I agree the answer works, I've always missed trusted Solutions like this for what they might leave out. I remember being in an introductory physics class and the teacher proving to us that solution to the differential equation generated by a spring, was the sine function because when you took the double derivative of the sine function you got another sine function in the opposite direction that only differ by a constant. At the time I thought well it works but is it the only thing that works? That seems wrong to me. Later that year I found out that was Not the Only Solution, there was the cosine function as well a the solutions that came from Eulers formula. Is there more? Maybe. I don't know, I haven't seen a proof that Euler's formula gives us the exhaustive set of answers for the spring equation.

So how do we know that the solution is all solutions for this functional equation?

[syedmdabid]

But it may be conditional identity, not an absolute identity ( always).

[wannabeactuary]

Let's try this approach as I am not convinced by uniqueness nor rearranging
f(1) = f(1)^2 / (2f(1)) = f(1)/2 ie f(1) = 0 just like ln(1) = 0


f(x)¬=0,
f(x^2) =f(x)/2
Note f'(x^2) = f'(x)2x and now I am stuck

[technolan]

if you set y=1 on the original functional equation, we will get f(x)=0 and this is not true since by the original functional equation we will get zeros in the denominator

[shacharh]

You skipped a lot of required reasoning because you have to assume from the start that f(1) is ill defined or that f is never zero.
Otherwise you either get something nonsensical or a function that's zero everywhere.

[mohamedb]

you proved existence now you have to prove uniqueness

[thomaslangbein]

f(0)=0 is definitely also a solution, if you exclude x=y and f(y)<>0. This should be allowed, as y=1/x must also be excluded, because 1 is not part of the domain. If you set x to 0 any function would work for f(y), with y<>0 and f(y)<>0. One could on the other hand also argue that with the restrictions f is not anymore generally valid, therefore not well defined => no solution at all.

[MichaelPolymhxanos]

Lets say xεR, y=1, and for text simplicity f(x)=fx and f(1)=f1. Then we get f(x*1)=fx=(fx*f1) / (fx+f1). Assuming fx+f1≠0 then this means fx*(fx+f1)=fx*f1 => fx^2 + fx*f1=fx*f1 => fx^2=0 => fx=0 for every xεR which is in logical condradiction with the fx+f1≠0. Because this contradiction occurs for every xεR, then for every xεR must be fx+f1=0, so f(x)=0

[tbg-brawlstars]

So isn't there like a systematic way of solving these problems ?
Like you just saw its like a logarithmic function
What if it's a property of a new function which we don't know yet