A Functional Equation from Türkiye 🇹🇷

[@MortezaSabzian-db1sl]

To a small generalization. If we have an equation like this; f(x)+f(u{1/(1-u^(-1)(x)})=v(x) where the function u^(-1)(x) is the inverse of u(x). Let u^(-1)(x)→x f(u(x))+f(u(1/(1-x)))=v(u(x)) and let f(u(x))=g(x) With the following few substitutions we arrive at a system of equations g(x)+g(1/(1-x))=v(u(x)) x→1/(1-x) g(1/(1-x))+g((x-1)/x)=v(u(1/(1-x))) x→1/(1-x) g((x-1)/x)+g(x)=v(u((x-1)/x)) As result we have; 1)g(x)+g(1/(1-x))=v(u(x)) 2)g(1/(1-x))+g((x-1)/x)=v(u(1/(1-x))) 3)g((x-1)/x)+g(x)=v(u((x-1)/x) We subtract equation 2 from 3 and add the result to equation 1; g(x)=1/2(v(u((x-1)/x))-v(u(1/(1-x)))+v(u(x)))=f(u(x)) So f(x)=..... For example u(x)=³√x v(x)=x³ f(x)=1/2((x³-1)/x³-1/(1-x³)+x³)

[@stephenshefsky5201]

Yes, that's a very nice generalization.

[@SyberMath]

nice!

[@MortezaSabzian-db1sl]

f(x)+f(1/3_√(1-x³))=x³ Let and set x³→x f(3_√x)+f(1/3√(1-x)=x Set and let f(3_√u)=g(u) g(x)+g(1/(1-x)=x This functional equation has been solved many times f(x)=(x⁹-x³+1)/2x³(x³-1) f(-1)=1/4 f(∞)=∞

[@terracottapie6872]

This is interesting. Never seen that g(x) functional equation.

[@stephenshefsky5201]

I derived the same general solution using the same substitution!

[@MortezaSabzian-db1sl]

@@stephenshefsky5201 Find my other comment and look, I got a more general look in that comment

[@scottleung9587]

I went down the route of the second method - like you said, I didn't get anywhere far.

[@yakupbuyankara5903]

1/4

[@giuseppepapari7419]

2:53. I am confused here: 1/ [ 1/ cube_root(1/2)] should be cube_root(1/2). Or do I miss something?

[@gabrielroy-manningham5734]

Yeah there is a small mistake

[@AllanPoeLover]

You are right. The method is good. But sometimes it need more patience.

[@BilmeMortaylı]

Lan Türkiye

[@paulkarch3318]

I tried this substitution for x. ((x^3 - 1)/x^3)^(1/3) and got f((1 - 1/x^3)^(1/3))) + f(x) = 1 - 1/x^3 but not sure if this leads anywhere.

[@yoav613]

Nice and easy😊