A Functional Equation | Can you solve?

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f(x)+f(1/(1-x))=x
#FunctionalEquations #Functions
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[misterdubity]

Do: Equation 1 PLUS Equation 3 MINUS Equation 2 then you get 2f(x) and get rid of the others

[flowingafterglow]

Wow, that was really confusing. Why not just add eq 1 and e3 and subtract eq 2, and you get directly to the expression for f(x).

f(x) = (x + 1/(x-1) + (x-1)/x)/2

[kinshuksinghania]

6:21 instead you should’ve negated only the 2nd equation and then added all three equations. We would’ve gotten 2f(x) directly that way

[ssarkar]

I love that you consider every functional equation as "nice".

[InnocentNeuron]

Any good textbook for functional equations that you will recommend? Thanks.

[shacharh]

This belongs in a class of functional equations where f is composed over some function g that is cylical in the sense that composing g over it self n times gives back x.
The methods to solve all the equations in this class is to substitute g^k(x) in the argument of f for all k<=n until you get n equations with n different arguments, and solve them as a system of linear equations.

[sudhirlele]

One can negate eq1 and eq3 and add all three or better still, negate only eq2 and add all of them.

[Roq-stone]

When I was finished with this one, I was like, “no way this is right!”
😂😂😂😂😂
I was wrong, it was right 😅

[mystychief]

Maybe easier to replace y=1/(1-x) so that x=1-1/y=(y-1)/y in eq.1 and then replace y with x again. The second time replace z=(x-1)/x so x=1/(1-z) and 1/(1-x)=(z-1)/z and then replace z with x again. The same 3 equations result, but this is easier to follow.

[andy_in_colorado]

I'm pretty sure you've done this problem before, or one very similar, but these are still fun.

[karmachoudhury]

Why don't you make video on number theory