Yes, I saw this in the Tompkins County Community College newsletter too. Here’s my solution: The equation has the form f(y)+f(g(y))=1/y where g(y)=1-1/y. The trick is that g(g(g(y)))=y. Using this we can write the following three equations: f(y)+f(g(y))=1-g(y) f(g(y))+f(g(g(y))=1-g(g(y)) f(g(g(y)))+f(y)=1-y Think of this as a system of linear equations in the variables f(y), f(g(y)), f(g(g(y))). From the solution we get f(y)=(1-y-g(y)+g(g(y)))/2=(1-y^2+y^3)/(2y-2y^2). -Veit

On Jun 24, 2017, at 5:46 AM, Guy Haworth <g.haworth@reading.ac.uk> wrote:

My college magazine includes this challenge:

Find a function f(x) such that for every real x not equal to 0 or 1, we have f(1/x) + f(1-x) = x

It's from Oxford BNC's Konstantin Ardakov.

I guess there's a clue here that 1/x and 1/(1-x) feature in the solution ... but how does one go about this?

Guy

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun