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I calculated the composition in both cases, F(X) and G(x), but I have no clue what an invariant is, and my lecturer has never mentioned. I tried to google but unsuccessfully.

A function F(x) is said to be invariant under another function G(x) if (F ∘ G)(x) = F(x), where ∘ denotes composition. Let a(x) = 1 - x and let b(x) = 1/x.

Calculate the function which is an invariant under both a(x) and b(x).

curiousStudents
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    It is defined in the problem. You need to find a function $F(x)$ such that, for all $x$ for which everything is defined, we have $F(1-x)=F(x)=F\left(\frac 1x\right)$. For instance, any constant works, though presumably they want a non-trivial solution if there is one. – lulu May 03 '20 at 13:28
  • Thanks for your comment, but is there a way, formula or something to follow, considering that I cannot find anything on the internet. – curiousStudents May 03 '20 at 13:56
  • this question is similar. – lulu May 03 '20 at 14:06

1 Answers1

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This problem hinges upon realizing that the set formed by the following $6$ transformations is closed under composition: $$x,\quad \frac 1x, \quad 1-x, \quad \frac 1{1-x},\quad \frac x{x-1},\quad \frac {x-1}x$$

(This is easily checked by direct calculation)

It follows that if we take any function and sum its translates over these six functions we will get a function invariant under both $x\mapsto \frac 1x$ and $x\mapsto (1-x)$ (and the other operations of course).

Taking $f(x)=x$ just returns the constant function $3$, but if we take $f(x)=x^2$ we get $$F(x)=x^2+(1-x)^2+\left(\frac 1x\right)^2+\left(\frac 1{1-x}\right)^2+\left(\frac x{x-1}\right)^2+\left(\frac {x-1}x\right)^2$$ and this function is a non-trivial example of what we are looking for.

lulu
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