Functions and Behudapanti

The equation $f(f(x)) = (x + x^2 + x^3 + \ldots) f(f(f(x))$ can be rewritten in the form $f(f(x)) = \frac{x}{1-x}f(f(f(x))$ , or, equivalently, in the form $(1-x)f(f(x)) = x f(f(f(x)).\:\:\:\:\:\:\:\:\:\:\:\:(1)$ Let us assume that the degree of the polynomial $f(x)$ is $n.$ Using that for any two polynomials $g(x)$ and $h(x)$ the degree of the polynomial $g(h(x))$ is the product of the degrees of $g(x)$ and $h(x),$ we obtain that the left side of (1) is a polynomial of degree $n^{2}+1$ and the right side is a polynomial of degree $n^{3}+1$ . Then $n^{2}+1=n^{3}+1$ . Therefore, $n=1$ or $n=0.$ If $n=0,$ then $f(x)$ will be a constant polynomial and the equation (1) will imply that $f(x)=0$ for all $x,$ but $f(x)$ must be a non-zero polynomial, so $n=1.$ It means that $f(x)=ax+b$ where $a \neq 0.$ Then formula (1) yields $-a^{2} x^2+(-a b-b+a^{2}) x +(ab+b) =a^{3} x^{2}+(a^{2} b+ab+b)x.$

Making corresponding coefficients equal, we get the following equations: $-a^{2}=a^{3},$ $-a b-b+a^{2}=a^{2}b+ab+b,$ $a b + b =0.$ Since $a\neq 0$ then we have to consider only solution of this system of equation, which is $a=-1,$ $b=1$ . Therefore, $f(x)=-x+1.$ Then $f^{2}(x)+x=(-x+1)^2+x=x^{2}-x+1=(x-\frac{1}{2})^2+\frac{3}{4}.$ The minimum value of this expression is $\frac{3}{4}=0.75$ , and it attained at $x=\frac{1}{2}.$