Let $ f(x) = x+1 $ and $g(x) = 2x$
Prove $f^2g = gf $ and determine $f^igf^jgf^k(x)$ explicitly as a function of x and in terms of i,j,k.
I got through the proof but I don't understand what the question is asking for, can someone tell me how I would go about solving this problem?
For every $x$ we have \begin{eqnarray} (f^2g)(x)&=&f(f(g(x)))=f(f(2x))=f(2x+1)=(2x+1)+1=2x+2=2(x+1)\\ &=&2f(x)=g(f(x))=(gf)(x), \end{eqnarray} i.e. $f^2g=gf$.
Notice that $$ f^2(x)=f(x+1)=x+1+1=x+2,\, f^3(x)=f(x+2)=x+2+1=x+3, $$ and therefore by indunction $f^i(x)=x+i$ for all $i\in \mathbb{Z}$. It follows that: \begin{eqnarray} (f^igf^jgf^k)(x)&=&f^i(g(f^j(g(f^k(x)))))=f^i(g(f^j(g(x+k))))=f^i(g(f^j(2(x+k))))\\ &=&f^i(g(f^j(2x+2k)))=f^i(g((2x+2k)+j))=f^i(g(2x+2k+j))\\ &=&f^i(2(x+2k+j))=f^i(4x+4k+2j)=4+4k+2j+i \end{eqnarray}