Solve for $Q$: $Q = (1-p) + p \cdot Q^{z-1} $, where $p\ \displaystyle\epsilon[0,1]$ and $z$ is positive integer.
My try:
$Q-1=p(Q^{z-1}-1)\implies Q-1=p(Q-1)(Q^{z-2}+Q^{z-3}+....+1)$ $\implies Q=1$ is one solution, how to get others, please help.
Solve for $Q$: $ Q = (1-p) + p \cdot Q^{z-1} $
0
$\begingroup$
algebra-precalculus
elementary-number-theory
factoring
-
0@Vim no nothing for $Q$. – 2017-01-30
-
0Related: [http://math.stackexchange.com/questions/2120739/how-does-one-solve-a-polynomial-with-one-arbitrary-power-term#comment4361485_2120739](http://math.stackexchange.com/questions/2120739/how-does-one-solve-a-polynomial-with-one-arbitrary-power-term#comment4361485_2120739) – 2017-01-30
-
0I suppose there is no closed form for $z-2>4$. – 2017-01-30
-
0$Q=1$ is always a solution. If $p=0$, then $Q=1$ is the only solution. If $p=1$, then $z=2$ so $Q$ can be any real number. If $Q\ne0$ then $Q[Q-(1-p)]=pQ^z$, perhaps something can be made of that? – 2017-02-01
-
0If $p\ne1$ and $z\in\{1,2\}$ then $Q=1$ is again the only solution. This leaves the cases when $p\ne1$ and $z>2$. – 2017-02-01