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I'm stuck on a particular part of a long question, so I will post the entire question and tell you guys where I'm stuck.

Q: The size of a locust population in each year is given in terms of the population in the previous year, according to the relation

$x_{t+1}=F(x_t)$, $F(x)=r(1-x^2)$, $r>0$.

where $t$ is the time measured in years, and $x_t$ is a dimensionless measure of puplation size.

(a) Sketch the graph of $F(x)$, ane hence (or otherwise) find ranges for the value of the initial population size for all $t>0$.

(b) Obtain the steady states of the model by finding the roots of a suitable monic quadratic $Q(x)$. Use linear stability analysis to describe the first bifurcation that occurs at a positive value $r=r_1$, which you should find.

(c) Write down the equation for period two solutions, and show that it can be rewritten in terms of $Q=Q(x)$ as $$r\left( (x-rQ)^2 -1 \right) +x=0.$$ Explain why the roots of $Q$ satisfy this equation, and hence show that for $r>r_1$ the solution with minimal period 2 alternates between the values $x_+$ and $x_-$, where $$x_{\pm}=\frac{1}{2r} \pm \sqrt{1-\frac{3}{4r^2}}.$$

(d) Give the condition for a pitchfork bifurcation in the doubled map, and apply it to find the value of $r_2>r_1$ such that the period two solution becomes unstable at $r=r_2$. Given that the next period doubling bifurcation happens at $r=r_3\approx 1.1696576163$, calculate an approximation to Feigenbaum's constant to 3 decimal places.

A: (a) Realistic population ranges for $x_0$ and $r$ I believe are $0 \leq x_0 \leq 1$ and $0<r\leq 1$. (b) I found the steady states by solving $F(x)=x$ and then the I also got that $$Q(x)=x^2+\frac{x}{r}-1.$$ Te first bifurcation also occurs at $r=r_1=\frac{\sqrt{3}}{2}$.

(c) I showed that $Q(x)=\frac{1}{r}(x-F(x))$ in order to show what they wanted, then also noted that the roots to $Q(x)$ satisfy $F(x)=x$, and solutions of that equation satisfy $F(F(x))=x$. So take a root of $Q(x)$, then we get $F(x^*)=x^*$, so $F(F(x^*))=F(x^*)=x^*$.

Edit: I just figured it out. It being in this form helps greatly because we can write $$r( (x-rQ)^2-1)+x=r(x^2-2xrQ+r^2Q^2-1)+x.$$ Then from here we get that $$r(x^2-2xrq+r^2Q^2-1)+x=rx^2-2xr^2Q+r^3Q^2-r+x.$$ Now we are almost there, we can then do $$rx^2-2xr^2Q+r^3Q^2-r+x=Q(r^3Q-2xr^2)+rQ.$$ So upon dividing by $Q$, as we want to eliminate the two steady state roots, we get the equation $$r^3Q-2r^2x+r=0.$$ Plug $Q$ back into that and simplify, you'll get a quadratic with the roots they wanted me to derive.

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With your definition of $Q(x)$, the solutions of $x=F(F(x))$ solve $$ x=r(1-r^2(1-x^2)^2)=r(1-(rQ(x)-x)^2), $$ or, equivalently, $$ (rQ(x)-x)^2=1-\frac{x}r=x^2-Q(x), $$ or, expanding the LHS, $$ Q(x)\cdot(r^2Q(x)-2rx+1)=0. $$ The solutions of period $2$ solve $r^2Q(x)-2rx+1=0$. Expanding $Q(x)$, this is $$ r^2x^2-rx+1-r^2=0, $$ hence the formulas for $x\pm$ when $r^2\geqslant3/4$. This solves (c).

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