To find the bifurcation values for $$x_{i+1}=f(x_i) = rx_i(1-x_i)$$first I set $rx(1-x) = 0$ and found the x values and then used the x values to find $r = 0$ and $r = 1$.
Do you think what I did here is correct? If not, can you help me find the mistakes here?
To study bifurcations of maps, begin by looking for fixed points. The logistic equation is
$$x_{i+1} = f(x_i) = rx_i(1-x_i)$$
so a fixed point satisfies
$$x = rx(1-x) \Rightarrow x(rx - r+1) = 0$$
which implies that there are fixed points at $x=0$ and $x=1-1/r$. To analyze bifurcations as $r$ varies, consider the linearization of the map around these fixed points $x_i=x^*+y_i$
$$y_{i+1} = f'(x^*)y_{i}$$
Now $f'(x)=r(1-2x)$ so linearizing about $x=0$ gives
$$y_{i+1} = ry_i$$
so there is a clear loss of stability when $r=1$ and a period-doubling bifurcation when $r=-1$.
Linearizing about $x=1-1/r$ gives
$$y_{i+1} = (2-r)y_i$$
so there is a bifurcation from instability to stability as $r$ passes through 1 from below, and then a period doubling bifurcation as $r$ passes through $3$.
That gives you three bifurcations, at $r=-1$, $r=1$ and $r=3$. To find further bifurcations you must analyze the second iteration of the map,
$$x_{i+2} = f(f(x_i))$$
and search for fixed points, which correspond to period-2 orbits. This isn't as hard as it sounds - you end up having to solve a quartic equation, but you already know two of the solutions, as any fixed point trivially satisfies this equation. That gives you further period-doubling bifurcations, from 2-cycles to 4-cycles. Finding the location of the next set of period-doubling bifurcations (from 4-cycles to 8-cycles) is hard.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{With the Logistic Map}\qquad x_{n + 1} = \mu x_{n}\pars{1 - x_{n}}\,,\qquad 0 < \mu < 4\tag{1} $$ you must study $\mu$-behaviour ( 'fixed points' ) of $\pars{1}$ and $\mu$-behaviour of \begin{align} x_{n + 1} &= \mu^{2}x_{n - 1}\pars{1 - x_{n - 1}} \bracks{1 - \mu x_{n - 1}\pars{1 - x_{n - 1}}} \\ x_{n + 1} &= \mu^{3}x_{n - 2}\pars{1 - x_{n - 2}}\bracks{1 - \mu x_{n - 2} \pars{1 - x_{n - 2}}}\braces{1 - \mu^{2} x_{n - 2}\pars{1 - x_{n - 2}} \bracks{1 - \mu x_{n - 2}\pars{1 - x_{n - 2}}}} \end{align} and so on.
