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We have the following equality:

$$ \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$

Some of the solutions I found:

  • $\mathrm{x} = 0$

  • Also for $\mathrm{a}=0$, every $\mathrm{x}$ is a solution I believe

I tried getting everything out of the brackets but that just gave a nasty equality which I couldn't solve.

Let's take $-1 \leq \mathrm{a} \leq 1$. I was also wondering if there is a way of knowing how many solutions this equality has beforehand? Or do we just have to look at the cases $\mathrm{a} = 1$, $\mathrm{a} = -1$, $\mathrm{a} \neq 0$ and $\mathrm{a} = 0$ individually to find every solution (i.e. both of the bounds, and for a (not) equal to 0)?

iostream007
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Prael
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1 Answers1

2

I've solve upto some extent $$ x = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$ $$ {a^2x(1-x)(1-ax(1-x))-x=0}\implies x=0$$ $$ {a^2(1-x)(1-ax(1-x))-1=0}$$ $$ {a^2(x-1)(1+ax(x-1))+1=0}$$ put x-1 =t $$ {a^2t(1+at(t+1))+1=0}$$ $$ {a^2t(1+at^2+at)+1=0}$$ $$a^3t^3+a^3t^2+a^2t+1=0$$ for this equation if we take a=1 $$t^3+t^2+t+1=0\implies t=-1,i,-i\implies x=0,1+i,1-i$$ so based on value of a there are different solution of t and x. solution in continuity based on @maming's comment

$$a^3t^3+a^3t^2+a^2t+1=0$$ $$a^3t^3+1+a^3t^2+a^2t=0$$ $${(at)}^3+1^3+a^3t^2+a^2t=0$$ $$(at+1)(a^2t^2-at+1)+a^2t(at+1)=0$$ $$(at+1)(a^2t^2-at+1+a^2t)\implies (at+1)=0\;,(a^2t^2-at+1+a^2t)=0$$ $$t=\frac{-1}{a}\implies \mathbf{x=1-\frac 1a}\;\;,a^2t^2+t(a^2-a)+1=0$$ $$t=\dfrac {-(a^2-a)\pm\sqrt{{(a^2-a)}^2-4\cdot a^2\cdot 1}}{2a^2}$$ $$t=\dfrac {a\left((1-a)\pm\sqrt{{(a-1)}^2-4}\right)}{2a^2}$$ $$t=\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}\implies x=1+\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}$$ $$x=\dfrac {(a+1)\pm\sqrt{{(a-1)}^2-4}}{2a}$$

iostream007
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