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We have the linear recurrence relation $$x_{n+1} = \dfrac{3}{2}x_n - 20$$ with $n = 0,1,2...$ and $a,b$ being constants. Does this equation have a fixed point? Does the equation have a period 2 (a period 2 solution $x_0,x_1$ is a solution where you get from $x_0$ to $x_1$ after 1 iteration and after the next iteration you get back to $x_0$) fixed point? Is the fixed point attractive?

I have absolutely no clue how to do this. My cheap school isn't willing to buy us books so we have to use horrible free pdfs from the internet and this one is particularly bad. The terminology is all over the place and as a result I have absolutely no clue on earth how to even begin this stuff. Can anyone at least point me in the correct direction?

tinlyx
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iEvenLift
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2 Answers2

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There is a fixed point at $x = 40$. Now put $y_n = x_n - 40$ for any $n$. Then you have $$ y_{n+1} + 40 = (3/2)(y_n + 40) - 40$$ so $$ y_{n+1} = {3\over 2} y_n.$$

The equilibrium at 40 is unstable and repels.

ncmathsadist
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  • Can you please explain why the equilibrium is unstable? – iEvenLift Jun 15 '13 at 20:21
  • Suppose $y_n > 0$. Then because of the recursive relation, $y_n\uparrow\infty$ as $n \to \infty$. So if an inital value $x_0 > 40$ is chosen, the $x_n$s increase to to $\infty$. Can you see what happens if $x_0 < 40$ is chosen? – ncmathsadist Jun 15 '13 at 21:39
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A fixed point for this recurrence relation is a value $x_n$ such that $x_{n+1}=x_n$. So you can try to solve $x=\frac{3}{2}x-20$, this will give you all the fixed points.

If a solution has period 2 then $x_n=x_{n+2}$ and $x_n\neq x_{n+1}$ so here you want to try to solve: $$ x_{n}=x_{n+2}=\frac{3}{2}x_{n+1}-20=\frac{3}{2}(\frac{3}{2}x_n-20)-20 $$ and check that any solution you get doesn't have period one.

A fixed point $x_0$ is attractive if for any $y$ which is close enough to $x_0$, the sequence $y=y_0,y_1,y_2,y_3,\ldots$ converges to $x_0$ (I can explain what this means if you don't know). You can test whether a fixed point is attractive if you define a function $f(x)=$the recurrence relation (in your case $f(x)=\frac{3}{2}x-20$) then if $f$ is (continuously) differentiable in a neighborhood of $x_0$ and $|f'(x_0)|<1$, your fixed point is attractive.

Note that this method only allows you to show that a point is attractive, I don't think it works to show a point is NOT attractive.

Also ncmathsadist's approach works.

James
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  • This is actually helpful. I assume this would just go on for a period 3,4... etc.? – iEvenLift Jun 15 '13 at 20:03
  • Indeed it does. – James Jun 15 '13 at 20:05
  • So is this correct; We have $x=40$ is a fixed point. If we try to find period 2 fixed points we get $x=40$ too, but that one doesn't satisfy the requirement $x_n \neq x_{n+1}$. So we don't have a period 2 solution. – iEvenLift Jun 15 '13 at 20:17
  • Also, I don't know how to show it's stability in this case. I would normally use $f'(40)$ and check if it's smaller, but the derivative of this function is just $\dfrac{3}{2}$ so I have no idea in this case. – iEvenLift Jun 15 '13 at 20:20
  • Then you have to be clever, look at ncmathsadist's post. He has a method which works. – James Jun 15 '13 at 20:25
  • Also, for the higher periods you can be a little smarter. Lets say we want to find all the period $n$ solutions. Using what I show you above, you repeatedly substitute $\frac{3}{2}x-20$ for $x$. Well however many times you substitute this linear relation, you are always going to get out a linear relation. I.e. your $n$th period stuff is always going to satisfy $x=ax+b$ for some $a$ and $b$.

    Now we know that our period one solution is going to work here, but we also know that an equation of the above type can have only one solution. So as we already have one, it can be the ONLY one.

     – James Jun 15 '13 at 20:31