[College Calc 01-02. Limits of Sequences] #05

Since we've had two problems for the price of one here, here are two solution methods for this kind of question.

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The first idea is to treat this as an iterated function system.

Let $f(x)=\frac14 x^2+\frac12 x -2$ .

The fixed points of this function are found by solving $x=f(x)$ ; that is $x=\frac14 x^2+\frac12 x -2$

There are two fixed points; $x=4$ and $x=-2$ . The important thing to consider is their stability. We have

$f'(x)=\frac12 x+\frac12$

so $f'(4)=\frac52$ and $f'(-2)=-\frac12$ .

A fixed point $x$ is only stable if $|f'(x)|\le 1$ ; hence $x=4$ is unstable and $x=-2$ is stable, and, in the limit, the iteration will tend to $x=-2$ for all starting points except $x=4$ and one other value (left as a bonus question, if you've read this far; all that matters here is that it's not $3$ ).

Hence the required limit is $\boxed {-2}$ .

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The original recurrence was $a_1=3;\;\;\;a_{n+1}=\frac12 a_n^2-4$

As observed elsewhere, this sequence doesn't converge; but it doesn't go off to infinity, or enter a cyclic pattern either.

Again, there are two fixed points, but using the method above, neither of them is stable.

The first few values are

and it doesn't really settle down after that.

Dividing both sides of the recurrence by $4$ , we find

$\left(\frac{a_{n+1}}{4}\right)=2\left(\frac{a_{n}}{4}\right)^2-1$

This form should look familiar; it's similar to the identity $\cos 2\theta \equiv 2\cos^2 \theta-1$

With this in mind, then, put $a_n=4\cos x_n$

We then have the very simple recurrence $x_{n+1}=2x_n$

When $n=1$ , $4\cos x_1=3$ ; so the full solution is $a_n=4\cos\left(2^{n-1} \cos^{-1} \frac34 \right)$

Although we proved above that $4$ and $-2$ are not stable fixed points, it's another nice exercise to show from this trig form that $a_n$ can't reach these values.

Let the limit (provided it exists) be $l$

Then $l=\dfrac {l^2}{2}-4\implies l=4,-2$

Since $a_n<0$ for $n\geq 3$ , therefore the required limit is $\boxed {-2}$ .

Obviously the sequence may not converge at all, but the question asks if it converges, what would be the limit? Let the limit be $\lambda$ ; then

$\begin{aligned} \lambda & = \frac 12 \lambda^2 - 4 \\ 2\lambda & = \lambda^2 - 8 \\ \lambda^2 - 2\lambda - 8 & = 0 \\ (\lambda - 4)(\lambda + 2) & = 0 \end{aligned}$

So if the limit exists, it can either be $4$ or $-2$ . But the actual values of $a_n$ are always less than $4$ (I don't know how to prove this). The graph show the first values of $a_n$ . Hence the limit is $\boxed{-2}$ .