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Given a sequence:

$$ x_{n+1} = \frac{1}{2-x_n} $$ Finding the general form of the sequence $x_n$ where $x_1 = {1 \over 2}$ and $n \in \mathbb N$

What I've done is expand the terms $n$ times and then go backwards in order to find some pattern, so:

$$ x_{n+1} = \frac{1}{2 - x_n} = \frac{1}{2-\frac{1}{2-x_{n-1}}} = \dots $$

Now wrapping from the bottom:

$$ \frac{1}{2 - {1 \over 2}} = {2 \over 3} \\ \frac{1}{2 - {2 \over 3}} = {3 \over 4} \\ \dots \\ {1 \over {2 - \frac{n-1}{n}}} = {n \over n+1} $$

So the pattern suggests that $x_n = {n \over n+1}$.

I'm not satisfied with that approach since i believe it's going to fail for more complicated sequences.

What approach or technique could i use to reach the same result in a more "analytical" way?

roman
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  • What do you mean by the "analytical" way? You see a pattern, now test it using induction. – user351797 Sep 18 '18 at 15:47
  • @Clclstdnt By that I meant any approach which would not require expanding the fraction and trying to find the pattern, I just couldn't find a better word for that. Perhaps it's better to say "symbolic" way. – roman Sep 18 '18 at 15:49
  • suppose $x_1 = \frac{p}{q}$? – John Joy Sep 18 '18 at 16:17

1 Answers1

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There are no general methods for solving $x_n=f(x_{n-1})$ where $f$ is a general function. The only way to go about it is compute $x_n$ for small values of $n$, hope to notice a pattern, then prove it by induction. If the problem is not possible by this method, you will not be expected to solve it.

However, there are some special classes which admit general solution methods. One example is $$ f(x) = \frac{ax+b}{cx+d} $$ The reason this works out nicely is that $x_n=f(f(\dots f(x)\dots))$, and function composition of functions resembling the above behaves exactly like matrix multiplication. To see this, compute the composition of $f$ with $g(x)=\frac{ex+f}{gx+h}$, rearrange it so it is of the form $\frac{px+q}{sx+t}$, and compare $p,q,s,t$ with the entries of matrix product $$ \begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}e&f\\g&h\end{bmatrix} $$

See wikipedia for more information.

This means that a closed form for $x_n$ can be found by putting the matrix of $f$ in Jordon form, because powers of a Jordon block matrix are easy to compute.

There are a number of other recurrence relations which can be solved exactly, like linear recurrence relations with constant coefficients like $x_n=2x_{n-1}-x_{n-2}$.

Mike Earnest
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