1

I am currently working on Newton's method for approximating roots to functions, and part of the chapter discusses other iterative functions. The basic definition is:

Consider a function $F$ and an initial number $x_0$. Define the subsequent numbers $x_n$ by the formula $x_n = F(x_{n-1})$. This process creates a list of numbers $x_0, x_1, x_2,..., x_n, ...$.

One of the questions I am looking at defines the iterative function by:

$x_{n+1} = x_n^2 + x_{n-2}$

And asks for $x_1$ and $x_2$ with a starting value of $x_0=0.6$.

So far, I have been OK with the definition - where the iterative function is defined by the previous term, but I am struggling with the function immediately above.

I have tried setting $n$ to 0, but this gives:

$x_1 = x_0^2 + x_{-2}$

Which does not seem plausible. If I set $n$ to be 2, then I get something more reasonable:

$x_3 = x_2^2+x_0$

But I have "jumped" past what I am looking for. I did consider allowing a negative subscript but treating it as the starting value:

$let\ n = 0: x_1 = x_0^2 + x_{-2}$

$let\ n=1: x_2 = x_1^2 + x_{-1}$

And calculating:

$x_1 = x_0^2 + x_0$

$x_2 = x_1^2 + x_0$

This approach got me close to the solution, but not to the solution. I am unsure how to handle this sort of function, as generating a new iteration of the function introduces more unknowns.

Any help would be greatly appreciated!

brocolliSally
  • 33
  • 2
    Your problem is indeed ill-defined : $x_0,x_1,x_2$ should have been given parameters, such that the recurrence relation $x_n = x_{n-1}^2 +x_{n-3}$ is valid for $n \ge 3$. – Abezhiko Jul 16 '23 at 17:15
  • Thanks for clarifying. I think I would still be stuck on how to proceed - but I did suspect there should be a parameter to prevent the subscript being negative :) – brocolliSally Jul 16 '23 at 17:27
  • Yes, as is, the problem is unsolvable. Alternatively, $x_1$ and $x_2$ could be determined from $x_0$ and two other values, e.g. $x_5$ and $x_{100}$, but three parameters have to be given beforehand in any case. – Abezhiko Jul 16 '23 at 17:42
  • Thanks! I did suspect that more information would be needed. I tried working out more iterations but each one adds a new unknown. I appreciate the input! – brocolliSally Jul 16 '23 at 18:46
  • Further iterations won't help, because the three informations (including $x_0$) have to be known from the start; you can't "produce" them, that is why this problem can't be solved as is because of lack of information. – Abezhiko Jul 17 '23 at 12:34
  • 1
    However, since you seem to consider initially recurrence relations involving only the previous term, hence the form $x_n = F(x_{n-1})$, I suspect that a typo has transformed $x_{n+1} = x_n^2 + x_n -2$ into $x_{n+1} = x_n^2 + x_{n-2}$. Then you would be able to compute straightforwardly $x_1$ and $x_2$ from $x_0$ only thanks to two steps of iteration. – Abezhiko Jul 17 '23 at 12:38
  • 1
    Yes, I can see that now, and you are correct - I was doing the problems on LibreText (https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/04%3A_Applications_of_Derivatives/4.09%3A_Newtons_Method/4.9E%3A_Exercises_for_Section_4.9), problem 14. I suspect you are correct - as soon as I tried $x_{n+1}=x_n^2+x_n-2$, I got the correct solutions. – brocolliSally Jul 17 '23 at 14:08

0 Answers0