Sequences
motivated by discussions on calculus forum

Suppose a sequence x(1), x(2), ... is defined like so:
x(n+1) = f(x(n))
>That's not usual, is it? I mean ... sequences defined like that?
Define "usual".
Anyway, let's consider such a sequence.

Graphically, we can:

  • Plot y = x and y = f(x)
  • Begin with some x(1)   ... for example, x(1) = 5    
  • The next term in the sequence, namely x(2) = f(x(1)), is obtained as follows:
    Move from the point (x(1), f(x(1)) horizontally to the line y = x then vertically to y = f(x).
  • The next term in the sequence, namely x(3) = f(x(2)), is obtained as follows:
    Move from the point (x(2), f(x(2)) horizontally to the line y = x then vertically to y = f(x).
  • The next term in the sequence, namely x(4) = f(x(3)), is obtained as follows:
    Move from the point (x(3), f(x(3)) horizontally to the line y = x then vertically to y = f(x).
  • The next term ...
>Okay, I get it. So where's x(n) going?
That depends upon f(x) and the initial x-value.
For the example above, f(x) = x - 1/x and, if we follow x(n), we eventually see something like this:  

>It don't look like it's gong anywhere in particular.
If it did approach some limit, L, then we'd expect L to satisfy L = f(L) ... that is: L = L - 1/L
... and that don't hardly got no solution.

Interesting things can happen depending upon the initial value of x.
For example:
     

Note, too, that x(n) will equal 0 if 0 = x - 1/x   and that has solution x = ±1.

Note:
There's an Excel spreadsheet to play with. Just click on the picture:

>Uh ... what's that big, long curve? Looks like a parabola.
Yes. See the graphs?
The blue points are the iterates x(n+1) = x(n) - 1/x(n) starting with x(0) = 40 and running to n = 800.
The red points are on the graph of x2 / 2 + n = 800.

>Uh ... they look the same.
Don't they?

>Is that exact?
No, but it's close for large x.

We can get a quick-and-dirty approximation like so:

x(n+1) = x(n) - 1 / x(n) ⇒ x(n+1) - x(n) = -1 / x(n) so (approximately!) dx/dn = -1/x which has solution: x2 / 2 + n = constant.

However, note that, if x(n)2 / 2 + n = C, then x(n+1)2 / 2 + (n+1) = { x(n) - 1/x(n) }2/2 + (n+1) = x(n)2 / 2 + n + 1/2x(n)2 = C + 1/2x(n)2 ≈ C for large x(n).
So, if x(n) lies on x2 / 2 + n = C, then so does x(n+1) ... very nearly (for large x-values).

Here are some other examples ... with different starting values:

Note:
If, for f(x), you stick in x - g(x)/g '(x) for some "reasonable" function g(x), the iterates are Newton iterates which (if you're lucky) converge to a root of g(x) = 0.

Iteration via   z(n+1) = F(z(n))   is even more fascinating if z = x + iy is a complex number. ... see Fractals            

That's a 2-dimensional ritual, generating a sequence of points (x(n), y(n)) via:
x(n+1) = f(x(n), y(n))
y(n+1) = g(x(n), y(n))


z(n+1) = z(n)2 - 3/4

See: Math Stuff