When you iterate a series, eg $\sin \theta -\theta =1$ you can use several different rearrangements to iterate. Some, however, come to a solution far faster than others; notably in this case $\theta =\sin \theta +1$ takes about 64 goes while a more complicated series takes nearer 4 (it involved squaring but my teacher took it in...)
I can see from drawing graphs that the closer a graph's gradient is to y=x the slower the iteration. However, this is not proportional but seems to go a little like a tan graph - suddenly zooming to infinity close to 1.
( Why, please? And what exactly is the relationship?)


An iteration of a curve f(x) must also depend on f''(x) but the relationship is not obvious. I realise it depends where you start but I'm hoping there is something else. The two curves I mentioned earlier didn't have hugely different gradients ( though quite diff like 0.7ish,0.2ish) & I started at the same point for both iterations.

Does anyone know anything to help or the answer?

Dear Alexandra,

Well, I can help with the first part of your question. I presume by ``iteration'' you mean that we are solving the equation x = f(x)
by choosing an initial value x₀ and then calculating x₁ = f(x₀ ), x₂ = f(x₁ ), and so on until we're close enough to the correct value of x.


Well, as you know we can draw a graph with the curves y = x and y= f(x) on it and follow the iteration on this graph. I'm going to assume we're close enough to the correct value of x that we can treat f(x) as a straight line with gradient g. Then if the error in x₀ is dx₀ , the error in x₁ will be g x dx₀ . (The easiest way to visualise this without drawing the graph is to imagine that f(x) = g x x, and then it's obvious that the error in x will be multiplied by g (which is less than 1) at each iteration.)


Now, what do we mean by the slowness of an iteration? One possible definition is the number of iterations required to reduce the error by a factor of N. We can calculate this number of iterations (call it n_i ) - it is the solution to the equation

gⁿ i = 1/N

So n_i = (log N) / (-log g).


This isn't quite a tan curve (it's 1/log g) but it does go to infinity at g = 1, as you mentioned.



> An iteration of a curve f(x) must also depend on f''(x) but the
> relationship is not obvious. I realise it depends where you
> start but I'm hoping there is something else. The two curves I
> mentioned earlier didn't have hugely different gradients (though
> quite diff like 0.7ish,0.2ish) & I started at the same point for
> both iterations.


log 0.7 = -0.36

log 0.2 = -1.61.



This is a ratio of 4.5, so I don't know how you got a ratio of 16 between the numbers of iterations you required for your two different methods. I suppose it is affected by f''(x), but I don't know exactly how. Iterations are fastest when f'(x) = 0 at the solution point (for example to calculate the square root of y by iteration, you use x = 1/2 * (x + y/x)), but that doesn't seem to apply in your case.

Eva.

Dear Eva,
Thank you very much for answering my question. When I used the exact values the log ratio thing worked.

Alexandra

PS - I just saw a really stupid error in my original question; please excuse!