How do I do iterations backwards?

Interesting question.

Let's say you were looking for a root of the equation

x^3 - 2x^2 - 14 = 0

The root is near x = 3.2919, but let's say that

somewhere in your forward iterations you had reached an

estimate of x = 3.30

I think you are asking whether it is possible to determine

from this what the previous estimate was, or whether there

are multiple paths that could've led exactly to x = 3.300000

For this case, let y = x^3 - 2x^2 - 14

so that dy/dx = 3x^2 - 4x

The "previous estimate" would've been

3.3 - dx

and the function value there would've been

-dy

so you would've had

dy/dx = 3(3.3-dx)^2 - 4(3.3 - dx)

and also

(3.3 - dx)^3 - 2(3.3 -dx)^2 - 14 = -dy.

Replacing the "dy" in the first of these equations, you have

(3.3 - dx)^3 - 2(3.3 - dx)^2 - 14 = 4 dx (3.3 - dx) - 3 dx (3.3 - dx)^2

35.937 - 32.67 dx + 9.9 (dx)^2 - (dx)^3 - 21.78 + 13.2 dx - 2 (dx)^2 - 14 =

= 13.2 dx - 4 (dx)^2 - 32.67 dx + 19.8 (dx)^2 - 3 (dx)^3

Collecting terms, this equation becomes

2 (dx)^3 - 7.9 (dx)^2 - 14 = 0

This equation has a unique solution, dx near 4.324334,

and it turns out NOT to make sense;

however, that's due to some algebra error on my part,

which you may be able to identify -- I'm not going to bother

to redo the algebra, because I can already see the general

answer to your question:

The answer to your question is that it IS possible to go backwards

and determine the previous estimate,

but the process of doing so will always involving finding the

root of some equation of the same degree as the one you

are trying to solve, so that undoing a single iteration becomes

a multi-iteration problem if you are doing manual calculations.