Firstly the approximation is only within the vicinity of some point. Move your point and you have to recalc.
It's for any function that can be represented as a power series. None of us are really meant to have a clue what that means, yet.
Unfortunately in my current haste I can't find the nice graphical resource I had on this. Add terms and you'll see the graph of the TS starts to accommodate itself to the graph of f better and better.
The main formula is for f(
x). However we use f(a+h) (or a-h) more. You can use x
0 for a here, too.
Try substituting a-h or a+h for a few of the
x's in the "standard formula" of p 569.
First term has no x. Next one .. (x - a) = (a + h - a) = h. Actually let's call that
+h...
No need to do more calcs. They all use that (x - a) term, basically.
OK. Now imagine doing x = a - h .. Your h now has a negative sign.
-h
And I think maybe I'll just leave this part all hanging like that, since it's that sign of h that turns out to be what makes terms cancel when you step out your central difference approximation of eg. f'
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OK and I might as well mention those formulas while I'm here.
All you do is take this "abbreviated TS", and solve for the derivative you want.
f(x) = f(a) + x f'(a) + k f''(xi) ... (truncation at the last)
Solve for f', using just the non-error terms before the truncation.
f'(a) = - (f(a))/ x. (The details are wrong, of course; I'm just trying to show that the principle is fairly simple).
(Or I hope it is, otherwise I'm in trouble).
