Calculating e^x-1 on classic HPs

[Dieter]

(01-12-2016 01:41 AM)Thomas Klemm Wrote:  Brilliant!
[...proof...]

Thank you very much. Indeed the idea behind this is rather trivial. I'd like to explain this method with an example.
Assume x = 3,141592654 E-6 and a 10-digit calculator.

Evaluating u = e^x with 10 digit precision yields
  u = 1,000003142

Accordingly, e^x–1 is evaluated as
u-1 = 0,000003142

But that's not e^x or e^x-1. The correct values are
e^x   = 1,0000031415975888...
e^x–1 = 0,0000031415975888...

The 10-digit result we got actually is not e^x but e^ln u = e^{3,141995064E-6} (resp. this minus one). Simply because only e^ln u exactly equals u:
e^ln u = u = 1,000003142000000...
e^ln u – 1 = 0,000003142000000...

So we got an exact result for an argument that is slightly off.
Precisely, we want the result for an argument that it is off by
dx = x – ln u

Now the rest is easy. With the first terms of a Taylor series we get
f(x+dx) ≈ f(x) + dx · f'(x)

and thus:
e^x-1 ≈ e^(ln u) - 1  +  (x - ln u) · d[e^x–1]/dx
     = (u - 1)  +  (x - ln u) · e^x

...and since u agrees with e^x to machine accuracy we finally get:

e^x-1 ≈ (u - 1) + (x - ln u) · u

That's it.

Dieter