Looking for TVM formulas

[Dieter]

(04-03-2014 01:10 AM)Manolo Sobrino Wrote:  

(04-02-2014 07:40 PM)Dieter Wrote:  ln 0,9 < z < ln 2 is all that's required. ;-)
With terms up to n = 9 the result has 11 valid digits.

You're right, lower bound is wrong for 11 digits. Yet I get 11 up to -0.38. Where did you get the analytical expression for the bounds? (yeah, I'm asking for "here be dragons" stuff )

Simple: the basic idea of a dedicated e^x—1 function is better accuracy than a simple e^x minus one.

For x > 0 this is true if the result is less than one. And e^x—1 < 1 means that x is less than ln 2.

For x < 0, e^x—1 is something between –1 and 0. Results within [–1; –0,1] have full accuracy even without a dedicated e^x—1 function. Which is required only for results > –0,1. Which means that x > ln 0,9.

So we only have to consider the interval from ln 0,9 to ln 2. Using the contiued fraction method up to n=9 this yields a largest error (at ln 2) of 3,6 units in the 12th place, i.e. 11 valid digits.

In all other cases a simple e^x minus one will return exact results, so the approximation is not required.

For more details, visit Casio's Keisan calculator, set the accuracy to 22 digits and copy the following code into the "Expression" window:

Code:

approx = z/(1 - z/(2 + z/(3 - z/(2 + z/(5 - z/(2 + z/(7 - z/(2 + z/(9 - z/2)))))))));
exact = e^z-1;
err = approx - exact;
shift = 10^(-(1+int(log(abs(approx)))));
print(exact, approx, err*shift);

Press "Execute", and then enter -1, 0.02, 101 at the now appearing "Variable" input line below. You will get a table for z = –1 to +1 with the true value, the approximation and the error, expressed in units of the nth significant digit (i.e. 3,5E-14 means the approximation is off by 3,5 units in the 14th digit).

Dieter