Calculating e^x1 on classic HPs

01152016, 06:46 PM
(This post was last modified: 01152016 07:13 PM by Dieter.)
Post: #31




RE: Calculating e^x1 on classic HPs
(01152016 06:07 PM)Gerson W. Barbosa Wrote: I'm currently running this on an ordinary 12C and Dieter's program on a 12C+, so I cannot compare speeds, but Dieter's should be slightly faster because it requires no tests. Hm, seems you forgot an essential test here: if x is so small that e^{x} rounds to 1 (here –5 E–11 < x < 5 E–10), the term (u–1)/ln u becomes 0/0... #) In such a case the program should return x. This additional test would also handle the x=0 case that's now tested in the first line. BTW, W. Kahan's method is supposed to calculate (e^{x}–1)/x. Even if this quotient is evaluated exactly, i.e. for small x very close to 1, there is always a potential error of half an ULP or 5 E–10 on a tendigit device. This translates to a possible error of up to 5 ULP in the final result, i.e. after the multiplication with x. Example: x=9 E–9. The exact quotient is 1,000000045... which on a tendigit calculator is rounded to 1,000000005. This indeed is the correctly rounded tendigit value. But the final result is returned as 9,000000045 E–9 instead of 9,0000000405000...E–9, i.e. there is an error of +4,5 ULP. Dieter 

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