Calculating e^x1 on classic HPs

02022019, 07:22 PM
Post: #36




RE: Calculating e^x1 on classic HPs
(01162016 12:40 PM)Gerson W. Barbosa Wrote: \[e^{x}1=\frac{e^{x}e^{x}}{e^{x}+1}\] (01162016 01:32 PM)Dieter Wrote: ... there are two peaks at +4 and –4 ULP that also showed up in earlier tests. I wonder where these come from. Any idea? My guess is that error goes at opposite direction, thus make things worse. exp(x) = 1/exp(x), so if exp(x) error is positive, exp(x) is negative. This made numerator even bigger. Worse, denominator shrink. Errors reinforcing each other, making bad peaks !  This is very different than Kahan's calculating (exp(x)1)/x via (u1)/log(u) For u close to 1, both numerator and denominator error goes the same way. A bonus is that the size of error also similar size, cancelling each other. Although calculated values of u1 and log(u) are not accurate, the ratio is very good. 

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