RE: HP48-HP50G : Error functions ERF(z), FRESNEL integrals, PROBIT, etc.
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11-14-2023, 11:24 PM
Post: #28
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RE: RE: HP48-HP50G : Error functions ERF(x or z), ERFC (x or z) & iERF/iERFc —>x
If x=10^(-9)
I try to improve the value found with my HP50G : ERF(10^(-9)) : 000000001128. Now with that value, .000000001128, I calculate the inverse iERF(000000001128) :9.9911511335E-10. It means that ERF (9.9911511335E-10) gives also 000000001128. With Taylor ERF (9.9911511335E-10 +8.8488665E-13) = .000000001128 +f'×h, with h=8.8488665E-13 Correct? And f'= 2/sqrt(pi)*exp(-x²). But as mentioned in on HP50G, exp (very small) =1 And exp (-very small) = 1/1 = Constant on HP50G. So that f'=2/sqrt(pi)*exp(-x²) =2/sqrt(pi)*1 = 1.12837916709. And f'×h = 1.12837916709 × 8.8488665E-13 = 9.98487661096E-13 And new found ERF = 000000001128 +f'×h= 000000001128 + 9.98487661096E-13= 1.12899848766E-9... ≠ 1.128379167! I suppose that I understood all wrong. Could you help me point the errors in order to get your very good solution value? By the way, is there a way (if possible an easy one!), with the HP50G, to get an approximate value of Exp(-1EE-18), which is less than 1? Thanks for your comprehension and patience. As you may have noticed, I am a dummy in maths and in many other topics. Regards |
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