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RE: HP48-HP50G : Error functions ERF(z), FRESNEL integrals, PROBIT, etc.
11-14-2023, 11:24 PM
Post: #28
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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RE: RE: HP48-HP50G : Error functions ERF(x or z), ERFC (x or z) & iERF/iERFc —>x - Gil - 11-14-2023 11:24 PM



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