HP Prime bug for approximate(erfc(z)), z as (0+b*i), or erf(b*i)

[Albert Chan]

(11-15-2023 08:23 PM)parisse Wrote:  There is a warning of low accuracy if re(z) is small : I must indeed stop the fraction expansion and
since we do not have multiprecision floats there are some wrong digits.

Throwing more precisions may not help.
Problem is more to do with how deep "fraction expansion" default we pick.

Let n = "last" numerator --> erfc(z, n=1) = exp(-z*z)/√(pi) / (z + (1/2)/(z + 1/z))

Let z = 3.16845945394*I, precisions = 1000 digits, results rounded to integer.

erfc(z, n=0.5) = -4292 I
erfc(z, n=1.0) = -4317 I
erfc(z, n=1.5) = -4323 I
erfc(z, n=2.0) = -4324 I
erfc(z, n=2.5) = -4325 I
erfc(z, n=3.0) = -4325 I
erfc(z, n=3.5) = -4326 I
erfc(z, n=4.0) = -4311 I
erfc(z, n=4.5) = -4324 I
erfc(z, n=5.0) = -4325 I
...
erfc(z, n=19.0) = -4324 I
erfc(z, n=19.5) = -4326 I
erfc(z, n=20.0) = 58957691655 I
erfc(z, n=20.5) = -4325 I
erfc(z, n=21.0) = -4326 I
...

For z=3.16845945394*I, n=20 is really bad.
But changing n will not help. It would just shift problematic z somewhere else.

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More precisions can't help because CF formula denominator is close to 0
But, erfc(z) is continuous ... something is going to break.

1 / ±0 = ±∞

Cas> erfc(3.16845945394*i)      → +58608047158.3*i
Cas> erfc(3.16845945395*i)      → −14620291004.7*i

CF Denomaintor = x * p(x²)/q(x²), rational polynomial with all positive terms
--> erfc CF formula for non-zero real argument will not get divide-by-0 issue.

(04-06-2024 11:04 PM)Albert Chan Wrote:  denominator may get to 0 only if re(z)=0