HP Prime: built-in inverse function of erf?

[Gil]

I don't know if there is on the HP PRIME built-in inverse function
for error function erf.

I have a probability .8

And I want its iERF: —>.906193802438.

[Albert Chan]

> ierfc(x):=normald_icdf(x/2)/-sqrt(2)
> ierf(x):=ierfc(1-x)
> ierf(0.8)

0.906193802437

[Gil]

As, in the HP PRIME, erf function admits as argument a complex number, it would be nice to have a built-in function inv_erf that would also admit as argument a complex number.

[Albert Chan]

Hi, Gil

I think erf inverse function is defined only for real x within ±1, otherwise we get infinite solutions.

> fsolve(erf(x)=2, x=i)            → 0.869417565326+1.34059481853*i
> fsolve(erf(x)=2, x=1+3i)      → 1.88472598633+2.28183639616*i
> fsolve(erf(x)=2, x=2+4i)      → 2.55833225411+2.9104037265*i
> fsolve(erf(x)=2, x=3+5i)      → 3.55858803599+3.85919595309*i

> fsolve(erf(x)=i, x=i)            → 0.731697153468*i
> fsolve(erf(x)=i, x=1+3i)      → 1.52285714842+2.04084734052*i
> fsolve(erf(x)=i, x=2+4i)      → 2.87881932565+3.2674470432*i
> fsolve(erf(x)=i, x=3+5i)      → 3.79852906513+4.13183559565*i

erf(z) = 2 --> erf(conj(z)) = conj(2) = 2-0i      → erfinv(2+0i) ≠ erfinv(2-0i)
erf(z) = i --> erf(-conj(z)) = -conj(i) = -0+i      → erfinv(0+i) ≠ erfinv(-0+i)

If we don't care about ±0, we doubled solutions for both erfinv(2) and erfinv(i)

[Gil]

"I think erf inverse function is defined only for real x within ±1, otherwise we get infinite solutions."

Not quite so as I understood from Wikipedia, in article Error Function, Properties for Abs (z=a+ib) < 1 :

The inverse error function is usually defined with domain (−1,1), and it is restricted to this domain in many computer algebra systems. However, it can be extended to the disk |z| < 1 of the complex plane, using the Maclaurin series[9]

Then
Inv
ERF

(
(.8,-.2)
)
—>(.801172930851,-.341012371796)

And ERF

(
(.801172930851,-.341012371796)
)
—> (.799999999998,-.200000000002)

Or
Inv
ERF

(
(0.,.85)
) —>
(0.,.649264402037

And ERF

(
(0.,.649264402037)
)
—> (0.,.849999999995)

It's what I developed for the HP50G in the directory program ERROR.9f for the error functions.

[Albert Chan]

(12-15-2023 01:06 PM)Gil Wrote:  Not quite so as I understood from Wikipedia, in article Error Function, Properties for Abs (z=a+ib) < 1 :

The inverse error function is usually defined with domain (−1,1), and it is restricted to this domain in many computer algebra systems. However, it can be extended to the disk |z| < 1 of the complex plane, using the Maclaurin series[9]

Then
Inv
ERF

(
(.8,-.2)
)
—>(.801172930851,-.341012371796)

Yes, we can define Maclaurin series solution as the "official" solution.
But, this is *not* properties of |z|<1. We still have infinite solutions.

> fsolve(erf(x)=0.8-0.2i, x=0.8-0.2i)  → 0.801172930852-0.341012371793*i
> fsolve(erf(x)=0.8-0.2i, x=1+3i)      → 1.53047592451+1.56695328912*i
> fsolve(erf(x)=0.8-0.2i, x=2+4i)      → 2.87803425787+3.00313505077*i
> fsolve(erf(x)=0.8-0.2i, x=3+5i)      → 3.7982932108+3.92873494629*i
> erf(Ans)                                        → 0.8-0.2i

[Gil]

Yes, sorry, you are right again, Albert.

For the time being I will leave the special solution sol_z given with the Taylor's development, with apparently abs (sol_z) < 1.

[Gil]

By the way, your solver seems to accept a complex unknown variable.

Is it the solver of the HP PRIME?

I use ONLY the HP50G EMU48.

[Albert Chan]

Examples were run on HP Prime emulator, 2.1.14181 (2018 10 16)
We don't really need solver for this. Plain newton's method work well.

> a := 0.8-0.2i
> Ans - (f:=erf(Ans)-a) / (2/sqrt(pi)*exp(-Ans^2))

0.817846322321-0.344502692613*i
0.800996592256-0.340824015653*i
0.801172956998-0.341012320073*i
0.801172930852-0.341012371793*i

Or, speed up with Halley's method

> a
> Ans - (f:=erf(Ans)-a) / (2/sqrt(pi)*exp(-Ans^2)+Ans*f)

0.800849610571-0.342519806279*i
0.801172932494-0.34101237069*i
0.801172930852-0.341012371793*i

[Gil]

In view of the simplicity of the above regarding the error function, I will of course change my inverse subroutine for INVERSE_ERF in and use Albert Chan's suggestion with Newton's method.