MaplePrimes - answers and comments on Question, Weirdness with the inverse Erf function

Statistics

acer — Wed, 10 Mar 2010 05:08:26 Z

Maybe you could use the Statistics package, with high working precision.

> restart:with(Statistics):

> X:=RandomVariable('Normal'(0,1)):

> evalf[50](2*CDF(X,1)-1);
             0.68268949213708589717046509126407584495582593345319
 
> evalf[50](Quantile(X,(%+1)/2));
             0.99999999999999999999999999999999999999999999999997

> evalf[50](Quantile(X,CDF(X,1)));
             0.99999999999999999999999999999999999999999999999996

Did I do that right?

acer

there is some trap

Axel Vogt — Wed, 10 Mar 2010 15:34:00 Z

There is some numerical trap for such extreme tasks

Note that your formula for the cdfN is only correct at the left wing

Then (as implicitely said by acer) you need high(er) precision

And finally - since the curve is extremely flat - you may wish to use all possible
strength of fsolve (look up the parameters in the help)

And provide a rough guess for a solution

I do not have Maple at hand now, but take asymptotics, convert to polynom and
solve to have a start value.

Or you can try to solve

 ln( erfc(x/sqrt(2)) ) = ln( y )

Edited to give the example:

> Digits:=18;
> 
> erfc(x/sqrt(2)); asympt(%,x, 2): convert(%, polynom);
> y=simplify(%) assuming 0 < x;
> solve(%, x);
> simplify(%) assuming 0 < y; 
> eval(%, y= 1e-18); 
> x0:=evalf(%);
> 
> fsolve(erfc(x/sqrt(2))= 1e-18, x=x0);

            8.83510978817539579

> eval(erfc(x/sqrt(2)), x=%);
> evalf(%);

         .999999999999999835e-18

> erfc(x/sqrt(2))= 1e-20; map(ln, %); fsolve(%, x=0 .. 20);

           9.33604484923406004

> eval(erfc(x/sqrt(2)), x=%);
> evalf(%);

         .999999999999999887e-20

inverse erf

Alex Smith — Wed, 10 Mar 2010 20:30:00 Z

This seems to work:

Digits:=200;

inv_erf := x->fsolve(erf(y/sqrt(2))=x,y,fulldigits);

inv_erf(1-10^(-20));

9.336...

If you take the square root of 2 out of the above defintion of inv_erf, the results are consistent with what WolframAlpa gives with InverseErf.

Need more digits, own solver, numerical analysis

outrider — Fri, 10 Aug 2018 05:54:04 Z

The actual value of inv_erf(1-10^-20)=6.60158062235514256151639163241871..., I believe. I used my own solver, a cubic version of Newton iteration called a Schroeder series, but with an extra step that helps convergence go faster. I did 9 iteration steps. Actually, I used Digits:=5000. This tok less than 5 sec on my high-frequency Apple MacBook Air

However, it took MAPLE a time of .42 sec to evaluate erf(6.60158062235), comiing very close to the desired value

* * * erf(x)=1-10^-20 * * *

The actual value found was

* * * erf(x)=0.999999999999999999990000000000648... (from 5,000 dits) * * *

and the last digit rounded down to '5' was very close. But even then, with Digits:=5000,.MAPLE was not able to calculate erf(x) correctely for the longer value of x above stated to 32 decimals. It did seem to help when I used fewer decismals, like the 11-decimal argument mentioned.

Much easier way

Carl Love — Fri, 10 Aug 2018 09:41:36 Z

Representing (1 - 1e-18) requires 18 significant digits, whereas representing 1e-18 requires only 1 significant digit. So I find the following way much easier than all the others presented in this thread, requiring no increase in Digits from its default. I apply the following obvious identity:

Quantile(Normal(mu,sigma), 1-x) = mu - sigma*Quantile(Normal(0,1), x).

Thus

Quantile(Normal(0,1), 1 - 1e-18) = -Statistics:-Quantile(Normal(0,1), 1e-18);
8.75729034878321

fsolve fails

vv — Sat, 11 Aug 2018 06:37:58 Z

It's a pity that fsolve fails for the equation erf(x) = 1 - 10^(-20) when Digits = 5000.
It works when Digits = 4000.
The equation can be solved via

Digits:=5000:
x1:=RootFinding:-Analytic(erf(x) = 1 - 10^(-20), 6-I/10 .. 7+I/10);

or even directly with the good old Newton's method:

restart;
y1:=1-10^(-20):
Digits:=5100:
a:=6.;h:=1. : N:=0:
K:=evalf(sqrt(Pi)/2):
while  abs(h)>1e-5000 do
h:=(erf(a)-y1)*exp(a^2)*K;
a:=a-h:  N:=N+1;
od:  N; a;