about inverse_erf

[Buck Golemon]

1) Sage seems unable to reduce `erf(x) == erf(y)` to `x == y`. How can I help this along?

solve(erf(x) == erf(y), x)[0].simplify_full()

Actual output: x == inverse_erf(erf(y))
Expected output: x == y

I had expected that sage would trivially reduce `inverse_erf(erf(y))` to `y`.

2) This output references 'inverse_erf', which doesn't seem to be importable t from anywhere in sage. Am I correct?

---

My concrete problem is re-deriving the formula for the normal-distribution cdf. I get a good solution from sage, but fail in showing that it's equivalent to a known solution because:

var('x sigma mu')
assume(sigma > 0)
eq3 = (-erf((sqrt(2)*mu - sqrt(2)*x)/(2*sigma)) == -erf((sqrt(2)*(mu - x))/(2*sigma)))
bool(eq3)

Actual output: False
Expected output: True

However this quite similar formula works fine:

eq3 = (-erf(sqrt(2)*mu - sqrt(2)*x) == -erf(sqrt(2)*(mu - x)))
bool(eq3)

Output: True

---
Include:
Platform (CPU) -- x86_64
Operating System -- Ubuntu 13.10
Exact version of Sage (command: "version()") -- 'Sage Version 5.13, Release Date: 2013-12-15'

[JamesHDavenport]

erf, as a function C->C, is not 1:1 (see 7.13(i) of DLMF), so this "simplification" would be incorrect.
I do not know how to tell Sage that you want real-valued functions/variables, when of course it would be correct to do the simplification.

[JamesHDavenport]

Furthermore, DLMF 7.17 only defines the inverse error function on the real line (in fact (-1,1))
I do not recall ever seeing a discussion of the complex inverse error function. Strecok (1968) shows that it satisfies y''=2yy'y',
but this is nonlinear, so the methodology of the paper below doesn't help.

Chyzak,F., Davenport,J.H., Koutschan,C. & Salvy,B.,
On Kahan's Rules for Determining Branch Cuts.
Proc. SYNSAC 2012 (ed. D. Wang et al.), IEEE Computer Society Press, Los Alamitos, CA, pp. 47-51

On Friday, 27 December 2013 22:40:40 UTC, Buck Golemon wrote:

[Buck Golemon]

Thanks. 

If I understand you, the problems lie in the complex domain, where I was only thinking of the real numbers.

Can I not do something to the effect of assume(x, 'real') ?

[Buck Golemon]

Yes, I can, but it doesn't have the intended (or any) effect:

sage: assume(x, 'real')

sage: assume(y, 'real')

sage: assumptions()

[x is real, y is real]

sage: solve(erf(x) == erf(y), x)

[x == inverse_erf(erf(y))]

[Buck Golemon]

I've found here:

http://mathworld.wolfram.com/InverseErf.html

(2)

with the identity holding for 

Is this a bit of information that can be added (by me?) to sage?

[JamesHDavenport]

In fact you don't really need MathWorld: erf is continuous monotone R->(-1,1), so must have an inverse function (-1,1)->R.
How you tell Sage this needs a Sage expert.

[Buck Golemon]

So I've succeeded in telling maxima how to simplify this, but it doesn't translate through to sage:

sage: print maxima.eval('''                                                               

declare([x, y], real)                                                                     

solve(erf(x) = erf(y), x)                                                                 

''')

done

[x=inverse_erf(erf(y))]

sage: print maxima.eval('''                                                     

matchdeclare (xx, lambda ([e], featurep (e, real)));

tellsimpafter (inverse_erf (erf (xx)), xx);

matchdeclare (yy, lambda ([e], -1 < e and e < 1));

tellsimpafter (erf (inverse_erf (yy)), yy);

''')

done

[inverse_erfrule1,?simp\-inverse\-erf]

done

[erfrule1,?simp\-erf]

sage: print maxima.eval('''

solve(erf(x) = erf(y), x)

''')

[x=y]

sage: var('x y')

(x, y)

sage: assume(x, y, 'real')

sage: assumptions()

[x is real, y is real]

sage: solve(erf(x) == erf(y), x)

[x == inverse_erf(erf(y))]

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[Ivan Andrus]

I think this might be because "the maxima used by the calculus package is different than the one in the interactive interpreter."  That's from sage.calculus.calculus?

Try the following:

sage: from sage.calculus.calculus import maxima as sm

sage: sm.eval('''matchdeclare (xx, lambda ([e], featurep (e, real)));

tellsimpafter (inverse_erf (erf (xx)), xx);

matchdeclare (yy, lambda ([e], -1 < e and e < 1));

tellsimpafter (erf (inverse_erf (yy)), yy);''');

sage: solve(erf(x) == erf(y), x)

[x == inverse_erf(erf(y))]

sage: assume(x, y, 'real')

sage: solve(erf(x) == erf(y), x)

[x == y]

-Ivan

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[Buck Golemon]

Yes, thanks!

That helped.

Sage will now reduce those erf functions, sometimes.

This equation will solve x as `x == r1`:

forget()

var('x mu sigma')

assume(sigma>0)

assume(x, mu, sigma, 'real')

from sage.calculus.calculus import maxima

maxima.eval('''matchdeclare(xx, lambda ([e], featurep (e, real)));\ntellsimpafter (inverse_erf (erf (xx)), xx);\nmatchdeclare (yy, lambda ([e], -1 < e and e < 1));\ntellsimpafter (erf (inverse_erf (yy)), yy);''')

demo = (-erf( sqrt(2)*(mu - x)) == -erf( sqrt(2)*mu - sqrt(2)*x))

show(demo)

show(solve(demo, x))

show(solve(solve(demo, x), x))

show(bool(demo))

1) Is it bad that I need to call solve() twice here?

2) What is r1?

This quite similar equation however refuses to solve and has boolean value False:

eqn = (1/2*erf(-(sqrt(2)*mu - sqrt(2)*x)/(2*sigma)) + 1/2 == 1/2*erf(-sqrt(2)*(mu - x)/(2*sigma)) + 1/2)

show(eqn)

show(solve(eqn, x))

show(solve(solve(eqn, x), x))

show(bool(eqn))

Actual output: False

Expected output: True

3) What prevents sage from knowing these arguments are real and using the above user-defined simpification?

[Ivan Andrus]

I'm afraid I'm in over my head here.  Though I do know that bool(eqn) doesn't necessarily mean what you think it means.

-Ivan