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. On Saturday, 28 December 2013 19:46:49 UTC, Buck Golemon wrote: > > I've found here: > http://mathworld.wolfram.com/InverseErf.html > > [image: erf^(-1)(erf(x))][image: =][image: x,] > (2) > > with the identity holding for [image: x in R] > > Is this a bit of information that can be added (by me?) to sage? > > > On Saturday, December 28, 2013 11:32:02 AM UTC-8, Buck Golemon wrote: >> >> 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))] >> >> >> On Saturday, December 28, 2013 11:27:09 AM UTC-8, Buck Golemon wrote: >>> >>> 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') ? >>> >>> On Saturday, December 28, 2013 10:07:41 AM UTC-8, JamesHDavenport wrote: >>>> >>>> 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. >>>> >>>> On Friday, 27 December 2013 22:40:40 UTC, Buck Golemon wrote: >>>>> >>>>> 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' >>>>> >>>>>
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