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?
On Mon, Dec 30, 2013 at 9:36 AM, Ivan Andrus <darthand...@gmail.com> wrote:
> 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
>
> On Dec 30, 2013, at 10:08 AM, Buck Golemon <buck.2...@gmail.com> wrote:
>
> 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))]
>
>
>
> On Sun, Dec 29, 2013 at 9:32 AM, JamesHDavenport <j.h.davenp...@bath.ac.uk
> > wrote:
>
>> 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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