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bool returns false with arcsin(x) and 2*arctan(x/(1+sqrt(1-x^2)))

asked 14 years ago

Shu gravatar image

updated 11 years ago

tmonteil gravatar image

The two expression should be equal but when I write bool(arcsin(x) == 2*arctan(x/(1+sqrt(1-x^2)))) it returns false.

Any clue - Why it does that? Any get around?

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answered 14 years ago

DSM gravatar image

updated 14 years ago

I can answer the "why it does that" question, anyway: Sage follows Maxima's conventions on equality testing. That is, it only returns True if it's convinced the equality holds. "False" here translates as "False or unknown".

I actually couldn't find a good workaround-- I'm not sure what the right way to fake new rewrite rules is. I guess you could confirm that the lhs and rhs agree via something like this:

sage: eq = arcsin(x) == 2*arctan(x/(sqrt(-x*x+1)+1))
sage: bool(eq)
False
sage: # but..
sage: bool(eq.subs(x=0))
True
sage: bool(diff(eq, x))
True

And if you had several cases where the "agree at one point and derivatives equal" trick works, you could put it in a function, maybe with something like this:

Warning: untested code! Will probably break when given unexpected input!


def tryharder_bool(some_expr):
    # hack hack hack
    if some_expr: 
        return True
    if type(some_expr) == Expression and some_expr.is_relational():
        v = some_expr.variables()
        if len(v) == 1:
            if some_expr.subs({v[0]:0}) and diff(some_expr, v[0]):
                return True
    return False

giving

sage: bool(eq)
False
sage: tryharder_bool(eq)
True

UPDATE:

For some of the other cases mentioned in the comments, playing around with the trig_* functions and/or converting to exponential form works:

sage: eq1 = arcsin(x) == 2*arctan(x/(sqrt(-x*x+1)+1))
sage: eq2 = sin(x) == 2*sin(x/2)*cos(x/2) 
sage: eq3 = sin(2*x) == 2*sin(x)*cos(x)
sage: eq4 = tan(x/2) == sin(x)/(1+cos(x))
sage: eqs = eq1, eq2, eq3, eq4
sage: [bool(q) for q in eqs]
[False, False, True, False]
sage: [bool(q.trig_reduce()) for q in eqs]
[False, True, True, False]
sage: [bool(q.trig_expand()) for q in eqs]
[False, False, True, False]
sage: [bool(q.trig_expand(half_angles=True)) for q in eqs]
[False, False, True, True]
sage: 
sage: 
sage: eq2.maxima_methods().exponentialize()
1/2*I*e^(-I*x) - 1/2*I*e^(I*x) == 1/2*(I*e^(-1/2*I*x) - I*e^(1/2*I*x))*(e^(-1/2*I*x) + e^(1/2*I*x))
sage: eq1.maxima_methods().exponentialize()
arcsin(x) == 2*arctan(x/(sqrt(-x^2 + 1) + 1))
sage: [bool(q.maxima_methods().exponentialize()) for q in eqs]
[False, True, True, True]
sage: 
sage: def trig_xeq_check(eq):
....:         return bool(eq) or bool(eq.maxima_methods().exponentialize()) or (bool(eq.subs(x=0)) and bool(diff(eq,x)))
....: 
sage: 
sage: [trig_xeq_check(q) for q in eqs]
[True, True, True, True]
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Comments

thanks for your answer. I liked you workaround. Just one thing, to find the point at which it agree can I use y=find_root(eq, -1000, 1000) and use that y.

Shu gravatar imageShu ( 14 years ago )

If the two expressions are really equal then they should agree everywhere, so it doesn't matter what point you choose. I'd avoid looking for a root, though: (1) it's slower, (2) it's generally a good idea to stay symbolic (i.e. not floating-point) as long as possible, and (3) there might not even be a root to find..

DSM gravatar imageDSM ( 14 years ago )

What if the two functions are not defined at the point I choose. e.g. in the above equation if I use bool(eq.subs(x=2)) It will return "False"

Shu gravatar imageShu ( 14 years ago )

Then depending on the function, it could crash, give you "nan == nan" which is False, or maybe even return True (if the function is undefined but evaluating the expression gave Infinity on both sides, for example). I'm afraid you're going to have to consider the functions involved.

DSM gravatar imageDSM ( 14 years ago )

Do you have any idea to work around for bool( sin(x) == 2*sin(x/2)*cos(x/2) ) ? The differentiation equality check also fails in sage for this one.

Shu gravatar imageShu ( 14 years ago )

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Asked: 14 years ago

Seen: 1,947 times

Last updated: Nov 17 '10