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problems with symbolic integration and then numerical evaluating

asked 2010-08-24 12:08:05 -0500

maldun gravatar image

updated 2010-08-24 12:08:48 -0500

William Stein gravatar image

can anyone explain this:

sage: integrate(legendre_P(64,x)*sin((1+x)*pi/2),x,-1,1).n()
1.16508247725542e79

from approximation one know's that the legendre coefficients converge exponentially to zero and not to infinity!

and indeed with mpmath I get a better answer:

sage: import sage.libs.mpmath.all as mpmath
sage: mpmath.call(mpmath.quad,lambda x: mpmath.legendre(64,x)*mpmath.sin(pi/2*(x+1)),[-1,1])
-5.04684703543649e-25

Is there an overhead happening, when I numerically evaluate large rationals or something???

Thanks in advance, maldun

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answered 2010-08-24 12:15:55 -0500

William Stein gravatar image

If you look at the output of

integrate(legendre_P(64,x)*sin((1+x)*pi/2),x,-1,1)

you'll see that it is a massive expression with large rationals, powers of pi, etc. If expr is any Sage symbolic expression, then expr.n(bits) works by evaluating all the "leaves" of the expression to the given bits of precision, then doing the arithmetic to evaluate the expression. Thus large roundoff and cancellation can and sometimes will occur. Indeed, that is exactly what is happening here. You can get around this by increasing the precision sufficiently. If you want to be sure of the result, you can use interval arithmetic. Here's how:

sage: a = integrate(legendre_P(64,x)*sin((1+x)*pi/2),x,-1,1)
sage: a.n(prec=300)
51516.6651093...60
sage: a.n(prec=500)
0
sage: a.n(prec=2000)
3.3204630...841332374315143e-97
sage: RealIntervalField(53)( a )
0.?e81
sage: RealIntervalField(200)( a )
0.?e37
sage: RealIntervalField(500)( a )
0.?e-54
sage: RealIntervalField(1000)( a )
3.32046...71175?e-97

For example, the lines

sage: RealIntervalField(500)( a )
0.?e-54

tell you that if evaluate the expression using 500 bits of precision, it's definitely 0.000000... (at least 53 zeros). Internally, this is done by evaluating with 500 bits of precision and rounding down and also evaluating with 500 bits and rounding up, and tracking the resulting "interval".

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Comments

that's an interesting behavior. Thanks for your quick response! maldun ( 2010-08-24 12:23:47 -0500 )edit
This is a great example that should make it into some doc/FAQ for those new to computer math software in general... which I will someday write... kcrisman ( 2010-08-26 02:48:47 -0500 )edit

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Asked: 2010-08-24 12:08:05 -0500

Seen: 636 times

Last updated: Aug 24 '10