[ quantlib-Patches-3047358 ] fixed gaussian orthogonal polynomial

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Patches item #3047358, was opened at 2010-08-17 19:25
Message generated for change (Comment added) made by renorm
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Category: None
Group: None
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: renorm (renorm)
Assigned to: Nobody/Anonymous (nobody)
Summary: fixed gaussian orthogonal polynomial

Initial Comment:
double with zero comparison  issue

Only fixed function are included

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>Comment By: renorm (renorm)
Date: 2010-09-14 15:44

Message:
Looks like "double with zero comparison" was meant for something. Let's
leave as it for now.

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Comment By: Luigi Ballabio (lballabio)
Date: 2010-09-01 10:07

Message:
With your patch, two cases in the test suite fail.  May you investigate?

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Comment By: renorm (renorm)
Date: 2010-08-18 09:08

Message:
the whole file with fixes

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[Bojan Nikolic]

This probably doesn't work because the logic was reversed here:

,----
|         if (std::fabs(denom)<QL_EPSILON) {
|             if (std::fabs(num)<QL_EPSILON) {
|                 QL_FAIL("can't compute b_k for jacobi integration\n");
`----

I think it should be something like

,----
|         if (std::fabs(denom)<QL_EPSILON) {
|             if (std::fabs(num)>QL_EPSILON) {
|                 QL_FAIL("can't compute b_k for jacobi integration\n");
`----

This test exists because l'Hopital rule doesn't work if the nominator is
finite, so an error is signalled if denom is close to zero while num is
_not_ close to zero.

Best,
Bojan

--
Bojan Nikolic          ||          http://www.bnikolic.co.uk/ql/

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[Kakhkhor Abdijalilov]

denom and num aren't continues. Looks like "if (!denom)" was meant to
deal with the case when denom is exactly zero because of particular
choice of i, alpha and beta. In that case num must be exactly zero
too.

For example, if i=0, alpha=0 and beta=0, then denom is bitwise 0. In
that case num must be bitwise zero too. It will work if the test "if
(!denom)" is meant to be true only for a particular values of i, alpha
and beta.

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