From: SourceForge.net <no...@so...> - 2004-04-26 12:09:29
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Bugs item #942261, was opened at 2004-04-26 05:09 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: zeta(-1) returns 1/12 instead of -1/12 Initial Comment: The Riemann zeta function at -1 takes the value -1/12 (see (69) in http://mathworld.wolfram.com/RiemannZetaFunction.html, or use Cauchy's residue theorem on the representation of zeta(z) as an integral around the Hankel contour). However, in maxima, the call zeta(-1) returns 1/12. Thanks, Jamie Walker ja...@sa... ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 |
From: SourceForge.net <no...@so...> - 2004-04-26 17:48:50
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Bugs item #942261, was opened at 2004-04-26 07:09 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: zeta(-1) returns 1/12 instead of -1/12 Initial Comment: The Riemann zeta function at -1 takes the value -1/12 (see (69) in http://mathworld.wolfram.com/RiemannZetaFunction.html, or use Cauchy's residue theorem on the representation of zeta(z) as an integral around the Hankel contour). However, in maxima, the call zeta(-1) returns 1/12. Thanks, Jamie Walker ja...@sa... ---------------------------------------------------------------------- >Comment By: Barton Willis (willisbl) Date: 2004-04-26 12:48 Message: Logged In: YES user_id=895922 Attached is a fix for zeta. The code could be improved in several ways. For one, zeta should simplify instead of evaluate. Also zeta should work for floating point arguments. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 |
From: SourceForge.net <no...@so...> - 2004-04-27 13:19:15
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Bugs item #942261, was opened at 2004-04-26 07:09 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: zeta(-1) returns 1/12 instead of -1/12 Initial Comment: The Riemann zeta function at -1 takes the value -1/12 (see (69) in http://mathworld.wolfram.com/RiemannZetaFunction.html, or use Cauchy's residue theorem on the representation of zeta(z) as an integral around the Hankel contour). However, in maxima, the call zeta(-1) returns 1/12. Thanks, Jamie Walker ja...@sa... ---------------------------------------------------------------------- >Comment By: Barton Willis (willisbl) Date: 2004-04-27 08:19 Message: Logged In: YES user_id=895922 I've attached tests for the zeta function. Barton ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2004-04-26 12:48 Message: Logged In: YES user_id=895922 Attached is a fix for zeta. The code could be improved in several ways. For one, zeta should simplify instead of evaluate. Also zeta should work for floating point arguments. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 |
From: SourceForge.net <no...@so...> - 2004-12-31 19:37:55
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Bugs item #942261, was opened at 2004-04-26 06:09 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 >Category: Lisp Core Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: zeta(-1) returns 1/12 instead of -1/12 Initial Comment: The Riemann zeta function at -1 takes the value -1/12 (see (69) in http://mathworld.wolfram.com/RiemannZetaFunction.html, or use Cauchy's residue theorem on the representation of zeta(z) as an integral around the Hankel contour). However, in maxima, the call zeta(-1) returns 1/12. Thanks, Jamie Walker ja...@sa... ---------------------------------------------------------------------- >Comment By: Robert Dodier (robert_dodier) Date: 2004-12-31 12:37 Message: Logged In: YES user_id=501686 Footnote -- There are also two implementations of the zeta fcn, bzeta and bfzeta, in share/numeric/bffac.mac. These return bfloat (big float) values. bzeta(-1,10) and bfzeta(-1,10) both return - 8.333333333333333B-2 which is -1/12 as a bfloat. The description of bzeta in doc/info/Number.texi states that bzeta is obsolete and bfzeta should be used instead. ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2004-04-27 07:19 Message: Logged In: YES user_id=895922 I've attached tests for the zeta function. Barton ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2004-04-26 11:48 Message: Logged In: YES user_id=895922 Attached is a fix for zeta. The code could be improved in several ways. For one, zeta should simplify instead of evaluate. Also zeta should work for floating point arguments. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=942261&group_id=4933 |