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From: Barton W. <wi...@un...> - 2025-05-18 21:25:29
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Hi Francisco, Thanks for the bug report. I made a ticket for this bug; see https://sourceforge.net/p/maxima/bugs/4559/ --Barton ________________________________ From: Francisco Carbajo <fra...@up...> Sent: Friday, May 16, 2025 11:34 AM To: max...@li... <max...@li...> Subject: [Maxima-discuss] Errors in atan and atan2 limits Caution: Non-NU Email Hi all. Some simple limits of the functions atan and atan2 are not calculated correctly (checked in Maxima 5.47.0 https://urldefense.com/v3/__https://maxima.sourceforge.io__;!!PvXuogZ4sRB2p-tU!H3rBpqWu89AdVKCnh3h0gMCL0X34bkdOtxOA9h4YkzyHvdUd4YEuRH7qezexOZfLpRDWnTpXDz_Ar_5qf2KAZyyU1A$ using Lisp SBCL 2.2.9.debian). For example (%i1) limit(atan(p/e),e,0,plus); %pi (%o1) --- 2 (%i2) limit(atan2(p,e),e,0,plus); %pi (%o2) --- 2 ignoring the sign of p. If p is declared as a negative number the answer is correct (%i5) assume(p<0); (%o5) [p < 0] (%i6) limit(atan(p/e),e,0,plus); %pi (%o6) - --- 2 (%i7) limit(atan2(p,e),e,0,plus); %pi (%o7) - --- 2 (%i8) To avoid this error just change p to a sum of two variables p+q (%i2) limit(atan((p+q)/e),e,0,plus); (%o2) atan2(q + p, 0) (%i3) limit(atan2((p+q),e),e,0,plus); (%o3) atan2(q + p, 0) (%i4) which is correct, shouldn't the answer be limit(atan(p/e),e,0,plus)=atan2(p,0)? Thank you very much _______________________________________________ Maxima-discuss mailing list Max...@li... https://urldefense.com/v3/__https://lists.sourceforge.net/lists/listinfo/maxima-discuss__;!!PvXuogZ4sRB2p-tU!H3rBpqWu89AdVKCnh3h0gMCL0X34bkdOtxOA9h4YkzyHvdUd4YEuRH7qezexOZfLpRDWnTpXDz_Ar_5qf2LPoUIHuw$ |