Crash when Solve for Integral with 15C

09042017, 02:37 AM
(This post was last modified: 09052017 08:19 PM by Gerson W. Barbosa.)
Post: #61




RE: Crash when Solve for Integral with 15C
(09032017 06:47 PM)Gerson W. Barbosa Wrote: Only the first 22 digits are correct. 600+ terms would be necessary for 100 digits, but a clean 50g will run out of memory after around 380 terms. EXPANDing inside the loop is slower at first, but becomes faster as the number of terms increases, not to mention a lesser memory requirement. Still, not a recommended method for the task (727 seconds on the emulator). 600 « 0 DUP ROT FOR n n 3 * 1 + DUP n ! * 6 ROT ^ SWAP / + EXPAND NEXT →STR "/" " " SREPL DROP "'" " " SREPL DROP OBJ→ SWAP 12 ALOG * SWAP IQUOT →STR DUPDUP SIZE DUP 11  SWAP SUB "." OVER + SREPL DROP » EVAL => 59639380918976775196293204202056658471019738405733913 533679285187391995176580099542319971838.280912881192 PS: A stackonly version hasn't helped, at least on the emulator, mostly due to my clumsy stack handling: 6 600 « DUP2 3 * 1 + DUP UNROT OVER 3 ^ UNROT ^ 4 PICK ! 1 6 PICK START DUP2 6 PICK * / 7 ROLL + EXPAND 6 ROLLD 5 PICK / UNROT OVER / ROT 5 ROLL 1  5 ROLLD 4 ROLL 3  4 ROLLD NEXT 5 DROPN →STR "/" " " SREPL DROP "'" " " SREPL DROP OBJ→ SWAP 12 ALOG * SWAP IQUOT →STR DUPDUP SIZE DUP 11  SWAP SUB "." OVER + SREPL DROP » EVAL => 59639380918976775196293204202056658471019738405733913 533679285187391995176580099542319971838.280912881192 (1293 seconds) PPS: It's better to sum the terms in ascending order, otherwise the timeconsuming expansions of long expressions are done since the beginning. Execution time down to 769 seconds despite more command inside the loop. Stack handling optimization might make it faster than the first version. 6 601 « OVER 3 ^ ROT 1 0 1 1 6 ROLLD 6 ROLLD 6 ROLLD DUP 8 ROLL START DUP2 6 PICK * / 7 ROLL + EXPAND 6 ROLLD 5 PICK * UNROT OVER * ROT 5 ROLL 1 + 5 ROLLD 4 ROLL 3 + 4 ROLLD NEXT 5 DROPN →STR "/" " " SREPL DROP "'" " " SREPL DROP OBJ→ SWAP 12 ALOG * SWAP IQUOT →STR DUPDUP SIZE DUP 11  SWAP SUB "." OVER + SREPL DROP » EVAL => 59639380918976775196293204202056658471019738405733913 533679285187391995176580099542319971838.280912881192 The arguments are Y: upper limit of integration X: number of terms (n1, in the previous version) PPPS: A version using local variables isn't any faster: 6 601 « 0 UNROT OVER 3 ^ ROT 1 DUPDUP → k b d f n « 1 SWAP START b d f * / + EXPAND k 'b' STO* 3 'd' STO+ n 'f' STO* 1 'n' STO+ NEXT » DUP TYPE 28  NOT SWAP →STR SWAP { "'" SWAP + "/1'" + } IFT "/" " " SREPL DROP "'" " " SREPL DROP OBJ→ SWAP 12 ALOG * SWAP IQUOT →STR "#" + DUPDUP SIZE DUP 12  SWAP SUB "." OVER + SREPL DROP "#" "" SREPL DROP » EVAL => 59639380918976775196293204202056658471019738405733913 533679285187391995176580099542319971838.280912881192 786 seconds on the emulator. 

09042017, 07:03 AM
Post: #62




RE: Crash when Solve for Integral with 15C
(09012017 11:51 AM)emece67 Wrote: \(5.963938091897677519629320420205\underline{665}×10^{91}\) in a wp34s after 2'20''. I used for this an wp34s that runs the new double exponential integration method. Can anybody with a wp34s still using the Romberg method repeat this calculation abd report the results? César  Information must flow. 

09042017, 09:30 AM
(This post was last modified: 09042017 09:35 AM by Pekis.)
Post: #63




RE: Crash when Solve for Integral with 15C
Hello,
After u=x^3 substitution, is this the correct form ? Integ(e^x^3 dx, 0, 6) = Integ(e^u/(3*u^(2/3)), 0, 216) u=x^3 => dx=du/(3*x^2) x=u^(1/3) => 3*x^2=3*u^(2/3) => dx=du/(3*u^(2/3)) Thanks 

09042017, 10:19 AM
Post: #64




RE: Crash when Solve for Integral with 15C
(09042017 09:30 AM)Pekis Wrote: Hello, Yes, it is. Elsewhere in this thread I'd wrongly written Integ(e^x^3 dx, 0, 6) = Integ(e^u/u^(2/3), 0, 216) neglecting the 1/3 constant in my own handwritten note: Integ(e^x^3 dx, 0, 6) = 1/3* Integ(e^u/u^(2/3), 0, 216) Sorry for the lack of attention and thanks for the correction. Gerson. 

09062017, 11:15 AM
Post: #65




RE: Crash when Solve for Integral with 15C
Found another problem to solve for integral this time HP 15C gave error message as error 0 have to do with the 'Complex Results from Real Numbers'
HP Prime gave answer as 3.64899739786 E 2 The answer should be 0.03662 The integral equation attached as jpg. Here is the program: LBL A ENTER SQR LSTx 1  + X<>Y LN 1/x  RTN P/R Gamo 

09062017, 12:10 PM
Post: #66




RE: Crash when Solve for Integral with 15C
(09062017 11:15 AM)Gamo Wrote: Found another problem to solve for integral this time HP 15C gave error message as error 0 have to do with the 'Complex Results from Real Numbers' Wolfram Alpha's result agrees with Prime: the answer is exactly 2ɣlog(4) which is ≈ 0.03648997397857652.... ɣ is the EulerMascheroni constant. X<> c Joe 

09062017, 12:27 PM
(This post was last modified: 09062017 12:53 PM by Dieter.)
Post: #67




RE: Crash when Solve for Integral with 15C
(09062017 11:15 AM)Gamo Wrote: Found another problem to solve for integral this time HP 15C gave error message as error 0 have to do with the 'Complex Results from Real Numbers' Where should a complex result appear here? Hmhmhm... you do not happen to have user mode on, have you? Check the keycode for the SQRT and 1/x commands. What do you see? 11 and 15 (correct) or 32 11 and 32 15? The latter is GSB A resp. GSB E due to user mode. (09062017 11:15 AM)Gamo Wrote: The answer should be 0.03662 And that's what the 15C returns. (09062017 11:15 AM)Gamo Wrote: Here is the program: Here is another error. This plus should be a divide. With a correct program you will get the correct result: FIX 5 0 [ENTER] 1 [∫yx] [A] => 0,03662 Edit: this is not the correct result, for more details see the posts below and the mentioned HPJournal article. Dieter 

09062017, 12:27 PM
Post: #68




RE: Crash when Solve for Integral with 15C
(09062017 11:15 AM)Gamo Wrote: Here is the program: The ENTER on line 2 is not necessary as the stack is filled with the x value. On line 7 it should be a division and not an addition. With the following program: Code: 00142.21.11 LBL A FIX 9 0 ENTER 1 f x A returns: 0,036489974 on the HP 15C Windows Emulator. 

09062017, 12:29 PM
Post: #69




RE: Crash when Solve for Integral with 15C
(09062017 11:15 AM)Gamo Wrote: Found another problem to solve for integral this time HP 15C gave error message as error 0 have to do with the 'Complex Results from Real Numbers' The '+' sign in your listing should be a '/'. More about this integral at http://www.hpl.hp.com/hpjournal/pdfs/Iss...98008.pdf starting at page 23. Gerson. 

09062017, 12:45 PM
(This post was last modified: 09062017 12:55 PM by Dieter.)
Post: #70




RE: Crash when Solve for Integral with 15C
(09062017 12:10 PM)Joe Horn Wrote:(09062017 11:15 AM)Gamo Wrote: ... FTR: The WP34s integration routine seems to perform better here: FIX 5: ∫ ≈ 0,03649 ALL 3: ∫ ≈ 0,03648997398363982 The latter is at least correct in all displayed digits. ;) (09062017 12:29 PM)Gerson W. Barbosa Wrote: More about this integral at ... Ah, now I know why this integral looked so familiar! It was mentioned in the HPJournal article on the thennew 34C and its integrate function (which seems to be more or less the same in the 15C). The indeed wrong answer in FIX 5 mode is explained there, just as the correct result 0,0364900 in FIX 7 mode: "The answers are wrong but the calculator is right". ;) Again, the error is caused by problems with the numeric evaluation of the integrand which has an very sharp spike near x=0, and it is shown how this can be avoided and the correct result can be obtained. Dieter 

09062017, 01:02 PM
Post: #71




RE: Crash when Solve for Integral with 15C
(09062017 12:29 PM)Gerson W. Barbosa Wrote: More about this integral at It's interesting that the program provided in Dr Kahan's article ignores the fact that the stack is filled with the value of the parameter when the function is called. The first two lines are: Code: LBL A Begin with u in the X register instead of: Code: LBL A Begin with u in the X, Y, Z and T registers 

09062017, 01:06 PM
Post: #72




RE: Crash when Solve for Integral with 15C
(09062017 01:02 PM)Didier Lachieze Wrote: It's interesting that the program provided in Dr Kahan's article ignores the fact that the stack is filled with the value of the parameter when the function is called. Right, the initial ENTER is not required as Solve and Integrate fill the stack with x. But the ENTER allows to calculate the function for different x (without having to manually fill the stack beforehand) in order to get an impression of its shape and the problems following from this (cf. fig. 10 a...c). Dieter 

09062017, 01:56 PM
(This post was last modified: 09062017 01:58 PM by Gamo.)
Post: #73




RE: Crash when Solve for Integral with 15C
Right that was from HP Journal magazine I try that formula and miss enter the last program from  and suppose to be divide. No wonder it gave out as Error 0
Gamo 

09062017, 02:02 PM
Post: #74




RE: Crash when Solve for Integral with 15C
(09062017 12:45 PM)Dieter Wrote:(09062017 12:10 PM)Joe Horn Wrote: Wolfram Alpha's result agrees with Prime: the answer is exactly 2ɣlog(4) which is ≈ 0.03648997397857652.... The last version of the wp34s integration routine returns: FIX 5: \(\int\approx0.03649\) in 4'' ALL 3: \(\int\approx0.03648997397857\underline{909}\) in 9'' DBLMODE: \(\int\approx0.036489973978576520559023667001244\underline{13}\) in 24'' César  Information must flow. 

09062017, 08:57 PM
Post: #75




RE: Crash when Solve for Integral with 15C  
09072017, 09:24 AM
Post: #76




RE: Crash when Solve for Integral with 15C
(09062017 08:57 PM)Dieter Wrote:(09062017 02:02 PM)emece67 Wrote: The last version of the wp34s integration routine returns: ... The new method is the doubleexponential one. It was uploaded to the svn repository some months ago. César  Information must flow. 

09082017, 02:57 PM
Post: #77




RE: Crash when Solve for Integral with 15C
ALL
I found the book L’Hôpital's Analyse des infiniments petits … … to be VERY interesting & EDIFYING with respect to the underlying foundation/basis of the Calculus. from the preface: "The Analyse consists of ten chapters, which L’Hôpital called “sections.” We consider it to have three parts. The first part, an introduction to the differential calculus, consists of the first four chapters: 1. In which we give the Rules of this calculus. 2. Use of the differential calculus for finding the Tangents of all kinds of curved lines. 3. Use of the differential calculus for finding the greatest and the least ordinates, to which are reduced questions De maximis & minimis. 4. Use of the differential calculus for finding inflection points and cusps. Taken together, these chapters provide a thorough introduction to the differential calculus in about 70 pages. The next five chapters are devoted to what can only be described as an advanced text on differential geometry, motivated in part by what were then cuttingedge research problems in optics and other fields. The final chapter is mildly polemical, demonstrating the superiority of Leibniz’ new calculus, when compared to the methods of René Descartes (1596–1660) and Johann van Waveren Hudde (1628–1704)." BEST! SlideRule 

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