jma@beach.cis.ufl.edu (John 'Vlad' Adams) (09/27/90)
Ok, forgive the novice question, but we all started with our first HP at one point... Is there a way to evaluate an integral sans limits on the 48SX? -- John M. Adams --**-- Professional Student on the six-year plan! /// Internet: jma@beach.cis.ufl.edu -or- vladimir@maple.circa.ufl.edu /// "We'll always be together, together in electric dreams" Tangerine Dream \\V// Sysop of The Beachside. FIDOnet 1:3612/557. 904-492-2305 (Florida) \X/
b3300876@rick.cs.ubc.ca (george kai yee chow) (09/28/90)
In article <24605@uflorida.cis.ufl.EDU> jma@beach.cis.ufl.edu (John 'Vlad' Adams) writes: >Ok, forgive the novice question, but we all started with our >first HP at one point... Is there a way to evaluate >an integral sans limits on the 48SX? Nope. And a browse thru the manual seems to support that. I was sorta disappoint since it really would be too much hassle. The best I've manage to do is to toss in symbolic bounds and let the 48 do the evaluate at. >-- >John M. Adams --**-- Professional Student on the six-year plan! /// >Internet: jma@beach.cis.ufl.edu -or- vladimir@maple.circa.ufl.edu /// >"We'll always be together, together in electric dreams" Tangerine Dream \\V// >Sysop of The Beachside. FIDOnet 1:3612/557. 904-492-2305 (Florida) \X/
davisp@skybridge.SCL.CWRU.Edu (Palmer Davis) (09/28/90)
In article <24605@uflorida.cis.ufl.EDU> jma@beach.cis.ufl.edu (John 'Vlad' Adams) writes: >Ok, forgive the novice question, but we all started with our >first HP at one point... Is there a way to evaluate >an integral sans limits on the 48SX? Sure... just use zero and the variable you're integrating with respect to as your "limits." -- PTD -- -- Palmer T. Davis | davisp@scl.cwru.edu -OR- ptd2@po.cwru.edu Case Western Reserve University | {att,sun,decvax,uunet}!cwjcc!skybridge!davisp ------------------------------------------------------+------------------------ Wake up and smell the cat food in your bank account. | Life is short.
dj1l+@andrew.cmu.edu (Demian A. Johnston) (09/28/90)
Yes; Just give the integral symbolic limits. For example if you are integrating with respect to "X" do sometyhing like giving the integral limits from 0 to "x" it's somewhere in the users manual. Demian J.
edp@jareth.enet.dec.com (Eric Postpischil (Always mount a scratch monkey.)) (09/28/90)
In article <24605@uflorida.cis.ufl.EDU>, jma@beach.cis.ufl.edu (John 'Vlad' Adams) writes: >Ok, forgive the novice question, but we all started with our >first HP at one point... Is there a way to evaluate >an integral sans limits on the 48SX? Not directly, but you can evaluate an integral with symbolic limits. So you can evaluate with the upper limit symbolic and the lower limit a constant. You could then either live with the constant expression in the result or remove it with some manipulation. -- edp
louxj@jacobs.CS.ORST.EDU (John W. Loux) (09/29/90)
>Ok, forgive the novice question, but we all started with our >first HP at one point... Is there a way to evaluate >an integral sans limits on the 48SX? I have no problem with a novice question, it's the novice answers that bother me, especially since the answer has been posted once before. First off, simply supplying 0 for one of the limits in definite integration does not work because all functions of zero do not return zero --- and some will burp and die on you. Second off, yes it is possible to do indefinite integration using the 48, but it's not particularly direct or documented. Consider: S X dX --- i.e., the integral of X with respect to X. First, create the expression using arbitrary, symbollic limits: S(A,B,X,X) EVAL'uate this to get the first integration step: 'X^2/2|(X=B)-(X^2/2|(X-A))' Split this expression using OBJ-> and drop the bottom three levels: 'X^2/2|(X=B)' You can now either evaluate this expression to replace X with B or use OBJ-> again to split the expression and drop the bottom three stack levels again. The result is either 'B^2/2' or 'X^2/2'. You can substitute X for B in the first expression in one of several ways (viz., 'X' 'B' STO EVAL) and add 'C' if you like ('C' +). This is the general procedure for expressions that the 48 can integrate symbollically. While we're at it, I'm still waiting for someone to respond to the posting on how one goes about symbollically integrating e^((y^2)/2) symbollically. The numeric solution is straight-forward: simply supply numeric limits, build the expression and use ->NUM to get the numeric result. I hope I've helped some, John
rrd@hpfinote.HP.COM (Ray Depew x2419) (09/29/90)
> Ok, forgive the novice question, but we all started with our > first HP at one point... Is there a way to evaluate > an integral sans limits on the 48SX? > John M. Adams --**-- Professional Student on the six-year plan! Two solutions: 1) Fake it, by setting the limits at -MAXR and MAXR. MAXR (MaxReal) is the largest number the 48 will handle. If you're curious to see what MAXR is, press MAXR ->NUM and you will see 9.999999999999E499 on Level 1. 2) Wlodek M-J, in his massive work on the HP-41, had a beautiful workaround for indefinite integrals. I gave the book to someone else (darn it!), so I don't remember what it was. But it was an elegant mathematical manipulation, and I used it several times on my 41. Perhaps someone else with the book would be good enough to post it here for our benefit? Regards Ray Depew HP Colorado IC Division rrd@hpfitst1.hp.com
madler@piglet.caltech.edu (Mark Adler) (09/29/90)
I responded to the e^(y^2/2) integration question privately, but it seems others are curious ... First off, what he really wanted was e^(-y^2/2), which is rather different for real y, and also quite useful, being the gaussian distribution (up to normalization). There is no symbolic answer that can be expressed as a rational function of trig and/or log/exp functions (except for special values of the integration bounds like -infinity, 0, and infinity). What is done when you have a useful, unsolvable integral is "solve it by naming it". The integral is given a name (the error function or complementary error function, erf(x) or erfc()), and then you have tables and subroutines to evaluate it. In fact, the HP has a function that'll do it: UTPN (upper tail probability normal) in MATH/PROB. Given the arguments 0 1 x, UTPN will return the integral from x to infinity of e^(-t^2/2)/sqrt(2pi) dt. The sqrt(2pi) is to make the result one for x equal to -infinity. The complementary error function (erfc()) is a little different, but a simple transformation of UTPN will give it. Naturally, UTPN is much faster (and probably more accurate) than doing the integral numerically on the HP-48. Mark Adler madler@piglet.caltech.edu
flinton@eagle.wesleyan.edu (10/04/90)
In article <19080015@hpfinote.HP.COM>, rrd@hpfinote.HP.COM (Ray Depew x2419) answers the question of John M. Adams: >> Ok, forgive the novice question, but we all started with our >> first HP at one point... Is there a way to evaluate >> an integral sans limits on the 48SX? A recent book gives another answer (of sorts). The book is "Calculus Activities" by Dennis Pence (PWS - Kent Publishing Co., Boston, 1990). In section 4 of Chapter 9 (pp. 172-175) you find the author's approach to plotting an indefinite integral over a given range. Sorry, no closed forms available -- just plots of acceptable approximations. Better than nothing, though, I hope. -- Fred
ags@seaman.cc.purdue.edu (Dave Seaman) (10/05/90)
In article <1990Oct4.005626.33899@eagle.wesleyan.edu> flinton@eagle.wesleyan.edu writes: >In section 4 of Chapter 9 (pp. 172-175) you find the author's approach >to plotting an indefinite integral over a given range. If you mean an "indefinite integral" in the sense that most people have been using the term in this newsgroup (i.e. an integral with no specified limits, yielding an answer that includes a constant of integration), then the plot is exceedingly simple. It is all black. In some quarters, the term "indefinite integral" means an integral of the form F(x) = (integral from a to x) f(t) dt where a is a constant. For some reason, I get the feeling that that is not what people have been asking for when they say they want to do indefinite integrals on the HP48. -- Dave Seaman ags@seaman.cc.purdue.edu