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[MUG] solve

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[MUG] solve
Author: Ruth Plathe    Posted: 09/10/2000 13:06:01 GDT
>> From: "Ruth Plathe" I have 4 equations: > level0:=(-In*sigma02*N0)+(N2/tau2)+(N5/tau5)+(N6/tau6)+((1/2)*k*N2^2 > )=0; level2:=(In*sigma02*N0)-(N2/tau2)-(In*sigma25*N2)-k*N2^2=0; > level5:=(In*sigma25*N2)-N5/tau5-(In*sigma56*N5)+((1/2)*k*N2^2)=0; > level6:=(In*sigma56*N5)-(N6/tau6)=0; With the following constants > sigma02:=10*10^(-21); > sigma25:=2*10^(-21); > sigma56:=2*10^(-21); > tau2:=2.5*10^(-3); > tau5:=1*10^(-3); > tau6:=0.72*10^(-3); > k:=0.4*10^(-18); I wish to get expressions for N0,N2,N5,N6 in terms of In This appears to be impossible - I get all sorts of recursion errors or it just gives me N2=N2 which is pretty useless. Does anyone have any tips on how to solve this? Ruth Plathe Postgraduate Research Student Centre for Imaging and Applied Optics, Swinburne University Ph: 9214 5680 (office) 9214 5686 (lab) Fax: 9214 5840

[MUG] when does maple 'give up' during a solve?
Author: Tlovie    Posted: 12/10/2000 19:03:50 GDT
>> From: I have a maple worksheet that involves solving a system of non-linear polynomial equations. Maple attempts to find the closed form solution to these equations, but never arrives at a solution before using all of my 128MB of ram. My question is, since maple hasn't stopped the computation on it's own accord, is it actually making progress at finding a solution(s), or is it likely that no matter how much memory I make available, it will never stop the computation? The equations I'm trying to solve are here, in case any insights can be had from looking at them. S:=matrix([[1,-s1,-s2],[-s1,1,-s3],[-s2,-s3,1]]); C:=matrix([[c11,c12,c13],[c12,1,c23],[c13,c23,c33]]); R:=collect(linalg[det](evalm(lambda^2*matrix([[1,0,0],[0,1,0],[0,0,1]])+ lambda*C+S)),lambda); T:=expand((lambda+(linalg[det](S))^(1/6))^6); eq1:=coeff(R,lambda,1)=coeff(T,lambda,1); eq2:=coeff(R,lambda,2)=coeff(T,lambda,2); eq3:=coeff(R,lambda,3)=coeff(T,lambda,3); eq4:=coeff(R,lambda,4)=coeff(T,lambda,4); eq5:=coeff(R,lambda,5)=coeff(T,lambda,5); solve({eq1,eq2,eq3,eq4,eq5},{c11,c12,c13,c23,c33}); Any comments would be appreciated. Tom Lovie.

[MUG] Re: solve
Author: Maple Group    Posted: 13/10/2000 16:17:18 GDT
>> From: Maple Group | >> From: "Ruth Plathe" | | I have 4 equations: | | > level0:=(-In*sigma02*N0)+(N2/tau2)+(N5/tau5)+(N6/tau6)+((1/2)*k*N2^2 | > )=0; level2:=(In*sigma02*N0)-(N2/tau2)-(In*sigma25*N2)-k*N2^2=0; | > level5:=(In*sigma25*N2)-N5/tau5-(In*sigma56*N5)+((1/2)*k*N2^2)=0; | > level6:=(In*sigma56*N5)-(N6/tau6)=0; | | With the following constants | | > sigma02:=10*10^(-21); | > sigma25:=2*10^(-21); | > sigma56:=2*10^(-21); | > tau2:=2.5*10^(-3); | > tau5:=1*10^(-3); | > tau6:=0.72*10^(-3); | > k:=0.4*10^(-18); | | I wish to get expressions for N0,N2,N5,N6 in terms of In -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Gaston Gonnet Date: Tue, 10 Oct 2000 15:19:15 +0200 (MET DST) To: Subject: solve Your system is indeterminate in one of the N variables. If you solve it with exact constants (instead of floating point numbers) you get N2 (200000000000000000000000 + In + 200 N2) {N2 = N2, N0 = 1/5 -------------------------------------------, In In N2 (In + 100 N2) N6 = 9/6250000000000000000000000 -----------------------------, 500000000000000000000000 + In N2 (In + 100 N2) N5 = -----------------------------} 500000000000000000000000 + In which means that for any arbitrary value of N2 there is a solution. Gaston Gonnet. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Stanley J Houghton Date: Tue, 10 Oct 2000 14:49:59 +0100 To: Subject: solve Ruth Unfortunately, your four equations are only three! If you add the first three they equal the level 6 equation. Thus you can only solve for three variables and hence the N2=N2 result. Hope it helps Stan -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 10 Oct 2000 17:28:37 +0100 From: "Dr Francis J. Wright" To: Subject: solve It is impossible because your equations are not independent. The values of the constants are largely irrelevant, although the disparity in their values would make me very worried about numerical instability if I had to do a numerical solution. Here is some analysis using Maple 6. I set up the equations as above, but not the constants. Then > ii := 0,2,5,6: solve({level||ii},{N||ii}); N2 (1 + In sigma25 tau2 + k N2 tau2) {N0 = ------------------------------------, In sigma02 tau2 N2 tau5 (2 In sigma25 + k N2) N5 = 1/2 -----------------------------, 1 + In sigma56 tau5 In sigma56 N2 tau5 (2 In sigma25 + k N2) tau6 N6 = 1/2 ---------------------------------------------, 1 + In sigma56 tau5 N2 = N2} This implies that the equations are not independent, which is easily confirmed by hand: > level0 := level0+level6; # eliminate N6 level0 := N2 N5 2 -In sigma02 N0 + ---- + ---- + 1/2 k N2 + In sigma56 N5 = 0 tau2 tau5 > level0 := level0+level2; # eliminate N0 N5 2 level0 := ---- - 1/2 k N2 + In sigma56 N5 - In sigma25 N2 = 0 tau5 > level0+level5; # not independent! 0 = 0 The best you can get is a 2-parameter family of solutions. In the light of this, I suggest you need to reconsider the model that led you to try to solve these equations. Francis -- Dr Francis J. Wright | mailto: School of Mathematical Sciences | tel: (020) 7882 5453 (direct) Queen Mary, University of London | fax: (020) 8981 9587 (dept.) Mile End Road, London E1 4NS, UK | http://centaur.maths.qmw.ac.uk/ -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Tue, 10 Oct 2000 10:06:32 -0700 (PDT) From: Robert Israel To: Subject: solve Maple (Release 6.1) quickly gives me the following: > solve({level0,level2,level5,level6},{N0,N2,N5,N6}); 24 N2 (.2000000000 10 + In + 200. N2) {N0 = .2000000000 ------------------------------------, In -23 In N2 (In + 100. N2) N6 = .1440000000 10 ---------------------, 24 .5000000000 10 + In N2 (In + 100. N2) N5 = ---------------------, N2 = N2} 24 .5000000000 10 + In N2=N2 is not "pretty useless", it's telling you an important fact: your four equations do not have a unique solution, and the variable N2 can be taken as arbitrary. In fact, you might note that one of your equations is redundant, as the sum of the equations is 0=0. Robert Israel Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 11 Oct 2000 10:03:30 +1000 To: From: (Bill Whiten) Subject: solve The equations are not independent. level0 can be obtained from the other equations. Using Maple Vr5.1: level0:=(-In*sigma02*N0)+(N2/tau2)+(N5/tau5)+(N6/tau6)+((1/2)*k*N2^2); level2:=(In*sigma02*N0)-(N2/tau2)-(In*sigma25*N2)-k*N2^2; level5:=(In*sigma25*N2)-N5/tau5-(In*sigma56*N5)+((1/2)*k*N2^2); level6:=(In*sigma56*N5)-(N6/tau6); subs(N6=solve(level6,N6),N5=solve(level5,N5),N0=solve(level2,N0),level0); simplify(%); 0 Regards, ----------- Bill Whiten, Julius Kruttschnitt Mineral Research Centre, The University Of Queensland, Tel: int +61 7 3365 5888 Isles Rd, Indooroopilly, Fax: int +61 7 3365 5999 Brisbane Qld 4068, AUSTRALIA. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 11 Oct 2000 02:13:52 +0200 From: Helmut Kahovec To: Subject: solve Hello Ruth, Well, I suggest the following approach to solve your equations: > solve( > {seq(cat(level,i),i=[0,2,5,6])}, > {seq(cat(N,i),i=[0,2,5,6])} > ); The last result is NULL. Thus solve() did not succeed. However, since your equations are homogeneous in the N-variables we can convert them into a linear system of equations for the variables x0=N0/N2, x2=N2, x5=N5/N2, and x6=N6/N2: > map(expand@eval,[seq(cat(level,i)/N2,i=[0,2,5,6])]); > eqns:=op(subs({N0=x0*N2,N5=x5*N2,N6=x6*N2},N2=x2,%)); > vars:=seq(cat(x,i),i=[0,2,5,6]); The matrix A of the linear system is > A:=linalg[genmatrix]({eqns},[vars],'b'); [ 1 ] [ 0 0 In sigma56 - ----] [ tau6] [ ] [ 1 1 ] [-In sigma02 1/2 k ---- ---- ] A := [ tau5 tau6 ] [ ] [In sigma02 -k 0 0 ] [ ] [ 1 ] [ 0 1/2 k - ---- - In sigma56 0 ] [ tau5 ] with the right-hand side b > eval(b); [ 1 1 ] [0, - ----, ---- + In sigma25, -In sigma25] [ tau2 tau2 ] and the rank > linalg[rank](A); 3 The rank of A is less than 4, which probably is the reason why solve() has not been able to do its job. Now we solve the linear system for x0, x2, x5, and x6 > linalg[linsolve](A,b,'r',t); and convert the solution into a list: > convert(%,list); [- (-2 tau2 t[1] - 2 tau2 t[1] In sigma56 tau5 - tau5 + tau2 In sigma25 tau5)/(tau5 In sigma02 tau2), -t[1] - t[1] In sigma56 tau5 + In sigma25 tau5 -2 ----------------------------------------------, t[1], k tau5 In sigma56 t[1] tau6] Since x2=N2 we have: > N2=op(2,%); -t[1] - t[1] In sigma56 tau5 + In sigma25 tau5 N2 = -2 ---------------------------------------------- k tau5 Rewriting the solutions in terms of N0, N2, N5, and N6 again yields > subs(%,applyop(u->u*N2,{1,3,4},%%)); > sols:={op(zip((u,v)->u=v,[seq(cat('N',i),i=[0,2,5,6])],%))}; %1 sols := {N2 = -2 ------, N0 = 2 ((-2 tau2 t[1] k tau5 - 2 tau2 t[1] In sigma56 tau5 - tau5 + tau2 In sigma25 tau5) 2 t[1] %1 %1)/(tau5 In sigma02 tau2 k), N5 = -2 -------, k tau5 In sigma56 t[1] tau6 %1 N6 = -2 -----------------------} k tau5 %1 := -t[1] - t[1] In sigma56 tau5 + In sigma25 tau5 These solutions do indeed satisfy the original equations: > simplify(subs(sols,eval([seq(cat('level',i),i=[0,2,5,6])]))); [0 = 0, 0 = 0, 0 = 0, 0 = 0] Finally, we introduce the parameters as equations (relations) > params:= > sigma02=10*10^(-21), > sigma25=2*10^(-21), > sigma56=2*10^(-21), > tau2=2.5*10^(-3), > tau5=1*10^(-3), > tau6=0.72*10^(-3), > k=0.4*10^(-18); and get the desired numerical values of N0, N2, N5, and N6 in terms of the parameter In: > map( > print@(u->collect(u,t[1]))@expand, > normal(subs({params},sols),expanded) > ): 22 N2 = (.5000000000 10 + .01000000000 In) t[1] - .01000000000 In / 46 1 22 \ N0 = |.1000000000 10 ---- + .4000000000 10 + .004000000000 In| \ In / 2 t[1] + / 45 \ | .2000000000 10 22| |-.006000000000 In + ---------------- - .2600000000 10 | \ In / 21 t[1] - .4000000000 10 + .002000000000 In 22 2 N5 = (.5000000000 10 + .01000000000 In) t[1] - .01000000000 t[1] In -25 2 2 N6 = (.007200000000 In + .1440000000 10 In ) t[1] -25 2 - .1440000000 10 t[1] In t[1] is an arbitrary parameter introduced by the fact that the rank of the matrix A is less than 4. Thus we get an infinite number of solutions for N0, N2, N5, and N6. With kind regards, Helmut -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 11 Oct 2000 16:50:46 +0100 From: Barsuhn Subject: solve To: Dear Ruth, I inserted three statements into the code you supplied. The first defining a variable eqsys for your system of equations, the seconds allows you to inspect your equation system after specifying various constants. The third is the solve-statement. > eqsys:={level0,level2,level5,level6}; > eqsys; > solve(eqsys,{N0,N2,N5,N6}); I obtain N2=N2, too. This shows that your four equations are not independent and thus not sufficient to solve uniquely for N0,N2,N5,N6. You could choose an arbitrary N2 and in any case would obtain a solution of your system of four equations. You can check this by substituting the solutions found into your system of equations. >subs(%,eqsys);simplify(%); Don't worry that simplify ends up with the set {0.=0}, i.e. with only one identity. The other three identities have been erased as a set consists only of different elements. Probably you should look for a further equation for your variables. All the best Jurgen -- ------------------- Prof. Dr. Jurgen Barsuhn Fachhochschule Bielefeld University of Applied Sciences Fachbereich Elektrotechnik und Informationstechnik Wilhelm-Bertelsmann-Str. 10 D-33602 Bielefeld ----------- -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Thu, 12 Oct 2000 14:20:37 -0400 (EDT) From: Robert K Wright To: Subject: solve > solset:= solve({level0,level2,level5,level6},{N0,N2,N5,N6}); 24 N2 (.2000000000 10 + In + 200. N2) solset := {N0 = .2000000000 ------------------------------------, In N2 (In + 100. N2) N5 = ---------------------, 24 .5000000000 10 + In -23 In N2 (In + 100. N2) N6 = .1440000000 10 ---------------------, N2 = N2} 24 .5000000000 10 + In > #Note N0,N5,N6 come out in terms of N2 and In. So apparently > #the four equations you give are not independent, and have > #a 1-parameter family of solutions. > #Below, substitute into the system of equations > #to check that they are really solved: > subs(solset,{level0,level2,level5,level6}); -20 24 {.2000000000 10 N2 (.2000000000 10 + In + 200. N2) - 400.0000000 N2 - 1/500000000000000000000 In N2 -18 2 - .4000000000 10 N2 = 0, -20 24 -.2000000000 10 N2 (.2000000000 10 + In + 200. N2) N2 (In + 100. N2) + 400.0000000 N2 + 1000 --------------------- 24 .5000000000 10 + In -20 In N2 (In + 100. N2) + .2000000000 10 --------------------- 24 .5000000000 10 + In -18 2 + .2000000000 10 N2 = 0, 1/500000000000000000000 In N2 N2 (In + 100. N2) - 1000 --------------------- 24 .5000000000 10 + In In N2 (In + 100. N2) - 1/500000000000000000000 --------------------- 24 .5000000000 10 + In -18 2 + .2000000000 10 N2 = 0, 0 = 0} > simplify(%); {0 = 0} > ************************ R. K. Wright Math/Stat Department 802-656-3429 University of Vermont

[MUG] fsolve and dsolve,numeric
Author: Douglas B Meade    Posted: 17/10/2000 22:13:02 GDT
>> From: "Douglas B. Meade" Dear MUG, A few weeks ago I posted a question about using the results from numeric dsolve in fsolve. I sent the same request to Maple Support. It took a few iterations, but they did provide a very nice solution to my problem. The full message from Technical Support is attached below. The basic idea is to use the usual wrapper to avoid evaluation of the solution for non-numeric arguments. The key is to use the method=taylorseries option to numeric dsolve to coerce Maple into returning a result that is compatible with fsolve. With this modification, all seems to work well -- at least on this type of problem. Doug -------- Original Message -------- Subject: Incident=334494 fsolve help Date: Mon, 18 Sep 2000 17:45:41 -0400 From: George Terlecki To: CC: Dear Dr. Meade, I was able to confirmed that this bug was fixed and the fix will be probably available in next full release of Maple. I have also a workaround presented below. > ff := dsolve({diff(x(t),t)=x(t),x(0)=2}, {x(t)}, type=numeric, output=listprocedure, method=taylorseries): > fx0 := subs(ff,x(t)); fx0 := proc(t) ... end proc # This next procedure allows for an unevaluated return, which is necc. for fsolve with dsolve/numeric > fx := proc(x) if not type(evalf(x),`numeric`) then 'procname'(x); else fx0(x); fi; end: > fx(0); 2. > fx(1); 5.436563657 > fsolve(fx(x)=3,x=0.1..1); .4054651078 > fsolve(fx(x)=4,x=0.1..1); .6931471800 Please let me know if you have any further questions with regards to this matter. Best Regards, George Terlecki Technical Support Analyst

[MUG] Re: when does maple give up during a solve?
Author: Maple Group    Posted: 18/10/2000 22:58:48 GDT
>> From: Maple Group >> From: | I have a maple worksheet that involves solving a system of non-linear | polynomial equations. Maple attempts to find the closed form solution to | these equations, but never arrives at a solution before using all of my | 128MB of ram. My question is, since maple hasn't stopped the computation | on it's own accord, is it actually making progress at finding a | solution(s), or is it likely that no matter how much memory I make | available, it will never stop the computation? -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Jason Schattman To: Subject: when does maple 'give up' during a solve? Date: Fri, 13 Oct 2000 11:46:43 -0400 It looks like the equations you're solving are of degree 6, and I believe there is, in general, no closed form formula for roots of polynomials (or systems thereof) higher than degree 4. You may need to use fsolve(), which looks for numerical solutions rather than symbolic ones. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Sat, 14 Oct 2000 09:56:23 -0700 (PDT) From: "Jason C. Leach" To: Subject: when does maple 'give up' during a solve? hi, Have a look at: ?startup ?kernelops With them you can set some options like, I think, CPU and Mem usage. If you are on a Unix system, you can do this do your proces also. j. ...................... ..... Jason C. Leach ... University College of the Cariboo. .. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Venkatramanan, Subramanian" To: Subject: when does maple 'give up' during a solve? Date: Sat, 14 Oct 2000 16:28:40 -0400 You can add your virtual memory by adding let's say 1GB of your C or D drive. I have had experiences with Maple taking more than 128 MB to solve, especially if its in symbolic form. But, it looks like your set of equations may not have exact solution. You probably want to try numerical solution. Venkat Venkat R. Subramanian Ph.D. Candidate Center for Electrochemical Engineering Department of Chemical Engineering University of South Carolina Office: (803) 576 5634 Home: (803) 254 6255 http://www.engr.sc.edu/grad/venkatra "It sure makes the day long when you get to work on time". -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Colin J Christopher" To: Date: Mon, 16 Oct 2000 13:08:49 GMT Subject: when does maple 'give up' during a solve? If there is no solution then Maple sometimes returns nothing, which is a bit disconcerting. However, there should be a solution here. I took u^6 = det(S) as a 6th equation, and substituted u instead of det(S)^(1/6) in the equations - otherwise the equations are just too complicated for Maple to solve if they involve 6th roots The new equations are now just polynomials. Solve returned nothing (it did terminate though). I then tried a grobner basis calculation, but run out of memory or something. The real problem is probably that it is best to work with the matrices themselves rather than get maple to work with the coefficients since there seems to be a large number of equations that way, and makes things too hard for Maple to deal with. also you miss out on any symmetry present. If the matrices are real then (since they are symmetric) you can find roots etc (I.e. let S = U^6) and factorize the quadratic in lambda in R by completing the square. R then becomes a product of two determinants of the form det(lam*Id + A_i) and T becomes (lam + det(U))^6. So, A_i have all their eigenvalues = det(U) and so must be of the form det(U)*Id. The only solutions therefore are C=2*det(U)*Id, D=det(U)^2*Id. Regards, Colin -------------------------------------------- Colin Christopher, Department of Mathematics and Statistics, University of Plymouth, Plymouth PL2 3AJ, UK. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Mon, 16 Oct 2000 11:18:46 -0700 (PDT) From: Robert Israel To: Subject: when does maple 'give up' during a solve? Your equations, converted into a polynomial system, are quite complicated. For example, equation 1 alone (after taking the 6th power of both sides, and subtracting one side from the other) becomes an irreducible polynomial of 10933 terms, of degree 12 in each of the s's and 6 in each of the c's. And in the course of the solution process, the amount of complication tends to rise enormously. In principle, "solve" should terminate in a finite amount of time after using a finite amount of memory. However, I would bet that these finite amounts would be astronomically large. Even with all the memory available in the world, don't expect to get an answer in this life. Robert Israel Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2

[MUG] slow series(solve(..)) quirk in Maple VI, V.X>3
Author: Desmond Johnston    Posted: 20/10/2000 15:24:28 GDT
>> From: Desmond Johnston The following command: z:=series(solve((c2-1)*g*(3*(c2-1)*z-1)**2-z+(c2*z-3*c2*(c2-1)**2*z**3)*(3*(c2-1 )*z-1)**2=0,z),g=0,40+1); for reverting a power series of an implicitly defined function- runs ten times faster or more on similar equipment with MapleV.3 compared with later versions. The rot sets in when the order in g reaches 30+1 or so. (The answer for Maple V.3 agrees with later versions). Playing around with stacklimit and gcfreq doesn't speed things up for V.X>3 or VI. Can anyone enlighten me as to what is going on, and how it might be avoided? Des Johnston Maths Department Heriot-Watt University Edinburgh

[MUG] dsolve numeric and analytic solution
Author: CT    Posted: 27/10/2000 11:01:51 GDT
>> From: CT Hello, We have solve a differential equation with "dsolve numeric" and we think that our solution could be approximate by a function (1/x^3 when x-> infinite). Do you know maple's functions doing such a kind of comparison ? Thank you for your help, Christine Universite Louis Pasteur. Strasbourg

[MUG] How to solve this equation by Maple?
Author: Meiers, Jerald    Posted: 31/10/2000 13:41:03 GMT
>> From: "Meiers, Jerald" What are the steps (or is it really possible?) to arrive at the solution "theta = 2*lambda" for the following equation of unknown 'theta', by Maple? eqn:=tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta) = 0; (Of course, we need to suppose we don't know the solution beforehand.) Thanks.

[MUG] Re: dsolve numeric and analytic solution
Author: Maple Group    Posted: 03/11/2000 13:34:08 GMT
>> From: Maple Group > >>> From: CT > > We have solve a differential equation with "dsolve numeric" and we >think that our solution could be approximate by a function (1/x^3 when >x-> infinite). Do you know maple's functions doing such a kind of >comparison ? -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Chris Eilbeck Date: Tue, 31 Oct 2000 11:08:54 GMT To: Subject: dsolve numeric and analytic solution try sticking in an inverse power series y = a[1]/x+a[2]/x^2 +a[3]/x^3 + ... into the differential equation and trying solving for the coefficients. It's a common technique in mathematics, most books on o.d.e.'s will have a chapter on this. ---------------------------------------------------------------------- Chris Eilbeck email: Department of Mathematics, Fax: +44 (0)131 451 3249 Heriot-Watt University, Phone: +44 (0)131 451 3220 Edinburgh EH14 4AS, Scotland. WWW: http://www.ma.hw.ac.uk/~chris/ ---------------------------------------------------------------------- -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Willard, Daniel Dr DUSA-OR" To: Subject: dsolve numeric and analytic solution Date: Tue, 31 Oct 2000 08:03:21 -0500 Try: > limit(1/x^3-[your solution], x=infinity);

[MUG] Re: How to solve this equation by Maple?
Author: Maple Group    Posted: 03/11/2000 13:35:43 GMT
>> From: Maple Group > >> From: "Meiers, Jerald" > > What are the steps (or is it really possible?) to arrive at the solution > "theta = 2*lambda" for the following equation of unknown 'theta', > by Maple? > > eqn:=tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta) = 0; > > (Of course, we need to suppose we don't know the solution beforehand.) > > Thanks. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 01 Nov 2000 18:21:18 +0100 From: Adri van der Meer To: Subject: How to solve this equation by Maple? > s := solve(eqn,theta): > convert(combine(simplify(convert(s,sincos))),tan); arctan(tan(2 lambda)) and, of course only > simplify(%,symbolic); gives the desired answer. -- A. van der Meer Dept. Applied Mathematics University of Twente Phone +31 (53) 4893427 P.O. Box 217 Fax +31 (53) 4894824 7500 AE Enschede -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Thu, 2 Nov 2000 03:01:22 -0500 (EST) From: Carl DeVore To: Subject: How to solve this equation by Maple? The trick is to use differentiation to verify that the complicated trig expression can be reduced in ways that Maple apparently does not know about. Here it is explicitly: > f:= (lambda,theta) -> tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta): > solve(f(lambda,theta)=0,theta); tan(lambda) -arctan(2 ----------------) 2 tan(lambda) - 1 > simplify(diff(%,lambda)); 2 > ans:= dsolve(diff(theta(lambda),lambda)=%); ans := theta(lambda) = 2 lambda + _C1 > solve(f(lambda,rhs(ans))=0); {_C1 = 0, lambda = lambda} Therefore, the integration constant is 0, and thus the answer is theta = 2 * lambda. Carl Devore University of Delaware -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Thu, 02 Nov 2000 09:43:42 +0100 (CET) From: To: Subject: How to solve this equation by Maple? With > solve( { convert(eqn, exp ) }, theta ); you get {theta = 2*lambda} E. Elbraechter ---------------------------------- E-Mail: Date: 02-Nov-2000 Time: 09:38:00 ---------------------------------- -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Thu, 02 Nov 2000 13:24:46 +0000 From: "Dr Francis J. Wright" To: Subject: How to solve this equation by Maple? The following (using Maple 6) is a bit tortuous because combine seems to work only on the sincos form and not on the tan form, thereby requiring conversion to and from sincos form. Otherwise, it would be fairly straightforward. > eqn:=tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta); 2 eqn := tan(lambda) sin(theta) + 2 tan(lambda) cos(theta) - sin(theta) > solve(eqn, theta); tan(lambda) -arctan(2 ----------------) 2 tan(lambda) - 1 > convert(%, sincos); sin(lambda) -arctan(2 ------------------------------) / 2 \ |sin(lambda) | cos(lambda) |------------ - 1| | 2 | \cos(lambda) / > simplify(%); sin(lambda) cos(lambda) arctan(2 -----------------------) 2 2 cos(lambda) - 1 > combine(%, trig); sin(2 lambda) arctan(-------------) cos(2 lambda) > convert(%, tan); arctan(tan(2 lambda)) > simplify(%, symbolic); 2 lambda Francis -- Dr Francis J. Wright | mailto: School of Mathematical Sciences | tel: (020) 7882 5453 (direct) Queen Mary, University of London | fax: (020) 8981 9587 (dept.) Mile End Road, London E1 4NS, UK | http://centaur.maths.qmw.ac.uk/ -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Thu, 2 Nov 2000 10:48:05 -0600 (CST) From: Andrzej Pindor To: Subject: How to solve this equation by Maple? |\^/| Maple 6 (SUN SPARC SOLARIS) ._|\| |/|_. Copyright (c) 2000 by Waterloo Maple Inc. \ MAPLE / All rights reserved. Maple is a registered trademark of <____ ____> Waterloo Maple Inc. | Type ? for help. > leqn:=tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta); 2 leqn := tan(lambda) sin(theta) + 2 tan(lambda) cos(theta) - sin(theta) > cleqn:=convert(leqn,tan); cleqn := 2 tan(lambda) tan(1/2 theta) tan(lambda) (1 - %1) tan(1/2 theta) 2 --------------------------- + 2 -------------------- - 2 -------------- 1 + %1 1 + %1 1 + %1 2 %1 := tan(1/2 theta) > subs(tan(1/2*theta)=y,cleqn); 2 2 tan(lambda) y tan(lambda) (1 - y ) y 2 -------------- + 2 -------------------- - 2 ------ 2 2 2 1 + y 1 + y 1 + y > solve(%=0,y); 1 tan(lambda), - ----------- tan(lambda) > solve(tan(1/2*theta)=tan(lambda),theta); bytes used=1000700, alloc=851812, time=0.31 2 lambda > solve(tan(1/2*theta)=-1/tan(lambda),theta); 1 -2 arctan(-----------) tan(lambda) Regards Andrzej -- Dr. Andrzej Pindor The foolish reject what they see and not what University of Toronto think; the wise reject what they think and not Information Commons what they see. Huang Po Phone: (416) 978-5045 Fax: (416) 978-7705

[MUG] Re: How to solve this equation by Maple?
Author: Maple Group    Posted: 07/11/2000 14:41:40 GMT
>> From: Maple Group | >> From: "Meiers, Jerald" | | What are the steps (or is it really possible?) to arrive at the solution | "theta = 2*lambda" for the following equation of unknown 'theta', | by Maple? | -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "peter lindsay" To: Subject: How to solve this equation by Maple? Date: Sat, 4 Nov 2000 12:20:26 -0000 tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta) can be simplified to (sec(lambda))^2 * sin(2*lambda-theta); after that the solution is easy. I dont know how to simplify tan(lambda)^2*sin(theta)+2*tan(lambda)*cos(theta)-sin(theta) in maple. Peter Lindsay -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Sat, 4 Nov 2000 19:38:34 -0500 (EST) From: David Vaughan To: Subject: How to solve this equation by Maple? The solution depends on what domains for lambda and theta are chosen. For example, if theta is any value that makes sine 0 then lambda is anything that makes tangent 0. If we assume these kinds of cases have been dealt with, then use the quadratic formula to solve for tangent(lambda) in terms of functions of theta, and some identities to get the result. No calculus is required. DCVaughan Department of Mathematics Wilfrid Laurier University

[MUG] solve
Author: Bill Moss    Posted: Thu, 07 Feb 2002 13:37:23 -0500
>> From: Bill Moss /> This worked in Maple 6 > eqn1 := c1*cos(1/2*(k/m)^(1/2)*L/V)-c2*sin(1/2*(k/m)^(1/2)*L/V) = 0; > eqn2 := c1*sin(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)+c2*cos(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)-k*L/(m*Pi^2*V^2-k*L^2)*Pi*V = 0; > solve({eqn1,eqn2},{c1,c2}); but does not in Maple 7. The output is null. Bill Moss Professor Mathematical Sciences Clemson University Clemson, SC 29634-0975 864.656.5225 http://www.math.clemson.edu/~bmoss

[MUG] Re: solve
Author: Carl Devore    Posted: Wed, 13 Feb 2002 15:27:29 -0500
>> From: Carl Devore /> On Thu, 7 Feb 2002, Bill Moss wrote: > This worked in Maple 6 > > eqn1 := c1*cos(1/2*(k/m)^(1/2)*L/V)-c2*sin(1/2*(k/m)^(1/2)*L/V) = 0; > > eqn2 := > c1*sin(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)+c2*cos(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)-k*L/(m*Pi^2*V^2-k*L^2)*Pi*V > = 0; > > solve({eqn1,eqn2},{c1,c2}); > but does not in Maple 7. The output is null. You should "declare" all the variables for solve for your first attempt: > solve({eq1,eq2}, {c1,c2,k,L,V,M}) or > solve({eq1, eq2}, indets({eq1, eq2}, And(name, Not(constant)))); ...and then prune away the unneeded variables. You probably suspected that whatever solution there was would be true for all k, L, V, m. But in fact there are some weird solutions that fix the value of k and give a c1 that is not 0 and does not depend on k.

[MUG] Re: solve
Author: Maple User Group    Posted: Thu, 21 Feb 2002 09:30:26 -0500
>> From: Maple User Group /> -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 20 Feb 2002 17:06:36 +0100 From: Preben Alsholm /> To: /> Subject: solve The problem does seem to be a genuine bug in solve. Take a look at the following simplified examples: restart; > eqn1 := c1*cos(w)-c2*sin(w) = 0: > eqn2 := c1*sin(w)*sqrt(b)+c2*cos(w)*sqrt(b)-a= 0: > solve({eqn1,eqn2},{c1,c2});#No answer > > restart; > eqn1 := c1*cos(w)-c2*sin(w) = 0: > eqn2 := c1*sin(w)*b+c2*cos(w)*b-a= 0: > solve({eqn1,eqn2},{c1,c2}); > a cos(w) a sin(w) {c2 = --------, c1 = --------} b b > restart; > eqn1 := c1*cw-c2*sw = 0: > eqn2 := c1*sin(w)*sqrt(b)+c2*cos(w)*sqrt(b)-a= 0: > solve({eqn1,eqn2},{c1,c2}); #No answer > > restart; > eqn1 := c1*cw-c2*sw = 0: > eqn2 := c1*sin(w)*sqrt(2)+c2*cos(w)*sqrt(2)-a= 0: > solve({eqn1,eqn2},{c1,c2}); > 1/2 1/2 a 2 sw a cw 2 {c1 = 1/2 ---------------------, c2 = 1/2 ---------------------} sin(w) sw + cos(w) cw sin(w) sw + cos(w) cw > restart; > eqn1 := c1*cw-c2*sw = 0: > eqn2 := c1*sw*sqrt(b)+c2*cw*sqrt(b)-a= 0: > solve({eqn1,eqn2},{c1,c2}); a cw a sw {c2 = ----------------, c1 = ----------------} 1/2 2 2 1/2 2 2 b (sw + cw ) b (sw + cw ) The combination of sin and cos with sqrt of a symbolic b in one of the two equations seems to have a bad effect. -- Preben Alsholm Institut for matematik (Department of Mathematics) DTU (Technical University of Denmark) -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Willard, Daniel Dr DUSA-OR" /> To: /> Subject: solve Date: Wed, 20 Feb 2002 11:29:26 -0500 It works, of course, in V 5.1. -----Original Message----- From: Carl Devore /> Sent: Wednesday, February 13, 2002 3:27 PM To: /> Subject: [MUG] Re: solve >> From: Carl Devore /> On Thu, 7 Feb 2002, Bill Moss wrote: > This worked in Maple 6 > > eqn1 := c1*cos(1/2*(k/m)^(1/2)*L/V)-c2*sin(1/2*(k/m)^(1/2)*L/V) = 0; > > eqn2 := > c1*sin(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)+c2*cos(1/2*(k/m)^(1/2)*L/V)*sqrt(k/m)- k*L/(m*Pi^2*V^2-k*L^2)*Pi*V > = 0; > > solve({eqn1,eqn2},{c1,c2}); > but does not in Maple 7. The output is null. You should "declare" all the variables for solve for your first attempt: > solve({eq1,eq2}, {c1,c2,k,L,V,M}) or > solve({eq1, eq2}, indets({eq1, eq2}, And(name, Not(constant)))); ...and then prune away the unneeded variables. You probably suspected that whatever solution there was would be true for all k, L, V, m. But in fact there are some weird solutions that fix the value of k and give a c1 that is not 0 and does not depend on k.

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