Robert Israel

6457 Reputation

21 Badges

15 years, 183 days
University of British Columbia
Associate Professor Emeritus
North York, Ontario, Canada

MaplePrimes Activity


These are answers submitted by Robert Israel

The Maple share library was replaced (as of Maple 6) by the on-line Maple Application Center. However, I can't find IntSolve there. In any case, I think IntSolve could handle a single integral equation (of certain types) but not a system of integral equations. If you give us an example of the type of system you're trying to solve, we might have some ideas of how to solve it.
You mean like this? > imagefile := cat(kernelopts(datadir), "/help/ImageTools/fjords.jpg"); with(ImageTools): img := Read(imagefile): Preview(img); Preview(Rotate(img,90));
You might try Import in the ExcelTools package (in Maple 11, Standard Worksheet). You can simply say e.g. > A:= ExcelTools[Import]("c:/myfolder/myfile.xls");
I don't know why your h needs abs twice: since abs(Omega*beta) >= 0, abs(Omega*beta)^(3/2) >= 0 as well. Anyway, you can try something like this: implicitplot3d(piecewise(h(beta,Omega) >0, F(beta,Omega,delta),undefined),beta=-1..1,Omega=-20 .. 20,delta=-6..6,axes=box);
Try something like this:
sys:= {seq(D(x[i])(t)=x[i](t) - 2*x[5-i](t),i=1..4), seq(x[i](0)=i,i=1..4)}: sol:= dsolve(sys,[seq(x[i](t),i=1..4)],numeric, output=listprocedure); S:= subs(sol,add(x[i](t),i=1..4)); e:= 1.5: plot(S,1 .. e); fsolve(S(t) = t);
An error seems to be occurring somewhere in the statement YP := ... in procedure p1. Actually you shouldn't assign a value to YP at all, just to the entries of YP. YP comes in to the procedure p1 as an array(1 .. N), and you're throwing that out and replacing it by an array(1 .. N-1). And then you try to assign a value to YP[N], which would not exist if YP was an array(1 .. N-1). I don't understand what you're trying to do in p1, but I suspect that what you should be doing is something like for i from 1 to N-1 do YP[i] := ... end do; YP[N] := ...
Do you mean something like this? plot([piecewise(t <=1,t,2-t^2), piecewise(t <=1,t^2,2-t),t=0..2]);
roots returns the roots of a polynomial in the rationals or an algebraic number field. In the case of x^2 + x + 1, you need your field to contain a square root of -3, so roots(x^2 + x + 1,I*sqrt(3)) would work. If you don't know or care what field the roots should be in, you should probably stick with solve and fsolve.
As Jacques noted, if dsolve can't find a closed-form solution it probably doesn't exist. You might be able to use a numeric solution or a series solution. See the help pages ?dsolve,numeric and ?dsolve,series.
Can you give us an example where this occurs?
Yes, this should be correct. Of course to actually get "true" or "false" rather than an equation, you need to apply evalb to it.
To get explicit solutions for quartics, set _EnvExplicit := true before using solve. Alternatively, you can try convert(..., radical) on the RootOf expressions. Warning: these explicit results tend to be very complicated. Personally, I find the RootOf notation to be much nicer. There's really not much you can do with the radical solution that you can't do with the RootOf.
OK, I see the problem. This version should work in any version of Maple from Maple 8 on: PlotFeasibleRegion:= proc(S::{set(`<=`), list(`<=`)}, xr:: name = range, yr:: name = range) # S is a set or list of non-strict inequalities # xr and yr of the form x = a..b, y = c..d # where x and y are the axis variables # The usual plot options, such as colour=..., can be specified # after these local SS, pts, i, j, n, x, y, a, b, c, d, R; x:= lhs(xr); y:= lhs(yr); a:= op(1,rhs(xr)); b:= op(2,rhs(xr)); c:= op(1,rhs(yr)); d:= op(2,rhs(yr)); SS:= convert(S,set) union {x >= a, x <= b, y >= c, y <= d}; n:= nops(SS); pts:= {seq(seq(solve({convert(SS[i], equality), convert(SS[j], equality)}), j=i+1 .. n), i=1..n-1)}; pts:= select(type,pts,set(name=realcons)); pts:= select(p -> andmap(is,subs(p,SS)), pts); pts:= map(subs, pts, [x,y]); R:=simplex[convexhull](pts); if R = {} then return NULL end if; plots[polygonplot](R,args[4..-1]) end proc; Cheers, Robert
For example, with some made-up data: > X:= [$0..9]; Z:= [1.85, 1.54, 1.35, 1.23, 1.17, 1.16, 1.17, 1.22, 1.29, 1.38]; S:=dsolve((D@@2)(y)(x)=a*D(y)(x)+b*y(x)+c); res:= [seq(eval(rhs(S),x=X[j])-Z[j],j=1..10)]; Optimization[LSSolve](res); [.26471056267204e-4, [a = -.226946305957795996, b = .542887865054803298e-1, c = -.290795758320473961e-1, _C1 = 1.09634198339014442, _C2 = .217404308130228680]] ... and to see how well it fits: > yfit:= subs(%[2],rhs(S)); P1:= plot(yfit, x = 0 .. 9); P2:= plot([seq([X[i],Z[i]],i=1..10)], style=point,symbol=circle, colour=black): plots[display](P1,P2); This is a case where the roots a/2 (+/-) sqrt(a^2+4b)/2 happen to be real. I don't think you'd be able to use the same form in a case where they are complex: LSSolve wouldn't like running into complex numbers. In that case you'd want to use the other form of the solution, involving sin and cos. In either case, you may also need to restrict the ranges of a and b in the LSSolve command to avoid complex numbers. Cheers, Robert
An alternative to Jacques's solution is to use print in your procedure to produce any plots before the end of the procedure. For example Energy:=proc(n) . . . print(plots[listdensityplot](...)); . . . plot(...) end proc;
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