MaplePrimes Commons General Technical Discussions

The primary forum for technical discussions.
What is the latest version of Maple 10 available? ( as in 10.x)
i am trying to solve: L_fire_7 := 0.5*trans_7 * p *((2 * h * c^2 / (lambda_7)^5) * (1 / (exp(h*c/(k*lambda_7*T)) -1))) = L7; L_fire_21 := 0.83*trans_21 * p *((2 * h * c^2 / (lambda_21)^5) * (1 / (exp(h*c/(k*lambda_21*T)) -1))) = L21; for T and p using PolynomialSystem. the result contains _X. i have no idea what that symbol means and can't find anything in the manual nor on the web. any help is highly appreciated! (also in regard to how to solve this better than using PolynomialSystem). -florian
I have written a relatively large simulation in Maple10 and it is literally taking forever to run for any decent size and time. I have been trying to convert to C to run in C, which became a serious problem, then I tried to use "compile" which again was very problematic. They seem to work only when the code is written in a C-like format anyway, in other words to use of lists and very fussy with arrays and procedures within procedures seems to create problems. I just wanted to ask if there is something clever I can do, to convert my procedure (there are about 20 of them) and if there is certain rigorous formatting I can do without having to literary re-write my whole code, as that is the way it seems now.
Hello , I have been trying to plot the solution to a boundary layer problem using Maple via the following programme: >BVP:={0.1*diff(y(x),[x$2]) + (x^0.5)*diff(y(x),x) - y(x) = 0, y(0)=0,y(1)=exp(2)}; >J:=dsolve(BVP,y(x), type=numeric,method=bvp[midrich]); >plots[odeplot](J,[x,y(x)], 0..1,color=blue,style=line,symbol=circle, symbolsize=12); However,before I can write the instructions to plot, I get an error message saying: "Error,(in dsolve/numeric,bvp) unable to achieve requested accuracy of 0.1e-5 with maximum 128 point mesh (was able to get 0.17e-4), consider increasing "maxmesh" or using larger "abserr" ."
I'm running Maple 10.02, Mac OS X.4.3, and Java 1.5, and when I give Maple an inverse laplace transform with a number, it returns the correct answer. But when I give it the same thing but replace the number with an arbitrary constant (i.e., I replaced 4 with k) it doesn't seem to be able to figure it out. Any suggestions on this?

> Maple Equation

I had been following a thread on comp.soft-sys.math.maple titled: Maple missing solutions?

In, short the OP was missing solutions to a set of equations involving sin and cos.

A respondent suggested converting to exp, which seems to have worked very well in that case.

For no particular reason I decided to try this with another equation and got some

results which I cannot explain. As follows:

> S1:=[solve(sin(x)=sin(1/x))];

A student of mine recently asked what algorithm Maple uses to calculate Eigenvalues. So, I tried diving into Maple Procedures. For example: >showstat(`Eigenvalues`); LinearAlgebra:-Eigenvalues := proc(A) .. [error catchers ommitted] .. 17 LinearAlgebra:-LA_Main:-Eigenvalues(`if` ... end proc How should I interpret that last line? Can I get Maple to show me what that procedure does? So far I've no luck cracking it. Also, is this the best/easiest way of going about learning what types of algorithms Maple uses. I understand that some (/many/all) of them may be proprietary and the persons at Waterloo may not want the general public knowing them, but even an answer as simple as "Accelerated Newton-Jacobi algorithm" would be good.
Hi, I have plotted the following chart using maple 10 on windows xp > plot(((x^2)+x-4)/(x-2),x=-4..6,y=-2..10); X being -4, 2 Which command needs to be executed to find the local extremity? Thanks in advance for any help… Robert
Another hour, another possible problem. Consider the following Maple statements:

> f := table([index=m]);
f := table([index = m])
> g := copy(eval(f));
g := table([index = m])
> g[index] := one;
g[index] := one
> h := g;
h := g
> f[index], h[index];
m, one
> g := copy(eval(h));
g := table([index = one])
> g[index] := two;
g[index] := two
> l := g;
l := g
> f[index], h[index], l[index];
m, two, two

So, after the first copy, the second table h has value one for index, while f still has value m for index. Yet, after the second copy, not only the third table l has value two for index, but also the second table h now has value two for index, while I have explicitly used copy to make a copy of h. Does this indicate a problem with the copy function?

a colleague just noticed the following (erroneous) behavior in Maple 10 :

> a := Psi(2, 2+2*I);
a := Psi(2, 2 + 2 I)
> printf("%+.6e\n", Re(a));
Error, (in fprintf) number expected for floating point format
> evalb(Im(a) <> 0);

The behavior is correct in Maple 9.5 :

> a := Psi(2, 2+2*I);
a := Psi(2, 2 + 2 I)
> printf("%+.6e\n", Re(a));
> evalb(Im(a) <> 0);

-- Regards,

Yet another question (after scouring the Maple documentation and coming up empty-handed): The following commands: with(student) intercept(y=sin(x), y=0) Result in this output: {y=0, x=0} How can I return the other intercepts with the x-axis? For instance, the first positive and first negative non-zero intercepts? (i.e. the "first" x-intercepts traveling left and right away from the y-axis?) Thanks as always, Bryan
Quick question: in maple , sqrt(4) returns 2. Why not plus/minus 2, since -2 is also a valid square root of 4? Thanks! Bryan
anyone has done significant development with these two?
Additionally to my previous post: First example returns (mathematically) wrong result,
eval(diff(v(z), z), [v = (x->x*H), z = H])
Equivalent works fine (just because multiplier "a" leads to implicit conversion of diff to D).
eval(diff(v(a*z), z), [v = (x->x*H), z = H, a = 1])
And the most exiting example (I think, that result can't be predicted by Maple developers also ):)))
eval(diff(v(x, y), x, y), {x = H, y = H})
         (D[1,1](v))(H,H)+2 (D[1,2](v))(H,H)+(D[2,2](v))(H,H)
Functional analog works, of course, without any errors.

Create new document (or worksheet) with content below and execute it step by step: > restart; > Eval(diff(v(z),z),z=H)==eval(diff(v(z),z),z=H); Eval(diff(v(z),z),z=H)=diff(v(H),H) > v:=z->z*H; v:=z->z*H > value((1)); H=2H As you can see, result of eval, which in this case is equivalent to subs(z=H,diff(v(z),z)), at the right side of first equation leads to wrong final result :( This...

First 71 72 73 74 75 76 Page 73 of 76