x11 := Vector([0.208408965651696e-3, -0.157194487523421e-2, -0.294739401402979e-2, 0.788206708183853e-2, 0.499394753201753e-2, 0.191468321959759e-3, 0.504980449104750e-2, 0.222150494088535e-2, 0.132091821964287e-2, 0.161118434883258e-2, -0.281236534046873e-2, -0.398055875132037e-2, -0.111753680372819e-1, 0.588868146012489e-2, -0.354191562612469e-2, 0.984082837373291e-3, -0.116041186868374e-1, 0.603027845850267e-3, -0.448778128168742e-2, -0.127561485214862e-1, -0.412027655195339e-2, 0.379387381798949e-2, -0.602550446997765e-2, -0.605986284736216e-2, -0.751396992404410e-2, 0.633613424008655e-2, -0.677581832613623e-2]):
y11 := Vector([ -21321.9719565717, 231.709204951251, 1527.92905167191, -32.8508507060675, 54.9408176234139, -99.4222178124229, -675.771433486265, 42.0838668074923, -12559.3183308951, 5.21412214166344*10^5, 1110.50031772203, 3.67149699000155, -108.543878970269, -8.48861069398811, -521.810552387313, 26.4792411876883, -8.32240296737599, -1085.40982521906, -44.1390030597906, -203.891397612798, -56.3746416571417, -218.205643256096, -178.991498697065, -42.2468018350386, .328546922634921, -1883.18308996621, 111.747881085748]):
z11 := Vector([ 1549.88755331800, -329.861725802688, 8.54200301129155, -283.381775745327, -54.5469129127573, 1875.94875597129, -16.2230517860850, 6084.82381954832, 1146.15489803104, -456.460512914647, 104.533252701641, 16.3998365630734, 11.5710907832054, -175.370276462696, 33.8045539958636, 2029.50029336951, 1387.92643570857, 9.54717543291120, -1999.09590358328, 29.7628085078953, 2.58210333216737*10^6, 57.7969622731082, -6.42551196941394, -8549.23677077892, -49.0081775323244, -72.5156360537114, 183.539911458475]):

a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t) + k4*u(t);
b1 := Diff(y1(t),t) = k5*x1(t)+ k6*y1(t)+ k7*z1(t) + k8*u(t);
c1 := Diff(z1(t),t) = k9*x1(t)+ k10*y1(t)+ k11*z1(t) + k12*u(t);
d1 := Diff(u(t), t) = 0;
ICS:=x1(0)=x11[1],y1(0)=y11[1],z1(0)=z11[1];
solL:=dsolve({a1,b1,c1,d1,ICS}, numeric, method=rkf45, parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
ans:=proc(p1,p2,p3) solL(parameters=[a1=p1,b1=p2,c1=p3]); end proc:
FitParams:=Statistics:-NonlinearFit(ans, x11, y11, z11, x1, y1, z1);

Error, (in Statistics:-NonlinearFit) unexpected parameters: Vector(27, {(1) = 1549.88755331800, (2) = -329.861725802688, (3) = 8.54200301129155, (4) = -283.381775745327, (5) = -54.5469129127573, (6) = 1875.94875597129, (7) = -16.2230517860850, (8) = 6084.82381954832, (9) = 1146.15489803104, (10) = -456.460512914647, (11) = 104.533252701641, (12) = 16.3998365630734, (13) = 11.5710907832054, (14) = -175.370276462696, (15) = 33.8045539958636, (16) = 2029.50029336951, (17) = 1387.92643570857, (18) = 9.54717543291120, (19) = -1999.09590358328, (20) = 29.7628085078953, (21) = 2582103.332, (22) = 57.7969622731082, (23) = -6.42551196941394, (24) = -...

So I am working on doing some trajectory simulations in Maple using standard Newton's Laws, some force expressions, and initial conditions.

Anyway, the numerical solution works fine if I let the initial conditions I specified (for z=-1) be actually for z=-0.9. To illustrate, when I give an initial condition like this:

x(-1) = x_0, D(x)(-1) = xd_0, Vz(-1) = v_0

the results don't make any sense. However, when using the same x_0, xd_0, and v_0 and I give initial conditions like this:

x(-.9) = x_0, D(x)(-.9) = xd_0, Vz(-0.9) = v_0,

the solutions at least make a bit of sense.

What's weird is that, when I let z -> 0.93 or so, the solution changes discontinuously. And this shouldn't happen. The initial conditions were calculated for and should work for z = -1. I don't understand why they aren't.

Here is my Maple document. ics1 are the problem.

dsolve_field_traject.mw

Do you guys have any idea what could be going on?

Hello, everyone!

Last week I’ve encountered problems with integration of Maple 17 in Microsoft Office Excel 2013. The Maplesoft note on the point (http://www.maplesoft.com/support/faqs/detail.aspx?sid=32651) offers some ways of fixing it up, though I’ve run all of them the problem is the same:

While the connection is established, after entering the formula “=Maple(“x+x”)”, the Excel returns “Critical Error in Formula”

Before contacting the Maplesoft Technical Support, I want to ask here whether someone had the same case and managed to solve it.

Many thanks in advance.

the question is as follow:

1)receive two integers p and q

2)declare two local p1 and q1 and give them intial values and q

3)check if p o q are equal or less to zero print works only with positive integers

4)while p1 not equal to q1 then p1-a1 otherwise q1-p1

5)whenever p1=q1 we have the GCD

note:must use procedure and call it for different values of p and q after the procedure is written

-by following the instruction above this is what i got

GCD:=proc(p,q)

local p1,q1;

p1:=p;

q1:=q;

if p<=0 OR Q<=0 then 'works only with positive integers'
else while p1<>q1 do if q1<p1 then p1-q1 else q1-p1

end if;

end do;

end if;

end proc;

but when I call two integers eg:p=2, q=6 -> GCD(2,6) maple just freeze...evaluating....forever. is it because i got the procedure wrong etc? it would be helpful if anyone can help me with this. thanks

A square has 36 sub-squares in it. How to Number each sub squares from 1 to 36, to make the sum of vertical, the sum of horizontal, and the sum of cross line are the same .Describe in general if possible.

Bonjour

Comment se calcul le résidue d'une fonction f(x,y) ?

Si par exemple f(x,y)=1/x+1/y+5/(x*y), alors le residue de f en (0,0) est défini comme étant (1/2*Pi)^2*int(f(x,y)) avec intégrale est sur une sphère qui contient l'origine.

Est ce qu'il est égal au coefficient 5 de 1/(x*y) dans le developpement de laurent en deux variables ??

Merci d'avance,

Gérard.

with(LinearAlgebra):

n := 31;

h := 0;

c1 := Array(1 .. 100);

c2 := Array(1 .. 100);

c3 := Array(1 .. 100);

for tt from 0 to n-4 do

c1[tt+1] := ...

c2[tt+1] := ...

c3[tt+1] := ...

od;

plot3d([c1[x],c2[x],c3[x]], x=1..27, y=1..27);

Does N variables caylay table have N permutation group so that can generate N functions?

for exmaple 3 variables cayley table have 3 permutation group, for 1, it has a permutation group , for 2 has a permutation group etc.

then does it mean that it has 3 functions, do it need to composite 3 functions in order to get a function belong to this cayley table?

1 1 1

1 2 2

1 2 3

Hello, everyone!

Last day I’ve discovered a strange behavior of the “optimization” tool:

Mostly it copes with the task – but there are cases (the data are the same) when the tool doesn’t return the opt. value: instead it returns the parameter name.
BUT if some of the previous calculations are re-evaluated (and the same input values are received) the “optimization” tools works quite well.

The question is: how to tackle the problem?

P.S. The way to force re-evaluation of previous functions in the body of this procedure (if the optimization failed) doesn’t help: it returns the optimizing parameter name once more.

Windows 8 (64-bit)
Maple 17.00

Hi, I am trying to plot these two curves:

I tried:

with(plots);  A := Array(1..2)  A[1] :=plot (0.199563349672261+0.0178636902277546 x^1.14406289706794-0.0182070811144750 x^(1.13867380551454),x=50..2050,  color=red);  A[2]  := plot(0.298910542599302+0.0117459591500434 x^1.00390277106937-0.0137065176395662 x^0.970667551759677, x = 50..2050, color = blue);  display (A);

and I got:

Error, (in plot) unexpected option: .298910542599302+0.117459591500434e-1*x^1.00390277106937-0.137065176395662e-1*x^.970667551759677

so I tried

so I am not sure how I would graph these two functions...

 is there two cayley tables when generate every modular lattice such as (x v y) ^ y = y

 how to generate these two cayley table for join and meet with axiom  (x v y) ^ y = y

Ive been given a task to write a fuction that will print true/false to if a given line will intersect a triangle given the co-ordinates of the vertices. 
I have a general idea on what needs to be done mathmaticly, but i have poor maple skills and the syntax is killing me.
I would appreciate some help in this regard.

Thanks.

Ok so my problem is that I need to write a function which produces the three inequalities defining a triangle given the 3 points that form its vertices, then with the 4th point supplied to the function I need the function to print false if the 4th point satisfies less than 2 of these inequalities and true if it satisfies more than 2 of these inequalities. Can anybody help me with this? On a previous post I obtained part of the function that produces the triangle, but the 2 inequalities part was not solved.

Thanks for your time

How to show the code in file with extension .lib and edit it?

Hello

Am trying to convert a vector projection to matrix type output.

Attached is a sample worksheet.

Transformation_matri.mw