@Markiyan Hirnyk 

First try, i change to 

result1 := Optimization:-Minimize([ans>=0, ans<=0],initialpoint=[.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003], feasibilitytolerance=0.01);

Error, (in Optimization:-Minimize) objective function must be an algebraic expression or procedure

Second try, i change to use ans for >=0, ans2 <=0

ans:=proc(k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12) sol(parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
add((X(tim[i])-x11[i])^2,i=1..N)+add((Y(tim[i])-y11[i])^2,i=1..N)+add((Z(tim[i])-z11[i])^2,i=1..N)+add((U(tim[i])-u11[i])^2>=0,i=1..N)
end proc;
ans2:=proc(k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12) sol(parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
add((X(tim[i])-x11[i])^2,i=1..N)+add((Y(tim[i])-y11[i])^2,i=1..N)+add((Z(tim[i])-z11[i])^2,i=1..N)+add((U(tim[i])-u11[i])^2<=0,i=1..N)
end proc;
ans(.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003);
result1 := Optimization:-Minimize([ans, ans2],initialpoint=[.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003], feasibilitytolerance=0.01);

Error, (in Optimization:-Minimize) objective function must be an algebraic expression or procedure

x11 := [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 := [ -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 := [ 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]:
u11 := [7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7,8,7];
a1 := Diff(x1(t),t) = k1*x1(t)+ k2*y1(t)+ k3*z1(t)+k4*u1(t);
b1 := Diff(y1(t),t) = k5*x1(t)+ k6*y1(t)+ k7*z1(t)+k8*u1(t);
c1 := Diff(z1(t),t) = k9*x1(t)+ k10*y1(t)+ k11*z1(t)+k12*u1(t);
d1 := Diff(u1(t),t) = 0;
ICS:=x1(1)=x11[1],y1(1)=y11[1],z1(1)=z11[1],u1(1)=u11[1];
sol:=dsolve({a1,b1,c1,d1, a2,b2,c2,d2,ICS}, numeric, method=rkf45, parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12],output=listprocedure);
X,Y,Z,U:=op(subs(sol,[x1(t),y1(t),z1(t),u1(t)]));
tim := [seq(n, n=1..27)];
N:=nops(tim):
ans:=proc(k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12) sol(parameters=[k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12]);
add((X(tim[i])-x11[i])^2,i=1..N)+add((Y(tim[i])-y11[i])^2,i=1..N)+add((Z(tim[i])-z11[i])^2,i=1..N)+add((U(tim[i])-u11[i])^2,i=1..N)
end proc;
ans(.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003);
result1 := Optimization:-Minimize([ans>=0, ans<=0],initialpoint=[.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003], feasibilitytolerance=0.01);

hi,

     there is a common  differential equation in my maple note,the solution of the eq. can be expressed by

associated Legendre function(s),but i get a result by hypergeometric representation.how i can translate the later into a  single Legendre fun？

 Thank you in advance  

Download question_12.19.mw

Is there a way to play animations in maplets?

I can send an animation to a plotter, but don't know how to play it.  

Thanks, Rollie

I have a linear space spanned by the column vectors of:

I want to know its exact intersection of the first quadrant in 16 dimensional space (meaning Sum(a[i]*e[i]),i=1..16), how could I accomplish it? The output could possibly be the vectors defining the convex cone in higher dimensional space...

I have an ipad air 16G running ios 7.0.4 and downloaded the MaplePlayer APP.  t seems to crash on several of the routines for example, "Approximaing Sphere" and "Linear System Tutor". The app was last updated in 2011.  Do you have plans to any upgrades plan in the near future?

Hi MaplePrimers,

I'm trying to solve a system of algebraic equations using 'solve' [float].  I'd prefer to use 'solve' over 'fsolve', as 'solve' solves my system in about 0.05s, whereas fsolve takes about 5 seconds.  I need to solve the system repeatedly at a different points, so time is important.  I don't know why there is such a large difference in time ... 

I have a few piecewise functions of order 3 to 5.  It solves fine with the other (piecewise) equations, but adding one piecewise function which gives me an error while trying to solve:

Error, (in RootOf) _Z occurs but is not the dependent variable.

I think this is due to solve finding multiple solutions.  Is there a way to limit solve to only real solutions?

Thanks in advance!

Hello, I have quite a complex thing to solve, but would simplify it here. I would like to solve the unknown in a matrix, how can I use the 'solve' function? For example

A:=Matrix(3,1,4)
B:=Matrix(3,1,[2,7,9])
The relation between them is A=B*y

How can I use the below function?
solve(A=B*y,y）

I have a programme which examine how well a particular set of data fits a theoretical function.

In fact, I faced a problem : the programme returns only one solution and I guess in an arbitrary way.

I would like to find a way to define an interval in which the programme seeks solutions .

How can I do so?

with (Optimization)
[ImportMPS, Interactive, LPSolve, LSSolve, Maximize, Minimize, NLPSolve,QPSolve];

data := [[2.954488^2, 1.644900e-5], [3.132092^2,1.614900e-5], [3.307416^2, 1.594200e-5], [3.471311^2, 1.550700e-5], [3.559775^2, 1.450200e-5], [3.669332^2, 1.499400e-5], [3.825572^2, 1.476900e-5], [3.962449^2, 4.133000e-8], [4.200714^2, 1.320900e-5],
[4.434636^2, 1.433400e-5], [4.638319^2, 1.259100e-5], [4.832908^2, 1.258500e-5], [5.078484^2,
1.216200e-5], [5.315167^2, 1.164300e-5], [5.662155^2, 1.131000e-5], [5.916080^2, 1.082400e-5],
[6.208865^2, 1.054800e-5], [6.526868^2, 1.002600e-5], [6.880407^2, 1.006200e-5], [7.243618^2, 9.594000e-6], [7.607233^2,
9.288000e-6], [7.916439^2, 8.958000e-6], [8.320457^2, 8.664000e-6], [8.721812^2, 8.439000e-6], [9.007774^2, 8.325000e-6], [8.721812^2, 8.439000e-6], [9.007774^2,8.325000e-6], [9.393083^2, 7.878000e-6], [9.668506^2, 7.755000e-6], [9.988994^2,7.623000e-6], [10.40192^2, 7.367000e-6], [10.94532^2, 6.928000e-6], [11.38244^2,6.812000e-6], [11.85200^2, 6.720000e-6], [12.18811^2, 6.422000e-6], [12.67281^2, 6.403000e-6], [12.96341^2,6.514000e-6], [13.49185^2, 6.032000e-6], [13.76590^2, 6.103000e-6], [14.4072^2,6.143000e-6], [14.45476^2, 6.095000e-6], [14.76313^2, 5.758000e-6], [15.09868^2,6.965000e-6]]:

f:= x -> abs(((2*(c*exp(-b*1.5e-6)/(2*150*(c^2-((1+I)*(sqrt(3.14*x/8.5e-5)))^2)))*(2*(((1-I)*c/(2*(sqrt(3.14*x/8.5e-5))))-((0.026/150)*sqrt(8.5e-5/2e-5)))*exp(-c*0.5e-3)+((1+((0.026/150)*sqrt(8.5e-5/2e-5)))*(1-((1-I)*c/(2*(sqrt(3.14*x/8.5e-5)))))*exp(((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3))-((1-((0.026/150)*sqrt(8.5e-5/2e-5)))*(1+((1-I)*c/(2*(sqrt(3.14*x/8.5e-5)))))*exp(-((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3)))+(b/(2*a*(b^2-((1+I)*(sqrt(3.14*x/9e-6)))^2)*((2*(1-((0.026/150)*sqrt(8.5e-5/2e-5)))*(1+((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150)))*exp(-((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3-b*1.5e-6))-(2*(1+((0.026/150)*sqrt(8.5e-5/2e-5)))*(1-((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150)))*exp(((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3-b*1.5e-6))-((1-((0.026/150)*sqrt(8.5e-5/2e-5)))*(1-((a/150)*sqrt(8.5e-5/9e-6)))*(1-((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(-((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3+((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))-((1-((0.026/150)*sqrt(8.5e-5/2e-5)))*(1+((a/150)*sqrt(8.5e-5/9e-6)))*(1+((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(-((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3-((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))+((1+((0.026/150)*sqrt(8.5e-5/2e-5)))*(1+((a/150)*sqrt(8.5e-5/9e-6)))*(1-((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3+((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))+((1+((0.026/150)*sqrt(8.5e-5/2e-5)))*(1-((a/150)*sqrt(8.5e-5/9e-6)))*(1+((1-I)*b*a/(2*(sqrt(3.14*x/8.5e-5))*150))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3-((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))))/(((1-((0.026/150)*sqrt(8.5e-5/2e-5)))*exp(-((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3))*(((1-((a/150)*sqrt(8.5e-5/9e-6)))*(1+((0.026/150)*sqrt(8.5e-5/2e-5))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))+((1+((a/150)*sqrt(8.5e-5/9e-6)))*(1-((0.026/150)*sqrt(8.5e-5/2e-5))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(-((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6)))-((1+((0.026/150)*sqrt(8.5e-5/2e-5)))*exp(((1+I)*(sqrt(3.14*x/8.5e-5)))*0.5e-3))*(((1+((a/150)*sqrt(8.5e-5/9e-6)))*(1+((0.026/150)*sqrt(8.5e-5/2e-5))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6))+((1-((a/150)*sqrt(8.5e-5/9e-6)))*(1-((0.026/150)*sqrt(8.5e-5/2e-5))/((a/150)*sqrt(8.5e-5/9e-6)))*exp(-((1+I)*(sqrt(3.14*x/9e-6)))*1.5e-6)))))))*((1+I)*(sqrt(3.14*x/2e-5)))*exp(-190e-6*((1+I)*(sqrt(3.14*x/2e-5))))):

residuals := map(p -> (f(p[1])-p[2]), data):
R:= LSSolve(residuals);

¿Que tipos de programas ayudan en  el desarrollo de problemas geometricos?

¿como puedo resolver problemas en maple17?  

I have a TextArea component on the worksheet. Is it possible to create on the worksheet some number of Sliders, where the number of sliders is defined by the number entered in the TextArea?

Hello, how can i solve these equations

eq[1]:=a'[1](t)*p[1](x)+b*a[1](t)*p[1](x)

eq[2]:=a'[2](t)*p[2](x)+b*a[2](t)*p[2](x)

eq[3]:=a'[3](t)*p[3](x)+b*a[3](t)*p[3](x)

eq[4]:=c[1]*a[1](t)+c[2]a[2](t)+c[3]a[3](t)-q(t)

eq[5]:=d[1]*a[1](t)+d[2]a[2](t)+d[3]a[3](t)-p(t),

where p[i](x) , c[i] , d[i] , q(t) and p(t) are known functions?

Why does the following statement not evaluate, or better yet, how can I make it do so?

A:=value(floor(p)) assuming p>0,p<1,p::real;

or

A:=simplify(floor(p)) assuming p>0,p<1,p::real;

or any one of a lot of different attempts along the above lines, all of which seem (to me) that they should yield

A:=0

rather than

A:=floor(p)

which is what I get.

Thanks in advance

I have theoretically 3(could eventually be more) layers with an incident wave with a wave equation for that wave.

It refracts into the 2nd layer from the first and now has a 2nd wave equation, then from the 2nd into the 3rd layer with a 3rd wave equation.

All the wave equations are of the form, Psi(z) = A_1psi_1(z) + B_1psi_2(z); this is just a general solution where psi_1&2 are linearly independant solutions that make up the general equation above and A_1 and B_1 are constant coefficients that would be A_2,B_2 and A_3,B_3 for the 2nd and 3rd layers respectively.

Transfer matrix method gives A_1,B_1 in terms of A_2,B_2(as it transfers from layer 1 to 2 they equate under boundary conditions so you can solve the simultaneous equations for results). You create a matrix of these results and multiply it with the respective matrix of the 2nd layer to 3rd layer to give you the overall transfer matrix from one side of the system to the other.

I think something to do with transfer function but not sure how to use it or set up the problem. 

Thanks in advance for any pointers.