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Hi,

I am trying to get the linearized model of a multibody system which has multiple inputs and multiple outputs. I initially used the "linearization template" of the MapleSim for an inverted pendulum, it worked fine. However, when I tried to get the linearized model for the double inverted pendulum, it gave this error message "problem at initial time: system is underdetermined; cannot solve for the following variables: 'Main.DFPSubsys1inst.eM_MyPendulum1_R1(t)'". 

See the attached worksheet: myDet.mw

Why does the determinant of A  contain  x, x^3, x^5, x^7 and x^9 ?

Maple15/Linux/Fedora16_x86_64

Is there any command in Maple 15 to linearize an PDE non-linear? Or or there is a package or help file to do so?

I'm handling with Non-linear PDEs in my work and I would like to solve them by these methods.

Thank You

Washington Inacio

Hello, I'm Tam and I had a problem with Conditional of Jacobi Iteration.

So, the proplem is:

>restart;

>with(LinearAlgebra):

A:=Matrix([[5,1,-1],[0,4,2],[-1,2,-7]]);

And I want to check the condition

with the maple program.

Fisrt, I typed

b:=sum(abs('A'[1,j]),j=1..ColumnDimension(A)); and I got b:=7

How to solve this problem in maple?
Decide which of three lines given are tangent to the graph of f(x) at given points.
The function to define is f(x)=2x^3-4x^2+3x-5
The possible tangent lines are:
y = x - 5
y = 2x - 5
y = 3x - 5

1. What is/are the zero(es) for this function? In other words, what is the solution for f(x)= 0

2.  What axes values (x- and y- ranges) are appropriate for this graph?

3. When plotting each linear function with f(x...

I tryto solve a linear program of big dimension using LPSolve function. My problem is that is returns a solution but I am not sure if it is unique or not, and for me it is really important.

I saw that in the new version there are options for LPSolve: method=interiorpoint or activeset. If I understood correctly, interiorpoint gives a solution in the middle of the interval and the activepoint - in the corner. My idea was to run LPSolve twice using those two methods. If...

Hi,

I have 5 docs with datas (x,y), like this one : BEF1txt

I want to put the 5 graphs side by side (parallels linears fonctions in a 3D graph). So I added the z data, z=1 in this case, and BEF2 got z=2, etc... But I am not able to make the 3D plot with the 5 linears fonction side by side.

May you help me please?

Thanks

everywhere i tried to find out how to solve system of linear equations like A*X=F, whre A is a matrix and X and F are vectors, the solution is X:=linsolve(A, F),

but in maple 14 linsolve doesn't work, so are there any analogues for this linsolve function in maple 14 or maybe another simple way to solve this system?

thanks

The coefficient of determination also known as R^2 tells how good a fit is. If R^2=1 the fit is perfect an if R^2=0 it's useless. But Maple don't have a native function to calculate R^2. I seached and found this: 
http://www.mapleprimes.com/questions/40171-Determining-An-Equations-Accuracy

But it only describe how to calculate R^2 on a 

Can one find an integral representation of the solution of the following system of PDEs?

pde1 := diff(f(x,y),x) + A(x) * ( p11* f(x,y) + p12 + p13*x ) 
= p14 * B(y)/B(x) * diff(f(x,y),x);

pde2 := diff(f(x,y),y) + A(y) * ( p21* f(x,y) + p22 + p23*y ) 
= p24 * B(x)/B(y) * diff(f(x,y),y);

f(x,y) is the unknown function,
A(x) and B(x) are known functions,
p11, p12, etc. are real parameters.

If p14=0 and p24=0, the right-hand...

How to solve the linear system by using maple

2y' + xy =2  with y(0)=1

Test.mw

If I use a binary {0,1} linear program the cardinal constraint ie control the
number of 1's (long positions in portfolio) works beautiful. See attached worksheet.

The problem starts when I convert the problem into a constrained integer {-1,0,1}
linear problem where -1= short position, 0=no position and 1=long position.

In the first example when we add the constraint  add(w[i], i = 1 .. NC) = 4

I can solve a [0,1] linear programming problem by using:

LPSolve(Ob, con, assume = binary);


I managed to solve [-1,1] linear programming problem by using the transformation
2*w[i]-1 ie   2*0-1=-1 and 2*1-1=1 as follows:

LPSolve(add(PP[i]*(2*w[i]-1), i = 1 .. NC)-add(Draw[i]*(2*w[i]-1), i = 1 .. NC), {add(w[i], i = 1 .. NC) = PS}, maximize = true, assume = binary);


Now my question is how can I solve a [-1,0,1] problem?
I tried to use:

Let f:Rⁿ  -> R^m a linear function i need help about a proc about matrix of the linear map Df(x),

our input will be n,m and f

such that,f:R²->R³ ,f(x,y,z)->(x+y,y+2z,x+y+z) and i wish resulting output with derivative matrix with respect to the standard bases in R² and R³

Dears,  I have a system of linear equation A*X=B   where A matrix 607x400  & X vector  400x1 and B vector 400x1

where A not square matrix. To sove,  I do

(A^T)*A*X=(A^T)*B,

Let   D= (A^T)*A        D  square matrix

then

D*X=(A^T)*B,

the solution is

X= (D^-1)*(A^T)*B.

could any one help me to do this by maple and cut the resluts in...