i have to compute lie algebra condition by using maple 15, but currently i have code in maple 5.,

i already try to run in maple 15 but it say 'error,unable to match delimeters'
i try to find error, but i cant find it..
the coding are here...

check[lie]:=proc(A,n)

local  i,j,k,l,m;  

for i from 1 by 1 to n do

for j from 1 by 1 to n do  

for k  from 1 by 1 to n do    

if A[i,i,k]<>0 then  

RETURN ('Input is NOT a Lie algebra (',i,i,k,')=',A[i,i,k], 'is not zero');  

elif A[i,j,k]+A[j,i,k]<>0 then  

RETURN ('Input is NOT a Lie algebra,(',i,j,k,')+(',j,i,k,')=',A[i,j,k]+A[j,i,k],'is not zero');  

else  

for 1 from 1 by 1 to n do

if  

simplify(sum(A[i,j,m]*A[m,l,k]+A[j,l,m]*A[m,i,k]+A[l,i,m]*A[m,j,k],   m=1..n))<>0  

then  

RETURN('Input is NOT a Lie algebra---the Jac(',i,j,l,') is not zero');    

fi;  

od;  

fi;  

od;  

od;  

od;  

print('Yes,input IS a Lie algebra');  

end:

can anyone help me here? Thank You..

Here, i attached the result in printscreen

Hi,

I ues the optimization tool in Maple2015.

In my experience, Maple optimizatiom tool just shows the optimum values, not a history of the optimizatiom process.

I want to know how I can obtain the history of the optimization process, such as values of the objective, constraints, and tolerance at every iteration.

Is there anyone who knows how to obtain the history data of the optimization process?

Hi everyone,

I am trying to solve the equation of heat tranfer, time dependent, with particular Initial and boundary conditions but I am stuck by technical problems both in getting an analytical solution and a numerical one.

The equation

the equation.

I defined a and b numerically. domain is : and I defined surf_power numerically.

The initial condition is : , T0 defined numerically

The boundary condition is : , because it has a shperical symetry.

To me, it looks like a well posed problem. Does it look fine ?

Problem in analytical solution :

It doesn't accept the boundary condition so I only input the initial condition and it actually gives me back an expression that can be evaluated but it never does : I can't reduce it more than an expression of fourier which I can't eval. The solution :
The solution calculated in (0,0). I was hoping T0...

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ?

The numerical solution problems :

Sometimes it tells me that my boundary condition is equivalent  to 0 = 0, and I don't see why. Some other times it tells me I only gave 1 boundary/initial condition even if I wrote both. Here is what I wrote for example :

(because it kept asking me to add these two options : 'time' and 'range')

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ? I must at least have syntax problems because even if I keep reading the Help, it's been a long time since I used Maple.

Thank very much for any indication you could give me !

Simon

I want to refer to the first element in a list such as x:=[1,2,3] as x[0] not x[1].

How to do it?

 my answer are 143.2030067,143.2030339,143.2030610

but, it is wrong. can anyone tell me where is the wrong part?

please!!

I am attempting to solve a system of second order ODEs. I place conditions on the solutions and use the solve command to figure the correct constants for the general solutions of the ODEs; however, the conditions do not appear to hold after I substitute the constants back into the general solutions. Any help would be greatly appreciated. Here's the code and an explanation:

First some constants

> A := 1; B := 9/10;
> j := 1-1/B;

 This is our homogeneous odes. I will give the general solutions of the inhomogeneous system momentarily 

> eqnv1 := diff(v1(x), `$`(x, 2)) = (1-1/(j+1))*v1(x)+v2(x)/(j+1);
> eqnv2 := diff(v2(x), `$`(x, 2)) = -v1(x)/(A*(j+1))+(B/A+1/(A*(j+1)))*v2(x);

Next we get the general solution of this sytem of odes.

> soln := dsolve([eqnv1, eqnv2])

Next we have our solutions of the inhomogeneous problem1. Basically solution v1neg, v2neg on [0,xi] and v1pos, v2pos on [xi,1]. We will assume v1,v2 are C^1 across xi; however, the location of xi is not known at this time so they must remain split.

> v1neg := op([1, 2], soln)-1;
> v2neg := op([2, 2], soln)-1/B;
> v1pos := op([1, 2], soln)+1;
> v2pos := op([2, 2], soln)+1/B;

There's probably a better way to do this, but I relabeled the constants:

> v1negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v1neg);
> v2negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v2neg);
>
> v1posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v1pos);
> v2posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v2pos);

Next we have eight conditions the solutions must satisfy. Namely v1, v2 are C^1 across xi and v1',v2' are 0 at {0,1}.

> syscon1 := subs(x = xi, v1negc) = subs(x = xi, v1posc);
> syscon2 := subs(x = xi, v2negc) = subs(x = xi, v2posc);
> syscon3 := subs(x = xi, diff(v1negc, x)) = subs(x = xi, diff(v1posc, x));
> syscon4 := subs(x = xi, diff(v2negc, x)) = subs(x = xi, diff(v2posc, x));
> syscon5 := subs(x = 0, diff(v1negc, x)) = 0;
> syscon6 := subs(x = 0, diff(v2negc, x)) = 0;
> syscon7 := subs(x = 1, diff(v1posc, x)) = 0;
> syscon8 := subs(x = 1, diff(v2posc, x)) = 0;

We solve to get the constants for the solutions.

> constants := simplify(evalf(solve({syscon1, syscon2, syscon3, syscon4, syscon5, syscon6, syscon7, syscon8}, {a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8]})));
>

We substitute the values for the constants.

> a[1] := op([1, 2], constants); a[2] := op([2, 2], constants); a[3] := op([3, 2], constants); a[4] := op([4, 2], constants); a[5] := op([5, 2], constants); a[6] := op([6, 2], constants); a[7] := op([7, 2], constants); a[8] := op([8, 2], constants);

Lastly we try to verify that the conditions from earlier hold:

> evalf(subs(xi = .2, subs(x = xi, v1negc-v1posc)));
-1.7597825261536669519
> evalf(subs(xi = .2, subs(x = xi, v2negc-v2posc)));
-1.8936659961101033997
> evalf(subs([x = 0, xi = .2], diff(v1negc, x)));
-0.38633519704430619686

They should hold for any xi, but they don't appear to. All of these should be 0. For a large xi, the numbers get very large so I was thinking perhaps roundoff error, but even when I do an exact solution and then evalf just at the end, I still have large error so I'm not sure what the problem is. Sorry for the long question. Thanks so much for the help.

Hello!

Vectors are defined by a lenth and an angle.

The arrow command wants me to give it a starting point and an endpoint in coordinates, for example:

with(LinearAlgebra);

with(plots);

gu := arrow([5, 10], [15, 20], .2, .4, .1, color = green)

I'm dealing with vectors from electrical circuits, so the information I have would be the lenth of the vector, and an angle given in degrees (which I could convert to radians if needed for the plot).

The goal is to plot all the vectors in the circuit on the same display so that it can be seen how they are in relation to each other, and to visualise the solution. This means I would need to find out the endpoint of the arrows to start the next arrow at that location with its angle in reference to the same axis. Does it make sence? I'm not sure about the wording of this.

Here's a picture of a calculated vector lenth and angle: http://imgur.com/hJtNeGg

So, the -45 degrees would need to be in reference to the y-axis for example, then the next vector can be placed at lets say perpendicular in reference to the vector sol_1, starting at its endpoint.

The only other way I have to draw this, excluding doing it by hand, is using the object tools from microsoft word, and to be honset I dislike using word for math stuff. I would much rather learn the syntax for Maple to do this.

I hope someone knows what I mean, Please ask if somethig is unclear,

Thanks!

 Obtain the tri-stimulus XYZ values from the CIE Color matching functions.

 Show the gamut of maximum chroma for the standard observer model with a D65 Illuminant.

 Approximate the white point of a Planckian source and compare to D65.

 Translate the maximum chroma gamut in xy to Lab (CIE L*a*b*) for perceived gamut (Violet and Magenta come together)

 Map the RGB color cube of fully saturated color into Lab and compare to perceivable colors.

10/6/15  Initial Document

•12/28/15 Improve RGB gamut with more data points: Procedures added for RGB to Lab: Wavlength Colors now based on CIEDE2000 model for Lab.                   

 Here is the latest version of this document, the MSL_data must be in a directory set in the mw file;

MSL_data.xlsx    Vision_RGB_Gamut.mw

how to plot exp(f)*erf(g) where f and g are complex function

(-6.328281880*10^(-49)-1.071713312*10^(-47)*I)*exp(-5996.664400+2.000000000*10^14*del+(1.547600000*10^16*I)*del)*(erf(11.+77.38000000*I+1.000000000*10^14*del)-1.*erf(-4.+77.38000000*I+1.000000000*10^14*del))

del from -100e-15 to 100e-15

In this paper we will demonstrate the many differences of implementation in the modeling of mechanical systems using embedded components through Maplesoft. The mechanical systems are used for different tasks and therefore have different structure in its design; as to the nature of the used functional elements placed on them, they vary greatly. This diversity is reflected in approaches and practices in modeling.

The following cases focus on mechanical components of the units manufacturing and processing machines. We can generate graphs for analysis using different dynamic pair ametros; all in real-time considerations in its manufacturing costs from the equations of conservation of energy.
Therefore modeling with Maplesoft ensures the smooth optimum performance in mechanical systems, highlighting the sustainability criteria for other areas of engineering.

XXXIII_Coloquio_SMP_2015.pdf

XXXIII_Coloquio_UNASAM_2015.mw

(in spanish)

L.AraujoC.

I am creating a plot but the numbers on the horizontal axis overlap.  Is there a way to change the directions of the numbers displayed under the horizontal axis (tickmark labels) from horizontal to vertical?

The picture below shows how the numbers are bunched together under the horizontal axis.  I'd like them to be displayed vertical

Thank you

1. Take a group for example. First we can set up a group G by

G:=<<a,b>|<a²=1,b³=1,(a.b)²=1>>. Actually,G=S₃. So how to simplify a long production,

such as "a.b.b.a.b.b.a.b"?

2. How to define a finitely presented algebra over some field, such as the enveloping algebra

or quantum group of a Lie algebra? And moreover how to do the similar computation about 

simplifying a long production?

Hi

Anyone could help me in solving the following system of equations to get constants C1, C2, C3 and C4. MALPE give me this "soution may have been lost".  The MAPLE sheet is also attached.

Download solution_lost.mw

Hi all,

I have this system

> system1D := H = alpha*gamma[2, 2]*d[2, 1]-beta*d[1, 2]*gamma[1, 2]^2-gamma*d[1, 2]*gamma[2, 1]^2+alpha*gamma[2, 2]^2*d[2, 2]-beta*d[2, 2]*gamma[2, 2]^2-gamma*d[2, 2]*gamma[2, 2]^2, E = alpha*gamma[2, 1]*d[1, 1]-beta*d[1, 2]*gamma[1, 1]-gamma*d[1, 1]*gamma[2, 1]+alpha*gamma[2, 1]^2*d[1, 2]-beta*d[2, 2]*gamma[2, 1]-gamma*d[2, 1]*gamma[2, 2], B = alpha*gamma[1, 1]*d[2, 1]-beta*d[1, 1]*gamma[1, 1]^2-gamma*d[1, 1]*gamma[1, 1]^2+alpha*gamma[1, 1]^2*d[2, 2]-beta*d[2, 1]*gamma[2, 1]^2-gamma*d[2, 1]*gamma[1, 2]^2, D = alpha*gamma[1, 2]*d[2, 1]-beta*d[1, 1]*gamma[1, 2]^2-gamma*d[1, 2]*gamma[1, 1]^2+alpha*gamma[1, 2]^2*d[2, 2]-beta*d[2, 1]*gamma[2, 2]^2-gamma*d[2, 2]*gamma[1, 2]^2, A = alpha*gamma[1, 1]*d[1, 1]-beta*d[1, 1]*gamma[1, 1]-gamma*d[1, 1]*gamma[1, 1]+alpha*gamma[1, 1]^2*d[1, 2]-beta*d[2, 1]*gamma[2, 1]-gamma*d[2, 1]*gamma[1, 2], C = alpha*gamma[1, 2]*d[1, 1]-beta*d[1, 1]*gamma[1, 2]-gamma*d[1, 2]*gamma[1, 1]+alpha*gamma[1, 2]^2*d[1, 2]-beta*d[2, 1]*gamma[2, 2]-gamma*d[2, 2]*gamma[1, 2], F = alpha*gamma[2, 1]*d[2, 1]-beta*d[1, 2]*gamma[1, 1]^2-gamma*d[1, 1]*gamma[2, 1]^2+alpha*gamma[2, 1]^2*d[2, 2]-beta*d[2, 2]*gamma[2, 1]^2-gamma*d[2, 1]*gamma[2, 2]^2, G = alpha*gamma[2, 2]*d[1, 1]-beta*d[1, 2]*gamma[1, 2]-gamma*d[1, 2]*gamma[2, 1]+alpha*gamma[2, 2]^2*d[1, 2]-beta*d[2, 2]*gamma[2, 2]-gamma*d[2, 2]*gamma[2, 2], H = alpha*delta[2, 2]*d[2, 1]-beta*d[1, 2]*delta[1, 2]^2-gamma*d[1, 2]*delta[2, 1]^2+alpha*delta[2, 2]^2*d[2, 2]-beta*d[2, 2]*delta[2, 2]^2-gamma*d[2, 2]*delta[2, 2]^2, E = alpha*delta[2, 1]*d[1, 1]-beta*d[1, 2]*delta[1, 1]-gamma*d[1, 1]*delta[2, 1]+alpha*delta[2, 1]^2*d[1, 2]-beta*d[2, 2]*delta[2, 1]-gamma*d[2, 1]*delta[2, 2], B = alpha*delta[1, 1]*d[2, 1]-beta*d[1, 1]*delta[1, 1]^2-gamma*d[1, 1]*delta[1, 1]^2+alpha*delta[1, 1]^2*d[2, 2]-beta*d[2, 1]*delta[2, 1]^2-gamma*d[2, 1]*delta[1, 2]^2, D = alpha*delta[1, 2]*d[2, 1]-beta*d[1, 1]*delta[1, 2]^2-gamma*d[1, 2]*delta[1, 1]^2+alpha*delta[1, 2]^2*d[2, 2]-beta*d[2, 1]*delta[2, 2]^2-gamma*d[2, 2]*delta[1, 2]^2, A = alpha*delta[1, 1]*d[1, 1]-beta*d[1, 1]*delta[1, 1]-gamma*d[1, 1]*delta[1, 1]+alpha*delta[1, 1]^2*d[1, 2]-beta*d[2, 1]*delta[2, 1]-gamma*d[2, 1]*delta[1, 2], C = alpha*delta[1, 2]*d[1, 1]-beta*d[1, 1]*delta[1, 2]-gamma*d[1, 2]*delta[1, 1]+alpha*delta[1, 2]^2*d[1, 2]-beta*d[2, 1]*delta[2, 2]-gamma*d[2, 2]*delta[1, 2], F = alpha*delta[2, 1]*d[2, 1]-beta*d[1, 2]*delta[1, 1]^2-gamma*d[1, 1]*delta[2, 1]^2+alpha*delta[2, 1]^2*d[2, 2]-beta*d[2, 2]*delta[2, 1]^2-gamma*d[2, 1]*delta[2, 2]^2, G = alpha*delta[2, 2]*d[1, 1]-beta*d[1, 2]*delta[1, 2]-gamma*d[1, 2]*delta[2, 1]+alpha*delta[2, 2]^2*d[1, 2]-beta*d[2, 2]*delta[2, 2]-gamma*d[2, 2]*delta[2, 2];


> subs({A = 0, B = 0, C = 0, D = 0, E = 0, F = 0, G = 0, H = 0, delta[1, 1] = 1, delta[1, 2] = 0, delta[2, 1] = 0, delta[2, 2] = 0, gamma[1, 1] = 1, gamma[1, 2] = 0, gamma[2, 1] = 0, gamma[2, 2] = 0, delta[1, 1]^2 = 0, delta[1, 2]^2 = 0, delta[2, 1]^2 = 1, delta[2, 2]^2 = 0, gamma[1, 1]^2 = 0, gamma[1, 2]^2 = 1, gamma[2, 1]^2 = 0, gamma[2, 2]^2 = 0}, {system1D});

The problem is: there is any simple way to use command "subs" when some expression such that delta[1,1]=1, gamma[1,1]=1, gamma[1,2]^2=1 have value and others are zero.

Can someone please advice and help me on this?

thanks

witribm

Hello to you all,

I have this DE

but when I try to change the variable

I get this

>algsubs(r(t) = 1/u(t), diff(diff(r(t), t), t)-k/r(t) = -K/r(t)^2);

Why is it not done for the rest of the terms.  Is there a more easy way to do it.  Take into account that I cannot use dsolve because it is in an course in integral.

Thank in advance for your help.

--------------------------------------
Mario Lemelin

Maple 2015 Ubuntu 14.04 - 64 bits
Maple 2015 Win 10 Pro -  64 bits
messagerie : mario.lemelin@cgocable.ca
téléphone :  (819) 376-0987