Cn ={ 1, n = 0 ,                                                  }

      {Xn−1[sum of] k=0   C(k)C(n−1−k) , otherwise.  }

looking to complete the definition of catalan so that catalan(n) returns Cn whenever n is a non-negative integer. usin g the definition above...any help appreciated

catalan:=
proc(n::TYPE)
description "Print the n'th Catalan number.";
option remember;
---MORE STUFF HERE---
end proc; # catalan

I am working on something where I need to use a stack. In the Maple documentation I noticed there are two different stack implementations. There is "stack" and "SimpleStack". I am curious if anyone has any experience using either of these and if there is any advantage to one over the other. Any suggestions or tips are appreciated.

Thanks,

Steven

Hello everybody

In the attached file, as you can see, 'u' is function of 'thet1(t)' explicitly, But when I want to differentiate from 'u' w.r.t 'thet1(t)', maple returns zero as the answer. How is it possible?

Thanks in advance

d.mw

I have an ODE with L(x,y), which includes partial derivatives of L(x,y) and coefficients y,_y1 and _y1^2. I want to split this equation into seperate equations of these coefficients. So I can solve for the unknown variable L(x,y).

ODE:=diff(L(x,y),x,x)+2*_y1(x)*diff(L(x,y),x,y)+_y1(x)^2*diff(L(x,y),y,y)+y(x)/x*diff(L(x,y),x)+(y(x)*_y1(x)/x+_y1(x))*diff(L(x,y),y)-y(x)/x^2*L(x,y)=0;

Could someone help me? 

I am currently trying to use the coeffs command but not really getting anywhere.

Thank you.

I a working with circuits, and was wondering whether or not it might be possible to shorten the proces of calculating parallel resistors. My idea is that using some symbol, such as || would find the equivalent resistor.

My idea is based on the fact that CrossProduct can also be done using &x command. Like so.

CrossProduct(a,b) = a &x b

So for resistors it would look something like:

Parallel(a,b) = a || b

For those interested the function for parallel resistors would be:

Parallel := (a,b) -> 1/(1/a+1/b);

the binomial coefficient  n k  can be defined recursively as follows for all nonnegative integers n, k:

(n)  = {0,      k>0

(k)  = {1       k=0, k=n

         {(n-1)+(n-1), otherwise.

          (k-1)   (k)

I need to complete a deinition of binom so that m so that binom(n,k) returns  n k  for all n greater than 0, and k greater than or equal to 0 using the definition of the binomial above..Any help appreciated..

binom:=
proc(n::TYPE1,k::TYPE2)
description "Compute a binomial coefficient";
option remember;
---MORE STUFF HERE---
end proc; # binom

I can calculate the quotient and remainder (r1) of a polinom (p1) divided by another (p2).

Unfortunately, I can't figure out how to divide r1 with p2, so that I get negative powers as well.

For example:

p1 = x^2+3

p2 = x+2

(x^2+3):(x+2) = x, and the remainder is: (-2x+3)

-----------

(-2x+3):(x+2) = -2, and the remainder is: (-1)

-----------

(-1):(x+2) = -x^-1, and the remainder is: (2x^-1)

-----------

(2x^-1):(x+2) = 2x^-2

...

As a result I get:

p1/p2 = x + -2 - x^-1 + 2x^-2 and so on...

How can I achieve this?

How does one explain this? 

Is this not a valid way for creating a floating-point matrix?
Otherwise, how do I explain the wrong result here?

-- Regards,

Franky

I am having trouble removing assumptions that are stored within expresssions.

Example code:

assume(l1>0): # this assumptions later helps to find a solution for a geometric problem with two four-bar-linkages

a := sqrt(l1);

save a, "test.m";

restart;

read "test.m"

a; # the assumptions are stored within the saved data

l1:='l1'; # try to remove the assumption

a; # assumption in a still existing

subs({l1=2}, a); # nothing happens: I can not access l1 any more

subs({l1~=2}, a); # This does not work either, nothing changes in a

So my question is: How do I remove the assumption within a stored expression?

My main problem lies in the handling of the expression with assumptions. At some point, I want to generate Matlab code, and the codegen-command gives me:

Warning, the following variable name replacements were made: l1~ -> cg

Hi all

I have  a simple question in export as eps format in Maple. i want the converted eps file with no box, but when i use export as eps by right click on plotted figure in Maple and then select " Export As--> Encapsulated PostScript(EPS)...", the below mentioned box(as in sample attached picture) will appear. How can I remove it? The Sample picture is attached. 

I will be grateful of any guidance.

Thanks in advance

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I'm trying to find lypunov exponent for this  system of ODEs. I know I need to take the Jacobian but not sure if it's possible the way it's currently defined. If anyone could provide insight it would be much appreciated.

Eqns:= diff(omega(t),t)=-(G*MSat*beta^(2)*(xH(t)*sin(theta(t))-yH(t)* cos(theta(t)))*(xH(t)*cos(theta(t))+yH(t)*sin(theta(t))))/((xH(t)^(2)+yH(t)^(2))^(2.5)),diff(theta(t),t)=omega(t), diff(xH(t),t)=vxH(t),diff(vxH(t),t)=-(G*M*xH(t))/((xH(t)^(2)+yH(t)^(2))^(1.5)),diff(yH(t),t)=vyH(t),diff(vyH(t),t)=-(G*M*yH(t))/((xH(t)^(2)+yH(t)^(2))^(1.5)): ;

ICs := omega(0) = omega0, theta(0) = theta0, xH(0) = a*(1+e), yH(0) = 0, vxH(0) = 0, vyH(0) = sqrt(G*M*(1-e)/(a*(1+e)));

I want the exponent for omega, I  procedure that takes some initial conditions, changing just w0, and computes the long term value of omega. This plots a sort of bifurcation diagram. I'd like an estimate of the exponent to compare what I see. 

thanks for the help

ft

Hello,

I want to compute powers of the Laplace-beltrami operator in some surface (e.g. cylinder):

https://en.wikipedia.org/wiki/Laplace_operator#Coordinate_expressions

and evaluate it for a given scalar function on the surface (e.g. a function on the cylinder).  I'm wondering if I only have to write the laplace-beltrami operator and then compute a simple power of it, or if I have to take care and compute the powers of the Laplace-beltrami operator first and then evaluate at the function on the cylinder. If the second option is the correct one (I think so), how can I compute that with maple? There is some trick way to compute powers of differential operators?

Thanks,

Fernando

Variable exists but is not shown when using save/read with extension .m

My example maple code is

 

a:=b^2:

save a, "test1.m"

restart:

read "test1.m"

a;

after the read command, I can access the variable, but it is not shown under "Variables".

This lead to some confusion when debugging the worksheet. Can I change this somehow?

Using the input type file format is not a solution, since then reading takes forever for complicated expressions.

Further, in the read command documentation it says "This functionality is not intended for end users" for saving the file as .m. What does that mean?

I am a Mathematica user trying to make the switch to Maple, so first of all I apologise if I am making a stupid mistake here. The complete script is attached below.

I have the following line in my script:

    b := Trace(map(diff, HermitianTranspose(W), coords[1]) . map(diff, W, coords[1]))

W is a matrix defined earlier in the script. I can provide the context if it would help but I'm not sure if it's relevant to the issue I'm having. The problem is that

    simplify(b) assuming real, r>0

gives a vastly different result to

    simplify(Trace(map(diff, HermitianTranspose(W), coords[1]) . map(diff, W, coords[1]))) assuming real, r>0

How can this be? What can I do to make Maple simplify the expression correctly?

conifold_metric.mw

Hello, i am doing some schmidt-analysis on a stirling engine, but my question is rather simple. I have the measured presure P at a given time T as a 229x2 matrix, i also have a function, V__total(T), for the total volume of the engine at a given time T. 

I then go on to create a pointplot PV, which is rougly the shape of a potato. I now want to find the area enclosed by this point plot, is there any way? 

I do something like:

> DATA:=ImportMatrix(filepath,skiplines=1); %import data from .txt file, skip header line. 

>P:=DeleteColumn(DATA,1); %Isolate presure column

>T:=DeleteColumn(DATA,2); %Isolate time coliumn

>V:=V__total~(T); %Generate volume vector as a function of time T

>pointplot(<P|V>,connect=true);

Is there any wat to finde the area enclosed by the curve/