The "ExcelTools" seems can work based on Excel 2003 format,which can only support 2^16 rows.Excel 2010 format cantains much more rows and columns. So I want to know when maple can support Excel 2010?

Hello,

I am trying to solve the coupled system of PDEs. Maple throws below error(highlighyted in bold letters) when I try to solve it using 'pdsolve'. Attached is the maple worksheet(BatteryPDEEquationsBackUp8thJuly13_constantCurrentInput.mw) having complete code. Please help. 

> pds := pdsolve(BatSys, IBC, numeric, variables, time = t, range = 0 .. L[N]+L[sep]+L[P]);

Error, (in pdsolve/indexlex) invalid input: a string/name list is expected for sort method `string`...

-(1/2*I)*(I*Pi*ln(2)-I*Pi*ln(g+4)-2*ln(2)^2+2*ln(2)*ln(g+4)-2*ln(2)*ln(-1/2*I)+ln(g+4)*ln(-1/2*I)-ln(g+4)*ln(2*I)+ln(-1/2*I)*ln(4-g)+ln(4-g)*ln(2*I))/Pi

why doesnt maple simplify this expression any further?

I tried radnormal and simplify.

but he complains when using combine saying:

Error, (in combine/ln) second argument must be a maple type

Why is that?

Yesterday I’ve updated to the new Version 17.01 and now I’ve the problem that I cannot enter  ‚}‘ and another problem is an exponent e. g. x^2. To enter it I have to write x^22. The first number is not shown in the worksheet.

I hope someone can help me to solve this problem.

what is the normalization constant N for this wave function x range from  x=o and x=L ?

Eigenvectors(A, C) can be used to solve the eigenvalue problem:
A . x = lambda . C . x

if a new term " lambda^2 . D . x " is added to the right hand side, where D is a new matrix, is there a simple way to solve the new eigenvalue problem:

A . x = lambda . C . x + lambda^2 . D . x

Thank you!

Simple Maple procedure definition is like this:

f := proc (x, y) sin(x+y)-exp(x)*y end proc:

How can I use indexed parameter declarations, say:

f := proc (x[1], y[2]) sin(x[1]+y[2])-exp(x[1])*y[2] end proc:

where x and y are arrays of unknowns. It seems that this does not work in Maple 13.

Originally, I want to solve a set of n nonlinear equations with n unknowns W[n] using fsolve(), and the output is a set of equations, say:

Hello,

I have some strange problems with the function "Equal" in LinearAlgebra, which should allow me to compare matrices.

First of all, this works without any problems:

> restart; with(LinearAlgebra);
> A := Matrix(2, 2, [[0, 0], [0, 0]]);
> Equal(A^2, A);

Maple correctly returns "true" since here A is equal to its square.

But:

> restart; with(LinearAlgebra);
> A := Matrix([[a, 0], [0, 0]]);

hi guys,i have a program when i run it after 3or4 minutes maple give this :

execution stopped:memory allocation failed.please see ?alloc for more detail.

when i ok it then give this :

the kernel has been shut down.further computations cannot be performed.

what do you think ?

thank you very much

Are you in or near Berkeley, California?

Are you experienced with Maplesoft for academics?

Can you train a student to use it with the tutorials and your own knowledge?

This student is located at University of California, Berkeley and needs an immediate hands on private trainer.

He will be taking Calculus 1A next semester and needs to learn to use this program this summer.

He is a student who uses an eye control computer without the use of his hands and needs this

Hi,

I'm using Maple 14 and I'm trying to solve a system of recurrence relations with initial conditions (a Vector Autoregressive model). The system is

const:=Vector[column]([0.009421681, -0.0005441856]):
lagY:=Vector[column]([0.796519372, 0.0179147112]):
lagZ:=Vector[column]([0.133049059, 0.5240695764]):
A:=<<const>|<lagY>|<lagZ>>;

eq1:=y(t) = A[1, 1] + A[1, 2]*y(t - 1) + A[1, 3]*z(t - 1);
eq2:=z(t) = A[2, 1] + A[2, 2]*y(t - 1) + A[2, 3]*z(t - 1);

Hi,

i wrote the following code

restart:
zet:=0.5:
rhop:=3880;
rhobf:=998.2;
cp:=773;
cbf:=4182;

eq1 := diff(u(eta), eta, eta)+0.133762025280e-2/(0.993000000000e-3+0.388362300000e-1*phi(eta)+.530162700000*phi(eta)^2)+(1/(eta-1)+(0.388362300000e-1+1.06032540000*phi(eta))*(diff(phi(eta), eta))/(0.993000000000e-3+0.388362300000e-1*phi(eta)+.530162700000*phi(eta)^2))*(diff(u(eta), eta));
eq2 := diff(T(eta), eta, eta)+(.63267672*((2.66666666666*(-1.1752324*10^6*phi(eta)+4.1744724*10^6))*u(eta)/p2+7.04876575828*(diff(phi(eta), eta))-(1.58058605349*(.597+4.45959*phi(eta)))*(diff(T(eta), eta))/(1-eta)))/(.597+4.45959*phi(eta));
eq3 := diff(phi(eta), eta)-5.00000000000*phi(eta)*(diff(T(eta), eta));
p:=proc(pp2) if not type([pp2],list(numeric)) then return 'procname(_passed)' end if:

res := dsolve({eq1=0,subs(p2=pp2,eq2)=0,eq3=0,u(0)=0,u(1-zet)=0,phi(0)=0.008,T(0)=0,D(T)(0)=1}, numeric);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)]));
res(parameters=[pp2]);
evalf(2/(1-zet^2)*Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*( F1(eta)*rhop*cp+(1-F1(eta))*rhobf*cbf )/(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))-pp2
end proc;

fsolve(p(pp2)=0,pp2=(20000)..(8000000));

it has an error and i cannot find that. would you please help me?

thanks

Hello everyone,

I want to know, how we can plot residual errors of the numerical scheme offered with dsolve command? This plot should show that the errors are in admissible range which indicates that the numerical scheme offers high accuracy.

For example, we have a bvp

dsolve({diff(y(x),x$2)-2*y(x)=0,y(0)=1.2,y(1)=0.9}, numeric,method=bvp[midrich]);

How to plot "Residual Error vs x" in this specific case?

 Thanks