Hello everybody,

I would like to assign 349 different variable not already existing (w[i], i from 1 to 349) to 349 expressions from a list. Unfortunately I do not know how to write it. When I try anything like a:=w[33]->listevnouveau[33], it does not work.

With this a, I want to solve this : solve(listedd[33]*w[33]=a,w[33]); which is only possible by assigning variable to the expression.

Regard, Charbo

Is there a direct command to give the (first) index in a list,
for which the entry coincides with some given value (without
coding a loop, to run through the list)?

---

PS: when I enter that question I automatically get an info,
that similar questions exists. But clicking on those links do
not give usefull infos, they just lead to a page of old threads
(Page 529 of 569 for the first link ...)

Hi all

I am trying to draw a 3d plot from the data that I have created in excel, how do I import from excel to maple and then draw the 3d plot using array?

below is the link to my dataset in excel

Book1.xls

your help is appreciated

What Should I do to get the correct result-which is obviously not something like my answer. 

> assume(a::(AndProp(NonZero, constant, real)));
> about(a);
Originally a, renamed a~:
  is assumed to be: AndProp(real,constant,Non(0))

> is(a, constant);
                                    true
> is(a = 0);
                                    false
> is(a^2 > 0);
                                    true
> assume(p::real, 0 < p and p < infinity...

How I can say that an object,say k, is a constant(it may be positive or negative) inside assume()?

assume(k::??should I say)

pls also tell me what kind of help pages I should see for this kind of "type inside assume()" problems. 

Thanks.

I'm working on a mecanic's problem and I need to solve a differential equation:

I want to determine x(t) and plot it... 

This is the way I do it: 

with(Physics):with(combinat):
comn := choose([u, v, mu, nu], 2):

def := seq(seq(seq(Physics[`.`](Dagger(Ket(aa, k)), Ket(bb, m)) = Physics[`.`](Ket(bb, m), Dagger(Ket(aa, k))), `in`(aa, comn[i][1])), `in`(bb, comn[i][2])), i = 1 .. nops(comn)), seq(seq(seq(Dagger(Physics[`.`](Dagger(Ket(aa, k)), Ket(bb, m)) = Physics[`.`](Ket(bb, m), Dagger(Ket(aa, k)))), `in`(aa, comn[i][1])), `in`(bb, comn[i][2])), i = 1 .. nops(comn));
Setup(def):

d1 := Physics[`.`...

I was recently browsing the Wolfram website, in particular the Online Integrator, and was very impressed that they are exposing a part of their product for free public use. I also noticed that they offer the option of submitting previously unknown anti-derivatives to them:

(From the FAQ page, http://integrals.wolfram.com/about/faq/#8)
“If you think you have a...

I have three different matrices, say A(3x6),B(3x6) and C(3x12) with the size of each matrix mentioned within parenthesis. I want to define a fourth matrix D (6x12) so that D=[A,B;C] (in Matlab notation). Can anyone please tell me how to define this D in Maple? Maple help seems not that helpful at least for beginners like me. Thank you in advance.

what is the command to change transfer matrix to state space?

When I enter the Greek Capital letter "Xi" (looking like three uneven horizontal lines) it comes up fine on screen but in print preview there is a space?  Lower case "xi" is fine.  Same thing happens in both Maple 14 and 15.

Hi everyone,

Is there a way that Maple can return a list of variables in a function? For example, if I have something like:
f:=(x,y)->x^2+y^2;
I would like Maple to return:
{x,y}

I found the indets command, but it does not seem to work for functions. For example:
f:=x^2+y^2; indets(f);  returns {x,y} but     f:=(x,y)->x^2+y^2; indets(f);    returns {f}

Any help would be appreciated.

I am looking for an way to do clifford and geometric algebra calculations with Maple.

Robert Israel helped me to plot  F2 vs k 

http://www.mapleprimes.com/questions/120380-Explicate-Plotting

I tried to modify his code to get reults for F2 vs S from equation A1, but i am still facing problems with it

The maple code is

restart:with(MultiSeries):with(plots):

parvalues:={C=2.02,H_liq=8.74,R=22.11,R1=0.0006};...

Hi,

I encountered the following problem in Maple 14

expr1:=sqrt(a)

assume(a>0)

expr2:=sqrt(a)

additionally(a>1)

subs(a=2,expr1) #gives correct answer of sqrt(2)

subs(a=2,expr2) #does not substitute a=2 in expr2

It seems the last statement with expr2 does not work as expected, I wondered why this behaviour is so, because it does work with expr1. The added assumption does not invalidate the first...