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These are replies submitted by sarra



Thank you.

@Rouben Rostamian  


The code is modified to clarify some idea. The product * is replaced by &x  I.e  x*y become x &x y..

e[0]&x e[1]=e[n-1] assuming  1<n;
e[0]&x e[0]=e[2];
e[1]&x e[1]=e[n-1] assuming  1<n: :
e[2]&x e[1]=e[n] assuming  1<n:
for i from 1 to n-1 do
e[i]&x e[0]=e[i+1];
end do;
 The rest of product of two vectors e[k] &x e[kk]=0 ( as you see the symbol product is remplaced by &x to avoid any problem) Attached the code





Thank you Acer.

I tried your method for the Hankel and its derivative but I have the same solution  and In practice we have the same solution.

ee := argument(HankelH1(v, x)):    
Harg := convert(asympt(convert(ee,arctan),x,6),polynom):

eeD := argument(diff(HankelH1(v, x),x)):    
HargD := convert(asympt(convert(ee,arctan),x,6),polynom):
I have Phin=thetan?? 

his is surprising as e solution


Great. Thanks.

I try to use the same code if I want to compute an asymptotic of the argument of  Hankel function, I tried like 

with(MultiSeries, asympt):
H:=convert(asympt(argument(HankelH1(v, x)),x,6),polynom):

But I get the following error:

Error, (in MultiSeries:-multiseries) unable to compute series

If there any solution of this problem.



Hello, Thanks for your remark, I changed in the code the star * by &x I think it will be more clear as you mentioned  before.

But, How can I use   applyrule  and mentioned that the function f is linear so that we can expand and make some simplification

Many thinks


Yes that I want.

@Rouben Rostamian  


Thank you for your comment and remarks.

The set of vectors ( e(0), ...., e(n))  is basis of some algebra... and not a vector components like v=(1,1,2) or something else.

The problem I have a set of vectors and we postulate that they satisfy some conditions.

for example e(0)*e(1)=e(n-1)  and this set of vectors is not necessary commutative.

The final goal is to prove that is a basis of Algebra and I need some computation using maple to verify my theoretical results.

So, The set of vectors  satisfy the condition proposed ( we postulate these conditions) and the product  * is not the usual product of two vectors.

I would like for example any help in maple how when one compute  x*y  there is some product of two vectors ( not the usual product of course), I would like to get a simplification in this quantitie using the definition of  the vector and what they satisfy as condition ( defined in the first of the worksheet).

This is the code modified just a litte modification was added to improve the code.





Thank you for your remark and your help. I tried and I get real function  in the case where I compute an asymptotic expansion for Hankel function.

But, in the case where  i try to get an asymptotic for the derivative of hankel i get a complex function.

for example:

with(MultiSeries, asympt):

asympt(abs(HankelH1(v, x)),x, 6):
eval(%, O=0):

I get this solution: it's a real function. But, if I remplace the Hankel funciton by diff( HankelH1(v, x)),x) in the obove line I get a complex function. WHy????


@Rouben Rostamian  


Thank you for your remaks. In the attached code I define the product of two vectors.
But when I compute the produt of x*y,  there is no simplification


Thank you for your answer

I tried using the command asympt but I think abs( Hankel) give me a complexe number. please, the attached code


@Rouben Rostamian  

Thank you for your remarks.


First: # The vectors e(i) satify the folowing conditions
    e(0)*e(1)=e(n-1) assuming  1<n;

"*" denotes product of two vectors. 

The set {e(0),e(1),e(2),...,e(n)} represent a set of n+1 vectors.

e(0): its a name of the first vectors

e(1) name of second vectors

So, I say that the product of first vector by second vectors I obtain the last but one vector e(0)*e(1)=e(n-1)

2) You have:

    for i from 4 to n do
What do you expect this to do?  You haven't said what n is.

for i from 4 to n do
end do:

In this loop a define a function f, like  f(e(4)) =(n+1)*e(4);  f(e(5)) =(n+2)*e(5); etc...
 the n will be  fixed later in my work.

for example, If I give n =5; Il woild like the previous loop dispay:

f(e(4)) =6*e(4);  f(e(5)) =7*e(5)









@Preben Alsholm 

Thank you for your remark.

Sorry, Ni is the function f the code is corrected

f:=x->2*sqrt(3)*a1*a2*(sum(pochhammer(1/3,k)*3^k*x^(3*k)/(3*k)! ,k=0..infinity)*sum(pochhammer(2/3,k)*3^k*x^(3*k+2)/(3*k+2)!  ,k=0..infinity)-sum(pochhammer(2/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity)*sum(pochhammer(1/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity));



yes we can. Than you.



thank you very much Acer.


Thank you acer for your answer and thanks for your remark.



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