Items tagged with asymptotic


In some cases, maple is able to return a series expansion when line A is called but fails to do so when line B is called, and in the help page for asympt it defines them to be identical procedures:

A: asympt(f,z)


this has generally occured when ever the function f is of the form:

where g(z) has an asymptotic expansion and when asympt(f,z) is called, maple provides this expansion divided by the exponential function, however line B as prescribed above returns an error. 


So, my question is, which line in showstat(asympt) is responsible for catching the error which line B as above encounters?

Hey guys,

I have the following occurence:




gives different results...the last one however seems to be the correct one...

What is happening here?


Dear all;

I open this good discussion, and hope can get a nice and strong idea in this domain of approximation of Hankel funciton and order truncation of infinite series. Thanks for all idea, can improve the discussion. 

Using the asymptotics of Hankel function for large argument and large orders ( both together) and
find   an order of truncation N of the obove series so that we can ensure an error bound  of epsilon( epsilon very small given).  abs(sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-infinity..infinity)-sum ( c[m]*HankelH1(m,x)*exp(I*m*theta), m=-N..N))<epsilon.

A first idea come in mind: the series converge, so that the general terms of this series converge to zero, and in particularity,  abs(c[m]* HankelH1(m,x))<1:
then abs(c[m])<1/ abs(HankelH1(m,x)).
So we can ensure an error bound of epsilon on the coefficient c[m] by imposing  (HankelH1(m,x))<epsilon  this lead to abs(c[m])<epsilon.
I consider the case where m and x are very large, I can suppose for example m=x*(1+zeta), with 0<zeta<1. zeta parameter. So that our truncation N depend on zeta.
And then how can I find and approximation of the inverse of Hankel function for large argument and large order. using m=x*(1+zeta). I think this give us N the truncation order.
I hope get a good discussion in this subjet.
Of course maybe there are other strong idea to compute the truncation series.
I get the following error in the code:

Error, (in MultiSeries:-multiseries) unable to expand with respect to parameter




###### Code### and error

assume(0 <= x);

assume(0 <= zeta<1);


sum(c[m]*HankelH1(m, x)*exp(I*m*theta),m=-infinity..infinity);


MultiSeries:-asympt(%,x, 4);

eval(%, O=0);



How to find asymptotic behaviour of a function.

For example at infinity

sinh(x) behaves as 1/2*exp(x)

1/sinh(x)  behaves as 2*exp(-x)

exp(-x)*(exp(-x)+1) behaves as exp(-x)

so that it works with a more complex expression.

Greetings to all.

I would like to share a brief observation concerning my experiences with the Euler-Maclaurin summation routine in Maple 17 (X86 64 LINUX). The following Math StackExchange Link shows how to compute a certain Euler-MacLaurin type asymptotic expansion using highly unorthodox divergent series summation techniques. The result that was obtained matches the output from eulermac which is definitely good to know. What follows is the output from said routine.

> eulermac(1/(1+k/n),k=0..n,18);
     1       929569        3202291        691                O(1)
O(- ---) - ----------- + ----------- - --------- + 1/1048576 ----
     19             15            17          11              19
    n      2097152 n     1048576 n     32768 n               n

        174611      5461        31       |      1           17        1
     - -------- + --------- + ------- +  |   ------- dk - ------- + ------
             19          13         9    |   1 + k/n            7        5
       6600 n     65536 n     4096 n    /                 4096 n    256 n

         1       1
     - ------ + ---- + 3/4
            3   16 n
       128 n

While I realize that this is good enough for most purposes I have two minor issues.

  • One could certainly evaluate the integral without leaving it to the user to force evaluation with the AllSolutions option. One can and should make use of what is known about n and k. In particular one can check whether there are singularities on the integration path because we know the range of k/n.
  • Why are there two order terms for the order of the remainder term? There should be at most one and a coefficient times an O(1) term makes little sense as the coefficient would be absorbed.

You might want to fix these so that the output looks a bit more professional which does enter into play when potential future users decide on what CAS to commit to. Other than that it is a very useful routine even for certain harmonic sum computations where one can use Euler-Maclaurin to verify results.

Best regards,

Marko Riedel

Dear friends,

I wonder how I would go about calculating the asymptotic expansion of

sum(5^j/j, j=1..m+1)?

The motivation for this calculation can be found here. The correct answer is

5/4 5^(m+1)/(m+1).

The classic asympt and the one from multiseries both fail on this one.


Marko Riedel

In Arfken(Mathematical methods for physicists,5-th edition,page 483),the asymptotic form of the Hankel function is approximated as



Is there any simple/direct way in Maple(using HankelH1(),or otherwise) to achieve this?I don't want to assign numerical values to t or s.

I'm using Maple 15.  It seems to me this worked in some previous version...

Consider the Lambert W function, y=LambertW(0,x) ... I want Maple to tell me the asymptotics for it,

something like this:


But I don't get that now.  Is my memory faulty that I got it in the past?

Maple 15:



not very useful...






I am interested in finding the asymptofic constant (Big O(1/n^(2m+2)) for the following expansion


series((1+1/n)^((1/2)/(sum(1/((2*k+1)*(2*n+1)^(2*k+1)), k = 0 .. m))), n = infinity, 10)


Upon using the preceding command in the maple i get


Error, (in asympt) unable to compute series

Page 1 of 1