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1. Consider the following fast convergent series:


As expected, the sum of the series is obtained very fast (with any precision), same results for S1 and S2.

2. Now change the series to a very slowly convergent one:


evalf(S1) is computed also extremely fast, because the acceleration algorithm works here perfectly.
But evalf(S2) demonstrates a bug:

Error, (in evalf/Sum1) invalid input: `evalf/Sum/infinite` expects its 2nd argument, ix, to be of type name, but received ...

3. Let us take another series:


Now evalf(S1) does not evaluate numerically and evalf(S2) ==> same error.
Note that I do not know whether this series is convergent or not, but the same thing happens for the obviously convergent series


(because it converges slowly (but absolutely) and the acceleration fails).
I would be interested to know a method to approximate (in Maple) the sum of such series.

Edit. Now I know that the mentioned series 

converges (but note that Leibniz' test cannot be used).

Before version 2016 Maple was incredibly good at evaluating an infinite power series and returning a simple function, e.g. 1/(3x+2).  Now version 2016 just returns the input sum expression with no change.  Is there some new command to get the old results?

Hello, I have a somewhat math and Maple question I'm hoping some can help with.

I have this curve,


and if I solve this system (numerically);

sol:= solve([curve=0,diff(curve,beta)=0],[alpha,beta]);

I get 6 special points (8 actually but two are critical). So I'll refer to them by subscript "i".

If I then do a coordinate transformation by;



and convert the 6 coordinates and curve, I'll get everything in terms of (u,v) coordinates.


soluv:=map(ln,sol);  (this is just pseudo - I don't know how to do it this way)

So now the 6 points are referred to by (u_i,v_i).

Next, I want to expand this curve locally around these six points, using the following (where "z" is the local coordinate);

u -> u_i + z^2

v -> v_i + sum(a_j*z^j,j=1..n)

where n is reasonable, though around 15.

curve3(i):=subs(u=z^2 + cat(Ubp,i),curve2);


Here, I'm not sure of the pros/cons of cat() vs a[i,j]....


Anyway, I'll then have an equation in terms of only the local coordinate "z". If I then solve each coefficient of "z", at each order, I should then be able to determine the power series v(z). I reason that since the curve is initially equal to zero, that every non-zero power of z will have a coefficient/equation (in terms of unknowns a[i,j]) that should be equal to zero.

This is analagous to solving differential equations with power series...


However, I'm a little lost in implementing this,

I am currently trying, for instance,

l1:= series(curve4(1,10),z,0,9);

e1:= seq(coeff(l1,z,i),i=1..4):



I believe I am doing something wrong though bcause every odd power is zero.


Thank you a lot for any suggestions and/or help,



Ubp and Vbp are those 6 points - I just kept them as symbols initially because I was still getting odd-powered coefficients as zero, regardless of their actual values. Hence a little bit of the "math" side of the problem...

I want to make sense of the expression

Int(t^2/ln(t)*exp(-t), t=0..infinity);

The denominator vanishes at t=1.  The singularity at t=1 is not integrable.  I want to see whether the integral is defined in the sense of Cauchy principal value.  Thus, I let

K := Int(t^2/ln(t)*exp(-t), t=0..1-a) + Int(t^2/ln(t)*exp(-t), t=1+a..infinity);

and wish to see whether the following limit exists:

limit(K, a=0, right);

Maple cannot evaluate this.  Nor can I.  Alternatively, we may try:

series(K, a=0);


series(K, a=0) assuming a>0, a<1;

In both cases Maple says that it is unable to compute the series.

So my question is: Does the Cauchy principal value exist, and can Maple help one to determine that?


@Markiyan Hirnyk 


Thanks for helping.

(1) But how to find period using code?

(2) how did you determine n=29 ?

(3) I don't need sequence of numbers, but would like to extract the coefficients of exp(..) terms. For example in the follwoing expression, how to extract the coefficients: (note that the expression is missing one more term and i don't know how to fix?)



You, I, and others like us, are the beneficiaries of decades of software evolution.

From its genesis as a research project at the University of Waterloo in the early 80s, Maple has continually evolved to meet the challenges of technical computing.

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);



I am having difficulty helping someone series expand an eigenvector solution.  I can expand the eigenvalues easily but get a numeric exception divide by zero when I attempt to expand a component of an eigenvector.  Mathematica seems to have no problem solving this problem.  Any help would be appreciated.






assume(varepsilon > 0)

H := Matrix(3, 3, {(1, 1) = 0, (1, 2) = -epsilon, (1, 3) = epsilon, (2, 1) = -epsilon, (2, 2) = 2-2*epsilon, (2, 3) = 0, (3, 1) = epsilon, (3, 2) = 0, (3, 3) = 2+2*epsilon})

Matrix(%id = 18446744078100429630)



evals, evecs := Eigenvectors(H):

e1 := convert(simplify(series(evals[1], varepsilon = 0, 4)), polynom)



e2 := convert(simplify(series(evals[2], varepsilon = 0, 4)), polynom)



e3 := convert(simplify(series(evals[3], varepsilon = 0, 4)), polynom)



simplify(series(evecs[1][1], epsilon = 0, 4))

Error, (in simplify/sqrt/local) numeric exception: division by zero





I am trying to extract the coefficients of z from its series expansion. In two cases I succeed in finding the coefficients, but in the last one I fail to get the correct coefficients. Some garbage value is obtained. What is the reason behind this? I have attached my maple program.


When I execute the command


and then refer to the equation in a new execution group using a equation label (CTRL-L on Windows), the equation is shown in Maple 18, but in Maple 2015 I get an error message: 'Error, missing operator or ';'. Using the % instead does work for both versions.

Is this intended behaviour or a bug in Maple 2015?




I need some help to compute the series approximation of the modulus and argument of hankel function for large x. The code display

 Error, (in asympt) unable to compute series

Thanks  for helping me.

#We define the hankel function as
#HankelH1(v,x) = BesselJ(v,x) + I*BesselY(v,x), where BesselJ and BesselY are bessel function.
#In this question the parameter "v " is  fixed. "

# Code

HankelH1(v, x);
# The modulus of Hankel function
Mn:=x->abs(HankelH1(v, x));
thetan:=x->argument(HankelH1(v, x));
phin:=x->argument(diff(HankelH1(v, x),x));
# Compute series
series(Nn(x),x=infinity, 7);
# I define the following function

# Series approximation
series(f(x),x=infinity, 7);



Good evening,

I am trying to solve the following sum equation:


sum((factorial(n)/factorial(2*n))^n, n = 1 .. infinity)


but I require the step-by-step solution as to how it is done, but can't seem to find that option. 


If anone could help me out, that would be great.


Kind regards,


I want to obtain the taylor series of a function say sin(x) at x=0 up to infinity. I mean that I don't want a trauncated series. I tried using "series" and "taylor" but they just give the truncated series.

let γ be the root 

i have to apply taylor series on f(x) and then do some substitution like (helped by a member of Mapleprime)

taylor(f(x), x = gamma, 8);
f(x[n]) := subs([x-gamma = e[n], f(gamma) = 0, seq(((D@@k)(f))(gamma) = factorial(k)*c[k]*(D(f))(gamma), k = 1 .. 1000)], %)

then find the derivative of result from above output

i do

b := diff((x[n]), e[n])

basically i have to find the value of newton method which is


here we substitute xn=γ and D(f)(xn)=b

and then want to apply f on yn

there are to problem which i face 

1  f(xn)/D(f)(xn) is not in simplified form i-e O(e[n]^8) and O(e[n]^7) is appeared in numerator and denominator respectively. how we get the simplified result.

2 wht step should i do to find f(yn)

plx help me to do this 

thanx in advance

Hi all.

Assume that we have partitioned [0,a], into N equidistant subintervals and in each subinterval we have M sets of poly nomials of the following form:

where Tm(t)=tm( namely Taylor Series) and tf is a(final point)
for Example with N=4, M=3 we have:

now we want to approximate a function, asy f(t), in this interval with following form:

Assume that we have

where '"." means common derivetive. How can we do the later integran in right way?

Note that t is unknown

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

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