## Computing a complex residue...

Greetings to all.

With the following matter I am betting on there being a simple mistake on my part due to fatigue owing to a challenging session of intense computing. The following link at Math.Stackexchange.Com points to a computation involving complex residues. Consult the link for additional details.

I usually verify my computations with Maple, I did the same this time. Thereby I happened on a curious phenomenon which I have documented below. Please study the session data provided, I believe it speaks for itself.

```user@host:~\$ math
Mathematica 10.0 for Linux x86 (64-bit)

In[1]:= Residue[z^2/(z^4 + 2*z^2 + 2)^2, {z, 2^(1/4)*Exp[3*Pi*I/8]}]

1/8      1/4
(-1)    ((-1)    + Sqrt[2])
Out[1]= ----------------------------
1/4      1/4           3
16 2    ((-1)    - Sqrt[2])

In[2]:= N[%]

Out[2]= 0.117223 - 0.0083308 I

In[3]:=
user@host:~\$ maple
|\^/|     Maple 18 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2014
<____ ____>  Waterloo Maple Inc.
|       Type ? for help.
> residue(z^2/(z^4 + 2*z^2 + 2)^2, z=2^(1/4)*exp(3*Pi*I/8));
0

> quit
memory used=0.9MB, alloc=8.3MB, time=0.07
user@host:~\$
```

I am looking forward to learning what the correct syntax is to get the residue in this case and I hope I can assist other users who might have run into the same problem. I will cancel the question should it turn out to be trivial and of little potential use to the community.

Best regards,

Marko Riedel

Post Scriptum. Being a programmer myself I would be curious to learn more about the algorithms that are deployed here and how and why they did not succeed.

## New features in Mathematica 10.3.0

Maple

Mathematica 10.3.0 was announced yesterday. This is the 6th release of Mathematica 10 during 16 months. I wonder its  MathematicaFunctionData and   FindFormula . At first sight, the former is an analog of FunctionAdvisor of Maple, but the latter isn't any analog. Also compare the outputs of

Residue[Binomial[n,k],{n,-j}]

(-1)^j/(j!*k!*(-j-k)!)

and

>`assuming`([residue(binomial(n, k), n = -j)], [integer, j > 0]);

residue(binomial(n, k), n = -j)
Let us wait for Maple 2016.

## it possible teach maple not simplify some values...

hi  for example to calculate the following

residue((Psi(-z)+Eulergamma)^2*h(z), z = 2)

gives

3*h(2)+(D(h))(2)

but it possible to write

as( Psi(2)+Eulergamma(z))*h(2)+(D(h))(2)

so that

and Psi(z)+Eulergamma== harmonicNumber(z-1)

the result must be

harmonicNumber(2)*h(2)+(D(h))(2)

it is possible that Maple gives explit form of the values function avoid to calculate automatic.

thanks

## Comment se calcule le residue d'une fonction de de...

Bonjour

Comment se calcul le résidue d'une fonction f(x,y) ?

Si par exemple f(x,y)=1/x+1/y+5/(x*y), alors le residue de f en (0,0) est défini comme étant (1/2*Pi)^2*int(f(x,y)) avec intégrale est sur une sphère qui contient l'origine.

Est ce qu'il est égal au coefficient 5 de 1/(x*y) dans le developpement de laurent en deux variables ??

Merci d'avance,

Gérard.

## How to calculate contour integral with Maple?...

How to calculate the integral of 1/sqrt(4*z^2+4z+3) over the unit circle counterclockwise for each branch of the integrand? Of course, with Maple.

## Mellin Transform.

by: Maple

Greetings to all. I will keep this brief and to the point. I would like to point out a certain deficit in the integral transform package. I have recently been calculating some Mellin transforms at this link and the base functions are of the following type.

`g := (p, q) -> 1/(x+p)/(x+(p*q-1)/q);`

Now to see the deficit here are some Mellin transforms that...

## residues of a function related to digital sums

by: Maple

I will post here the output of two commands that has me a bit worried. Of course it is entirely possible that I am overlooking something very simple, which is why I am writing to your site.

`> residue(1/(2^s-1), s=2*Pi*I*3/log(2));                                           0`

## beta function vs. sine

by: Maple

Dear friends,

I am reporting with a brief comment concerning the integral int(1/(1+x^a), x=0..infinity) with a>=2 a real number. This was evaluated here.

Now Maple 15 (X86 64 LINUX) will quite happily compute this in its most simple form involving the sine when a is not a positive integer or a rational number. If it is, however, a beta function term results,...

## possible symbolic integration bug

by:

Dear friends,

I will present a series of commands for you to ponder that should speak for themselves.

```> q := n -> int(1/(1-2^n*exp(I*theta*n))^2*2*I*exp(I*theta), theta=0..2*Pi);
2 Pi
/
|          2 I exp(theta I)
q := n ->  |      ------------------------ dtheta
|            n                2
/       (1 - 2  exp(theta n I...```

## an integral of the hyperbolic cosine

by:

Dear friends,

I was working on the integral of cosh(ax)/cosh(x) from zero to infinity with "a" being a rational number p/q where p<q and p-q being odd. I was quite shocked to discover that Maple 15 (X86 64 LINUX)cannot do this integral while Mathematica gets it right every time. I certainly hope this will be rectified soon.

For those who might be interested the calculation of the integral

## residue at pole...

How to find the residue of   {e^(az)}/z(z+b)^m   at the pole -b of order m

## why does it still not work ?...

I'm sorry if I'm duplicating the information here. I tried to post a comment but it doesn't seem to be appearing.

When I change = to := it does not work. Maple says

Error, invalid input: MultiSeries:-series expects its 2nd argument, eqn, to be of type {name, name = algebraic}, but received 1.*I

## help evaluating an integrand - integrating in u st...

Hello,

I have a huge expression involving psi and phi called A_1 which I paste below.  I am trying to integrate it using the residue theorem in the variable phi. I make a substitution in u as below for sin(k*phi)  and ask for the singularities of each operand at u. But when I get my final result it still has the variable u in it. If its evaluated the integral in u then it should not return u in the result. Evidently it’s done something...

## help needed - evaluating an integrand...

Hi guys I am trying to evaluate  an integral of the form

A_0 := (1/(2*Pi)+Q*cos(k*phi))^2/(1+(1/(2*Pi)+Q*cos(k*phi))^2);

In this case I specify a value for Q=15 as given below and since maple cannot evaluate the integral directly I have written code to evaluate the integral using the residue theorem. So for each singularity I calculate the residue at that point and multiply the sum of the residues by 2*Pi*I as in the code below. However the...

## without multiseries and selecting poles with posit...

Okay so I can see how to get the poles. But I have some further questions.

1) For beta AND Q both GREATER THAN 0 WHICH poles have positive imaginary part and is there a PROGRAMMATIC way of selecting these ones?

2) How do I calculate the residue at these poles? The approach given by Alec above seems useful and is much...

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