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division of two series

lemelinm's picture
Posted on 2008-10-10 13:26 By lemelinm (Maple Rating 3 224)
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Hi all,

While working with the method of Frobenius to solve a second-order homogeneous equation,  I stumble at a point where I have to divide two series.

You will find in the attach worksheet what I mean.

Frobenius division.mw

Any help would be appreciate.

Thanks in advance

Mario

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Robert Israel's picture

division

Comment on 2008-10-10 14:00 from Robert Israel (Maple Rating 4 1431)

Do you mean this?

> series(S, x);

-1/x+1-(1/4)*x+(1/128)*x^3-(21/1024)*x^4-(109/12288)*x^5+O(x^6)

 Or perhaps this?

> convert(S, FormalPowerSeries, x);

-1/x+Sum(Sum(2*x^k/alpha^(k+1), _alpha = RootOf(3072-1536*_Z+576*_Z^2-160*_Z^3+35*_Z^4)), k = 0 .. infinity)

 

lemelinm's picture

It was so simple but...

Comment on 2008-10-10 14:46 from lemelinm (Maple Rating 3 224)

 the second, FormalPowerSeries, give a result that have no sense at all.  What about the two summations sign.  And there is no roots for _Z.  Do you have at hand an example that yould give the summand of a serie.

But anyway, you solve my problem.  Thanks Robert!

Axel Vogt's picture

stay cool, it makes

Comment on 2008-10-10 15:17 from Axel Vogt (Maple Rating 4 948)

it says: for all the 4 roots alpha of the denominator you have to sum up (after substracting 1/x), so it writes your S as -1/x + A/B1 + A/B2, where A is your numerator (of degree 4) and B1,B2 are quadrics with complex roots only (coming from a partial fraction or as displayed: it uses A/b1+A/b2+a/b3+A/b4). Note that your S is rational (of degree = [4,5]) and thus has no global series due to singularities

Robert Israel's picture

Double summation

Comment on 2008-10-10 15:47 from Robert Israel (Maple Rating 4 1431)

What Maple's result means is this. The polynomial 3072-1536*_Z+576*_Z^2-160*_Z^3+35*_Z^4
has four complex roots: let's call them alpha_1 to alpha_4. The coefficient of x^k for k from 0 to
infinity is sum_{j=1}^4 2/(alpha_j)^(k+1).

The radius of convergence of the series is approximately 3.029725460, which is the distance from 0 to the closest root of the polynomial.

Axel Vogt's picture

variant of this

Comment on 2008-10-10 16:24 from Axel Vogt (Maple Rating 4 948)

normal( S+1/x ) is a rational function of degree [3,4] where the
denominator B is as above. You can enforce a representation as
hypergeom by 1/t; convert(%,hypergeom, include=all); eval(%,t=B);

Then by multiplying with the numerator you get a representation
for S+1/x as polynom*hypergeom.

The hypergeom converges for abs(1-t) < 1 and gives your a series,
and beyond you have to use some of the linear transforms given in
Abramowitz&Stegun.

May be some re-scaling makes it a bit better, but the problem with
the poles will remain.

lemelinm's picture

I will digest this

Comment on 2008-10-10 16:06 from lemelinm (Maple Rating 3 224)

 and if I am still stuck, I'll get back to you.  Thanks for the trouble.

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