       Re: Series Expansion

• To: mathgroup at smc.vnet.net
• Subject: [mg100476] Re: [mg100457] Series Expansion
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 5 Jun 2009 03:02:04 -0400 (EDT)
• Reply-to: hanlonr at cox.net

```
It won't work with arbitrary series coefficients (also, would need to be a[n] vice a_n and b[n] vice b_n). Using simple examples,

fm1[a_, x_] = 1 + Sum[a^n*x^n, {n, 1, M}] // Simplify

(a^(M + 1)*x^(M + 1) - 1)/(a*x - 1)

fm2[b_, x_] = 1 + Sum[b^n*x^n, {n, 1, M + 2}] // Simplify

(b^(M + 3)*x^(M + 3) - 1)/(b*x - 1)

fm[a_, b_, x_] = (fm1[a, x]/fm2[b, x])

((b*x - 1)*(a^(M + 1)*x^(M + 1) - 1))/((a*x - 1)*(b^(M + 3)*x^(M + 3) - 1))

Series[fm[a, b, x], {x, 0, 5}] // Normal // Simplify

((a^5*x^5 + a^4*(x^4 - b*x^5) + a^3*(x^3 - b*x^4) + a^2*(x^2 - b*x^3) +
a*(x - b*x^2) - b*x + 1)*(a^(M + 1)*x^(M + 1) - 1))/
(b^(M + 3)*x^(M + 3) - 1)

Bob Hanlon

---- Ossama Kullie <okullie at chimie.u-strasbg.fr> wrote:

=============
Hi every body,

I have the following function fm which is a rational between to series :

fm1 = 1 + Sum[a_n*x^n, {n, 1, M}]
fm2 = 1 + Sum[b_n*x^n, {n, 1, M + 2}]
fm = (fm1/fm2) // Expand

Now I want to have aseries exansion of fm to some order:

Series[fm, {x, 0, 5}] // Expand
But I get the input nothing else? can somebody there give me some hit

1/(1 + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(2 + M\)]\(
\*SuperscriptBox[\(x\), \(n\)]\ b_n\)\)) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(M\)]\(
\*SuperscriptBox[\(x\), \(n\)]\ a_n\)\)/(1 + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(2 + M\)]\(
\*SuperscriptBox[\(x\), \(n\)]\ b_n\)\))

Best Regards,
O. Kullie

```

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