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
please and thanks in advance.
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