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MathGroup Archive 2007

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Series and order

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74738] Series and order
  • From: "Hugh" <h.g.d.goyder at cranfield.ac.uk>
  • Date: Tue, 3 Apr 2007 00:26:24 -0400 (EDT)

I wish to expand the expression e1, below, in a Taylor's series
expansion about z0 in the two variables z1 and z2 and include only
first order terms (c1, c2, a and b are coefficients). Terms in z1 and
z2 are of similar order so that products of their first order terms
are second order and should be discounted compared to first order
terms. Problem 1: how do I tell Series to do this? Problem 2: the
coefficients c1 and c2 are of similar order to z1-z0 and thus
additional terms can be set to zero as second order. How do I do
this?

Below I give a warm up problem for a function f[z1,z2] which just
seems to expand in terms of one variable and then in terms of the
other without regard for term order. If we can solve the first problem
then it may be possible to solve the second by expanding in terms of
the coefficients as well.

Thanks

Hugh Goyder


Series[f[z1, z2], {z2, z0, 1}, {z1, z0, 1}]


SeriesData[z2, z0, {SeriesData[z1, z0,
    {f[z0, z0], Derivative[1, 0][f][z0, z0]}, 0, 2, 1],
   SeriesData[z1, z0, {Derivative[0, 1][f][z0, z0],
     Derivative[1, 1][f][z0, z0]}, 0, 2, 1]}, 0, 2, 1]


e1 = (z0^2 - 2*I*c1*z0*z1 - z1^2)*(z0^2 - 2*I*c2*z0*z2 -
      z2^2) - 4*z1*z2*(I*a*z0 + b*Sqrt[z1*z2])^2;


Simplify[Normal[Series[e1, {z2, z0, 1}, {z1, z0, 1}]],
  {0 < z0}]


2*z0^2*((2 - I*a*b - 2*b^2)*z0^2 +
   (2*a^2 - 9*I*a*b - 8*b^2 - 2*(-I + c1)*(-I + c2))*z1*z2 +
   z0*((-2 + 3*I*a*b + 4*b^2 - 2*I*c1)*z1 +
     (-2 + 3*I*a*b + 4*b^2 - 2*I*c2)*z2))



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