Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What does this happen to my series expansion?

  1. Sep 4, 2009 #1
    I came across this really strange error when doing series expansions in Mathematica. Suppose I were to let,

    [tex]F(z) = \frac{(z+2)^2}{(z+1)^2} - 0.4[/tex]

    Now [tex]F(z) = 0[/tex] gives [tex]z \approx -3.72076, -1.61257[/tex]. Suppose we take the second value of [tex]z^* = -1.61257[/tex]. What is the series expansion of [tex]\sqrt{F(z)}[/tex] about this point?

    [tex]F(z) \sim 0 + F'(z^*)(z-z^*) + \ldots \approx 3.37088 (z-z^*) + \ldots[/tex]. Thus [tex]\sqrt{F(z)} \sim \sqrt{3.37088}\sqrt{z-z^*}[/tex].

    Now let's try doing this in Mathematica:

    Code (Text):

    moo = z /. Solve[disc[z] == 0, z]
    F[z_] = -0.4 + ((z+2)/(z+1))^2;
    Chop[Series[F[z], {z, moo[[2]], 4}]];
    Sqrt[%]
     
    This seems to give the right answer. The first two terms are:

    Code (Text):
    1.83599 Sqrt[z + 1.61257] + 3.43262 (z + 1.61257)^(3/2)
    Now here's the problem. I try:

    Code (Text):

    Series[Sqrt[F[z]], {z, moo[[2]], 4}]
     
    The output of the first three terms are,

    Code (Text):

    0. + 1.05367*10^-8 I) - (0. + 1.59959*10^8 I) (z + 1.61257) - (0. + 1.21417*10^24 I) (z + 1.61257)^2
     
    which is nonsensical.

    What is the problem? Does it have to do with the fact that we're using an approximation to a root?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted