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Maclaurin series for multivariable

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I got a few functions I need to expand to series using Maclaurin forumlas.

    2. Relevant equations

    http://mathworld.wolfram.com/MaclaurinSeries.html

    3. The attempt at a solution

    So here are the ones I managed to do:
    [tex] f= \sqrt{1-x^2-y^2} [/tex]
    writing it in another form:
    [tex] f= (1+(-x^2-y^2)^{\frac{1}{2}} [/tex]
    Then I use:
    [tex](1+x)^m = 1 + mx + ..[/tex]
    as a function of one variable where [tex]x = -x^2-y^2[/tex] and I get the correct answer.
    same goes for
    [tex]z=\frac{1}{1-x+2y}[/tex]
    and
    [tex]p=\ln(1+x+y)[/tex].
    Basically - I found that whenever there is no multiplication involved, I can just treat the two variables as one big variable and it works (according to my given answers).
    The problem comes when I got stuff like this:
    [tex]g=\frac{\cos{x}}{\cos{y}}[/tex]
    or
    [tex]v=e^{x}\cos{y}[/tex].
    Expanding each part and then dividing or multiplying (as you would you do if it was a true single var function) doesn't work. Expanding with taylor series from the start works - but the point is to use the maclaurin series.
    So where did I go wrong?
     
  2. jcsd
  3. Sep 11, 2009 #2

    HallsofIvy

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    ?? The MacLaurin series is a Taylor's series, just evaluated at x= 0.
     
  4. Sep 11, 2009 #3
    I know, but the MacLaurin series are given equations so you will not have to differentiate all over again. For example, for functions in the form of ex or siny. You just plug it in the formula.
    My question is - can it be done for functions like f(x,y) = exsiny? If yes - how? This way the differentiation process can be avoided.
     
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