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Expanding a series

  1. Oct 11, 2006 #1
    I have the series (cosx)^-1/2 (so it's square-rooted and on the bottom)

    And I need to get the first few terms. My idea has been to do this: [1 + (cosx - 1)]^-1/2

    But I have no idea how to take the binomial expansion of that. Do I:

    Take the BE first, substituting "cosx - 1" as just X or something, then after I get the first few terms of that, substitute in the series for "cosx - 1" back into X and adding everything up? The only reason I haven't tried this yet is because it seems like this would take an infinite amount of work... :yuck:
     
  2. jcsd
  3. Oct 11, 2006 #2

    quasar987

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    if you only need the first few terms, why no just differentiate (cosx)^-1/2 a few times?

    [tex]\frac{1}{\sqrt{\cos x}} = 1+ \frac{1}{2}\left[ \frac{d}{dx}\frac{1}{\sqrt{\cos x}}\right]_{x=0}x+...[/tex]
     
  4. Oct 11, 2006 #3
    Because your 1st derivative is missing a sinx (Unless I don't get your mathematical notation...), and the next derivatives only get more gruesome. This teacher keeps spouting "we're smart, but lazy", so I'm guessing there is a trick I am missing here.
     
  5. Oct 11, 2006 #4

    quasar987

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    it's not messy at all after you evaluate the derivative at x=0. All the cos become 1 and all the sines become 0.
     
  6. Oct 11, 2006 #5
    quasar, you are doing a MacLaurin series right?
     
  7. Oct 11, 2006 #6

    quasar987

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    Are you going to say that after a few derivatives, nasty sines come poping in the denominator?

    Just to be on the same side, make that a Taylor series centered on [itex]\pi/4[/itex] where both cos and sine are [itex]1/\sqrt{2}[/itex].
     
    Last edited: Oct 11, 2006
  8. Oct 11, 2006 #7
    Errr... when can I evaluate at x=0? Don't I need to keep it a variable until I am completely done deriving?
     
  9. Oct 11, 2006 #8

    quasar987

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    Yes. It's true you are lazy, lol.
     
  10. Oct 11, 2006 #9
    I tried this once already. I got to through 2 derivations before it became really ugly. And my f(0)'s were something like 1, 0, and 1/2. That only makes it 2 terms, really, unless I want to cheat and say it's 0x. But I probably can't. So if all the odd n terms are supposed to be 0, I have a lot of deriving to go. I said I'm lazy, but smart. Smart enough to know that there has to be a different approach. This isn't a calc class where we are learning this the first time and have to go through this just to know how it's done. This is math physics, so it's mainly just a review + more critical thinking.
     
  11. Oct 11, 2006 #10

    quasar987

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    Alright, so let's get back to your first idea.

    [tex](\cos x)^{-1/2} = (1+(\cos x - 1))^{-1/2} \equiv (1-X)^{-1/2}[/tex]

    Expand this in a binomial series, then resubstitute X--> cos(x)-1. You will have what you want: a series expansion for (cosx)^-1/2. It is not a power series, but it is a series expansion.

    It is frequent that calc classes do not cover the binomial series for non-positive integer exponents, but any physics class will assume that you know it. See

    http://en.wikipedia.org/wiki/Binomial_series

    The series you find will converge for all values of x such that |X(x)|<1. Find what that corresponds to in terms of x.
     
    Last edited: Oct 11, 2006
  12. Oct 11, 2006 #11
    I have to get a power series expansion (or whatever it's called, where it's 1 + x/2! etc). So would I then expand cosx - 1 and plug those results into each individual binomial expansion? That sounds like just as much work. But I guess I have to try something.
     
  13. Oct 11, 2006 #12

    quasar987

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    yeah, do that then. Since X = cos(x) - 1, and cos(x)'s power series's first term is 1, the 1's cancel and then the term of lowest degree in [itex]X^n[/itex] is [itex]x^n[/itex]. So if you decide to keep only the terms up to degree 2, you're left with the first two terms (in x) of X and the first term (in x) of X².
     
    Last edited: Oct 11, 2006
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