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Homework Help: Exploiting Geometric Series with Power Series for Taylors Series

  1. Sep 28, 2010 #1
    I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

    Find the Taylor series for....

    EXAMPLE 1:
    [tex]f(x) = \frac{1}{1- (x)}[/tex] around [tex]x = 2[/tex]

    Then,

    [tex] \frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} = \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}}[/tex] provided that [tex]| \frac{x+2}{3} | < 1[/tex] or, [tex] -5 < x < 1[/tex]


    Does this look correct?

    Now if I do the same question in another fashion...

    [tex]\frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}} [/tex] provided that [tex]| \frac{(-x - 2)}{3} | < 1[/tex] or, [tex] -5 < x < 1[/tex]

    They aren't equal, in this one I have a [tex](-1)^{n}[/tex] kicking around in my sum.

    What am I doing wrong here?
     
  2. jcsd
  3. Sep 28, 2010 #2
    So is the first one correct, and the 2nd one incorrect?

    EDIT: The post above this one was removed.
     
  4. Sep 28, 2010 #3

    vela

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    In the second series, you're expanding

    [tex]\frac{1}{1+x} = 1-x+x^2-x^3+\cdots[/tex]

    You forgot the alternating sign, which cancels with the factor of [itex](-1)^n[/tex].
     
  5. Sep 28, 2010 #4
    How do you see,

    [tex]\frac{1}{1+x}[/tex], I can never see it in my work.

    So is the 2nd one incorrect then, or correct and I'm just not understanding why?
     
  6. Sep 28, 2010 #5

    fzero

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    A series expansion around [tex]x=2[/tex] is an expansion in [tex]x-2[/tex]. Also, for the 2nd derivation, you'll find that the expansion of [tex]1/(1+u)[/tex] has a factor of [tex](-1)^n[/tex] that cancels out the one that you introduced.
     
  7. Sep 28, 2010 #6
    But the 2nd derivation is the same question as the first question, right?

    I'm not doing,

    [tex]1/(1+x)[/tex]

    I'm doing [tex]1/(1-x)[/tex]

    I'm sure you know this, and it is probably me who's misunderstanding something but I still can't see it.

    Is it just that the [tex](-1)^{n}[/tex] I introduced gets canceled, and that's it?
     
  8. Sep 28, 2010 #7

    fzero

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    The 1/(1+u) appears in the 3rd part of the 2nd derivation:

    Once you perform the proper expansion, everything falls into place.
     
  9. Sep 28, 2010 #8

    vela

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    When you do the expansion, you'd expanding a function of the form 1/(1+u), not 1/(1-u), where u=-(x+2)/3.
     
  10. Sep 28, 2010 #9
    So would this be correct,

    [tex]\frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} (\frac{-x-2}{3})^{n}[/tex]

    That would take care of the "+ sign", right?

    But then when I pull the other, [tex](-1)^{n}[/tex] out I'll get,

    [tex](1)^{n}[/tex] and my problems will be fixed, correct?
     
  11. Sep 28, 2010 #10

    fzero

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    Yes, both expressions agree.

    Are you meant to expand around x=2 or x=-2? Your OP says x=2, but your expansions are around x=-2. You have the proper radii of convergence, so I'm not sure if the first line of your example is a typo or not.
     
  12. Sep 28, 2010 #11
    It was suppose to be for x=2, there was no typo.

    I guess if I'm doing,

    [tex]\frac{1}{1 - x}[/tex] then I sub in, [tex] x + 2[/tex] or [tex] x - 2[/tex]?

    Similarly,

    if I'm doing,

    [tex]\frac{1}{1 + (-x)}[/tex] then I sub in, [tex] x + 2[/tex] or [tex] x - 2[/tex]?

    This should be the last bit of confusion I have to clear up!
     
  13. Sep 28, 2010 #12

    fzero

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    Around x=2 you will be expanding an expression of the form

    [tex]\frac{1}{1\pm a(x-2)}[/tex].

    The important thing is that x=2 is actually within the radius of convergence of your expansion.
     
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