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Binomial vs Geometric form for Taylor Series

  1. Dec 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Sorry if this is a dumb question, but say you have 1/(1-x)

    This is the form of the geometric series, and is simply, sum of, from n = 0 to infiniti, X^n. I am also trying to think in terms of Binomial Series (i.e. 1 + px + p(p-1)x/2!...p(p-1)(p-2)(p-(n-1) / n!).

    1/(1-x) is also equivalent to (1 - (-x) ) ^ -1 --> Here we can substitute p = -1, and x = -x into the original binomial series, and I believe it should also still work out to X^n.

    My question is when I should use one over the other. It took me a little bit to even get this connection, so I just want to make sure I'll be Ok when I see it on the test. 1/(1-x) is exactly in the geometric (sum) form, so I can see that, but once the functions start to change, I want to make sure I'm Ok.

    Thanks
     
  2. jcsd
  3. Dec 8, 2014 #2

    Stephen Tashi

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    Science Advisor

    What does "work out to X^n" mean?

    Perhaps your question is whether the substitutions you suggested will make the binomial series sum to [itex] \frac{1}{1+x} [/itex]

    There is such a thing as a "negative binomial series" http://mathworld.wolfram.com/NegativeBinomialSeries.html Is that infinite series what you mean by "binomial series"? Or does your notation indicate a binomial series with only a finite number of terms?
     
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