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Binomial series

  1. Feb 26, 2015 #1
    The binomial series ##(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + ...## only converges for ##|x| < 1## right?
    Is it true that writing ##(1 + x)^n## differently (i.e. ##x^n (1 + \frac{1}{x})^n##) extends the validity of this series to include values of ##x## such that ##|x| > 1##?
     
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  3. Feb 26, 2015 #2

    Svein

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    No. A binomial series is only defined for a given n, Whatever the value of x, there exists an N such that [itex]nx>1000 [/itex] for n>N.
    No.
     
  4. Feb 27, 2015 #3

    micromass

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    Yes. Although the series might converge for more values than just ##|x|<1##. Complete details are here: http://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence

    Yes, If ##|x|>1##, then your trick can be used. If ##|x|<1##, then your trick doesn't work, but the original series does of course. If ##|x|=1## then the situation is a bit annoying.
     
  5. Feb 27, 2015 #4
    Interesting. I once watched a video in which someone proves that the sum of all natural numbers is ##-\frac{1}{12}##, and in one of the steps, he substituted ##x = 1## into the binomial series. Apparently, anything is possible if you're Euler!
     
  6. Feb 27, 2015 #5

    micromass

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    Yes, that is valid, but only under conditions. For the standard convergence of series, plugging in ##x=1## is forbidden. But there are other definitions where series do not converge how we they usually do. Under those definitions, you do get ##-1/12##.
     
  7. Feb 27, 2015 #6
    Is it possible to approximate ##\sqrt{2}## using the binomial series?
     
  8. Feb 27, 2015 #7

    micromass

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    Yes, but not directly. What you can do (for example) is to use ##x=-1/2## to approximate ##1/\sqrt{2} = \frac{\sqrt{2}}{2}##. And then you can easily find ##\sqrt{2}##.

    Edit: I guess you can even do it directly, but convergence won't be very rapid.
     
    Last edited: Feb 27, 2015
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