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Infinite number of terms in the expansion (1+x)^n?

  1. Apr 11, 2009 #1
    hey,

    I'm having a good deal of difficulty understanding why the following expansion has an infinite number of terms within it:

    (1+x)n (|x|<1 where n is any real number)

    Would someone mind explaining this to me please?


    Thanks,
    Oscar
     
  2. jcsd
  3. Apr 11, 2009 #2
    If n was an integer there would be n terms, but how do you suppose someone expands [tex](1+x)^\sqrt{2}[/tex]?
     
  4. Apr 12, 2009 #3
    I think he is referring to Newton's form of the binominal theorem for the absolute value of n less than 1.
    Something like[tex] \sqrt{x+1}= x+\frac{1}{2x}-\frac{1}{8x^3}+-[/tex] In the case of x=5^2, we can expand like
    [tex]\sqrt{5^2+1} = 5+1/10-1/1000+-+ =5 +99/1000 =5.099 +-+[/tex] This is sometimes a convient way to get rough answers mentally.
     
    Last edited: Apr 12, 2009
  5. Apr 12, 2009 #4

    CRGreathouse

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    If you follow the (infinite) formula for integer n, you'll see that all but finitely many terms are equal to zero. If n is not an integer then the terms are all nonzero.
     
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