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

  • Thread starter 2^Oscar
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  • #1
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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
 

Answers and Replies

  • #2
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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
If n was an integer there would be n terms, but how do you suppose someone expands [tex](1+x)^\sqrt{2}[/tex]?
 
  • #3
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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:
  • #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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