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

1. Apr 11, 2009

2^Oscar

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. Apr 11, 2009

epkid08

If n was an integer there would be n terms, but how do you suppose someone expands $$(1+x)^\sqrt{2}$$?

3. Apr 12, 2009

robert Ihnot

I think he is referring to Newton's form of the binominal theorem for the absolute value of n less than 1.
Something like$$\sqrt{x+1}= x+\frac{1}{2x}-\frac{1}{8x^3}+-$$ In the case of x=5^2, we can expand like
$$\sqrt{5^2+1} = 5+1/10-1/1000+-+ =5 +99/1000 =5.099 +-+$$ This is sometimes a convient way to get rough answers mentally.

Last edited: Apr 12, 2009
4. Apr 12, 2009

CRGreathouse

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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