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

[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

 

[epkid08]

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 (1+x)^\sqrt{2}?

 

[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.

 

[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.