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

Click For Summary

Discussion Overview

The discussion revolves around the expansion of the expression (1+x)n for real values of n, particularly focusing on why this expansion contains an infinite number of terms when |x|<1. The scope includes theoretical aspects of binomial expansion and the implications of different values of n.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Oscar expresses confusion about the infinite number of terms in the expansion of (1+x)n for real n, particularly when |x|<1.
  • Oscar questions how to expand (1+x)√2, noting that if n were an integer, there would be a finite number of terms.
  • One participant references Newton's binomial theorem, suggesting that for non-integer n, the expansion results in all nonzero terms, contrasting with the case of integer n where only finitely many terms are nonzero.
  • Another participant mentions that following the formula for integer n leads to most terms being zero, implying a distinction in behavior between integer and non-integer values of n.

Areas of Agreement / Disagreement

Participants appear to have differing views on the nature of the expansion for integer versus non-integer n, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

The discussion does not resolve the specific mathematical steps or assumptions regarding the expansion for non-integer values of n, leaving some aspects of the topic unexplored.

2^Oscar
Messages
45
Reaction score
0
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
 
Mathematics news on Phys.org
2^Oscar said:
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]?
 
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:
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 15 ·
Replies
15
Views
4K