- #1
Alettix
- 177
- 11
Hello!
When studying binominal expansion:
## (a+b)^n = \sum_{k=0}^{n}{{n \choose k}a^{n-k}b^k} ##
in high school, we proved this formula with combinatorics considering that "you can choose either a or b each time you multiply with a binom". Probably, this is not a real mathematical proof at all, but at least we developed an understanding about the concept and found it logical.
Now I have often come across the statement that this formula is valid "for any value of n, whether positive, negative, integer or non-integer" and I have also used it when approximating things in physics. However, I lack an understanding of why the formula works for negative and/or non-integer exponents. How is even ##{n \choose k}## defined for non-integer or negative n?
I would be really happy if you could help me understand this, either with a real proof (hopefully not involving maths I haven't learned yet) or with a logic-argument such as that one I encountered in high school (or both!). Thank you in advance! :)
When studying binominal expansion:
## (a+b)^n = \sum_{k=0}^{n}{{n \choose k}a^{n-k}b^k} ##
in high school, we proved this formula with combinatorics considering that "you can choose either a or b each time you multiply with a binom". Probably, this is not a real mathematical proof at all, but at least we developed an understanding about the concept and found it logical.
Now I have often come across the statement that this formula is valid "for any value of n, whether positive, negative, integer or non-integer" and I have also used it when approximating things in physics. However, I lack an understanding of why the formula works for negative and/or non-integer exponents. How is even ##{n \choose k}## defined for non-integer or negative n?
I would be really happy if you could help me understand this, either with a real proof (hopefully not involving maths I haven't learned yet) or with a logic-argument such as that one I encountered in high school (or both!). Thank you in advance! :)