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B Binomial Expansion with Negative/Rational Powers

  1. Sep 5, 2016 #1
    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 statment 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! :)
     
  2. jcsd
  3. Sep 5, 2016 #2

    micromass

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    Have you encountered Taylor series before?
     
  4. Sep 5, 2016 #3
    Not in full depth I belive, but to a basic extent yes.
     
  5. Sep 5, 2016 #4

    micromass

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    Cool! Can you find the Taylor series of ##(1+x)^{1/2}## for me? Just find the first few terms and see if you notice a pattern.
     
  6. Sep 5, 2016 #5
    I believe the Taylor expansion at x = a should be:
    ##T_a = (1+a)^{1/2} + \frac{x-a}{2(1+x)^{1/2}} - \frac{(x-a)^2}{4(1+x)^{3/2}} = (1+a)^{1/2} + \frac{(1+a)^{1/2}}{2} - \frac{(1+a)^{1/2}}{4} ##

    I see that the factor ##(1+a)^{1/2}## is repeated, and that the denominators seem to follow ##2^k## (and also change sign?), but I am not yet sure that I have discovered the pattern we are looking for.
     
  7. Sep 5, 2016 #6

    micromass

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    Oh, but you should do a Taylor expansion around ##a=0##. What do you get then? You should get as coefficients a generalization of the binomial coefficient, whatever that generalization is.
     
  8. Sep 5, 2016 #7
    Oh wait, in my previous post, what comes after the second equivalence sign is totally wrong!

    So if I continue from the first equality with a = 0, we get:
    ## (1+x)^{1/2} = 1 + \frac{x}{2} - frac{x^2}{4} = 1 + \frac{1}{2} x +\frac{1}{2} (\frac{1}{2}-1) x^2 ##
    which is what one might expect from the binomial theorem if we write it like:
    ##(a+b)^n = a^n + na^{n-1}*b + \frac{n(n-1)}{2!} a^{n-2}*b^2+ ... +b^n ##
    instead of using the sum and ##{n \choose k}## notation.
     
  9. Sep 5, 2016 #8
    However, this implies futher questions! Because if I am not totally wrong, we will never reach ##b^n## if n is not a positive integer, which means that the binomial expansion is an infinite series and more of an approximation and not an exact formula if n is negative and/or rational. Am right? And if it is just an approximation, for which values of x (or a and b) is it valid? I remember from physics that ##(1+x)^n = 1+nx## require x<<1, but how does this change when more terms are included?

    Secondly, how can one genearilize what we have just done? Should I try to find the MacLaurin series for ##(a+b)^n##?
     
  10. Sep 5, 2016 #9

    micromass

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    OK, very good questions. It's true that it's an infinite series, and thus you can never write down an exact formula with finitely many terms.

    For the right formula, we put ##\binom{\alpha}{k} = \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!}##. Then we have
    [tex](1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2}x^2 + ... + \binom{\alpha}{n} x^n + ...[/tex]

    There are some constraints on this formula. First for all, we want ##\alpha\in \mathbb{C}##, so ##\alpha## can be any complex number. For ##x## however, the formula only holds in general for ##x\in (-1,1)## unless ##\alpha## is a positive integer (in which case it holds for any ##x##). More comprehensive conditions on ##x## can be found on wiki: https://en.wikipedia.org/wiki/Binomial_series

    Thus in general, we do not expand ##(a+b)^\alpha##, but only ##(1+x)^\alpha##. If you want to expand ##(a+b)^\alpha##, then you'll have to do it as follows:
    [tex](a+b)^\alpha = a^\alpha ( 1 + (b/a))^\alpha =a^\alpha \sum \binom{\alpha}{k}\left(\frac{b}{a}\right)^k = \sum \binom{\alpha}{k} b^k a^{\alpha-k}[/tex]
    PROVIDED THAT ##b/a\in (-1,1)##. If it is not, then we'll have to look at ##a/b##.
     
  11. Sep 6, 2016 #10
    Hmm, I am really sorry but I am not entierly sure I can follow you. Where does ##x\in (-1,1)## and ##b/a\in (-1,1)## come from? If we have ##x=1## that's just ##2^\alpha## and if ##x=-1## it's ##0^\alpha##, which doesn't make sense really so I must have missunderstood you.

    Also, Is ##{a \choose k}## defined for negative and/or non-integer ##a##? I cannot imaging choosing 5 elements out of -4, but maybe I am just stuck in the "combinatorics box".
     
  12. Sep 6, 2016 #11

    PeroK

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    It's best, perhaps, to take an example:

    ##(1 + x)^{-1} = 1 -x + x^2 -x^3 \dots##

    But, this is actually a geometric series, so you can see quite easily that it converges only when ##|x| < 1##

    This means the Taylor/Binomial expansion is only valid for a limited range of ##x \in (-1 , 1)##. This is called the Radius of Convergence (R), and applies generally to Taylor series. In some cases ##R = +\infty##, which means the expansion is valid for all ##x##. In other cases, like this one, R is finite.

    If you start with ##(a + b)^{-1}## and if, say, ##a > b##, then ##b/a < 1## so:

    ##(a + b)^{-1} = a^{-1}(1 + b/a)^{-1} = a^{-1}(1 - (b/a) + (b/a)^2 - (b/a)^3 \dots##

    And, if ##a < b##, then you would do:

    ##(a + b)^{-1} = b^{-1}(1 + a/b)^{-1} = b^{-1}(1 - (a/b) + (a/b)^2 - (a/b)^3 \dots##

    In order to get a convergent expansion in each case.
     
  13. Sep 6, 2016 #12

    PeroK

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    Yes, although I don't know if anyone would still view this as "combinatorial". It's just a generalisation, which has the same recursive pattern as the normal binomial coefficients.
     
  14. Sep 6, 2016 #13
    Oh, now I understand! So ##x \in (-1 , 1)## doesn't mean that x should be either ##1## or ##-1##, but rather that ## 1 \leq x \leq -1##. Otherwise the series would diverge and could not be approximated as a sum of low degree terms.
     
  15. Sep 6, 2016 #14
    However, Micromas said earlier that we do not generally expand ##(a+b)^n##, but rewrite it as ##a^n(1+b/a)^n##. Does this mean that the formula is not valid for the first case, or simply that we like to keep the structure of the second case?
     
  16. Sep 6, 2016 #15

    fresh_42

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    ##x \in (-1,1) ## actually means ##-1 < x < 1## and to obtain this you could either write ##a^n (1+b/a)^n## or ##b^n (1+ a/b)^n## because one of the two quotients will be in this range (unless ##a=b##).
     
  17. Sep 6, 2016 #16

    PeroK

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    I must confess, I've always reduced it to ##(1+x)^r## without thinking. Good question! Maybe @fresh_42 knows the answer?
     
  18. Sep 6, 2016 #17

    fresh_42

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    Sorry, I didn't get this. I only wanted to explain why we have to distinguish between ##a/b## and ##b/a##.

    Of course there is no need to rewrite ##(a+b)^n## for natural ## n##. In post #9 in which it has been used, ##n=\alpha## has been complex. But in any case ##(1+x)^{\alpha}## will do.
     
  19. Sep 6, 2016 #18

    PeroK

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    If we take the same example:

    ##(a + b)^{-1} = a^{-1} - a^{-2}b + a^{-3}b^2 - a^{-4}b^3 \dots = a^{-1}(1 - (b/a) + (b/a)^2 - (b/a)^3 \dots)##

    So, that's just another way of doing it when ##|a| > |b|##. I didn't know you could do that!
     
  20. Sep 6, 2016 #19

    fresh_42

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    I think I understood you now. Indeed a good question. ##a,b## real could help.

    Edit: IMO the basic question is, when does the closed formula ##(a+b)^{\alpha}## equals the expansion into a series? Therefore the quotient ##|a / b|## has to be less than ##1## (or vice versa). The closed formulas might be equal but this doesn't guarantee their formal expansions into a series are.
     
    Last edited: Sep 6, 2016
  21. Sep 7, 2016 #20
    So, if I understand you right the equality
    ## \sum_{k=0}^{\alpha}{{\alpha \choose k}a^{\alpha-k}b^k} =a^\alpha\sum_{k=0}^{\alpha}{{\alpha \choose k}1^{\alpha-k}(b/a)^k} ##
    does not hold for any value of ##\alpha##, but rather positive integers only?
     
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