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Binomial Expansion Question - fractional powers

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data

    My question is simple is there a formula for the bi/tri-nomial expansion of bi/tri-nomials raised to fractional powers. that is,

    [tex] (x^{2}+1)^{1/2} [/tex] or [tex] (x^{2}+x+1)^{1/2} [/tex]

    I know pascals triangle for integer exponents but i can't really find anything about fraction exponents. I also know that 1/2 or 1/3 are square or cube roots but in that form i don't see anyway of expanding them.

    thanks!
     
  2. jcsd
  3. May 3, 2010 #2

    rock.freak667

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    Homework Helper

    If you have a fractional or negative power, you have an infinite number of terms. Normally you'd expand it the usual way. But you work out nC1 and nC2 to get results such as:

    nC1 =n

    nC2= n(n-1)/2!


    and then you'd just substitute for n.
     
  4. May 3, 2010 #3
    I don't really understand what you are saying, can you use those rules to show me on one of the examples that I listed?
     
  5. May 3, 2010 #4
    Can someone please explain this to me?
     
  6. May 3, 2010 #5

    rock.freak667

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    The expansion of (a+b)n is as follows:

    [tex](a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...[/tex]

    If you insert in your calculator something like 1/2C1, it won't give you a number. But you can simplify nC1 from the definition of nCr

    [tex]\left( ^n _r \right) = \frac{n!}{(n-r)!r!}[/tex]

    [tex] \left( ^n _1 \right) = \frac{n!}{(n-1)!1!}[/tex]

    n!=n(n-1)(n-2)...3.2.1 = n(n-1)!

    You can similarly simplify nC2, in the same manner.


    [tex](a+b)^n = b^n + nb^{n-1}a + ... [/tex]
    Sp
     
  7. May 3, 2010 #6
    so does

    nC1 read n choose 1?

    I really have no experience with n choose k equations, thats why I am having a hard time understanding you.
     
  8. May 3, 2010 #7

    Mark44

    Staff: Mentor

    Yes, nC1 is read as "n choose 1." It also means the number of combinations of n things taken 1 at a time.

    It is also written as
    [tex]\left( \begin{array}{c} n \\ 1 \end{array} \right)[/tex]

    Take a look at this Wike article, especially the section on Newton's generalized binomial theorem - http://en.wikipedia.org/wiki/Binomial_theorem
     
  9. May 3, 2010 #8

    rock.freak667

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    Well yes they are essentially the same idea. Except you are not really 'choosing' in a binomial expansion.

    But you do understand the definitions of n! and nCr as I've typed above right?
     
  10. May 3, 2010 #9
    Ill take a shot at understanding it though, so in:

    [tex]
    (a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...
    [/tex]

    is:

    [tex] b^{n-1}a [/tex]

    then multiplied by:

    [tex]
    \left( ^n _1 \right) = \frac{n!}{(n-1)!1!}
    [/tex]

    is that how it goes? I got the pattern, just those n choose r phrases are a mystery to me at the moment

    or rather

    [tex] \left (^n _k \right) [/tex]

    become the binomial coefficients, is that right?
     
    Last edited: May 3, 2010
  11. May 3, 2010 #10

    Mark44

    Staff: Mentor

    Right. The snag here is that for "n choose k" expressions, both n and k are normally integer values, with n >= r. For your problem, n is going to be 1/2.
     
  12. May 3, 2010 #11
    ok, so i guess the only other question I have is how would you do factorials with fractions,

    what would

    [tex] \frac{1}{2}! [/tex] be?

    I know integer factorials are easy,

    [tex] 4! = 1*2*3*4 [/tex]...
     
  13. May 3, 2010 #12
    On my calculator I found a factorial button, at least I think that's what it is, if i put in:

    (1/2)! it gives 0.8862269255, does that sound right? It doesn't work out to a fraction but if i solve the equation:

    [tex] \left(^n _k \right) = \frac{n!}{(n-k)!k!} [/tex]

    for

    [tex] \left(^n _1 \right) [/tex]

    with

    [tex] n! = n(n-1)! [/tex]

    you get:

    [tex] \frac{\frac{1}{2}(-\frac{1}{2})!}{(-\frac{1}{2})!(1)!} = \frac {1}{2} [/tex]

    does it look like i have a hold on this here?
     
  14. May 3, 2010 #13
  15. May 3, 2010 #14
    ok then..... I really don't know if this is what I am suppose to use but would this be appropriate?

    [tex] \Gamma (n + \frac {1}{2}) = \frac {(2n-1)!!}{2^{n}} \sqrt{ \pi } = \left(^{n-\frac{1}{2}} _{\;\;n} \right) n! \sqrt{ \pi } [/tex]

    If that's the right equation would I use that were the normal binomial theorem calls for a factorial of a fraction?

    This is much more complicated that I had originally hoped.

    Also how do I know when to stop? With the normal theorem using whole integers there should be n+1 terms for a binomial raised to the n powers, but when n = 1/2 n+1 = 3/2 or 1+ 1/2 terms, which does make sense.
     
  16. May 3, 2010 #15

    Mark44

    Staff: Mentor

    As rock.freak667 said in post 2, with fractional or negative exponents, you get an infinite number of terms, unlike what happens when n is a positive integer.
     
  17. May 3, 2010 #16
    ah so this is hopeless? No way to get a finite number of terms? Basically its an infinite sum?
     
  18. May 3, 2010 #17

    rock.freak667

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    The only way to get a finite number of terms is if you neglect certain powers and higher. So if you need to approximate a square root for example, to a certain degree of accuracy, higher terms will become negligible.
     
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