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Any quick tricks to multiplying by powers?

  1. Dec 11, 2011 #1
    Hey guys

    Basically I want to evaluate ((1/3 +x/2 +x3/6))3

    But instead of doing the long hand is there any way off doing this quickly? Basically I was wanting to find the coefficient of x3?

    So ya any tricks because I seem to remember doing something like 3(1/3*1/2*1/6) +... can't remember..

    Thanks in advance!!
    Regards
    Tam
     
  2. jcsd
  3. Dec 11, 2011 #2

    phyzguy

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    What you want are called 'multinomial coefficients' - here's a link:
    http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients

    For your problem, if you consider this of the form (a+b+c)^3, there are two ways to get x^3, you can have a^2*c, or you can have b^3. The coefficients of these will be :

    Coef(a^2 c) = 3!/(2! 0! 1!) , and
    Coef(b^3) = 3!(0! 3! 0!)

    Hopefully from there you can work out the coefficient of x^3. If you do that, what do you get?
     
  4. Dec 11, 2011 #3
    Thanks.. Okay we need:Ʃn!/p!q!r! For apbqcr

    So does that give us the coeff of x3 = 3 + 1 = 4... doesn't seem right...

    Any help would be great!?

    Thanks
     
  5. Dec 11, 2011 #4

    phyzguy

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    No, you need:

    (1) c^3 + (3) a^2 c =

    (1) (x/2)^3 + (3) (1/3)^2 (x^3/6) = (What) x^3
     
  6. Dec 12, 2011 #5
    Okay using what you wrote I get 13/72. This is correct as I have checked with my answer.. However I am unsure of why you multiplied b3 by (1) and a2c by (3)?

    Thanks for you help
    Regards
    Tam
     
  7. Dec 12, 2011 #6

    phyzguy

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    These are the multinomial coefficients, n!/(p! q! r!), like you wrote in your earlier post,

    for a^2 c, n=3, p=2, q=0, r=1, so n!/(p! q! r!) = 3!/(2! 0! 1!) = 3

    for b^3, n=3, p=0 q=3, r=0, so n!/(p! q! r!) = 3!/(0! 3! 0!) = 1
     
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