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Math Help - binomial problem

  1. #1
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    binomial problem

    HI,
    I need help with this one (x^1/2 - y^1/2)^5.

    just need to know where to start.

    Thanks.
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  2. #2
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    Re: binomial problem

    Quote Originally Posted by anthonye View Post
    HI,
    I need help with this one (x^1/2 - y^1/2)^5.

    just need to know where to start.

    Thanks.
    the formula for a binomial expansion is probably a good place to start

    (x+y)^n=\sum_{k=0}^n \left(n \atop k\right)x^k y^{n-k}\mbox{ where }\left(n \atop k\right)=\frac{n!}{k!(n-k)!}

    note that (x-y)=(x+(-y))
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  3. #3
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    Re: binomial problem

    ok but what do I do with the powers on the variables?
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  4. #4
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    Re: binomial problem

    Quote Originally Posted by anthonye View Post
    ok but what do I do with the powers on the variables?
    (x^{1/2} - y^{1/2})^n = \left((x^{1/2}) + (-y^{1/2})\right)^n
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  5. #5
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    Re: binomial problem

    Just not seeing a method here?
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  6. #6
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    Re: binomial problem

    Go to the formula. Wherever you see an x, replace it by (x^{1/2}). Wherever you see a y, replace it by (-y^{1/2}).

    Example: x^k becomes (x^{1/2})^k
    Last edited by SlipEternal; December 24th 2013 at 10:18 AM.
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  7. #7
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    Re: binomial problem

    OK but wouldn't the first term be 5c0(x^1/2)^5.
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  8. #8
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    Re: binomial problem

    Yes, that is correct. (x^{1/2})^5 = x^{1/2}\cdot x^{1/2} \cdot x^{1/2} \cdot x^{1/2} \cdot x^{1/2} = x^{1/2+1/2+1/2+1/2+1/2} = x^{5\cdot 1/2} = x^{5/2}. In other words, (a^b)^c = a^{b\cdot c}.
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