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Math Help - Binomial Expansion

  1. #1
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    Binomial Expansion

    Hi

    Can anyone help me with expanding the following equation?

    [1-(1/x)]^1/4

    Thank you.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by dd86 View Post
    Hi

    Can anyone help me with expanding the following equation?

    [1-{1}/{x}]^1/4

    Thank you.
    (1-\frac{1}{x})^{1/4} = \Sigma_{n=-\infty}^0x^n (-1)^n {\frac{1}{4}\choose{-n}}<br />

    For understanding: Binomial Theorem for Rational n
    Last edited by Also sprach Zarathustra; July 10th 2010 at 08:05 PM.
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  3. #3
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    Sorry, I think one of the brackets have not been closed. I'm a bit confused cos of that
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  4. #4
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    The binomial series:

    (1 + x)^n = \sum_{k = 0}^{\infty}{n\choose{k}}x^k, where {n\choose{k}} = \frac{n(n-1)(n-2)\dots (n-k+1)}{k!}.


    So in your case:

    \left(1 - \frac{1}{x}\right)^{\frac{1}{4}} = \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\left(-\frac{1}{x}\right)^k

     = \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\frac{(-1)^k}{x^k}

     = 1 - \frac{1}{4x} - \frac{3}{32x^2} - \frac{7}{128x^3} - \frac{77}{2048x^4} + O\left[\left(\frac{1}{x}\right)^5\right].
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  5. #5
    MHF Contributor chisigma's Avatar
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    In the formula for the binomial expansion...

    \displaystyle (1+t)^{\alpha} = 1 + \binom {\alpha}{1}\ t + \binom {\alpha}{2}\ t^{2} + \binom {\alpha}{3}\ t^{3} + \dots (1)

    ... You set \alpha=\frac{1}{4} and t= -x^{-1} and obtain...

    \displaystyle (1-\frac{1}{x})^{\frac{1}{4}} = 1 - \frac{1}{4}\ x^{-1} - \frac{3}{2\ 4\ 4}\ x^{-2} - \frac{3\ 7}{6\ 4\ 4\ 4}\ x^{-3} - \dots (2)

    The expansion (2) converges for |x|>1 ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; July 10th 2010 at 08:19 PM. Reason: -1/4 instead of 1/4... sorry!...
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  6. #6
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    Quote Originally Posted by chisigma View Post
    In the formula for the binomial expansion...

    \displaystyle (1+t)^{\alpha} = 1 + \binom {\alpha}{1}\ t + \binom {\alpha}{2}\ t^{2} + \binom {\alpha}{3}\ t^{3} + \dots (1)

    ... You set \alpha=\frac{1}{4} and t= -x^{-1} and obtain...

    \displaystyle (1-\frac{1}{x})^{- \frac{1}{4}} = 1 + \frac{1}{4}\ x^{-1} + \frac{5}{2\ 4\ 4}\ x^{-2} + \frac{5\ 9}{6\ 4\ 4\ 4}\ x^{-3} + \dots (2)

    The expansion (2) converges for |x|>1 ...

    Kind regards

    \chi \sigma
    Sorry, there's something wrong with your calculations.

    For one thing, {\frac{1}{4}\choose{2}} = \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!} = \frac{-\frac{3}{16}}{2} = -\frac{3}{32} \neq \frac{5}{2\,4\,4}...

    Your calculation of the x^{-3} term is also incorrect.

    Also, all the coefficients except for the x^0 term should be negative.
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  7. #7
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    Thank you all for your invaluable help.

    I was confused when expanding the above equation as the textbook I am referring to stated different formulae where n is a non-positive integer and when x>1 or x<1.

    Can the expansion formula that is used when n is a positive integer be still used when n is a negative integer or when n is not an integer at all?
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Prove It View Post
    Sorry, there's something wrong with your calculations.

    For one thing, {\frac{1}{4}\choose{2}} = \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!} = \frac{-\frac{3}{16}}{2} = -\frac{3}{32} \neq \frac{5}{2\,4\,4}...

    Your calculation of the x^{-3} term is also incorrect.

    Also, all the coefficients except for the x^0 term should be negative.
    The expression...

    \displaystyle (1 - \frac{1}{x})^{-\frac {1}{4}} = 1 + \frac{1}{4}\ x^{-1} + \frac{5}{2\ 4\ 4}\ x^{-2} +  \frac{5\ 9}{6\ 4\ 4\ 4}\ x^{-3} + \dots (1)

    ... is perfectly correct... the only problem was that it was requested the expasion of (1-\frac{1}{x})^{\frac{1}{4}} and not of (1-\frac{1}{x})^{-\frac{1}{4}} ... fortunately when You have posted that was already been corrected...

    Kind regards

    \chi \sigma
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  9. #9
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    Quote Originally Posted by dd86 View Post
    Thank you all for your invaluable help.

    I was confused when expanding the above equation as the textbook I am referring to stated different formulae where n is a non-positive integer and when x>1 or x<1.

    Can the expansion formula that is used when n is a positive integer be still used when n is a negative integer or when n is not an integer at all?
    The formula is essentially the same, except you need to use an infinite series rather than finite, and you need to use the revised binomial coefficients.
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  10. #10
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    By revised binomial coefficients, you mean that I have to ensure that I substitute the correct variables in the expansion to get the correct result, right?
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  11. #11
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    I mean that for n as a positive integer, then {n\choose{k}} = \frac{n!}{k!(n-k)!}, and you can look it up on Pascal's Triangle.

    For n in general, {n\choose{k}} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!}.
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  12. #12
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    Thank you very much.
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