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Math Help - Applying Binomial Theroem

  1. #1
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    Applying Binomial Theroem

    Using the equation; F_{r}=12\frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{R_{0}+x}\right)^{13}-\left(\frac{R_{0}}{R_{0}+x}\right)^{7}\right]
    I must apply the binomial theroem to; (\frac{1}{R_{0}+x})^{13} And (\frac{1}{R_{0}+x})^7,

    In order to show that F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x
    Also, x is very small comapared to R_{0} so, \left| \frac{x}{R_{0}}\right|<1


    What Ive done so far is, \frac{1}{R^{13}_{0}}.\frac{1}{1+(\frac{x}{R_{0}})^  {13}}
    =\frac{1}{R^{13}_{0}}\left(1+(\frac{x}{R_{0}})\rig  )^{-13}\approx \frac{1}{R^{13}_{0}}\left(1+(-13)\frac{x}{R_{0}}\right)=\frac{1}{R^{13}_{0}}\lef  (1-13\frac{x}{R_{0}}\right)

    So, does? F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x=12\frac{U_  {0}}{R_{0}}\left[\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)-\frac{1}{R^{7}_{0}}\left(1-7\frac{x}{R_{0}}\right)\right]

    If so, could someone show me some working because I cant quite get to the right answer. Any help will be greatly appreciated, thanks.
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  2. #2
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    Re: Applying Binomial Theroem

    Quote Originally Posted by johnkash View Post
    Using the equation; F_{r}=12\frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{R_{0}+x}\right)^{13}-\left(\frac{R_{0}}{R_{0}+x}\right)^{7}\right]
    I must apply the binomial theroem to; (\frac{1}{R_{0}+x})^{13} And (\frac{1}{R_{0}+x})^7,

    In order to show that F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x
    Also, x is very small comapared to R_{0} so, \left| \frac{x}{R_{0}}\right|<1


    What Ive done so far is, \frac{1}{R^{13}_{0}}.\frac{1}{1+(\frac{x}{R_{0}})^  {13}}
    =\frac{1}{R^{13}_{0}}\left(1+(\frac{x}{R_{0}})\rig  )^{-13}\approx \frac{1}{R^{13}_{0}}\left(1+(-13)\frac{x}{R_{0}}\right)=\frac{1}{R^{13}_{0}}\lef  (1-13\frac{x}{R_{0}}\right)

    So, does? F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x=12\frac{U_  {0}}{R_{0}}\left[\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)-\frac{1}{R^{7}_{0}}\left(1-7\frac{x}{R_{0}}\right)\right]

    If so, could someone show me some working because I cant quite get to the right answer. Any help will be greatly appreciated, thanks.
    You have a bit of an algebra error

    \left( \frac{R_0}{R_0+x}\right)^{13} =\left( \frac{1}{1+\frac{x}{R_0}}\right)^{13}= (1+\frac{x}{R_0})^{-13}

    Then the binomial series is

    1-\frac{13}{R_0}x

    In both series you shouldn't have the factors of R_0 to the seventh and thirteenth power.
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  3. #3
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    Re: Applying Binomial Theroem

    So are you saying that they should be? \frac{1}{R_{0}}\left(1-\frac{13}{R_{0}}x\right) and \frac{1}{R_{0}}\left(1-\frac{7}{R_{0}}x\right)
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