Applying Binomial Theroem

**johnkash** · October 6th 2012, 05:08 PM

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.

**TheEmptySet** · October 6th 2012, 05:22 PM

Originally Posted by johnkash

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.

**johnkash** · October 6th 2012, 06:00 PM

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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