I'm trying to follow this work, and I can't figure out how they are simplifying the following expression.(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \frac{r}{r_1} = \left( 1 - \frac{d}{r} \cos \theta + \left( \frac{d}{2r} \right)^2 \right)^{-1/2} [/tex]

[tex] \frac{r}{r_1} \approxeq 1 - \frac{1}{2}\left( -\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right) + \frac{3/4}{2} \left(-\frac{d}{r} \cos \theta + \frac{d^2}{4r^2} \right)^2 [/tex]

This is with the assumption [itex] r \gg d [/itex].

Are they doing a taylor expansion?

Another related question.

The book performs a binomial expansion from,

[tex] \left( R^2 - \vec R \cdot \vec d + \frac{d^2}{4} \right)^{-3/2} \approxeq R^{-3}\left(1-\frac{\vec R \cdot \vec d}{R^2} \right) [/tex]

So a binomial expansion needs to be of the form,

[tex] (X+Y)^n [/tex] right?

So would X just be [itex] R^2 [/itex] and Y would be the scalar [itex] -\vec R \cdot \vec d + \frac{d^2}{/4} [/itex] ?

So does X and Y each have to be a function of one variable respectively? Or can they be any algebraic term? for example, could

[tex] X = R^2 \cos \theta [/tex]

[tex] Y = \sin (\phi+R) [/tex]

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# Homework Help: I need help with a simplification

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