- #1
FrogPad
- 810
- 0
I'm trying to follow this work, and I can't figure out how they are simplifying the following expression.
[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]
[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]
Last edited: