I need help with a simplification

  • Thread starter FrogPad
  • Start date
  • #1
809
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]
 
Last edited:

Answers and Replies

  • #2
809
0
I figured it out.

It's a binomial series where:
[tex] (X+Y)^n = \ldots [/tex]
[tex] X = 1 [/tex]
[tex] Y = \frac{d}{r} \left(-\cos \theta + \frac{4}{dr} \right) [/tex]


Follow up question. Why use the binomial expansion? Why not perform a taylor expansion?
 
  • #3
J77
1,083
1
Taylor series is just a special case of binomial where you have a "one plus" with a positive exponent.

(What you have can be called a negative binomial series.)
 

Related Threads on I need help with a simplification

Replies
1
Views
846
Replies
2
Views
730
Replies
3
Views
646
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
915
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
923
Top