Expansion of 1/(1+quantity) in Coulomb's Law

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Discussion Overview

The discussion revolves around the expansion of the expression \( \frac{1}{1+\frac{2dsin(\theta)}{r_1}} \) as presented in a lecture on Coulomb's Law. Participants explore the mathematical reasoning behind this expansion, specifically using Taylor polynomials, and examine the conditions under which the approximation holds.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant references a lecture that expands \( \frac{1}{1+\frac{2dsin(\theta)}{r_1}} \) into \( 1-\frac{2dsin(\theta)}{r_1} \) and seeks clarification on this process.
  • Another participant suggests using Taylor polynomials to derive the expansion, providing the general form of the series and noting that for small \( d \) or large \( r_1 \), the approximation is valid.
  • A question is raised regarding the restrictions on the variables \( d \), \( r_1 \), and \( \theta \), with an example provided that illustrates a potential discrepancy in the approximation when specific values are chosen.
  • Further clarification is sought on the conditions for the approximation, with emphasis on the requirement that \( d \sin(\theta) \) must be much smaller than \( r_1 \).
  • One participant challenges the assumption that \( \theta \) can be arbitrary, arguing that since \( \sin(\theta) \) is bounded, the key condition is that \( d \) must be much smaller than \( r_1 \).

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for the approximation to hold, particularly regarding the role of \( \theta \) and the relationship between \( d \) and \( r_1 \). There is no consensus on the exact restrictions or the validity of the approximation under various conditions.

Contextual Notes

Participants note that the approximation relies on the assumption that \( d \) is much smaller than \( r_1 \), but the discussion reveals uncertainty about how \( \theta \) influences this relationship and whether other conditions might apply.

yayscience
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I'm watching some lectures on electromagnetism from NPTEL on Youtube.
Lecture 3 on Coulomb's Law, at 29minutes:
http://www.youtube.com/watch?v=0A45kt2U3U8&feature=player_profilepage#t=1740s


The professor expands this:
\frac{1}{1+\frac{2dsin(\theta)}{r_1}}
into:
1-\frac{2dsin(\theta)}{r_1}

I'm not seeing this expansion. Could someone provide a hint (or a good resource for similar types of problems? Anything would be helpful.)

Thanks all!
 
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You can use taylor polynomials, if you're familiar with those.

\frac{1}{1+x} = f(x) = f(0) + f'(0) x + \frac{f''(0)x^2}{2!} + . . .

Notice that the nth derivative evaluated at 0 = (-1)^n. So

\frac{1}{1+x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + . . .

Now let x = 2dsin(ɵ)/r.

\frac{1}{1+\frac{2dsin\theta}{r_1}} = 1 - \frac{2dsin\theta}{r_1} + \frac{(\frac{2dsin\theta}{r_1})^2}{2!} - . . .

For small d or large r, the expansion the professor gives is a good approximation because the squared terms are very small.

If you haven't experienced Taylor Polynomials before, say so and I'll explain what I just did.
 
What's the restriction on d? r1? Theta?

I'm assuming that d and r1 are positive integers and theta is any angle. Algebraically speaking, they're not equivalent. Choosing d = 1, r1 = 1, and theta = pi/2, you get:

1/(1+2sin(pi/2)) = 1/3

and

1 - 2sin(pi/2) = -1

It looks like he just pulled that out of thin air, but what I think he's doing is using an approximation using the fact that d << r1.
 
The expression (1+x)n is well approximated by 1+nx -- provided that |x| is much much smaller than 1, i.e. |x|«1.

For the example at hand -- n=-1, x=2d(sinθ)/r1 -- this means the approximation holds if
d sinθ « r1

Loosely speaking, this means either d or sinθ must be small, or r1 must be large.
 
Redbelly98 said:
Loosely speaking, this means either d or sinθ must be small, or r1 must be large.

Couldn't θ be arbitary? sinθ is bounded between -1 and 1, so all you should really need is that d << r1.
 

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