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Expansion of 1/(1+quantity)

  1. Jul 4, 2011 #1
    I'm watching some lectures on electromagnetism from NPTEL on Youtube.
    Lecture 3 on Coulomb's Law, at 29minutes:

    The professor expands this:

    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!
  2. jcsd
  3. Jul 4, 2011 #2
    You can use taylor polynomials, if you're familiar with those.

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

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

    [tex]\frac{1}{1+x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + . . . [/tex]

    Now let x = 2dsin(ɵ)/r.

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

    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.
  4. Jul 4, 2011 #3


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    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


    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.
  5. Jul 5, 2011 #4


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    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.
  6. Jul 5, 2011 #5


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    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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