1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    http://www.youtube.com/watch?v=0A45kt2U3U8&feature=player_profilepage#t=1740s


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

    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

    gb7nash

    User Avatar
    Homework Helper

    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.
     
  5. Jul 5, 2011 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    gb7nash

    User Avatar
    Homework Helper

    Couldn't θ be arbitary? sinθ is bounded between -1 and 1, so all you should really need is that d << r1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Expansion of 1/(1+quantity)
  1. < or = to 1 (Replies: 5)

Loading...