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Taylor Series, Binomial Series, Third Order Optics

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that if cosΦ is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes Equation 4 for third-order optics. [Hint: Use the first two terms in the binomial series for [itex]ℓ^{-1}_o[/itex] and [itex]ℓ^{-1}_i[/itex]. Also, use Φ ≈ sinΦ.]

    2. Relevant equations
    Sorry that these don't look very nice. I'm pretty sure all the parentheses are necessary. I've been trying to learn as I go so I didn't know all the commands to make it look nicer. It's taken me well over 30 minutes to type this whole thread out (I was experimenting so yeah...). I'd recommend just looking at the screenshots from my pdf.

    Equation 1:
    [itex]n_1/ℓ_o + n_2/ℓ_i = (1/R)(n_2s_i/ℓ_i - n_1s_o/ℓ_o)[/itex]
    Equation 2 (comes in a pair):
    [itex]ℓ_o = √(R^2 + (s_o + R)^2 - 2R(s_o + R)cosΦ)[/itex]
    [itex]ℓ_i = √(R^2 + (s_i - R)^2 + 2R(s_i - R)cosΦ)[/itex]
    Equation 4:
    [itex]n_1/s_o + n_2/s_i = (n_2 - n_1)/R + h^2( (n_1/(2s_o))(1/s_o + 1/R)^2 + (n_2/(2s_i))(1/R - 1/s_i)^2)[/itex]

    L3dPkMm.png

    3. The attempt at a solution
    Third-degree Taylor polynomial for cosΦ is:
    [itex]1 - Φ^2/2![/itex]
    I plugged in Φ ≈ sinΦ into the third-degree polynomial:
    [itex]1 - sin^2Φ/2[/itex]
    Plugged that into Equation 2 and got:
    [itex]√(R^2sin^2Φ + Rs_osin^2Φ + s^2_o)[/itex]
    [itex]√(R^2sin^2Φ - Rs_isin^2Φ + s^2_i)[/itex]

    Then I plugged that into Equation 1. I don't know what to do after this. I know the hint says to use the first two terms in the binomial series for [itex]ℓ^{-1}_o[/itex] and [itex]ℓ^{-1}_i[/itex], but I have no clue how I would do that since it's not in the form where I could use the binomial series. I have tried manipulating it so it would be, but nothing has worked.
    Side Note:
    This is the first time I have posted on this forum so I apologize if I have broken any rules.
     
  2. jcsd
  3. Dec 4, 2013 #2
    You have not broken any rules that I know about. You will find the Latex gets easier to type with practice. I also spent forever on my first several attempts.

    Let's start with the binomial theorem. You have square roots in ##l_0## and ##l_1##. I'm sure the hint is to use the binomial theorem with n = 1/2 and then grab the first two terms.

    Why did you substitute sin ##\phi## for ##\phi##? They are close when ##\phi## is close to zero, but not necessarily elsewhere. I'm guessing that the purpose of the Taylor polynomial was to get rid of the trig term. Perhaps you have additional notes that suggest this substitution is the way to proceed?
     
  4. Dec 4, 2013 #3
    I used [itex]sin ϕ[/itex] for [itex]ϕ[/itex] because it says that in the hint as noted above.
    I don't think binomial theorem works. If it does, I can't see it.

    My confusion lies in two parts:
    1. How do you find the first two terms in the binomial series for [itex]ℓ^{-1}_o[/itex] and [itex]ℓ^{-1}_i[/itex]?
    2. Where does the h come from in the final equation?
     
  5. Dec 4, 2013 #4
    The binomial theorem says ##(a + b)^r = a^r + ra^{r-1}b## + stuff. If r is a positive integer, you are already familiar with it, and the sum terminates. If r is a not a positive integer, there is an extended binomial theorem which becomes an infinite series. Read more here :http://rutherglen.science.mq.edu.au/wchen/lnfycfolder/fyc22.pdf

    There is lots of literature about when the series does and does not converge, but you don't have to be concerned with this, since you were told to use the first two terms. In this case your r is going to be -1/2.

    However, the expressions for your l's are trinomials. Since it is ##\phi## you are concerned with, you could group the first two terms as your "a" and let the last term be your "b".

    As for the "h", it certainly isn't defined anywhere. I suspect it is some factor that comes up naturally in the course of proving these things, which one then bundles into the ##h^2##. You have several choices here:

    1)Ask your teacher what h is
    2)Look up the 3rd order optics equation somewhere and see what it says h is
    3)Plow ahead and if you wind up with a square factor in that place, call it ##h^2##

    Looking up the optics equation independently might be a good idea anyway. You may find a different formulation for it, which is easier to obtain, or some explanation that helps you.
     
  6. Dec 5, 2013 #5
    Thanks for the help. I'll give what you said a shot in the morning. At this point, I'm doing this for myself since we already turned it in. :P
    I don't like to give up on problems. I'll update this later if I run into anything.
     
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