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Simple Taylor expansion

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the quadratic approximation to the potential function?

    2. Relevant equations

    U(x) = U0((a/x)+(x/a))
    U0= 20

    3. The attempt at a solution

    This is just the last part of a question on my engineering homework, I never learned Taylor expansions before even though I have taken all the class prerequisites. So if you could just walk me through it that would be much appreciated.
  2. jcsd
  3. Sep 14, 2008 #2


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    Read this http://en.wikipedia.org/wiki/Taylor_series and try it out yourself. It's just a cookbook formula. You have to add various derivatives of U(x) at x=4 times powers of (x-4) up to quadratic. It's not that mysterious. We'll be glad to look at your efforts.
  4. Sep 14, 2008 #3
    So for my first try I got U=(1.25)x2-10x+60. I am pretty sure I remember something about only going out to the second order place to make it "quadratic" so I got.

    40 for the first place, 0 zero for the first order, and 1.25x2-10x+20 for the second order.

    Does this look right?
  5. Sep 14, 2008 #4


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    Yes, that's right. For most purposes you probably want to leave that in the form 40+5*(x-4)^2/4, since you are thinking of x's near 4.
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