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Power Series Expanded, Arfken 5.7.16

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    The behavior of a neutron losing energy by colliding elastically with nuclei of mass A is described by a parameter ξ1,
    ξ1 = 1 + [itex]\frac{(A-1)^2}{2A}[/itex]*ln[itex]\frac{A-1}{A+1}[/itex]

    An approximation, good for large A, is
    ξ2= [itex]\frac{2}{A+2/3}[/itex]

    Expand ξ1 and ξ2 in powers of A−1. Show that ξ2 agrees with ξ1 through (A−1)2. Find the difference in the coefficients of the (A−1)3 term.

    2. Relevant equations

    Taylor Expansion, see wiki page

    ln(1-x)= x+ [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+[itex]\frac{x^n}{n}[/itex]

    ln(1+x)= x- [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+(-1)^(n+1)[itex]\frac{(x^n)}{n}[/itex]

    3. The attempt at a solution

    Alright so since it is in powers of A−1 I decided to substitute x=A−1. I get:

    ξ1= 1+ [itex]\frac{x((1/x)-1)^2}{2}[/itex]* ln(1-x)-ln(1+x)

    I use the expansions from above and substitute.
    Now I do not know how to expand the entire thing from here.
     
    Last edited: Nov 13, 2011
  2. jcsd
  3. Nov 15, 2011 #2

    vela

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    Suggestion: Use LaTeX for the entire expression instead of for just bits and pieces.

    You should clean up your math a bit. Those expressions above aren't true because the series on the righthand sides have an infinite number of terms. They don't terminate with the nth term.
    Again, what you wrote is incorrect. Use parentheses (or square brackets, etc.)!
    [tex]\xi_1 = 1+\frac{x}{2}\left(\frac{1}{x}-1\right)^2[\log(1-x)-\log(1+x)][/tex]You just need to substitute the series in, combine terms, and then multiply everything out and simplify. In other words, do a bunch of algebra. Note that since you're only looking for the first few terms, those are the only ones you really need to worry about. Anything that results in higher powers of x you can ignore.
     
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