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Series expansion

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data

    I want to show/prove that

    √(y)√(1+y) - ln[√(y)+√(1+y)] = 2y^(3/2)/3 when y<<1 by series expansion.

    2. Relevant equations

    √(1+y) = 1+y/2 - (y^2)/8 + ....
    ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]

    3. The attempt at a solution
    I'm thinking I sub in the expansions above into the initial equation. My question then, is what do I do next?
  2. jcsd
  3. Jul 27, 2008 #2

    Gib Z

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    Homework Helper

    Use [tex]\log_e (1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} [/tex] and let [tex] x = \sqrt{y} + \sqrt{1+y} -1 [/tex]. The first two terms of that expansion should clean up very nicely with the [itex]\sqrt{y}\sqrt{1+y}[/itex] from the original question, only slightly hard part is expanding that cubic term. Good luck
  4. Jul 27, 2008 #3
    Cool, I understand this method, just wondering how I would go about it if I were to use the relevant equations though.
  5. Jul 27, 2008 #4

    Gib Z

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    Well the problem changes to showing [tex]\log_e \left( 1 + \sqrt{y} + \frac{y}{2} - \frac{y^2}{8} ... \right) \approx \sqrt{y} - \frac{y^{3/2}}{6} - \frac{y^{5/2}}{8} ...[/tex].

    Using the log (1+x) expansion to only a linear term doesnt work, so I might try using two terms, but that might take a while, Sorry I can't see anything better at the moment.
  6. Jul 27, 2008 #5
    that's ok, its just that those specific results are provided, so I'm eager to utilize them!
  7. Jul 28, 2008 #6
    After referring to a textbook, it seems a more appropriate way would be to let Q = √(y) + y/2 - y^2/8 with ln(1+Q) = Q - (Q^2)/2 + (Q^3)/3 - ..... [similiar to what was initially mentioned].

    I expand (without worrying about terms higher than 3/2) and am left with

    does this appear correct?
    pardon my forgetfulness, but can fractional numbers raised to powers be added?
  8. Jul 28, 2008 #7
    *to the top*

    help please!
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