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Finding a Taylor Series from a function and approximation of sums

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\mu = \frac{mM}{m+M}[/itex]

    a. Show that [itex]\mu = m[/itex]
    b. Express [itex]\mu[/itex] as m times a series in [itex]\frac{m}{M}[/itex]

    2. Relevant equations

    [itex]\mu = \frac{mM}{m+M}[/itex]

    3. The attempt at a solution

    I am having trouble seeing how to turn this into a series. How can I look at the given function differently and see a series in it?
     
  2. jcsd
  3. Feb 12, 2013 #2

    LCKurtz

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    I don't see how that can be. Is there something you aren't telling us?
    Maybe try writing$$
    \frac{mM}{m+M}= m\left( \frac M {M+m}\right) = m\left( \frac 1 {1+\frac m M}\right)$$and use long division.
     
  4. Feb 13, 2013 #3

    I like Serena

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    I'd go for writing it as a geometric series.
    $${1\over 1+x}=1-x+x^2-x^3+...$$
     
  5. Feb 13, 2013 #4

    HallsofIvy

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    If [itex]\mu= \frac{mM}{m+Mz}= m\frac{M}{m+ M}[/itex] then [itex]\mu= m[/itex] if and only if [itex]\frac{M}{m+ M}= 1[/itex] which leads to [itex]M= m+ M[/itex] and then [itex]m= 0[/itex].
     
  6. Feb 13, 2013 #5
    Apologies everyone, it is actually [itex]\mu \approx m[/itex].

    Also, I do see how I could turn LCKurtz suggestion into a geometric series: [itex]m * \frac{1}{1-(- \frac{m}{M})}[/itex] is the sum of the geometric series: [itex]\Sigma^{\infty}_{0} (-1)^nm(\frac{m}{M})^n[/itex]
     
  7. Feb 14, 2013 #6

    Redbelly98

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    Is there more that you haven't told us? Making this approximation depends on how large m and M are relative to each other. But the problem statement says nothing about their relative magnitude or what they are supposed to represent.
     
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