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Converting 1/(1-z) to a sum

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    So I'm checking my solutions to past question and there's one bit that throws me.
    1/(1+(z-1)) = Σ(-1)n(z-1)n (for 0<|z-1|<1)
    I don't know where the (-1)n factor came from. Is it just something that always happens that I didn't know about / forgot about, or is there some other reason that I'm missing?
    Thanks

    (2. Relevant equations)


    (3. The attempt at a solution)
     
  2. jcsd
  3. Dec 17, 2015 #2

    Mark44

    Staff: Mentor

    ##\frac 1 {1 + u} = 1 - u + u^2 - u^3 +- \dots (-1)^n u^n + \dots##
    BTW, the expression in your thread title is different from what you actually asked about.
     
  4. Dec 17, 2015 #3
    That just made me realize DX
    The expression 1/(1+(z-1)) is the same as 1/(1-(-(z-1)))
    and the general rule is: 1/(1-x) = Σxn
    That double minus is where the -1 factor came from.
    I think I understand it, now sure if I explained why I understand it well enough (so many brackets). Just missed out on the sign :D
     
  5. Dec 22, 2015 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It helps to know that the sum of a "geometric series", [itex]\sum_{n= 0}^\infty ar^n[/itex] is [itex]\frac{a}{1- r}[/itex]. That can be shown by looking a the finite series, [itex]S= \sum_{n=0}^N ar^n= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^N[/itex]. We can write that as [itex]S= a+ \left(ar+ ar^2+ \cdot\cdot\cdot+ ar^N\right)= a+ r\left(a+ ar+ \cdot\cdot\cdot+ ar^{N-1}\right)[/itex]. The quantity in the parentheses on the right is almost "S" again! We can make it so by adding [itex]ar^N[/itex] inside the parentheses and subtracting [itex]ar^{n+1}[/itex] outside the parentheses: [itex]S= a+ r\left(a+ ar+ \cdot\cdot\cdot+ ar^{N-1}\right)+ ar^N)- ar^{N+1}= a+ r\left(S\right)- ar^{N+1}[/itex]. Subtracting [itex]rS[/itex] from both sides, [itex]S- rS= (1- r)S= a- ar^{N+1}= a(1- r^{N+1})[/itex]. Finally divide both sides by 1- r to get [itex]S= \frac{a(1- r^{N+1})}{1- r}[/itex].

    As long as |r|< 1, we can take the limit as N goes to infinity. Then [itex]r^{N+1}[/itex] goes to 0 and [itex]\sum_{n= 0}^\infty ar^n= \frac{1}{1- r}[/itex].

    We can think of [itex]\frac{1}{1+ (z- 1)}[/itex] as [itex]\frac{1}{1- (1- z)}[/itex] and write that as a geometric series with a= 1 and r= 1- z: [itex]\sum_{n=0}^\infty (1- z)^n= \sum_{n=0}^\infty ((-1)(z- 1))^n= \sum_{n=0}^\infty (-1)^n (z- 1)^n[/itex].

    Essentially, the "[itex](-1)^n[/itex]" comes from the fact that this is [itex]\frac{1}{1+ (z- 1)}[/itex] rather than [itex]\frac{1}{1- (z-1)}[/itex] which would have power series [itex]\sum_{n=0}^\infty (z- 1)^n[/itex].
     
    Last edited: Dec 22, 2015
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