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Second approximation

  1. Dec 30, 2005 #1
    I have this algebraic term: (1+i)^-1 where i is very very small i<<<1

    if we used second approximation what would it equal to?

    thanks in advance!
     
  2. jcsd
  3. Dec 30, 2005 #2

    Tide

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    You can use the binomial expansion:

    [tex]\frac {1}{1+i} = 1 - i + i^2 + \cdot \cdot \cdot[/tex]

    so in first approximation you have 1 and in second approximation 1 - i etc.
     
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