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Homework Help: Big O notation

  1. Dec 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Taking e to be small, we have been given these two formulas to prove

    m1 = [tex]\frac{-1+\sqrt{1-4e}}{2e}[/tex] = [tex]\frac{-1+(1-2e+O(e^2))}{2e}[/tex] = -1+ O(e)

    m2= [tex]\frac{-1-\sqrt{1-4e}}{2e}[/tex] = [tex]\frac{-1-(1-2e+O(e^2))}{2e}[/tex] = -1+ O(1)

    2. Relevant equations

    3. The attempt at a solution

    Firstly, the second stage in each of these? I am assuming what they have done is say that
    (1 - 4e) = (1 - 2e + O(e2))2
    If you do this manually you get
    (1 - 2e - O(e2) - 2e + 4e2 + O(e3) + O(e2) - O(e3) + O(e4))
    Is this right - I am assuming O(e^2) means "some term of the order e^2

    This simplifies to
    (1 - 4e + O(e2) + O(e3) + O(e4)
    And as e is little the O(e2) can be neglected...however if this is the case then when both including O(e2) at all??

    Secondly (sorry)
    breaking down the first equation is ok:
    = -1/2e +1/2e - 2e/2e + O(e2)/2e
    = -1 + O(e)

    however breaking down the second equation
    = -1/2e -1/2e +2e/2e - O(e2)/2e
    = -1/e + 1 - O(e)

    Our tutor said that 1 - O(e) is the same as O(1) becuase O(e) is small; but if this is so, why doesnt the first equation become -1?
  2. jcsd
  3. Jan 4, 2009 #2
    have had another go at this, and still cannot work out the first bit - the random removal of O(e^2) in some places and not others :-(
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