Proving the Equivalence of Two Formulas with e as Small

[KateyLou]

Homework Statement

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

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

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

Homework Equations

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) because O(e) is small; but if this is so, why doesn't the first equation become -1?

 

[KateyLou]

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 :-(