# Big O notation

## 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)

## 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?