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## Homework Statement

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

m

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

m

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

^{2}))

^{2}

If you do this manually you get

(1 - 2e - O(e

^{2}) - 2e + 4e

^{2}+ O(e

^{3}) + O(e

^{2}) - O(e

^{3}) + O(e

^{4}))

Is this right - I am assuming O(e^2) means "some term of the order e^2

This simplifies to

(1 - 4e + O(e

^{2}) + O(e

^{3}) + O(e4)

And as e is little the O(e

^{2}) can be neglected...however if this is the case then when both including O(e

^{2}) at all??

Secondly (sorry)

breaking down the first equation is ok:

= -1/2e +1/2e - 2e/2e + O(e

^{2})/2e

= -1 + O(e)

however breaking down the second equation

= -1/2e -1/2e +2e/2e - O(e

^{2})/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?