Approximation to λ_+ in solution to exercise 3b (set 5)

Approximation to λ_+ in solution to exercise 3b (set 5)

by Michal Jan Tyszkiewicz -
Number of replies: 1
I don't understand how we make the approximation that

1/2 [1 - 2δ + sqrt(1 + 4δ - 4δ^2)] \approx 1/2 [1 - 2δ + (1 + 2δ - 6δ^2)]

I see that we could for instance expand sqrt(1 + 4δ - 4δ^2) with Taylor series around 0, but that would lead to 1 + 2δ - 4δ^2 + ..., (notice -4δ^2 not -6δ^2).

Thank you for your help
In reply to Michal Jan Tyszkiewicz

Re: Approximation to λ_+ in solution to exercise 3b (set 5)

by Olivier Lévêque -

Dear Michal,

There is a subtlety here (which by the way does not impact the main conclusion of the exercise): with Taylor expansion, you have, when x is small:

sqrt(1+x) ~= 1 + x/2 - x^2/8

so

sqrt(1+ax+bx^2) ~= 1 + (ax + bx^2)/2 - (ax+bx^2)^2/8 ~= 1+ (a/2) x + (b/2-a^2/8) x^2, and not 1 + (a/2) x + (b/2) x^2

(this mistake was by the way corrected by Alireza last year, as I had it wrong...)

All the best,

Olivier