Compute the first three terms of the Maclaurin series for sqrt(f) if the Maclaurin series for f begins -->

1+ax+bx^2+...

By the given Maclaurin series,
f(0) = 1
f '(0) = a
f ''(0)/2! = b ==> f ''(0) = 2b
------------------------
Now,
g(x) = [f(x)]^(1/2) ==> g(0) = [f(0)]^(1/2) = 1^(1/2) = 1.
g'(x) = (1/2) [f(x)]^(-1/2) * f '(x) ==> g'(0) = (1/2) [f(0)]^(-1/2) * f '(0) = a/2.
g''(x) = (1/2)(-1/2)[f(x)]^(-3/2) * [f '(x)]^2 + (1/2) [f(x)]^(-1/2) * f ''(x)
==> g''(0) = (-1/4) * 1 * a^2 + (1/2) * 1 * 2b = b - a^2/4.

So, [f(x)]^(1/2) = 1 + (a/2) x + ((b - a^2)/4) * x^2/2! + ...

I hope this helps!

we are given that f(x) = 1 + ax + bx^2 + ...
this means: f(0) = 1, f '(0) = a, (1/2) f ''(0) = b

Let g(x) = sqrt(f(x)). At x=0, we then have g(0) = sqrt(f(0)) = 1 (we assume that the positive square root was implied, as usual in this notation).
We also have: g^2 = f, and after a first differentiation:
2 g g' = f ', and at x=0 we have: 2 g(0) g '(0) = f '(0) = a, which gives us: g '(0) = a/2.

Differentiating again:
2(g ')^2 + 2g g '' = f '', and at x=0:
2 (g'(0))^2 + 2 g(0) g ''(0) = f ''(0) = 2b
(g'(0))^2 + g(0) g ''(0) = b
a^2 /4 + g ''(0) = b
g ''(0) = b - a^2 /4

So the first term of the maclaurin series for g(x) are:
g(x) = g(0) + g '(0) x + (1/2) g ''(0) x^2 + ...
= 1 + (a/2) x + (1/8) (4b - a^2) x^2 + ...