# Thread: [SOLVED] Maclaurin series for square root

1. ## [SOLVED] Maclaurin series for square root

Can anyone tell me where I messed up (not getting the correct answer):

What is the 4th order Maclaurin of: $f(x) = \sqrt{9 + x^2}$

$1 + u = 9 + x^2 \implies$

$\sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^2}{2} + O(x^3) \implies$

$\sqrt{9 + x^2} = 1 + \frac{8 + x^2}{2} - \frac{(8 + x^2)^2}{8} + O(x^6)$

EDIT: The substitution $u = \frac{x^2}{9}$ works better. Then I can just use $3\sqrt{1 + \frac{x^2}{9}}$

2. I'm getting

$3+\frac{x^{2}}{6}-\frac{x^{4}}{216}+\frac{x^{6}}{3888}-\frac{5x^{8}}{279936}+...............$

3. Yeah I got the correct answer after using a proper substitution.