How do you find the 1st 4 nonzero terms in the Taylor series expansion about x=0 for f(x)= sqrt(1+x)?

Jun 27, 2015

$f \left(x\right) \approx 1 + \frac{x}{2} - {x}^{2} / \left(8\right) + \frac{3 {x}^{3}}{48}$ (for x close to 0)

Explanation:

For a general function $f \left(x\right)$, we can do a Taylor series expansion about $x = 0$, (called the Maclaurin Series) by doing the following:

f(x) = f(0) + x*f'(0) + (x^2*f''(0))/(2!) + (x^3*f^((3))(0))/(3!) + ...

This can be written more concisely with summation notation as

f(x) = sum_(n=0)^(\oo)(x^n f^((n))(0))/(n!)

In this case, we have $f \left(x\right) = \sqrt{1 + x}$
Note that
$f ' \left(x\right) = \frac{1}{2 \sqrt{1 + x}}$

$f ' ' \left(x\right) = - \frac{1}{4 {\left(1 + x\right)}^{\frac{3}{2}}}$

${f}^{\left(3\right)} \left(x\right) = \frac{3}{8 {\left(1 + x\right)}^{\frac{5}{2}}}$

This gives:

$f \left(0\right) = 1$
$f ' \left(0\right) = \frac{1}{2}$
$f ' ' \left(0\right) = - \frac{1}{4}$
${f}^{\left(3\right)} \left(0\right) = \frac{3}{8}$

So the first 4 nonzero terms of the Taylor expansion about $x = 0$ for $f \left(x\right) = \sqrt{1 + x}$ are

$f \left(x\right) \approx 1 + \frac{x}{2} - {x}^{2} / \left(8\right) + \frac{3 {x}^{3}}{48}$

PS: I hope there are no errors. :)