# Question 0a436

Sep 30, 2015

Maclaurin expansion to the 4th term:

$1 + \frac{1}{2} x - \frac{1}{8} {x}^{2} + {x}^{3} / 16 + O \left({x}^{4}\right)$

And using this series to approximate $\sqrt{0.5}$: $0.773$ which is 95.9% accurate.

#### Explanation:

A Maclaurin series is just a Taylor series expansion of a function about 0:

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

In our case,

$f \left(x\right) = \sqrt{x + 1}$ and $f \left(0\right) = 1$

To get the other terms, of the series, we must compute derivatives of the function.

To get the first derivative, we should use the power rule

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

And to get the second term of the expansion, we evaluate the first derivative with $x = 0$ and multiply it by x.

$f ' \left(0\right) x = \frac{1}{2 {\left(0 + 1\right)}^{\frac{1}{2}}} x = \frac{1}{2} x$

To get the second derivative, we take the derivative of the first derivative.

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

To get the third term of the expansion, we evaluate the second derivative with $x = 0$, multiply it by ${x}^{2}$ and divide it by 2!

f''(0)x^2/(2!) = -1/4(0+1)^(-3/2)x^2/(2!) = -1/4*x^2/(2!) = -1/8x^2

To get the third derivative, we take the derivative of the second derivative.

${f}^{3} \left(x\right) = - \frac{1}{4} {\left(x + 1\right)}^{- \frac{3}{2}} \mathrm{dx} = \frac{3}{8} {\left(x + 1\right)}^{- \frac{5}{2}}$

To get the fourth term of the expansion, we evaluate the second derivative with $x = 0$, multiply it by ${x}^{3}$ and divide it by 3!

f^3(0)x^3/(3!) = 3/8(0+1)^(-5/2)x^3/(3!) = x^3/16#

Thus, the Maclaurin series up to the fourth term is

$1 + \frac{1}{2} x - \frac{1}{8} {x}^{2} + {x}^{3} / 16 + O \left({x}^{4}\right)$

Now, to approximate $\sqrt{0.5}$, we solve for $x$

$\sqrt{0.5} = \sqrt{x + 1}$

$0.5 = x + 1$

$- 0.5 = x$

and we plug $x = - 0.5 = - \frac{1}{2}$ into our series

$f \left(- \frac{1}{2}\right) = 1 + \frac{1}{2} \left(- \frac{1}{2}\right) - \frac{1}{8} {\left(- \frac{1}{2}\right)}^{2} + {\left(- \frac{1}{2}\right)}^{3} / 16 = 1 - \left(\frac{1}{4}\right) + \left(\frac{1}{32}\right) - \left(\frac{1}{128}\right) \cong 0.773$