# How do you use Taylor series to estimate the accuracy of approximation for f(x)=sqrt(x) with a=1 and n=3 with 0.9<=x<=1.1?

Dec 14, 2017

Taylor's Theorem guarantees such an estimate will be accurate to within about 0.00000565 over the whole interval $\left[0.9 , 1.1\right]$.

#### Explanation:

Let $f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$ so that $f ' \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}}$, $f ' ' \left(x\right) = - \frac{1}{4} {x}^{- \frac{3}{2}}$, and $f ' ' ' \left(x\right) = \frac{3}{8} {x}^{- \frac{5}{2}}$.

Since $f \left(1\right) = 1$, $f ' \left(1\right) = \frac{1}{2}$, $f ' ' \left(1\right) = - \frac{1}{4}$, and $f ' ' ' \left(1\right) = \frac{3}{8}$, the Taylor polynomial of degree $n = 3$ for $f$ centered at $a = 1$ is

P_{3}(x)=1+1/2 (x-1) - (1/4)/(2!) (x-1)^{2}+ (3/8)/(3!) (x-1)^{3}

$= 1 + \frac{1}{2} \left(x - 1\right) - \frac{1}{8} {\left(x - 1\right)}^{2} + \frac{1}{16} {\left(x - 1\right)}^{3}$.

If we let ${E}_{3} \left(x\right) = f \left(x\right) - {P}_{3} \left(x\right)$ be the error in using ${P}_{3} \left(x\right)$ to approximate $f \left(x\right)$ for $x \approx a = 1$, then Taylor's Theorem guarantees

|E_{3}(x)|=|E_{n}(x)|\leq M/((n+1)!) * |x-a|^{n+1}=M/(4!)|x-1|^{4}, where $| f ' ' ' ' \left(t\right) | \setminus \le q M$ for all $t$ between $x$ and $a = 1$.

Since $f ' ' ' ' \left(x\right) = - \frac{15}{16} {x}^{- \frac{7}{2}}$, we have $| f ' ' ' ' \left(t\right) | = \frac{15}{16} {t}^{- \frac{7}{2}}$, which is maximized for $t \setminus \in \left[0.9 , 1.1\right]$ at $t = 0.9$. Therefore, we can take $M = | f ' ' ' ' \left(0.9\right) | = \frac{15}{16} \cdot {0.9}^{- \frac{7}{2}} \approx 1.35557$.

Hence, rounding up gives $| {E}_{3} \left(x\right) | \setminus \le q \frac{1.3556}{24} \cdot {0.1}^{4} \approx 0.00000565$ for $x \in \left[0.9 , 1.1\right]$.

You can double-check this by noting that $\sqrt{.9} - {P}_{3} \left(.9\right) \approx - 0.00000420$ and $\sqrt{1.1} - {P}_{3} \left(1.1\right) \approx - 0.00000365$.