Question

How do you use Taylor series to estimate the accuracy of approximation for $f(x)=\sqrt{x}$ and $a=1$ and $n=3$ with $0.9\le x\le 1.1$

Answer 1

First of all we will differentiate the function $f(x)$ with respect to $x$ on both sides then we will keep differentiating the function with respect to $x$ to get the maximum value of the function and after that we will use Taylor's inequality theorem to obtain the required result.

Complete step-by-step solution:
The Taylor series was discovered by James Gregory a Scottish mathematician and was formally introduced by an English mathematician named Brook Taylor.
A Taylor series is an expansion of some function into an infinite sum of terms where each term has a larger exponent. You may also represent Taylor series in the form of functions of several variables. Taylor Series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point.
If the Taylor Series is centred at $0$then the series is known as the Maclaurin series.
In order to use an approximation method intelligently, we need to have an idea of how good our approximation is. That is, how big an error could we be making. The error is the difference between the approximation value and the exact answer. Notice, that we cannot know exactly how far off we are, without know the exact answer in the first place
We write a general ${{n}^{th}}$ degree polynomial $f(x)={{a}_{0}}+{{a}_{1}}(x-c)+{{a}_{2}}{{(x-c)}^{2}}+...............$
Where ${{a}_{0}},{{a}_{1}},...$ are coefficients on each polynomial term, whereas $c$ is a constant that represents where on $x-axis$ to start the approximation.
We have given $f(x)=\sqrt{x}$
Now differentiating $f(x)$ with respect to $x$ on both sides:
$\Rightarrow \dfrac{d}{dx}f(x)=\dfrac{d}{dx}\sqrt{x}$
$\Rightarrow f'(x)=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$
$\Rightarrow f'(x)=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}$
Now again differentiating $f'(x)$ with respect to $x$ on both sides:
$\Rightarrow \dfrac{d}{dx}f'(x)=\dfrac{d}{dx}(\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}})$
$\Rightarrow f''(x)=\dfrac{1}{2}\times (-\dfrac{1}{2}){{x}^{-\dfrac{1}{2}-1}}$
$\Rightarrow f''(x)=-\dfrac{1}{4}{{x}^{-\dfrac{3}{2}}}$
Now again differentiating $f''(x)$ with respect to $x$ on both sides:
$\Rightarrow \dfrac{d}{dx}f''(x)=\dfrac{d}{dx}(-\dfrac{1}{4}{{x}^{-\dfrac{3}{2}}})$
$\Rightarrow f'''(x)=-\dfrac{1}{4}\times (-\dfrac{3}{2}){{x}^{-\dfrac{3}{2}-1}}$
$\Rightarrow f'''(x)=\dfrac{3}{8}{{x}^{-\dfrac{5}{2}}}$
Now the fourth derivative will be negative. So we know that $f'''(x)$ will be decreasing over the interval
If we differentiate $f'''(x)$ with respect to $x$ on both sides:
$\Rightarrow \dfrac{d}{dx}f'''(x)=\dfrac{d}{dx}(\dfrac{3}{8}{{x}^{-\dfrac{5}{2}}})$
$\Rightarrow f''''(x)=\dfrac{3}{4}\times (-\dfrac{5}{2}){{x}^{-\dfrac{5}{2}-1}}$
$\Rightarrow f''''(x)=-\dfrac{15}{8}{{x}^{-\dfrac{7}{2}}}$
As fourth derivative $|f''''(t)|=\dfrac{15}{16}{{t}^{-\dfrac{7}{2}}}$ that is maximized for $t\in [0.9,1.1]$ hence we can take:
$\Rightarrow \left| f''''(0.9) \right|\le M$
$\Rightarrow \left| \dfrac{15}{16}\times {{(0.9)}^{-\dfrac{7}{2}}} \right|\le M$
$\Rightarrow 1.35557\le M$
By Taylor’s inequality, we have:
$\Rightarrow \left| {{R}_{n}}(x) \right|\le \dfrac{M}{(n+1)!}{{\left| x-a \right|}^{n+1}}$
Where $n=3$ and $a=1$
$\Rightarrow \left| {{R}_{n}}(x) \right|\le \dfrac{1.35557}{(3+1)!}{{\left| 1.1-1 \right|}^{3+1}}$
$\Rightarrow \left| {{R}_{n}}(x) \right|\le \dfrac{1.35557}{(4)!}{{(0.1)}^{4}}$
$\Rightarrow \left| {{R}_{n}}(x) \right|\le \dfrac{1.35557}{4\times 3\times 2\times 1}(0..0001)$
$\Rightarrow \left| {{R}_{n}}(x) \right|\le 0.000005648$
Hence $\left| {{R}_{n}}(x) \right|\le 0.000005648$ for $x\in [0.9,1.1]$

Note: We must remember that the sum of partial series can be used as an approximation of the whole series. Taylor Series is used in computer science, chemistry, calculus, physics and also in the power flow analysis of electrical power systems. It is a series that is used to create an estimation of a function.