Taylor-approximation of square root

Question

Can someone please guide me with this question: Use the Taylor polynomial of $$f(x)=\sqrt{1+x}$$ around $0$ to get an estimate for $\sqrt{2}$ correct to one decimal places (that is, remainder less than $0.05$). State clearly the polynomial you are using. First give the solution as a fraction, then use a calculator to write the solution with one decimal precision. Use a calculator to verify that the solution is correct.

Answer 1

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Indeed, the use of $\ds{\root{\vphantom{\large A}1 + x}}$ to evaluate approximately $\ds{\root{\vphantom{\large A}2} = \root{\vphantom{\large A}1 + \color{#f00}{1}}}$ $\ul{is\ not}$ very convenient because the $\ds{\root{\vphantom{\large A}1 + x}}$ Taylor expansion converges for $\ds{\verts{x} < 1}$. The following approach provides a simple evaluation which is 'fully' convergent ( it's more or less borrowed from the numerical stuff ):

\begin{align} \color{#f00}{\root{\vphantom{\large A}2}} & = \root{98 \over 49} = {1 \over 7}\root{\vphantom{\large A}{100 - 2}} = {10 \over 7}\root{{1 - {1 \over 50}}} \\[5mm] & = {10 \over 7}\bracks{% 1 - {1 \over 100} + \sum_{n = 2}^{\infty}{1/2 \choose n}\pars{-\,{1 \over 50}}^{n}} \\[5mm] & = {10 \over 7}\bracks{% 1 - {1 \over 100} + \sum_{n = 2}^{\infty}{n - 3/2 \choose n}\pars{1 \over 50}^{n}} \\[5mm] & \ds{=} {10 \over 7}\bracks{% 1 - {1 \over 100} + {1 \over 2500}\sum_{n = 0}^{\infty}{n + 1/2 \choose n + 2}\pars{1 \over 50}^{n}} \\[5mm] & = {10 \over 7} - {1 \over 70} + {1 \over 1750}\sum_{n = 0}^{\infty}{n + 1/2 \choose n + 2}\pars{1 \over 50}^{n} \tag{1} \end{align}

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$$ \begin{array}{c} \ds{\mbox{Numerical evaluation of}\ \root{2} = 1.41421356237309\ldots} \\[2mm] \pars{~\mbox{which can be compared with the following table}~} \end{array} $$

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From expression $\ds{\pars{1}}$: \begin{equation} \begin{array}{rclc}&&& \mbox{Error}\\[8mm] \hline \ds{10 \over 7} & \ds{\approx} & \ds{1.4\color{#f00}{2\ldots}} & \ds{1.0153\ \%} \\[3mm] \ds{99 \over 70} & \ds{\approx} & \ds{1.4142\color{#f00}{8\ldots}} & \ds{5.1019 \times 10^{-3}\ \%} \\[3mm] \ds{19\ 799 \over 14\ 000} & \ds{\approx} & \ds{1.41421\color{#f00}{4\ldots}} & \ds{5.1148 \times 10^{-5}\ \%} \\[3mm] \ds{1\ 979\ 889 \over 1\ 400\ 000} & \ds{\approx} & \ds{1.4142135\color{#f00}{7\ldots}}& \ds{6.4032 \times 10^{-7}\ \%} \\[3mm] \ds{22\ 627\ 417 \over 16\ 000\ 000} & \ds{\approx} & \ds{1.414213562\color{#f00}{5\ldots}}& \ds{8.9735 \times 10^{-9}\ \%} \\&&& \\ \hline \end{array} \end{equation}

Answer 2

The Taylor series for $f(x)=\sqrt{1+x}$ is given by

$$f(x)=1+\sum_{n=1}^\infty \frac{(-1)^{n+1} (2n)!x^n}{4^n(n!)^2(2n-1)}$$

Inasmuch as the series is an alternating series, the error of the $N-1$'th partial sum for $x=1$ is bounded by $\frac{(2N)!}{4^N(N!)^2(2N-1)}$. We want to find $N$ such that the error is less than $0.05$. Thus, we set

$$\frac{(2N)!}{4^N(N!)^2(2N-1)}<0.05$$

which is satisfied for $N\ge 4$. Therefore, the square root of $2$ can be approximated to one decimal place as

$$\sqrt{2}\approx 1+\frac12-\frac18+\frac1{16}=\frac{23}{16}\approx 1.4375$$

However, we can check numerically that for $N=3$, the approximation

$$\sqrt{1+x}\approx 1+\frac12-\frac18 \approx 1.375$$

is also within the prescribed tolerance of $0.05$.

Answer 3

Just for your curiosity.

Following Dr. MV's answer, let us consider the more general case where you would like a remainder less that $\epsilon$. Then, as Dr. MV wrote, you are looking for $n$ such that $$\frac{(2n)!}{4^n(n!)^2(2n-1)}<\epsilon\tag 1$$ For approximating the lhs, use Stirling approximation which is $$m!\sim \sqrt{2 \pi m} \left(\frac m e\right)^m$$ This makes $(1)$ much simpler since it becomes $$\frac{1}{\sqrt{\pi } \sqrt{n} (2 n-1)}<\epsilon\tag 2$$ and we can write $$\frac{1}{2n\sqrt{\pi } \sqrt{n} }<\frac{1}{\sqrt{\pi } \sqrt{n} (2 n-1)}<\epsilon\tag 3$$ from which $$n >\frac 1{(2 \sqrt{\pi}\epsilon)^{2/3}}\approx \frac{0.43}{\epsilon ^{2/3}}\tag 4$$ Since $n$ is an integer, then select $$n=\left\lceil \frac{0.43}{\epsilon ^{2/3}}\right\rceil\tag 5$$

For your case where $\epsilon=0.05$, the above formula would lead to $n=4$. To $\epsilon=0.005$ would correspond $n=15$. As Dr. MV mentioned, it is possible that one less term could be resuired. For the last considered remainder, summing up to $n=15$ leads to $$\frac{95064943}{67108864}\approx 1.41658$$ but $n=14$ leads to $$\frac{47365319}{33554432}\approx 1.41160$$ which is also within the prescribed tolerance of $0.005$.