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The question I am trying to solve is as follows:

  1. Expand $\sqrt{1+x}$ to two terms plus remainder. Estimate the remainder.
  2. Use the expression of Ex. 1 (discarding the remainder) to calculate $\sqrt{2}$ What is the degree of accuracy of the approximation?

Here, is my attempt at a solution:

  1. The first part seemed extremely simple as I just have to expand the function according to Taylor’s theorem: $$\sqrt{1+x} = 1 + \frac{x}{2} + R_1(x)$$ where $R_1$ can be expressed in Lagrange form: $$R_1(x) = -\frac{x^2}{8(1+\theta x)^{\frac{3}{2}}}, \theta \in [0,1]$$

The part I struggled to understand was “estimate the remainder”. To estimate the remainder doesn’t one need a value for $x$ or some range for $x$? In order to make it sense for me, I estimated the remainder at $x=1$ $$\frac{\sqrt{2}}{32} \lt |R_1(1)| \lt \frac{1}{8}$$

  1. Just continuing on for the moment, if we approximate $\sqrt{2}$ using the Taylor polynomial above we simply get $1.5$. The problem now is bounding the error term: $$|R_1(1)|=\frac{1}{8(1+\theta)^{\frac{3}{2}}} \leq \frac{1}{8} $$ since $\frac{1}{(1+\theta x)^{\frac{3}{2}}}$ achieves its maximum value when $\theta = 0$. Thus, the error for this approximation is at most $0.125$.

What perplexes me the most is the answer the textbook provides:

  1. The book claims that the expansion of $\sqrt{1+x}$ to two terms is: $$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{1}{4(1+\theta x)^{\frac{3}{2}}}$$ and that the remainder can be estimated as: $$-\frac{1}{4} \lt R_1 \lt - \frac{\sqrt{2}}{16}$$

I understand how they obtained the range for the remainder (simply substituting $\theta = 0$ and $\theta = 1$). However, I don’t understand how to obtain this expression for the remainder.

  1. The textbook simply stated the percentage error of the approximation is a little over $6$%. However, I don’t see how I could get this answer from the methods I know. The only way I could only think of was to calculate $\sqrt{2}$ to 4 or 5 decimal places and actually calculate the percentage error.

I feel a little embarrassed about posting this, as the question seems like to me to be very simple; however, I can’t think of any way to calculate the exact error of the approximation, as everything I have learned to this point is essentially calculating the “worse-case scenario” for the remainder.

I would appreciate any guidance on this.

EDIT:

I have found a way to calculate the value $\sqrt{2}$ using the Taylor expansion above by expressing $\sqrt{2}$ as $\frac{7}{5}\sqrt{1+\frac{1}{49}}$. However, I still don’t understand how the textbook obtained their expression for the remainder.

David Choi
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1 Answers1

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Taylor's Theorem says $$ \begin{align} f(x) &=f(0)+xf'(0)+\int_0^x(x-t)f''(x)\,\mathrm{d}t\tag{1a}\\ &=f(0)+xf'(0)+\frac{x^2}2f''(\xi)\tag{1b} \end{align} $$ for some $\xi$ between $0$ and $x$.

Applying $(1)$ to $f(x)=\sqrt{1+x}$ gives $$ \begin{align} \sqrt{1+x} &=1+\frac x2-\frac{x^2}{8(1+\theta x)^{3/2}}\tag2 \end{align} $$ for some $\theta\in[0,1]$, as in your answer.

As I said in chat, I think the book is in error.

robjohn
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