Using the binomial expansion to approximate a square root.

Question

Find the first four terms of the expansion $\sqrt{1-4x}$ in ascending powers of x. Hence, approximate $\sqrt{6}$ to four decimal places.

I've expanded it to form this: $1-2x-2x^2-4x^3$

When approximating the value of $\sqrt{6}$, bearing in mind that for the expansion to be valid , $\left|x\right|<\frac{1}{4}$, I took out a value of $\sqrt{10}$ to form $\sqrt{10}\sqrt{0.6}$. Then, letting $x = 0.1$, I substituted $x$ into the expansion, then multiplied by $\sqrt{10}$ to find an approximate value for $\sqrt{6}$. Is there anything invalid about this? Why is it better to take out a factor of $2.5$?

Answer 1

Here we look at some arguments why taking out a factor $2.5$ is a convenient choice.

Problem: We want to (manually) approximate $\sqrt{6}$ by using the first few terms of the binomial series expansion of \begin{align*} \sqrt{1-4x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{4}\\ &= 1-2x-2x^2-4x^3+\cdots\tag{1} \end{align*}

In order to apply (1) we are looking for a number $y$ with \begin{align*} \sqrt{1-4x}&=\sqrt{6y^2}=y\sqrt{6}\tag{2}\\ \color{blue}{\sqrt{6}}&\color{blue}{=\frac{1}{y}\sqrt{1-4x}} \end{align*}

We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root. We obtain from (2) \begin{align*} 1-4x&=6y^2\\ 4x&=1-6y^2\\ \color{blue}{x}&\color{blue}{=\frac{1}{4}-\frac{3}{2}y^2}\tag{3} \end{align*}

When looking for a nice $y$ which fulfills (3) there are some aspects to consider:

• We have to respect the radius of convergence $|x|<\frac{1}{4}$.

• Since we want to calculate an approximation of $\sqrt{6}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.

• Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{6}$.

We will see it's not hard to find values which have these properties.

We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that \begin{align*} \frac{3}{2}y^2 \end{align*} is close to $\frac{1}{4}$. So, $y^2$ is close to $\frac{1}{4}\cdot\frac{2}{3}=\frac{1}{6}$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $a,b$ so that
\begin{align*} y^2=\frac{a^2}{b^2}\approx \frac{1}{6} \end{align*} This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $6$ we might quickly come to $100$ and $16$. These are two small squares and we have $6\cdot 16=96$ close to $100$. That's all.

Now it's time to harvest. We choose $y^2=\frac{16}{100}=\frac{4}{25}$ resp. $y=\frac{2}{5}$. We obtain for $x$ from (3) \begin{align*} x=\frac{1}{4}-\frac{3}{2}y^2=\frac{1}{4}-\frac{3}{2}\cdot\frac{4}{25}=\frac{1}{100} \end{align*} We have now an appropriate value $x=\frac{1}{100}$ and we finally get from (1) the approximation: \begin{align*} \color{blue}{\sqrt{6}} \approx \frac{5}{2}\left(1-2\cdot 10^{-2}-2\cdot 10^{-4}- 4\cdot 10^{-4}\right)\color{blue}{=2.449\,4}9 \end{align*}

We have $\sqrt{6}=\color{blue}{2.449\,4}89\,7\ldots$ with an approximation error $\approx 2.572\times 10^{-7}$. This result is quite impressive when considering that we have used just four terms of the binomial series.

At this point it might be clear that factoring out $2.5=\frac{5}{2}=\frac{1}{y}$ is a good choice.

Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.