Approximating square roots using binomial expansion.

Personally, I wouldn't have done it that way. So here is how I would've done it:

Method 1:

$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$

which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.

$$1.375^2=1.890625$$

Obviously it approaches the correct value as you take more terms.

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Method 2:

This is called fixed-point iteration/Newton's method, and it basically goes like this:

$$x=\sqrt2\implies x^2=2$$

$$2x^2=2+x^2$$

Divide both sides by $2x$ and we get

$$x=\frac{2+x^2}{2x}$$

Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so

$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$

and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$. For example, with $x_0=1$,

$x_0=1$

$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$

$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$

$x_3=\dots=\frac{577}{408}=1.414215686$

And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.

We want to (manually) approximate $\sqrt{2}$ by using the first few terms of the binomial series expansion of \begin{align*} \sqrt{1-2x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-2x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{2}\\ &= 1-x-\frac{1}{2}x^2-\frac{1}{2}x^3+\cdots\tag{1} \end{align*} Here we look for a way to determine appropriate values of $x$ using the binomial expansion.

In order to apply (1) we are looking for a number $y$ with \begin{align*} \sqrt{1-2x}&=\sqrt{2y^2}=y\sqrt{2}\tag{2}\\ \color{blue}{\sqrt{2}}&\color{blue}{=\frac{1}{y}\sqrt{1-2x}} \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*} 1-2x&=2y^2\\ \color{blue}{x}&\color{blue}{=\frac{1}{2}-y^2}\tag{3} \end{align*}

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

• We have to respect the radius of convergence $|x|<\frac{1}{2}$.

• Since we want to calculate an approximation of $\sqrt{2}$ 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 $\sqrt{2}$.

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 $ y^2 $ is close to $\frac{1}{2}$. 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}{2} \end{align*} This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $2$ we might quickly come to $100$ and $49$. These are two small squares and we have $2\cdot 49=98$ close to $100$. That's all.

Now it's time to harvest. We choose $y^2=\frac{49}{100}$ resp. $\color{blue}{y=\frac{7}{10}}$. We obtain for $x$ from (3) \begin{align*} x=\frac{1}{2}-y^2=\frac{1}{2}-\frac{49}{100}=\frac{1}{100} \end{align*} We have now a nice value $\color{blue}{x=\frac{1}{100}}$ and we finally get from (1) the approximation: \begin{align*} \color{blue}{\sqrt{2}} \approx \frac{10}{7}\left(1- 10^{-2}-\frac{1}{2}\cdot 10^{-4}-\frac{1}{2}\cdot 10^{-6}\right)\color{blue}{=1.414\,213\,5}71\ldots \end{align*}

We have $\color{blue}{\sqrt{2}=1.414\,213\,5}62\ldots$ with an approximation error $\approx 9.055\times 10^{-9}$. This result is quite impressive when considering that we have used just four terms of the binomial series.

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.