Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.

Binomial expansion

$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $

Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$ = 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ... $$

The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$

And $$ (1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$ Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.

Is there an easy way to determine an appropriate value for $x$?


4 Answers 4


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.


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

Method 1:


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.


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

Method 2:

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

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


Divide both sides by $2x$ and we get


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


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





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


It is often easier to begin with a slightly different problem to simplify slightly. Stay with me and you will see why.

Instead of finding the square root of 2, look at the square root of 200 and note that:

200 = 14² + 4

This is part of a general approach that says if N = a² + b, where b << a, then we may write:

$$\sqrt N = a(1 + \frac b {a^2} ) =a[1 + \frac b {2a^2} - \frac {b^2} {8a^4} + \frac {b^3} {16a^6} - ... ] = a + \frac b {2a} - \frac {b^2} {8a^3} + ...$$

Now, if we use N = 200, a = 14, b = 4, we get the first two terms give:

$$ \sqrt 200 = 14\frac 1 7 $$

That is $ \sqrt 2 = 1.4142857...$ which is correct to 5 significant figures. By including the next term, we get $\sqrt 200 = 14 + \frac 1 7 - \frac 1 {1372}$. This gives the superb result:

$ \sqrt 2 = 1.414212827...$, which is correct to 6 significant figures.

APPENDIX - Another example of $\sqrt{13}$

Using the same approach and only the first two terms, we note that 36² = 1296. Thus, $\sqrt{1300} = 36\frac 1 {18} = $. This gives:

$\sqrt{13} = 3.60555 $, correct to 6 significant figures.


In fact you can take any two numbers which can be added to get 2 (not nesserly 0.01 but at least you should know the root of one of them

So for example $\sqrt{2} = {(1+1)^{1/2}}$

Know all what you need is to expand it using bio theorem and for 2 terms you ll get 1.5

$1+(0.5)\cdot 1= 1.5 $ Which is not too bad approximation for $\sqrt{2} = 1.4142135624 $

  • $\begingroup$ I have edited your answer to use MathJax, but it is still in need of improvement. What do you mean when you say $\sqrt{2} = \frac{(1+1)^1}{2}$? This is not an equality... Also, what is "bio theorem"? Do you mean the binomial theorem? Your answer is altogether hard to understand. $\endgroup$
    – Xander Henderson
    Commented Apr 8, 2018 at 21:58

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