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# Archimedes Numerical Roots

Archimedes estimated the value of $\pi$ by finding the perimeters of regular polygons inscribed in a circle and circumscribed around the circle. He managed to establish that $3\frac{10}{71} < \pi < 3\frac{1}{7}$.

Before he could find the perimeters of polygons he need to be able to calculate square roots. How did he calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between $1$ and $2$. How do I know this? Now calculate the average of $\frac{3}{2}$ and 2 (which is 1.75) - this is a second approximation to $\sqrt 3$. i.e. we are saying that a better approximation to $\sqrt 3$ is $$x_{n+1} = \frac{(\frac{3}{x_n} + x_n)}{2}$$ where $x_n$ is an approximation to $\sqrt 3$ .

We then repeat the process to find the new (third) approximation to $\sqrt{3}$ $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214... $$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?

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Archimedes estimated the value of $\pi$ by finding the perimeters of regular polygons inscribed in a circle and circumscribed around the circle. He managed to establish that $3\frac{10}{71} < \pi < 3\frac{1}{7}$.

Before he could find the perimeters of polygons he need to be able to calculate square roots. How did he calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between $1$ and $2$. How do I know this? Now calculate the average of $\frac{3}{2}$ and 2 (which is 1.75) - this is a second approximation to $\sqrt 3$. i.e. we are saying that a better approximation to $\sqrt 3$ is $$x_{n+1} = \frac{(\frac{3}{x_n} + x_n)}{2}$$ where $x_n$ is an approximation to $\sqrt 3$ .

We then repeat the process to find the new (third) approximation to $\sqrt{3}$ $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214... $$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?

Did you know ... ?

BBC News on 6 January 2010 reported that a computer scientist Fabrice Bellard claimed to have computed the mathematical constant pi to nearly 2.7 trillion digits, some 123 billion more than the previous record. He used a desktop computer to perform the calculation, taking a total of 131 days to complete and check the result. This version of pi takes over a terabyte of hard disk space to store.

Previous records were established using supercomputers, but Mr Bellard claims his method is 20 times more efficient. The prior record of about 2.6 trillion digits, set in August 2009 by Daisuke Takahashi at the University of Tsukuba in Japan, took just 29 hours.

BBC News on 6 January 2010 reported that a computer scientist Fabrice Bellard claimed to have computed the mathematical constant pi to nearly 2.7 trillion digits, some 123 billion more than the previous record. He used a desktop computer to perform the calculation, taking a total of 131 days to complete and check the result. This version of pi takes over a terabyte of hard disk space to store.

Previous records were established using supercomputers, but Mr Bellard claims his method is 20 times more efficient. The prior record of about 2.6 trillion digits, set in August 2009 by Daisuke Takahashi at the University of Tsukuba in Japan, took just 29 hours.

Find the value of sqrt(2+sqrt3)-sqrt(2-sqrt3)and then of cuberoot(2+sqrt5)+cuberoot(2-sqrt5).

Find the smallest numbers a, b, and c such that: a^2 = 2b^3 = 3c^5 What can you say about other solutions to this problem?