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A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

Weekly Challenge 16: Archimedes Numerical Roots

Stage: 5 Challenge Level: Challenge Level:1

A sequence of numbers $x_1, x_2, x_3, ... ,$ starts with $x_1 = 2$, and, if you know any term $x_n$, you can find the next term $x_n+1$ using the formula $x_{n=1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$.

Solution by Andaleeb of Woodhouse Sixthform College, London.

For the iteration $$x_{n+1} = \frac{1}{2} \left( x_n + \frac{3}{x_n} \right)$$ \begin{eqnarray} \\ x_1 &=& 2,\; x_2 = 1.75,\; x_2 = 1.732142857,\; x_4 = 1.73205081 \\ x_5 &=& 1.732050808,\; x_6 = 1.732050808,\; x_7 = 1.732050808 \\ x_8 &=& 1.732050808;\end{eqnarray} We notice that when $x_n = 1.732050808$, so is $x_{n+1}$. Squaring these terms we get $x_1^2 = 4, x_2^2 = 3.0625, ... , x_5^2 = 3$ and the rest of the other terms are the same!! This implies that when $x_n \approx \sqrt{3}$ so is $x_{n+1}$ and the values of $x_n$ tend to the limit $\sqrt{3}$. This special property can easily be proven. Assume that the limit exists, so $x_{n+1} = x_n = x$, then solve the equation $$X = \frac{1}{2}\left(X + \frac{3}{N} \right)$$ If we test it for $N = 3$, we see that $x_{29} = 1.44224957$, which is what the calculator gives for the cube root of 3. Testing it for $N = 8$, we get $x_1 = 2$, which is the right answer. By experimentation you can soon discover for yourself that it is not safe to assume that the same method works finding fourth roots using the iteration formula. $$x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x_n^3} \right)$$ There is work to do to show that the iteration $x_{n+1} = F(x_n)$ converges to a limit $L$ if and only if $-1 < F'(L) < 1$.

There was also a correct solution from Andrei Lazanu (School 205, Bucharest). The first part is very clear but I have tried to simplify his solution to the second part for inclusion here. Perhaps someone could improve on this for us. Thank you for your hard work Andrei.

First, I approximated 3 using the method given in the problem. I know that 3 is between 1 and 2 because $1^2 < (\sqrt 3) ^2 < 2^ 2 $ or $1 < 3 < 4$. I know that the approximation of $\sqrt 3$ correct to five decimal places is: $$\sqrt 3 \approx 1.73205$$.

Now I show each of the approximation steps: First approximation: $$\sqrt{3} \approx {2}$$.
Second approximation: $$\sqrt{3}\approx {{{3\over{2}} + 2} \over {2}} ={1.75}.$$
Third approximation: $$\sqrt{3} \approx {{{3\over{1.75}} + 1.75} \over {2}} = {1.732142857}.$$ Fourth approximation: $$\sqrt{3} \approx {{{3\over{1.732142857}} + 1.732142857} \over {2}} = {1.73205081}$$

So, four approximations are sufficient to approximate $\sqrt{3}$ correct to 5 decimal places.

You could think of the above as $$ \sqrt{a^2}\approx {{{a^2\over{n}} + n} \over {2}} ={m}$$ Where n is the approximation to the root of a 2 (that is "a") and m the next approximation. The first approximation (n) differs from a by k. I can therefore write n as a + k where k is numerically less than a (k could be negative). So I have the next approximation $$\quad = {{{a^2\over{a+k}} + a+k} \over {2}}$$

The next approximation = $${{{a^2\over{a+k}} + a+k} \over {2}}$$
But $${{{a^2\over{a+k}} + a+k} \over {2}} = {{2a^2 + 2ak + k^2} \over{2(a+k)}}$$ and $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{2a(a+k)+ k^2} \over{2(a+k)}}= {{2a(a+k)} \over{2(a+k)}} + {{k^2} \over{2(a+k)}} = a + {{k^2} \over{2(a+k)}}$$
While a is positive, $${{k^2} \over{2(a+k)}}$$must be positive as k is numerically less than a.

So $$a< {{{a^2\over{a+k}} + a+k} \over {2}}$$

But the same equation could be written as: $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{a^2+ (a+k)^2} \over{2(a+k)}}= {{(a+k)^2} \over{2(a+k)}} + {{a^2} \over{2(a+k)}} = {{a+k} \over{2}} + {{a^2}\over {2(a+k)}}$$

The following number is equal to a+k: $${{a+k} \over{2}} + {{a^2}\over {2(a+k)}} + {{2ak+k^2}\over{2(a+k)}} = {{(a+k)^2 + a^2 + 2ak + k^2}\over{2(a+k)}} = {{a^2 + k^2 + 2ak + a^2 + 2ak + k^2}\over{2(a+k)}} = {{2(a^2 + 2ak + k^2)}\over{2(a+k)}}={{(a+k)^2}\over{(a+k)}} = (a+k)$$

This means that $${{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$ From the two inequalities I obtain that: $$a< {{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$ This means that the solution obtained goes closer and closer at each step to the real value of whether k is positive or negative.