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I am attempting to resolve the following problem:

Find an approximation to $\sqrt{5}$ correct to an exactitude of $10^{-10}$ using the bisection algorithm.

From what I understand, $\sqrt{5}$ has to be placed in function of $x$ but I am not sure where to go from there.

Also, a function in Mathematica are given to do the calculations in which the function $f(x)$, $a$ and $b$ (from the interval $[a, b]$ where $f(a)$ and $f(b)$ have opposite signs), the tolerance and the number of iterations.

Omari Celestine
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    Are you supposed to do this by hand? Because an accuracy of $10^{-10}$ requires over $30$ bisections. That's just tedious and boring, and has very little educational value over, say, $10^{-2}$. – Arthur Jan 14 '19 at 16:33
  • Note that $\sqrt{5}$ is the solution to $x^2-5 = 0$. – D.B. Jan 14 '19 at 16:35
  • @Arthur Algorithms in Mathematica are given to solve the problem. – Omari Celestine Jan 14 '19 at 17:23
  • Bisection is definetely not an efficient way for approximating $\sqrt{5}$. If you exploit the convergents of $\frac{1+\sqrt{5}}{2}$, which are ratios of consecutive Fibonacci numbers, you have that $2\frac{F_{27}}{F_{26}}-1$ meets the wanted constraints. This can be through $\leq 15$ integer multiplications/ratios, see https://math.stackexchange.com/a/2081266/44121 – Jack D'Aurizio Jan 14 '19 at 18:04
  • @SimplyBeautifulArt It is not the same problem. – Omari Celestine Jan 14 '19 at 20:41

1 Answers1

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The bisection method is a method for finding the roots of a continuous function ie finding $ x $ such that $ f(x) $ equals zero. Thus, you need a function which has a zero equal to $ \sqrt{5} $. The function $x^2 - 5 = f(x)$ will do.

Then, select an initial interval $[a, b]$ which contains the root. and for which $f(a)$ and $f(b)$ have different signs. The interval $[2, 3]$ will do.

The bisection method converges by repeatedly cutting the length of this interval in half.

Here is pseudocode from wikipedia:

N ← 1
While N ≤ NMAX # limit iterations to prevent infinite loop
  c ← (a + b)/2 # new midpoint
  If f(c) = 0 or (b – a)/2 < TOL then # solution found
    Output(c)
    Stop
  EndIf
  N ← N + 1 # increment step counter
  If sign(f(c)) = sign(f(a)) then a ← c else b ← c # new interval
EndWhile
Output("Method failed.") # max number of steps exceeded

The difference between the midpoint of the $n-th$ interval and the root $c$ is given by $$|c_n - c| \leq \frac{|b-a|}{2^n}$$ which follows directly from the fact that the interval is halved each iteration. You can figure out the number of iterations necessary for the desired $\epsilon$ using this formula.

Klint Qinami
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