The bisection method in practice

1. Existence and uniqueness. $g(x)=\cos x-x$ is continuous with $g(0)=1>0$ and $g(1)=\cos1-1<0$, so a root exists by the intermediate value theorem. Moreover $g'(x)=-\sin x-1\le0$, and it vanishes only at isolated points, so $g$ is strictly decreasing and the root is unique.

2. Bisection. At each step compute the midpoint $m$ of $[a,b]$; keep $[a,m]$ if $g(a)$ and $g(m)$ have opposite signs, else keep $[m,b]$.

• Step 1: $[0.5,\ 1]$
• Step 2: $[0.5,\ 0.75]$
• Step 3: $[0.625,\ 0.75]$
• Step 4: $[0.6875,\ 0.75]$
• Step 5: $[0.71875,\ 0.75]$
• Step 6: $[0.734375,\ 0.75]$
• Step 7: $[0.734375,\ 0.7421875]$
• Step 8: $[0.73828125,\ 0.7421875]$
• Step 9: $[0.73828125,\ 0.740234375]$
• Step 10: $[0.73828125,\ 0.7392578125]$

The true value is $\alpha\approx0.7390851$, inside the last interval.

3. Number of steps. After $n$ steps the interval has length $2^{-n}$. We need $2^{-n}\le10^{-6}$, that is $n\ge6\ln10/\ln2\approx19.9$. So $20$ steps are enough.

Remark. The method is slow (about three steps per decimal digit) but it is guaranteed to converge as soon as the function is continuous and changes sign.