Zeros and the bisection method: trapping a root by halving

Zeros & bisection

How can you find a zero of f(x) = x³ − 3x + 1 to two decimal places without solving the equation?

Pick an interval where f changes sign, then halve it again and again, keeping the half that still changes sign. Watch [a, b] shrink around a root of x³ − 3x + 1, and see when the method cannot start.

Drag the ringed points on the board.

High school
Zeros of a function
Intermediate value theorem
Bisection method
Approximation to a given accuracy

a

0.0

b

1.0

n

b_n - a_n = \dfrac{b - a}{2^n}

$a_n$: 0.2500
$b_n$: 0.5000
$b_n - a_n$: 0.2500
$m = \tfrac{a_n + b_n}{2}$: 0.3750
$f(m)$: -0.0723

f(a) · f(b)< 0: a zero is trapped

Guided tour

1A sign change traps a zero
On $[0,1]$ : $f(0)=1>0$ and $f(1)=-1<0$ . The curve is unbroken, so it must cross the $x$ -axis somewhere in between. The midpoint is $m=0.5$ , where $f(0.5)=-0.375$ .
2Keep the half that still changes sign
$f(0.5)<0$ and $f(0)>0$ , so the zero is in $[0,0.5]$ ; throw the other half away. Two more halvings give $[0.25,0.375]$ : after $n$ steps the interval is $\frac{1}{2^n}$ as long, here $0.125$ .
3Stop when the interval is short enough
For accuracy $0.01$ we need $\frac{1}{2^n}<0.01$ , so $n=7$ : the interval is $[0.3438,0.3516]$ , only $0.0078$ long. Its midpoint $0.3477$ is within $0.004$ of the true zero $0.3473$ .
4No sign change, no guarantee
Now try $[-1,2]$ : $f(-1)=3$ and $f(2)=3$ have the same sign, yet the curve crosses the axis twice in between. The theorem only works one way: a sign change guarantees a zero, but no sign change proves nothing. Choose a smaller interval before you start.

Explanation

1
Zeros are where the graph meets the axis
A zero of $f$ is a real number $x_0$ with $f(x_0)=0$ . It is not a point: it is the $x$ -coordinate where the graph meets the $x$ -axis, and the same number is a root of the equation $f(x)=0$ . The cubic $x^3-3x+1$ has three zeros, near $-1.879$ , $0.347$ and $1.532$ , and none of them has a tidy formula.
2
The existence theorem
If $f$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$ , then $f$ has at least one zero in $(a,b)$ .
- It says at least one: on $[-2,2]$ the sign changes and there are three.
- The converse is false: on $[-1,2]$ the sign does not change, yet there are two.
- If $f$ is also monotonic on $[a,b]$ , the zero is unique.
3
The bisection method
- Choose $[a,b]$ with $f(a)\cdot f(b)<0$ .
- Take the midpoint $m=\frac{a+b}{2}$ and compute $f(m)$ . If $f(m)=0$ , you are done.
- If $f(a)\cdot f(m)<0$ , the zero is in $[a,m]$ : set $b=m$ . Otherwise set $a=m$ .
- Repeat until $b-a$ is smaller than the accuracy you need; any number in $[a,b]$ will do.
4
How many steps?
Each step halves the interval, so after $n$ steps its length is $\dfrac{b-a}{2^n}$ . For an interval of length $1$ and accuracy $0.01$ you need $2^n>100$ , so $n=7$ ; for $0.001$ , $n=10$ . Every step buys about $0.3$ decimal digits. That is slow compared with Newton's method, but bisection never fails once the sign change is there.

Check yourself

Q01

In which interval must $f(x)=\ln x+2x-6$ have a zero?

Q02

Starting from an interval of length $1$ , how many halvings make it shorter than $0.05$ ?

Q03

If $f(a)\cdot f(b)>0$ for a continuous $f$ , then on $(a,b)$ …

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