Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — locating a root and bisecting
Show that f(x)=x3−x−1 has a root between 1 and 2, and use two bisections to narrow it.
- 1f(1)=1−1−1=−1 and f(2)=8−2−1=5Opposite signs.
- 2f is a polynomial and so continuous, therefore a root lies in (1,2).Both conditions must be stated.
- 3Bisect: f(1.5)=0.875>0, so the sign change is now in (1,1.5).Keep the half where the sign still changes.
- 4Bisect again: f(1.25)=−0.297<0, so the root is in (1.25,1.5).
- 5Each step halves the width; the true root is 1.3247…Reliable but slow — three or four steps per decimal place.
Example 2 — Newton–Raphson
Apply the Newton–Raphson method to f(x)=x3−x−1 with x0=1.5, giving x2.
- 1xn+1=xn−f′(xn)f(xn)Follow the tangent to the axis.
- 2f′(x)=3x2−1
- 3f(1.5)=0.875 and f′(1.5)=5.75
- 4x1=1.5−5.750.875=1.3478
- 5f(1.3478)=0.1008, f′(1.3478)=4.4499
- 6x2=1.3478−4.44990.1008=1.3252Two steps give four correct figures; the root is 1.32472.
Example 3 — why one rearrangement works and another does not
The equation x3−x−1=0 can be rearranged as x=(x+1)1/3 or as x=x3−1. Explain why only the first converges.
- 1An iteration xn+1=g(xn) converges near a root when ∣g′(x)∣<1.The test to apply before iterating.
- 2For g(x)=(x+1)1/3: g′(x)=31(x+1)−2/3
- 3At the root x≈1.3247: g′≈0.19<1 ✓ so it converges.Terms: 1.2599, 1.3123, 1.3223, 1.3243, …
- 4For g(x)=x3−1: g′(x)=3x2
- 5At the root: g′≈5.3>1 ✗ so it diverges.From x0=1.3: 1.197, 0.715, −0.634, … running away.
- 6Same equation, opposite behaviour — the rearrangement matters.