Why a Local Condition Is Useful
Newton’s method uses the update \(x_{n+1}=x_n-f(x_n)/f'(x_n)\) to approximate a root of \(f\). Its success depends on where the iteration starts: a small step at one point does not by itself guarantee that later steps remain small or move toward a root.
A useful way to analyze the iteration is to regard one Newton step as a function of the current value. If that function shrinks distances throughout an interval around a root, then every starting value in the interval stays there, and the iterates approach the root. This tutorial establishes that sufficient condition and explains precisely what it does—and does not—guarantee.
The Newton Map and Contraction Condition
Let \(f\) be differentiable on an interval where \(f'(x)\ne 0\). The Newton iteration map is
Starting from \(x_0\), Newton’s method is the sequence defined by \(x_{n+1}=g(x_n)\), provided every iterate lies where \(g\) is defined. A root \(\alpha\) of \(f\) is a fixed point of \(g\) when \(f'(\alpha)\ne 0\): if \(f(\alpha)=0\), then \(g(\alpha)=\alpha\). Conversely, if \(g(x)=x\) and \(f'(x)\ne0\), then \(f(x)=0\).
For a differentiable map, a convenient sufficient test is a uniform derivative bound. If \(g\) is differentiable on an interval containing \(I\), and \(|g'(x)|\le q<1\) throughout \(I\), then the mean value theorem gives \(|g(x)-g(y)|\le q|x-y|\) for all \(x,y\in I\). The word uniform matters: the same \(q<1\) must work at every point in the interval.
Why the Condition Guarantees Convergence
The proof uses the fixed point and the contraction bound together. Comparing any point in \(I\) with \(\alpha\) shows that its image is no farther from \(\alpha\) than it was before. Applying the same argument repeatedly gives an error bound that tends to zero.
For \(x\in I\), the derivative bound implies \(|g(x)-g(\alpha)|\le q|x-\alpha|\). Since \(g(\alpha)=\alpha\), this is \(|g(x)-\alpha|\le q|x-\alpha|\).
Because \(q<1\), \(|g(x)-\alpha|\le |x-\alpha|\). If \(x\in I\), the point \(g(x)\) is therefore no farther from \(\alpha\) than \(x\) is. Since \(I\) is an interval containing \(\alpha\), \(g(x)\in I\).
For \(x_0\in I\), the first step gives \(|x_1-\alpha|\le q|x_0-\alpha|\). Applying the same inequality at each subsequent iterate gives \(|x_n-\alpha|\le q^n|x_0-\alpha|\).
Since \(0\le q<1\), \(q^n\to0\). Thus \(|x_n-\alpha|\to0\), which means \(x_n\to\alpha\). The iterates are defined throughout because they remain in \(I\).
This proves the theorem, including its convergence claim. The estimate is a geometric error bound: each step reduces the distance to the root by at least the factor \(q\), in the sense that \(|x_{n+1}-\alpha|\le q|x_n-\alpha|\). It is a sufficient condition, not a claim that every convergent Newton iteration must satisfy this particular bound.
Computing the Derivative of the Newton Map
Suppose \(f\) is twice differentiable and \(f'\ne0\) on the interval under consideration. Differentiating the Newton map using the quotient rule gives
This formula turns the contraction test into a condition on \(f\), \(f'\), and \(f''\). In particular, let \(\alpha\) be a simple root, meaning \(f(\alpha)=0\) and \(f'(\alpha)\ne0\). Then \(g(\alpha)=\alpha\) and \(g'(\alpha)=0\). If \(f\) is twice continuously differentiable near \(\alpha\), then \(g'\) is continuous near \(\alpha\), as long as \(f'\) stays nonzero there. Consequently, for any chosen \(q\) with \(0<q<1\), continuity allows an interval around \(\alpha\) small enough that \(|g'(x)|\le q\) throughout it. On a closed interval of this kind, the theorem applies.
This explains why a simple root has a local convergence guarantee: sufficiently close starting values lie in an interval on which the Newton map is a contraction. The interval can depend on the function and the root. This is a local statement, not a guarantee for arbitrary starting values.
Worked Applications
Worked Example: A Contraction Interval for \(x^2-2\)
Let \(f(x)=x^2-2\), whose positive root is \(\alpha=\sqrt{2}\), and take \(I=[1.3,1.5]\). On this interval \(f'(x)=2x\ne0\), and
For \(x\in I\), \(x^2\in[1.69,2.25]\), so \(|x^2-2|\le0.31\) and \(2x^2\ge3.38\). Therefore
Also \(1.3^2=1.69<2<2.25=1.5^2\), so \(\alpha\in I\). The theorem applies with \(q=0.1\). For example, starting at \(x_0=1.5=3/2\),
The next step is
Both displayed iterates lie in \([1.3,1.5]\). Substituting \(3/2\) and \(17/12\) into \(g(x)=(x^2+2)/(2x)\) gives \(17/12\) and \(577/408\), respectively, so the arithmetic agrees with the iteration. The theorem guarantees convergence from every starting value in this interval, not just from \(3/2\).
Worked Example: A Contraction Interval for \(x^3-1\)
Take \(f(x)=x^3-1\), with simple root \(\alpha=1\), and \(I=[0.9,1.1]\). Here \(f'(x)=3x^2\ne0\) on \(I\). The map and its derivative are
The derivative \(g'\) is increasing because \(g''(x)=2/x^4>0\). Its absolute value on \(I\) is therefore bounded by the larger of its endpoint absolute values. At \(x=0.9=9/10\),
since \(4(542)=2168<2187\). At \(x=1.1=11/10\),
since \(4(662)=2648<3993\). Thus \(|g'(x)|\le1/4\) on \(I\). Starting at \(x_0=0.9=9/10\), one step gives
This value lies in \(I\). Substitution into the map confirms the calculation: \(\bigl(2(9/10)+1/(9/10)^2\bigr)/3=(9/5+100/81)/3=1229/1215\). The theorem guarantees convergence to \(1\) for every starting value in \(I\), with \(|x_n-1|\le(1/4)^n|x_0-1|\).
Worked Example: Isolating the Negative Root of \(x^2-1\)
Let \(f(x)=x^2-1\), and consider its simple root \(\alpha=-1\) on \(I=[-1.2,-0.8]\). The interval excludes the other root \(1\), and \(f'(x)=2x\ne0\) on \(I\). We have
As \(x\) ranges over \(I\), \(x^2\) ranges from \(0.64\) to \(1.44\). The expression \(\frac12(1-1/s)\) increases as \(s>0\) increases. At the two endpoint values of \(s=x^2\), the absolute derivative values are
Hence \(|g'(x)|\le3/10\) throughout \(I\). Starting at \(x_0=-0.8=-4/5\),
For a further arithmetic check,
Both values are in \(I\); direct substitution in \(g(x)=(x^2+1)/(2x)\) gives the displayed iterates. The contraction theorem ensures that iterates starting anywhere in \(I\) converge to \(-1\), rather than to the other root.
A Common Mistake: Checking Only at the Root
Because \(g'(\alpha)=0\) at a simple root, it is tempting to conclude that the Newton map is a contraction on any interval around that root, or to use the derivative value at a starting point as if it bounded the derivative throughout the interval. Neither conclusion follows from a single derivative value. The condition needed in the theorem is a uniform bound on the whole interval.
For instance, in the \(x^2-2\) example, \(g'(1.5)=1/18\). A mistaken argument might say, “Since \(1/18<1\), the iteration is a contraction on \([1.3,1.5]\).” This checks only one point. The correction is to bound every derivative value on the interval: \(|g'(x)|\le31/338<0.1\) for all \(x\in[1.3,1.5]\). That uniform estimate justifies the conclusion.
What the Condition Does Not Say
The contraction test is a local sufficient condition. It guarantees convergence only for starting values in the interval where the derivative bound holds. It does not say that Newton’s method converges from every starting value, and it does not rule out convergence when no convenient contraction interval has been established. A chosen interval may also be too wide for the bound even when a smaller interval around the root works.
For practical use, the essential checks are: the root is fixed by the Newton map, the map is defined throughout the chosen interval, and one uniform derivative bound \(q<1\) holds there. Together, these give an invariant interval and a geometric error estimate. The condition explains local convergence without relying on a particular numerical starting value.
Check Your Understanding
Use the definitions and arguments above to answer these questions.
- What does it mean for a map to be a contraction on an interval, and why must its contraction factor be less than \(1\)?
- State the derivative bound that is sufficient for a differentiable map to be a contraction.
- Why does a simple root make the derivative of the Newton map equal to zero at the root?
- What does the estimate \(|x_n-\alpha|\le q^n|x_0-\alpha|\) tell you about iterates that start in the contraction interval?
- Why is checking \(|g'(x)|<1\) at just one point not enough to establish contraction on an interval?
