When Newton’s Method Does Not Behave as Expected
Newton’s method is often effective when the starting value is suitable and the function behaves well near the root being sought. But the existence of a root alone does not ensure that a Newton iteration will find it. The iterates may repeat in a cycle, make a large excursion away from a root, or grow without bound.
These behaviors are consequences of repeatedly applying the Newton update, not merely of making an arithmetic mistake. Studying them clarifies what the method does and does not guarantee: a local convergence result concerns starting values sufficiently close to a root under appropriate hypotheses; it is not a promise of global convergence from every starting value.
The Newton Iteration and Its Domain
The formula comes from replacing the graph near \(x_n\) by its tangent line. That line has equation \(y=f(x_n)+f'(x_n)(x-x_n)\). Setting \(y=0\) and solving for its horizontal intercept gives \(x=x_n-f(x_n)/f'(x_n)\), exactly the Newton update. The step therefore uses a local linear approximation to estimate a root. If that tangent-line intercept is far away, the next iterate can also be far away.
A sequence of Newton iterates is well-defined only as long as every iterate at which a step is taken has a nonzero derivative. A zero derivative is one reason a step can fail, but it is not the only possible problem: even when every step is defined, the resulting sequence need not converge to a root.
An Exact Two-Cycle
An oscillation can be an alternating pattern that approaches a root, or it can persist indefinitely. The following example gives an exact two-cycle: two points map to each other, so the iterates never settle down.
Worked Example: A Newton Iteration That Repeats Forever
Consider \(f(x)=x^3-2x+2\), starting from \(x_0=0\). First verify that a real root exists. Since \(f\) is a polynomial, it is continuous, and $$ f(-2)=(-2)^3-2(-2)+2=-8+4+2=-2, \qquad f(-1)=(-1)^3-2(-1)+2=-1+2+2=3. $$ Because the function changes sign between \(-2\) and \(-1\), the Intermediate Value Theorem guarantees a root in \((-2,-1)\).
Differentiate: $$ f'(x)=3x^2-2. $$ At \(x_0=0\), \(f'(0)=-2\ne0\), so the first step is defined: $$ x_1=0-\frac{f(0)}{f'(0)} =0-\frac{2}{-2}=1. $$ At \(x_1=1\), \(f'(1)=3(1)^2-2=1\ne0\), and $$ x_2=1-\frac{f(1)}{f'(1)} =1-\frac{1}{1}=0. $$ The same two computations now repeat: \(N_f(0)=1\) and \(N_f(1)=0\). Thus \(x_0,x_1,x_2,x_3,\ldots\) alternate between \(0\) and \(1\).
Check: Substituting \(0\) into the update gives \(0-2/(-2)=1\), and substituting \(1\) gives \(1-1/1=0\). Both derivatives are nonzero. The function has a real root, but this starting value produces a cycle rather than convergence to that root.
This example separates two different statements. “The function has a root” is a fact about \(f\). “The iterates from this starting value converge to a root” is a fact about repeatedly applying \(N_f\). The first does not imply the second.
Leaving a Root’s Neighborhood
An iterate can also leave a chosen neighborhood of a root and later return toward that root. A neighborhood here simply means an open interval containing the root. Leaving such an interval is not, by itself, proof of divergence; it shows that proximity at one step does not ensure proximity at the next.
Worked Example: A Large Excursion Before Convergence
Take \(f(x)=x^2-1\), which has the root \(r=1\). Its derivative is \(f'(x)=2x\). For \(x\ne0\), the Newton update simplifies as follows: $$ N_f(x)=x-\frac{x^2-1}{2x} =\frac{2x^2-(x^2-1)}{2x} =\frac{x^2+1}{2x} =\frac12\left(x+\frac1x\right). $$ Choose the neighborhood \(U=(0,2)\) of \(1\), and start at \(x_0=\frac14\), which lies in \(U\). The first step is $$ x_1=\frac14-\frac{(1/4)^2-1}{2(1/4)} =\frac14-\frac{-15/16}{1/2} =\frac14+\frac{15}{8} =\frac{17}{8}. $$ Since \(\frac{17}{8}>2\), the new iterate lies outside \(U\).
The next step is defined because \(x_1>0\): $$ x_2=\frac12\left(\frac{17}{8}+\frac{8}{17}\right) =\frac12\left(\frac{289}{136}+\frac{64}{136}\right) =\frac{353}{272}. $$ This value is greater than \(1\) and less than \(2\), since \(272<353<544\). In fact, all subsequent iterates converge to \(1\). To see why, for any \(x>1\), $$ N_f(x)-1 =\frac{x+1/x-2}{2} =\frac{(x-1)^2}{2x}>0, $$ and $$ x-N_f(x) =\frac{x-1/x}{2} =\frac{x^2-1}{2x}>0. $$ Therefore \(1<N_f(x)<x\). Since \(x_1>1\), the iterates from \(x_1\) onward decrease and remain bounded below by \(1\). They have a limit \(L\ge1\). Passing to the limit in \(x_{n+1}=\frac12(x_n+1/x_n)\) gives \(L=\frac12(L+1/L)\), so \(2L^2=L^2+1\), hence \(L^2=1\). The condition \(L\ge1\) forces \(L=1\).
Check: The update at \(x_0=\frac14\) gives \(\frac{17}{8}\), and the next update gives \(\frac{353}{272}\). Every iterate from \(x_0\) onward is positive, so \(f'(x_n)=2x_n\ne0\); the displayed updates are defined. The first step leaves \(U\), but the later iterates converge to the root \(1\).
The interval \(U\) was chosen to illustrate an excursion, not to claim that every starting point inside it fails. Local convergence results apply when the starting point is sufficiently close to the root under the required hypotheses. A chosen neighborhood need not be small enough to guarantee that every point in it has that behavior.
Divergence Despite a Simple Root
A stronger failure occurs when the iterates grow without bound. The next example has a root at which the derivative is nonzero, and its Newton steps remain defined along the entire divergent sequence.
Worked Example: Newton Iterates That Diverge to Infinity
Let $$ f(x)=\frac{x}{1+x^2}. $$ The denominator is positive for every real \(x\), so \(f\) is defined everywhere, and \(f(x)=0\) exactly when \(x=0\). This root is simple because \(f'(0)=1\). Differentiating gives $$ f'(x)=\frac{(1+x^2)-2x^2}{(1+x^2)^2} =\frac{1-x^2}{(1+x^2)^2}. $$ For \(x\ne\pm1\), substitution into the Newton formula yields $$ N_f(x) =x-\frac{x/(1+x^2)}{(1-x^2)/(1+x^2)^2} =x-\frac{x(1+x^2)}{1-x^2} =\frac{-2x^3}{1-x^2} =\frac{2x^3}{x^2-1}. $$
Start at \(x_0=2\). The derivative is nonzero there, and $$ x_1=N_f(2)=\frac{2(2)^3}{(2)^2-1} =\frac{16}{3}>2. $$ Whenever \(x_n>1\), the update is defined and satisfies $$ x_{n+1}=\frac{2x_n^3}{x_n^2-1}>2x_n, $$ because \(x_n^2-1>0\) and \(\frac{x_n^2}{x_n^2-1}>1\).
We now prove a lower bound by induction. The first iterate satisfies \(x_1>2^1x_0\). If \(x_n>2^n x_0\) for some \(n\ge1\), then \(x_n>2\), so the preceding inequality applies and $$ x_{n+1}>2x_n>2\cdot2^n x_0=2^{n+1}x_0. $$ Thus \(x_n>2^n x_0\) for every \(n\ge1\). Since \(x_0=2\), this lower bound grows without bound, and therefore \(x_n\) diverges to \(+\infty\).
Check: The first update is \(\frac{16}{3}\), consistent with the exact Newton map. For every \(n\ge1\), \(x_n>2\), so \(f'(x_n)=\frac{1-x_n^2}{(1+x_n^2)^2}\ne0\). No step fails because of a zero derivative; instead, the iterates move farther from the existing simple root at \(0\).
What These Examples Show
The three examples exhibit distinct outcomes: a repeating cycle, a temporary excursion followed by convergence, and divergence. Each behavior is consistent with the Newton formula. The method chooses the zero of a tangent line, and that intercept depends on the function and the current iterate. Unless conditions ensure good local behavior, a tangent line at one point gives no general guarantee about where later iterates will go.
When analyzing a Newton sequence, keep these questions separate:
- Is the step defined? Check that \(f'(x_n)\ne0\) whenever the update is used.
- Does a root exist? This is a property of the equation \(f(x)=0\), not of the iteration from a particular starting value.
- What is the sequence doing? Compute or bound the iterates, and distinguish convergence, a cycle, and unbounded growth.
- What can local theory conclude? A local guarantee concerns starting values within a suitable region and under its hypotheses; it does not automatically cover distant starting values.
A few computed steps can reveal a pattern, but a claim about every iterate needs justification. In the divergence example, the inequality \(x_{n+1}>2x_n\) and induction prove unbounded growth. In the cycle example, checking both updates proves that the same two values recur forever. These arguments establish the behavior rather than merely suggesting it from a short numerical list.
Common Mistake: Treating a Root as a Global Guarantee
A common error is to reason, “There is a root, so Newton’s method must converge to it.” The two-cycle example directly disproves that inference: the function has a real root, yet the iterates from \(x_0=0\) alternate between \(0\) and \(1\). Another error is to stop checking the derivative after the first update. A sequence is not a valid Newton iteration past any stage where the required derivative is zero. In the divergence example, this check is satisfied at every iterate, which is why the growth argument describes a genuine Newton sequence.
It is also important not to label every excursion as divergence. In the \(x^2-1\) example, \(x_1\) leaves the chosen interval \(U\), but the later sequence is decreasing and converges to \(1\). A large step indicates that the iterate has moved; it does not, on its own, determine what all future iterates will do.
Check Your Understanding
Answer these questions using the iteration formula and the reasoning in the examples.
- What condition on \(f'(x_n)\) is required for the Newton step from \(x_n\) to be defined?
- For \(f(x)=x^3-2x+2\), what are the first two iterates from \(x_0=0\), and why do they repeat?
- In the example \(f(x)=x^2-1\), which iterate first leaves \(U=(0,2)\), and what happens afterward?
- For \(f(x)=\frac{x}{1+x^2}\), why does the bound \(x_{n+1}>2x_n\) imply divergence from \(x_0=2\)?
- Why does proving that a function has a root not prove that Newton’s method converges from an arbitrary starting value?
