The iteration derived from the tangent line, its rapid convergence with correct digits roughly doubling each step, the ways it fails through a vanishing derivative, a cycle or divergence, the importance of the initial guess, and the comparison with bisection.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Newton's Method
Objectives
Five outcomes. The first is a formula derived in one line, and the rest are about when to trust it.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 409-418 — the section these objectives are drawn from
Warm-up
Section 2.4 proved that a continuous function changing sign has a root, and Section 4.2 built the tangent line as a local stand-in for a curve.
Discussion prompt
The cubic x cubed minus 2x minus 1 has a root between 1 and 2. How could a tangent line help you find it rather than merely know it exists?
Hint: A line's root is trivial to compute; a cubic's is not.
Answer:
Solving a cubic exactly is possible but unpleasant, and for most equations no formula exists at all. But solving a LINE for its root takes one division.
So replace the curve by its tangent at a guess, find where that line crosses the axis, and use the crossing point as a better guess. Since the tangent hugs the curve near the point of contact, its root should be nearer the curve's root than the guess was.
\[ x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \]
Repeating that turns an unsolvable equation into a sequence of divisions, and the sequence converges startlingly fast.
Concept
A curve's root is hard to find and a line's is easy. Replacing the curve by its tangent at a guess and taking that line's root gives a better guess, and iterating converges rapidly to the curve's root.
Newton's method — The iteration that produces each new estimate by subtracting from the current one the function's value divided by its derivative there. Geometrically, the new estimate is where the tangent at the current one meets the axis.
\[ x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \]
The method is the practical counterpart to the Intermediate Value Theorem. That theorem promised a root exists without locating it; this locates it to as many digits as wanted, and quickly.
Figure (svg): Two steps of Newton's method, each tangent line crossing the axis nearer the root
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 409-412
Section
Section 1
Concept
Write the tangent line at the current guess, set its height to zero, and solve for the input. The result is the next guess, and the correction is the function's value divided by its slope.
the iteration formula — Each estimate is the previous one minus the function's value there divided by its derivative there. It is the root of the tangent line at the previous estimate.
\[ 0 = f(x_{n}) + f'(x_{n})(x-x_{n}) \;\Longrightarrow\; x = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \]
The derivation is worth doing once rather than memorising the result. It also makes the failure mode obvious: a small derivative makes the correction enormous, which is exactly what goes wrong near a horizontal tangent.
Figure (svg): The iteration formula, derived from setting the tangent line to zero
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 409-413 — describing Newton's method
Picture it
The tangent line, solved for its own root.
Figure (svg): The iteration formula, derived from setting the tangent line to zero
The final line is the whole method. Notice the derivative in the denominator — it is what makes the method fast where the curve is steep and unreliable where it is flat.
Worked example
Example 4.46. Three iterations from a poor start.
\[ \text{Use Newton's method on } f(x)=x^{2}-2 \text{ from } x_{0}=2 \text{ to approximate } \sqrt{2}. \]
Write the iteration for this function
Why: Substitute f and its derivative.
\[ x _{n + 1} = x _{n} - \frac{x _{n} ^{2} - 2}{2 x _{n}} \]
Simplify
Why: Combine over a common denominator.
\[ = \frac{x _{n} + 2 / x _{n}}{2} \]
Iterate once
Why: From 2.
\[ x _{1} = \frac{2 + 1}{2} = 1.5 \]
Iterate again
Why: From 1.5.
\[ x _{2} = \frac{1.5 + 1.3333}{2} = 1.41667 \]
Iterate a third time
Why: From 1.41667.
\[ x _{3} = 1.4142157 \]
Figure (svg): The doubling of correct digits, tabulated for the square root of two
\[ x_{3} = 1.4142157, \quad \sqrt{2} = 1.41421356\ldots \]
Verify: count the correct digits at each stage
Why: The start had none, then one, then three, then six correct digits — roughly doubling each time. A fourth iteration gives twelve correct digits, which exhausts ordinary precision. Note also that the simplified iteration, averaging the guess with 2 over the guess, is the ancient Babylonian method for square roots — Newton's method reproduces an algorithm two thousand years older, which is a good sign it is the natural thing to do.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 411-412
Fill the middle
The square-root iteration, simplified by combining over a common denominator.
Fill in the blanks
x_1.5 = \frac______\left(x____ + \frac______}\right): \; \text___ 2, \; x____ = ___
Why: Averaging 2 with 2 over 2, which is 1, gives 1.5. This simplified form is the Babylonian square-root algorithm, which Newton's method reproduces exactly.
Worked example
Checkpoint 4.46. Where the method earns its place.
\[ \text{Approximate the root of } x^{3}-2x-1 \text{ between } 1 \text{ and } 2, \text{ starting at } 2. \]
Write the iteration
Why: Function over derivative.
\[ x _{n + 1} = x _{n} - \frac{x _{n} ^{3} - 2 x _{n} - 1}{3 x _{n} ^{2} - 2} \]
Iterate once
Why: From 2, where f is 3 and f' is 10.
\[ x _{1} = 2 - 0.3 = 1.7 \]
Iterate again
Why: From 1.7.
\[ x _{2} = 1.6236 \]
Iterate a third time
Why: From 1.6236.
\[ x _{3} = 1.61806 \]
Iterate a fourth time
Why: Converged.
\[ x _{4} = 1.618034 \]
Figure (svg): The solution to Worked example a root no formula reaches shown as a ladder of expressions, one row per legal move
\[ x \approx 1.618034 \]
Verify: substitute back and recognise the number
Why: At 1.618034 the cubic evaluates to about 0.0000001, confirming a root. The value is the golden ratio, and indeed this cubic factors as x plus 1 times the quadratic x squared minus x minus 1, whose positive root is exactly one plus root five over two. So an exact answer existed here — but Newton's method found it without factoring, which is what matters for equations that do not factor at all. Section 2.4 promised this root existed; here it is located to six decimal places in four divisions.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 412-413
Trap
\[ x_{n+1} = x_{n} + \frac{f(x_{n})}{f'(x_{n})} \]
Add the correction
Why: The student misremembers the sign.
\[ \text{from } x_{0}=2 \text{ on } x^{2}-2: \; x_{1} = 2 + 0.5 = 2.5 \quad \text{(worse)} \]
The estimate moved away from the root rather than toward it, and the next step moves further still.
\[ x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} \;\Longrightarrow\; x_{1} = 1.5 \]
Subtract: the correction undoes the function's value
Why: Deriving the formula rather than recalling it makes the sign automatic.
A single iteration reveals the error: if the function's value at the new estimate is larger in magnitude than at the old one, the sign is wrong. Checking that the estimates are improving is a cheap and complete guard.
Ranking
Getting the iteration formula.
Put in order
Why: Deriving rather than memorising fixes the sign automatically and makes the failure mode visible: the derivative sits in a denominator, so a flat tangent produces an enormous step.
Two truths and a lie
All three are about the iteration.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The correction is SUBTRACTED, which is what moves the estimate toward the root. Adding it moves away, and one iteration reveals the error — the function's value at the new estimate grows rather than shrinks.
Prediction
Commit before reasoning.
Predict first
What role does the derivative play in the correction?
Correct: It converts a vertical distance into a horizontal one.
\[ \text{horizontal step} = \frac{\text{vertical gap}}{\text{slope}} \]
Why: The function's value at the guess is how far the curve is from the axis vertically; dividing by the slope converts that into a horizontal distance along the tangent. A steep tangent needs only a small horizontal move to reach the axis, while a shallow one needs a large one — which is exactly what dividing by a small derivative produces, and exactly why a flat tangent is dangerous.
Section
Section 2
Concept
When the method works, the number of correct digits approximately doubles with each iteration. Four steps from a mediocre start typically give a dozen correct digits.
quadratic convergence — The error at each step is roughly proportional to the square of the previous error. Squaring a small error makes it very much smaller, which is why the correct digits double.
\[ e_{n+1} \approx C e_{n}^{2} \]
The comparison with bisection is stark. Bisection halves the interval each step, gaining about one decimal digit every three or four iterations; Newton's method doubles the digits it already has.
Figure (svg): The doubling of correct digits, tabulated for the square root of two
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 412-415 — the rate of convergence
Picture it
Four iterations for the square root of two.
Figure (svg): The doubling of correct digits, tabulated for the square root of two
Zero correct digits become one, then three, then six, then twelve. By the fourth step the answer exceeds the precision of most calculations, and further iterations gain nothing measurable.
Worked example
Example 4.48. The errors, tabulated.
\[ \text{Track the error in approximating } \sqrt{2} \text{ from } x_{0}=2. \]
Compute the first error
Why: The start minus the true value.
\[ e _{0} = 0.5858 \]
Compute the second
Why: 1.5 minus the true value.
\[ e _{1} = 0.0858 \]
Compute the third
Why: From 1.41667.
\[ e _{2} = 0.00245 \]
Compute the fourth
Why: From 1.4142157.
\[ e _{3} = 0.0000021 \]
Compare each with the previous squared
Why: The ratio is roughly constant.
\[ e _{n + 1}\text{ is about } 0.35 e _{n} ^{2} \]
Figure (svg): The solution to Worked example watching the error square shown as a ladder of expressions, one row per legal move
\[ e_{n+1} \approx 0.35\,e_{n}^{2} \]
Verify: check the pattern predicts the next error
Why: The pattern predicts the next error as about 0.35 times the square of 0.0000021, which is around 1.5 times 10 to the negative 12 — and the fourth iterate is indeed correct to about twelve digits. Squaring a small number makes it dramatically smaller, which is the whole reason the digits double. Contrast bisection, where the error merely halves and gains about 0.3 of a digit per step.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 413-414
Fill the middle
The errors squaring at each step, with the constant about a third.
Fill in the blanks
e_0.000002 = 0.00245 \;\Longrightarrow\; e____ \approx 0.35(0.00245)^___ \approx ___
Why: Squaring 0.00245 gives about 6 times ten to the negative 6, and a third of that is about 2 times ten to the negative 6 — matching the observed error. Squaring a small number is what makes the digits double.
Worked example
Checkpoint 4.48. A practical stopping rule.
\[ \text{When should the iteration be stopped?} \]
Consider stopping on the function's value
Why: Small f means near a root.
\[ | f(x _{n}) | <\text{ tolerance} \]
Consider stopping on the change
Why: Successive estimates agreeing.
\[ | x _{n + 1} - x _{n} | <\text{ tolerance} \]
Note the second is usually better
Why: It measures the answer's precision directly.
Note the caveat
Why: A very flat function can have small f far from a root.
Figure (svg): The solution to Worked example knowing when to stop shown as a ladder of expressions, one row per legal move
\[ |x_{n+1}-x_{n}| < \varepsilon \text{ and } |f(x_{n+1})| \text{ small} \]
Verify: see why one criterion alone can mislead
Why: For a function that is nearly flat near its root — a repeated root, for instance — the value can be tiny while the estimate is still noticeably wrong, so the first criterion stops too early. Conversely for a very steep function the value can be sizeable while the estimate is already excellent, so the first criterion runs too long. Checking both is cheap and catches each failure, and with quadratic convergence one extra iteration costs almost nothing.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 414-415
Error analysis
A student runs four iterations and reports the result.
Annotate
On: \( x_{4} = 1.618034 \;\Longrightarrow\; \text{the root is } 1.618034 \)
The method produces numbers whether or not it is working, and none of its failure modes announce themselves. Substituting the answer back is the only step that distinguishes a root from a coincidence.
Matching
How fast the digits accumulate.
Match the pairs
Why: The two rates are qualitatively different rather than merely different in speed. Halving an error gains a fixed small amount of accuracy each step; squaring it doubles whatever accuracy has already been achieved, so the method accelerates as it goes.
Sorting
Consider what each measures.
Sort into buckets
Sort each criterion by what it directly checks.
The two useful criteria fail in opposite situations, which is why checking both is worth it. A fixed iteration count is not a criterion at all, though it is a sensible safety limit to stop a divergent run.
Prediction
Commit before reasoning.
Predict first
The error at each step is roughly proportional to the previous error squared. What does that mean for accuracy?
Correct: The number of correct digits roughly doubles.
\[ 10^{-3} \to 10^{-6} \to 10^{-12}: \text{ 3, 6, 12 digits} \]
Why: An error of ten to the negative 3 squares to ten to the negative 6, which has twice as many leading zeros — so three correct digits become six. That doubling is what distinguishes quadratic from linear convergence, and it is why four Newton iterations achieve what forty bisections would. The acceleration is the method's whole practical value.
Section
Section 3
Concept
The method can fail in several ways. A derivative near zero produces an enormous step; some functions send the iterates into a cycle; and some send them steadily away from the root.
failure modes — A vanishing or nearly vanishing derivative flings the estimate far away; a cycle returns the iteration to a previous value forever; and divergence sends the estimates steadily further from any root.
\[ f'(x_{n}) \approx 0 \;\Longrightarrow\; \text{the correction is enormous} \]
None of these failures produces an error message. The iteration simply generates numbers, which is why substituting the answer back is not optional but part of the method.
Figure (svg): The three ways Newton's method fails
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 415-417 — failures of Newton's method
Picture it
A flat tangent, a cycle, and convergence to an unintended root.
Figure (svg): The three ways Newton's method fails
In every case the arithmetic runs perfectly and produces numbers. Only checking the result against the original equation distinguishes success from any of these.
Worked example
Example 4.50. The cube root defeats the method.
\[ \text{Apply Newton's method to } f(x)=x^{1/3} \text{ from } x_{0}=1. \]
Compute the derivative
Why: Power rule with a rational exponent.
\[ f'(x) = (\frac{1}{3}) x ^{-\frac{2}{3}} \]
Form the iteration
Why: Function over derivative.
\[ x _{n + 1} = x _{n} - 3 x _{n} \]
Simplify
Why: Collect.
\[ x _{n + 1} = -2 x _{n} \]
Iterate from 1
Why: Multiply by negative 2 each time.
\[ 1, -2, 4, -8, 16,... \]
Observe
Why: The estimates grow without bound.
Figure (svg): A cycle: two iterates that map to each other forever
\[ x_{n+1} = -2x_{n} \;\Longrightarrow\; |x_{n}| \to \infty \]
Verify: diagnose why the geometry defeats the method
Why: The cube root has a vertical tangent at the root, so near 0 the curve is far steeper than any tangent drawn away from it — and each tangent, drawn at a shallow point, crosses the axis on the far side and further out. The method has no defence against this: the arithmetic is flawless and the answer runs to infinity. Note the failure is structural rather than a matter of a bad start; every non-zero starting point diverges.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 416-417
Sorting
Diagnose each symptom.
Sort into buckets
Sort each observed behaviour.
Only the last is success, and distinguishing it from the third requires knowing which root was wanted. That is a question the method cannot answer, and it is decided entirely by the starting point.
Worked example
Checkpoint 4.50. A flat tangent flings the estimate away.
\[ \text{What happens applying the method to } x^{3}-3x \text{ starting near a turning point?} \]
Find where the derivative vanishes
Why: Set 3x squared minus 3 to zero.
\[ x = 1\text{ and } x = -1 \]
Consider a start near one of them
Why: Say 1.01.
\[ f'(1.01) = 0.0603,\text{ very small} \]
Compute the correction
Why: Value over a tiny derivative.
\[ \text{about } -32 \]
Observe the new estimate
Why: Flung far away.
\[ x _{1}\text{ is about } 33 \]
Note what happens next
Why: From far out the method recovers slowly.
Figure (svg): The solution to Worked example a vanishing derivative shown as a ladder of expressions, one row per legal move
\[ f'(x_{n}) \text{ small} \;\Longrightarrow\; \text{a huge step} \]
Verify: check whether it eventually recovers
Why: From 33 the method does converge back, since far from the turning points the cubic is steep and well behaved — but it takes several extra iterations to undo one bad step. In other cases the flung estimate lands in a region attracted to a different root, or outside the domain entirely. Avoiding starts near horizontal tangents is the practical lesson, and a quick plot shows where they are.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 417-417
Trap
\[ x_{5} = 16 \text{ after five iterations} \]
Report 16 as the root
Why: The student trusts the arithmetic.
\[ f(16) = 16^{1/3} \approx 2.52 \ne 0 \]
The sequence was diverging, and the numbers looked exactly like a converging sequence would until they were checked.
\[ \text{substitute back: } |f(x_{n})| \text{ must be small} \]
Verify the answer against the original equation
Why: The method's failure modes are all silent.
Two additional signs are worth watching: successive estimates should be getting closer together, and the function's value should be shrinking. A sequence where the steps grow rather than shrink is diverging, whatever the numbers look like individually.
Fill the middle
The cube root's iteration, which simplifies to a multiplication.
Fill in the blanks
x_-2 = x____ - 3x____ = ___x____
Why: Each step multiplies the estimate by negative 2, so the magnitudes double and the sign alternates. The iterates run away from the root at 0 rather than toward it, from any non-zero start.
Two truths and a lie
All three are about failure.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The cube root has a root at the origin and Newton's method diverges from every non-zero start. Existence of a root is no guarantee of convergence, which is the sharpest contrast with bisection — that method always converges when a sign change is present.
Prediction
Commit before reasoning.
Predict first
The cube root has a root at 0 and Newton's method diverges from every start. Why?
Correct: It has a vertical tangent at the root, so tangents drawn elsewhere overshoot.
\[ x_{n+1} = -2x_{n}: \text{ overshoot by a factor of two, every time} \]
Why: The cube root is perfectly continuous and its derivative exists everywhere except at 0. The problem is geometric: the curve is nearly vertical near the root and nearly flat far from it, so a tangent drawn at any guess is far shallower than the curve and its root lands beyond the true one — by a factor of two, as the simplified iteration shows. The structural nature of that failure is why no choice of start helps.
Section
Section 4
Concept
A function with several roots has several possible outcomes, and the iteration converges to whichever root the starting point is attracted to. A sketch or a sign change is the usual way to choose sensibly.
the basin of attraction — The set of starting points from which the iteration converges to a particular root. For a function with several roots these sets can be intricate, and neighbouring starts can lead to different roots.
\[ x_{0} \text{ near a root and away from } f'=0 \]
The Intermediate Value Theorem is the natural companion. A sign change on an interval both guarantees a root and supplies a starting point in the right region.
Figure (svg): The importance of the initial guess, with two starting points on the same function
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 413-418 — choosing the initial approximation
Picture it
A cubic with three roots and two starting points.
Figure (svg): The importance of the initial guess, with two starting points on the same function
The start on the right converges to the nearby root; the one near the origin sits close to a turning point and is thrown elsewhere. A plot costs seconds and prevents both problems.
Worked example
Example 4.51. The Intermediate Value Theorem supplies the region.
\[ \text{Find a good starting point for the root of } x^{3}-2x-1 \text{ between } 1 \text{ and } 2. \]
Confirm a root exists there
Why: Evaluate at both ends.
\[ f(1) = -2, f(2) = 3: a\text{ sign change} \]
Check for horizontal tangents in the interval
Why: Solve the derivative.
\[ 3 x ^{2} - 2 = 0\text{ at about } 0.816,\text{ outside} \]
Conclude the interval is safe
Why: No flat tangents inside.
\[ \text{any start in } [1, 2]\text{ should work} \]
Pick a start
Why: Either endpoint, or the midpoint.
\[ x _{0} = 2 \]
Iterate
Why: Four steps.
\[ \text{converges to } 1.618034 \]
Figure (svg): The solution to Worked example choosing a start from a sign change shown as a ladder of expressions, one row per legal move
\[ x_{0} = 2 \;\longrightarrow\; 1.618034 \]
Verify: check what happens from a start outside the safe region
Why: Starting at 0.8, very near the horizontal tangent at 0.816, the derivative is about negative 0.08 and the first step throws the estimate to around negative 33. It does eventually recover, but several iterations are wasted. Checking for horizontal tangents inside the interval before starting costs one derivative computation and avoids that entirely — and the sign change from the Intermediate Value Theorem provides the interval in the first place.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 417-418
Sorting
Near a root, and away from horizontal tangents.
Sort into buckets
Sort each start for the cubic x^3 - 3x, whose turning points are at plus and minus 1.
The last is a borderline case: at 0.5 the derivative is negative 2.25, not tiny, but the start sits between two turning points in a region attracted unpredictably. When in doubt, a plot settles it faster than analysis.
Worked example
Checkpoint 4.51. The method finds a root, not the root.
\[ \text{For } x^{3}-2x, \text{ which root does the method find from } x_{0}=1.5 \text{ and from } x_{0}=0.3? \]
Identify the roots
Why: Factor.
\[ x(x ^{2} - 2):\text{ roots at } 0\text{ and plus or minus } \sqrt{2} \]
Start at 1.5
Why: Near the positive root and away from turning points.
\[ \text{converges to } 1.4142 \]
Examine the second start
Why: Compute the derivative there.
\[ f'(0.3) = -1.73,\text{ and } f(0.3) = -0.573 \]
Take one step
Why: The correction is about negative 0.33.
\[ x _{1}\text{ is about } -0.03 \]
Continue
Why: Now attracted to the root at 0.
\[ \text{converges to } 0 \]
Figure (svg): The importance of the initial guess, with two starting points on the same function
\[ x_{0}=1.5 \to \sqrt2; \quad x_{0}=0.3 \to 0 \]
Verify: notice that both runs succeeded
Why: Neither run failed: both converged to genuine roots, and substituting back confirms each. But they found different roots, so a student wanting the positive root and starting at 0.3 would get a correct answer to a different question. The method finds A root determined by the starting point, and knowing which one you want is the user's responsibility — a sketch settles it in seconds.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 418-418
Error analysis
A student seeks the root of a cubic near 1 and starts at a convenient value.
Annotate
On: \( f(x)=x^{3}-3x, \; x_{0}=1.01 \;\Longrightarrow\; x_{1} \approx 33 \)
Checking where the derivative vanishes before choosing a start costs one computation and avoids the single most common practical failure. Starting points near horizontal tangents should always be avoided.
Fill the middle
The cubic whose turning points must be avoided as starting values.
Fill in the blanks
f'(x) = 3x^\pm 1-3 = 0 \;\Longrightarrow\; \text___ x = ___
Why: The derivative vanishes at plus and minus 1, so starts near either produce enormous corrections. Computing these before choosing a start costs one line and prevents the commonest practical failure.
Ranking
Choosing a starting point sensibly.
Put in order
Why: Steps a and b are Section 2.4's theorem doing exactly the job it was built for: guaranteeing the search will succeed. Step c is the check that prevents the flat-tangent failure, and step e is the confirmation without which none of the failure modes would be visible.
Prediction
Commit before reasoning.
Predict first
A function has three roots. Which one does Newton's method find?
Correct: Whichever the starting point is attracted to, usually but not always the nearest.
\[ x_{0}=1.5 \to \sqrt2, \quad x_{0}=0.3 \to 0: \text{ both correct, different roots} \]
Why: For a start comfortably inside a steep region the nearest root is found, but a start near a turning point can be flung across to a distant one. The sets of starting points leading to each root can be surprisingly intricate — for some functions they interleave in a fractal pattern. In practice a plot and a start well away from horizontal tangents makes the outcome predictable.
Section
Section 5
Concept
Bisection halves an interval containing a sign change and always converges, but slowly. Newton's method converges very fast and can fail entirely. Serious root-finders use bisection to get close and then switch.
bisection — Repeatedly halving an interval on which the function changes sign, keeping the half where the sign change persists. It is guaranteed by the Intermediate Value Theorem and gains about one decimal digit every three or four steps.
\[ \text{bisection: } e_{n+1} = \tfrac{1}{2}e_{n}; \quad \text{Newton: } e_{n+1} \approx Ce_{n}^{2} \]
The two methods have exactly complementary weaknesses, which is why combining them is standard. Bisection cannot fail but is slow; Newton is fast but has no safety net.
Figure (svg): Newton's method against bisection, compared on speed and reliability
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 415-418 — iterative processes
Picture it
The two methods side by side.
Figure (svg): Newton's method against bisection, compared on speed and reliability
Every row is a trade-off, and the two columns are opposites in each. That is what makes the hybrid strategy at the bottom the standard approach in numerical software.
Worked example
How many steps each needs for six digits.
\[ \text{Compare the two methods for } \sqrt{2} \text{ to six decimal places.} \]
Count Newton's steps
Why: Digits double from a start of 2.
Set up bisection
Why: A sign change on the interval from 1 to 2.
\[ \text{initial width } 1 \]
Compute bisection's rate
Why: Each step halves the width.
\[ \text{need width below } 10 ^{-6} \]
Solve for the count
Why: Two to the n must exceed a million.
\[ \text{about } 20\text{ steps} \]
Compare
Why: Three against twenty.
Figure (svg): The solution to Worked example comparing the two on one root shown as a ladder of expressions, one row per legal move
\[ \text{Newton: } 3; \quad \text{bisection: } 20 \]
Verify: extend the comparison to twelve digits
Why: For twelve digits Newton needs one more step, four in total, because the digits double. Bisection needs about forty, twice as many as for six digits, because it gains a fixed amount per step. The gap widens as more precision is demanded, which is the practical meaning of quadratic against linear convergence. But bisection was guaranteed to succeed and Newton was not.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 416-417
Matching
Complementary strengths.
Match the pairs
Why: The derivative requirement matters when the function is known only as data or as a black box, where bisection works and Newton does not. Combining them takes the guarantee from one and the speed from the other.
Worked example
Checkpoint 4.51. Use each where it is strong.
\[ \text{Describe a reliable and fast root-finding strategy.} \]
Start with a sign change
Why: Guaranteed by the Intermediate Value Theorem.
Bisect a few times
Why: Safely narrow the interval.
Switch to Newton
Why: Now close to the root and away from flat tangents.
Guard the switch
Why: If a Newton step leaves the interval, reject it.
Confirm
Why: Substitute back.
Figure (svg): The solution to Worked example the hybrid strategy shown as a ladder of expressions, one row per legal move
\[ \text{bisect} \to \text{Newton, with a bracket guard} \]
Verify: see why the guard matters
Why: The guard is what makes the hybrid genuinely safe: any Newton step that would leave the bracketing interval is discarded in favour of a bisection step, so the interval always contains the root and the method cannot diverge. This is essentially how production root-finders such as Brent's method work, and it inherits bisection's guarantee together with Newton's speed.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 418-418
Trap
\[ \text{a root is needed inside an automatic control loop} \]
Use Newton's method for its speed
Why: The student optimises for iterations.
If the derivative happens to be small for some input, the method flings its estimate away and the loop returns nonsense with no warning.
\[ \text{bracket first, then Newton with a fallback} \]
Choose the method for the situation's demands, not only for speed
Why: Unattended computation needs a guarantee more than it needs a few saved iterations.
Twenty bisection steps take microseconds, so the speed advantage is often irrelevant while the reliability difference is not. The hybrid gives both, which is why it is what numerical libraries actually implement.
Fill the middle
Halving an interval of width 1 until it is below a millionth.
Fill in the blanks
2^20 > 10^___ \;\Longrightarrow\; n \approx ___
Why: Two to the twentieth is about 1.05 million, so twenty steps suffice. Newton needs three for the same precision, and the gap widens as more digits are demanded.
Sorting
Consider what the situation demands.
Sort into buckets
Sort each situation.
The third case is decisive: with data rather than a formula there is no derivative to compute, so Newton is unavailable whatever its speed. Choosing a method is about what the situation supplies as much as what it demands.
Prediction
Commit before reasoning.
Predict first
Why do numerical libraries use a hybrid rather than Newton alone?
Correct: Because Newton has no safety net and bracketing guarantees the root stays enclosed.
\[ \text{bracket maintained} \Rightarrow \text{cannot diverge}; \quad \text{Newton steps} \Rightarrow \text{fast} \]
Why: A hybrid rejects any Newton step that would leave the bracketing interval and substitutes a bisection step instead, so the interval always contains a root and divergence is impossible — while most steps are Newton steps and the convergence stays fast. Newton is far from slow and bisection is not more accurate; the difference is entirely about guarantees, and hybrids are more complicated to code, not less.
Comparison
Fill the blanks. The two columns are opposites in every row.
Comparison matrix
| Newton | Bisection | |
|---|---|---|
| Convergence | digits double each step | about one digit per 3-4 steps |
| Needs | the derivative | only a sign change |
| Guarantee | none: it can diverge or cycle | always converges |
| Best used | close to a root, away from flat tangents | to get safely close first |
The last row is the hybrid strategy in two cells: bisection establishes safety and Newton supplies speed once the estimate is close enough for it to be trustworthy.
Pattern
Given an equation whose roots cannot be found algebraically.
Step five is not optional. Every failure mode of the method is silent, so the iteration produces plausible numbers whether or not it is converging to anything.
Stewart, Calculus: Early Transcendentals 8e, §4.8 Newton's Method §4.8, pp. 345-349
Check
The iteration.
Check your understanding
For f(x) = x^2 - 2 starting at x0 = 2, what is x1?
Answer: A
Why: The correction is f(2)/f'(2) = 2/4 = 0.5, subtracted from 2.
Check
Convergence rate.
Check your understanding
About how many correct digits does Newton's method gain per step?
Answer: A
Why: The error is roughly proportional to the previous error squared, which doubles the leading zeros.
Check
Failure modes.
Check your understanding
Newton's method applied to the cube root from x0 = 1 gives 1, -2, 4, -8. What is happening?
Answer: A
Why: The iteration simplifies to multiplying by -2, so the magnitudes grow without bound.
Real world
A mortgage of 200 000 pounds is repaid over 25 years with monthly payments of 1200 pounds. The relationship between the monthly interest rate r and the payment cannot be solved for r algebraically.
Discussion prompt
Explain why no formula gives r, set up Newton's method for it, and say what precautions the lender's software must take.
Hint: The payment formula contains r both inside and outside a power.
Answer:
\[ P = L\cdot\frac{r(1+r)^{n}}{(1+r)^{n}-1}, \qquad n = 300 \]
The unknown r appears both as a factor and inside a power raised to the 300th, so no rearrangement isolates it — this is the same difficulty as the variable appearing in both base and exponent in Section 3.9, and it has no algebraic solution.
Newton's method applies directly. Define f of r as the computed payment minus 1200, differentiate, and iterate. From a sensible start such as 0.005, which is 6 percent a year, it converges in three or four steps to r about 0.004816, or roughly 5.78 percent annually.
The precautions matter. The software runs unattended on customer data, so every failure mode of the method is a failure the customer sees. It must bracket the rate between sensible bounds — say 0 and 2 percent monthly — reject any Newton step leaving that bracket in favour of a bisection step, cap the iteration count, and verify the answer by recomputing the payment.
That is precisely the hybrid strategy of the last idea, and the reason for it is the reason given there: three saved iterations are worth nothing, and a silent divergence quoting a nonsensical interest rate is worth a great deal. Speed is almost never the binding constraint in this kind of work; reliability always is.
Commit first
Answer, then rate your confidence honestly.
Predict first
Newton's method has run five iterations and produced a number. What must you do next?
Correct: Substitute it back to confirm.
\[ |f(x_{n})| \text{ small} \;\Longrightarrow\; \text{genuinely a root} \]
Why: Every failure mode of the method is silent: a diverging sequence, a cycle and a convergent run all produce numbers that look alike. Only substituting into the original equation distinguishes them, and it costs one evaluation. Running more iterations does not help a diverging sequence, and comparing with bisection is a reasonable but far more expensive check than simply evaluating the function once.
Explain it
They ran the method, got a sequence of numbers, and want to know whether it worked.
Discussion prompt
In four sentences or fewer, give them the two checks.
Hint: One check is about the sequence and one about the answer.
Answer:
First look at the steps: successive estimates should be getting closer together, not further apart. A sequence like 1, negative 2, 4, negative 8 is diverging however plausible the individual numbers look.
Then substitute the final estimate into the original equation. If the value is close to zero it is genuinely a root; if not, the method has converged to nothing or run away, and no amount of extra iterations will fix it. Both checks together cost one evaluation and catch every failure mode.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the formula, derive it from the tangent rather than recalling it — the sign then takes care of itself. For starting points, find where the derivative vanishes and stay away. For failure, check that the steps shrink and substitute the answer back. For choosing, ask whether reliability or speed is the binding constraint, and whether a derivative is even available. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a curve with a starting point, its tangent, and the tangent's root marked as the next estimate — then a second tangent from there. Beneath the picture, derive the iteration formula in four lines from the tangent equation. Below, tabulate four iterations for the square root of two from a start of 2, with a column for the number of correct digits, and write one sentence on why the digits double. In the middle of the page, draw the three failure pictures: a nearly flat tangent flinging an estimate away, the cube root diverging, and a cubic with three roots where two starts find different ones. Beside each write the diagnosis in a few words. In the lower half, make a two-column comparison of Newton and bisection covering speed, requirements and guarantees, and write the hybrid strategy beneath it in two lines. At the bottom, write the two checks that reveal whether a run succeeded. In a margin, write why the derivative sits in the denominator.
If your comparison table has Newton winning every row, look again — bisection wins on guarantees and on needing no derivative, and those are exactly the rows that decide which method production software uses.
Recap
Five things, and the last is knowing when not to use the method at all.
| If you see | Then |
|---|---|
| A root that resists algebra | Newton's method locates it |
| A derivative near zero at the start | Choose a different starting point |
| Steps that grow rather than shrink | The iteration is diverging |
| A converged answer | Substitute it back before reporting it |
| Several roots | The start decides which one you find |
| No derivative available | Bisection, not Newton |
| Unattended computation | Bracket, and guard every Newton step |
Section 4.10 closes the chapter by reversing the question. Instead of differentiating a function, it asks which function has a given derivative — and that reversal is what the whole of Chapter 5 is built on.
OpenStax Calculus Volume 1, §4.9 Newton's Method §4.9, pp. 409-418 — everything on these slides traces back here
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