Worked examples — Newton-Raphson method for root finding

Before anything, a reminder of the three symbols, in plain words, so nothing is assumed:

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The scenario matrix

Every worked example below is tagged with the cell of this matrix it exercises.

Cell	Situation	What could go wrong / be special	Example
A	Nice simple root, good start	Nothing — fast quadratic convergence	Ex 1
B	Negative starting guess / other quadrant	Sign of $f^{\prime }$ steers you the "other way"	Ex 2
C	Double root ($f^{\prime }=0$ at the root)	Convergence drops to linear	Ex 3
D	Degenerate step: $f^{\prime }(x_n)=0$ exactly	Division by zero — method breaks	Ex 4
E	Runaway / diverging start	Guess flies off to infinity	Ex 5
F	Two roots, start decides which	Basin of attraction	Ex 6
G	Real-world word problem	Set up $f$ from a story, mind the units	Ex 7
H	Exam-style twist (reciprocal / division-free)	Design $f$ so the update has no division	Ex 8

Notice the matrix covers both signs of the guess, the degenerate slope, both limiting behaviours (converge fast, diverge), and applied + exam flavours. Nothing is left uncovered.

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Cell A — the well-behaved simple root

Forecast: you start at $2$, the true root is $\approx 1.52$. Will $x_1$ land below $2$? By how much roughly — a small nudge or a big jump?

Steps

1. Evaluate at the guess: $f(2)=8-2-2=4$, $f^{\prime }(2)=12-1=11$. Why this step? The update needs both the current height and the current slope; we always compute them first.
2. Apply the formula: $x_1=2-\frac{4}{11}=2-0.363636\dots =1.636364$. Why this step? We ride the tangent from the point $(2,4)$ down to where it crosses the axis — that crossing is $x_1$. Look at the lavender tangent in the figure below dropping to the $x$-axis.
3. Repeat: $f(1.636364)=4.379-1.636-2=0.743$ (to 3 dp), $f^{\prime }(1.636364)=3(2.6767)-1=7.030$. Then $x_2=1.636364-\frac{0.743}{7.030}=1.530681$. Why this step? Feed the new guess back in — each pass uses a fresh tangent at the new point.
4. Once more: $x_3=1.521452$, and $x_4=1.521380$ — already 6 correct digits. Why this step? To watch the digits double per step (2 correct → 4 → 6): the quadratic-convergence signature the Taylor series argument predicts.

Verify: plug $x=1.521380$ into $f$: $1.521380^3-1.521380-2=3.52138-1.52138-2\approx 0.000$. Height is essentially zero ✅. And $1.636364<2$ as forecast — a modest first nudge, then it snaps in.

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Cell B — a negative guess in a different quadrant

Forecast: starting at $-1$, which root do we head for — $+\sqrt{2}$ or $-\sqrt{2}$? Watch the sign.

Steps

1. $f(-1)=1-2=-1$, $f^{\prime }(-1)=-2$. Why this step? On the left branch the slope is negative; that negative sign is exactly what redirects us toward the negative root.
2. $x_1=-1-\frac{-1}{-2}=-1-0.5=-1.5$. Why this step? Height over slope is $\frac{-1}{-2}=+0.5$; subtracting moves us left, toward $-1.414$. Never assume the jump goes right.
3. $x_2=\frac{1}{2}(-1.5+\frac{2}{-1.5})=\frac{1}{2}(-1.5-1.3333)=-1.416667$. Why this step? Same doubling-of-digits, just mirrored across zero.
4. $x_3=\frac{1}{2}(-1.416667-1.411765)=-1.414216$.

Verify: $(-1.414216{)}^2-2=2.000007-2\approx 0$ ✅. We landed on $-\sqrt{2}$, not $+\sqrt{2}$ — the starting sign chose the basin. Compare with Ex 6.

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Cell C — a double root, convergence slows to linear

Forecast: the parent promised quadratic speed. Here the digits will not double. Guess: does each step halve the remaining error instead?

Steps

1. $f(2)=1$, $f^{\prime }(2)=2$. So $x_1=2-\frac{1}{2}=1.5$. Error went $1\to 0.5$. Why this step? Ordinary update — but note the error only halved.
2. $f(1.5)=0.25$, $f^{\prime }(1.5)=1$. $x_2=1.5-0.25=1.25$. Error $0.5\to 0.25$. Why this step? Again exactly halved — this is linear convergence, one bit per step.
3. $f(1.25)=0.0625$, $f^{\prime }(1.25)=0.5$. $x_3=1.25-0.125=1.125$. Error $0.25\to 0.125$. Why this step? The pattern $x_{n+1}-1=\frac{1}{2}(x_n-1)$ holds forever — the quadratic guarantee failed because it assumed a simple root ($f^{\prime }(\text{root})\ne 0$).

Verify: the error sequence $1,0.5,0.25,0.125$ is exactly $2^{-n}$ — a geometric (linear) decay, confirming the parent's mistake-callout #3 ✅. Contrast Ex 1 where errors squared.

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Cell D — degenerate step: the slope is exactly zero

Forecast: what is the tangent's slope at $x=0$? If it's flat, where does a flat line cross the $x$-axis?

Steps

1. $f(0)=-1$, $f^{\prime }(0)=2\cdot 0=0$. Why this step? We must check $f^{\prime }$ before dividing (parent mistake #2). Here it is exactly $0$.
2. The update $x_1=0-\frac{-1}{0}$ is undefined — division by zero. Geometrically the tangent at $x=0$ is horizontal (see the flat coral line in the figure); a horizontal line never meets the $x$-axis. Why this step? This is the degenerate case the theory warns about — the recipe simply has no answer.

Verify: $f^{\prime }(0)=0$ exactly, so $f(x_0)/f^{\prime }(x_0)$ has a zero denominator — no finite $x_1$ exists ✅. The fix: perturb the start, e.g. $x_0=0.5$ gives $f^{\prime }=1\ne 0$ and converges fine to $1$. (Check: $x_1=0.5-\frac{-0.75}{1}=1.25$, then $x_2=1.025$, closing on $1$.)

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Cell E — a runaway start that diverges

Forecast: for large $x$, $\arctan x$ flattens toward $\frac{\pi }{2}$ — the slope $\frac{1}{1+x^2}$ becomes tiny. Tiny slope + not-tiny height means a huge jump. Guess: does $x_0=2$ move us closer to $0$ or hurl us further away?

Steps

1. $f(2)=\arctan 2=1.107149$, $f^{\prime }(2)=\frac{1}{5}=0.2$. Why this step? Notice the height ($1.11$) is large relative to the slope ($0.2$) — a warning sign.
2. $x_1=2-\frac{1.107149}{0.2}=2-5.535744=-3.535744$. Why this step? We overshoot zero and land further from the root than we started, on the other side. The nearly-flat tangent (parent mistake #2) launched us.
3. $f(-3.535744)=-1.295647$, $f^{\prime }(-3.535744)=\frac{1}{1+12.5015}=0.074066$. $x_2=-3.535744-\frac{-1.295647}{0.074066}=-3.535744+17.49$-ish $=+13.95$. Why this step? The distance is growing each step ($2\to 3.54\to 13.95$) — textbook divergence.

Verify: $|x_1|=3.5357>|x_0|=2$, and $|x_2|\approx 13.95>|x_1|$ ✅ — the guesses run away, confirming Newton-Raphson is only local. Starting instead at $x_0=0.5$ (small height, slope $0.8$) converges: $x_1=0.5-\frac{0.463648}{0.8}=-0.079560$, already close to $0$.

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Cell F — two roots, the start picks the basin

Forecast: start at $x_0=4$ (right of $0.5$) — do you reach $3$ or $-2$? Then start at $x_0=-1$ — same question, opposite side.

Steps

1. Right start $x_0=4$: $f(4)=16-4-6=6$, $f^{\prime }(4)=7$. $x_1=4-\frac{6}{7}=3.142857$. Why this step? We're to the right of the vertex, on the branch owned by root $3$.
2. $f(3.142857)=9.8776-3.1429-6=0.7347$, $f^{\prime }(3.142857)=5.285714$. $x_2=3.142857-0.139002=3.003855$. Why this step? Homing on $3$ — the right basin. Look at the figure: everything right of the dashed vertex flows to $3$.
3. Left start $x_0=-1$: $f(-1)=1+1-6=-4$, $f^{\prime }(-1)=-3$. $x_1=-1-\frac{-4}{-3}=-1-1.3333=-2.333333$. Why this step? Left of the vertex, so we fall into the left basin of root $-2$.
4. $f(-2.333333)=5.4444+2.3333-6=1.777778$, $f^{\prime }(-2.333333)=-5.666667$. $x_2=-2.333333-(-0.313725)=-2.019608$.

Verify: from $4$: $x_2=3.003855$ heads to $3$ ✅. From $-1$: $x_2=-2.019608$ heads to $-2$ ✅. The vertex $x=0.5$ ($f^{\prime }=0$) is the boundary between the two basins of attraction — links to Fixed-point iteration.

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Cell G — a real-world word problem (mind the units)

Forecast: zero interest ($x=1$) gives exactly $5000 (five deposits). We need $6000, so $r$ is a positive, smallish rate. Guess: is $r$ nearer 5% or 15%?

Steps

1. Start $x_0=1.1$ (a 10% trial): $f(1.1)=1.4641+1.331+1.21+1.1-5=0.1051$, $f^{\prime }(1.1)=5.324+3.63+2.2+1=12.154$. Why this step? Pick a sensible physical starting rate; compute height and slope in the same units (pure numbers now, since $x$ is a ratio).
2. $x_1=1.1-\frac{0.1051}{12.154}=1.1-0.008648=1.091352$. Why this step? One Newton jump lowers the growth factor slightly — a small correction because we started close.
3. $f(1.091352)=0.002$-ish, $f^{\prime }(1.091352)\approx 11.95$, giving $x_2=1.091342$. Why this step? Already converged; more passes barely move it — quadratic convergence at a simple root.

Verify: growth factor $x=1.091342\Rightarrow r=0.091342\approx 9.13\%$. Plug back: $1000\cdot \frac{1.091342^5-1}{0.091342}=1000\cdot \frac{1.548014-1}{0.091342}\approx 6000$ ✅. Units check: dollars in = dollars out; $r$ is a pure rate ✅.

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Cell H — an exam-style twist (division-free reciprocal)

Forecast: you must invent an $f$ whose root is $1/a$ and whose update cancels the division. What function has $1/a$ as a root but only involves $a$ multiplicatively?

Steps

1. Choose $f(x)=\frac{1}{x}-a$. Its root is where $\frac{1}{x}=a$, i.e. $x=\frac{1}{a}$ — exactly the target. Why this step? We design $f$ so its zero is the number we want.
2. Differentiate: $f^{\prime }(x)=-\frac{1}{x^2}$. Why this step? The update needs the slope; note it also contains a $1/x^2$ — we're about to make those cancel.
3. Update: $x_{n+1}=x_n-\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}=x_n+x_n^2\left(\frac{1}{x_n}-a\right)=x_n+x_n-a{x}_n^2=x_n(2-a{x}_n)$. Why this step? The two reciprocals cancel algebraically, leaving only multiplications and a subtraction — no division. That's the exam's punchline.
4. Iterate with $a=3$, $x_0=0.3$: $x_1=0.3(2-0.9)=0.3\cdot 1.1=0.33$; $x_2=0.33(2-0.99)=0.33\cdot 1.01=0.3333$; $x_3=0.3333(2-0.9999)=0.333333$. Why this step? Confirm it homes on $1/3=0.\overline{3}$ using multiplication alone.

Verify: $x_3=0.333333\approx \frac{1}{3}$; check $1/x_3=3.00000=a$ ✅. Every operation was a multiply or subtract — the division-free goal met ✅.

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Recall Self-test: match example to cell

Which example shows linear (not quadratic) convergence, and why? ::: Ex 3 — a double root, where $f^{\prime }(\text{root})=0$ so the Taylor guarantee fails. Which example has no valid $x_1$ at all? ::: Ex 4 — $f^{\prime }(0)=0$ exactly, so we divide by zero. In Ex 6, what value of $x$ separates the two basins of attraction? ::: The vertex $x=0.5$, where $f^{\prime }(x)=2x-1=0$. Why does Ex 8's update avoid division? ::: The $1/x$ from $f$ and the $1/x^2$ from $f^{\prime }$ cancel, leaving $x_{n+1}=x_n(2-a{x}_n)$.

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Connections

• Tangent line and linear approximation — every step here is one tangent, drawn and dropped to the axis.
• Taylor series — explains why Ex 1 doubles digits but Ex 3 doesn't.
• Derivatives — definition and rules — supplied $f^{\prime }$ in every example.
• Fixed-point iteration — the basin idea in Ex 6 is the fixed-point convergence picture.
• Bisection method — the safe fallback when Ex 4/Ex 5 misbehave (safeguarded Newton).
• Roots of polynomials — Ex 1, 6, 7 are all polynomial roots.

Concept Map