Worked examples — Newton-Raphson method for root finding

4.1.33 · D3 · Maths › Calculus I — Limits & Derivatives › Newton-Raphson method for root finding

Kuch bhi shuru karne se pehle, teen symbols ka reminder, plain words mein, taaki kuch assumed na ho:

---

Scenario matrix

Neeche har worked example us matrix ke cell ke saath tagged hai jise woh exercise karta hai.

Cell	Situation	Kya galat ho sakta hai / special kya hai	Example
A	Nice simple root, accha start	Kuch nahi — fast quadratic convergence	Ex 1
B	Negative starting guess / doosra quadrant	$f^{\prime }$ ka sign tumhe "doosri taraf" steer karta hai	Ex 2
C	Double root ($f^{\prime }=0$ root par)	Convergence linear ho jaati hai	Ex 3
D	Degenerate step: $f^{\prime }(x_n)=0$ exactly	Division by zero — method toot jaata hai	Ex 4
E	Runaway / diverging start	Guess infinity ki taraf ud jaata hai	Ex 5
F	Do roots, start decide karta hai kaun sa	Basin of attraction	Ex 6
G	Real-world word problem	Story se $f$ set up karo, units dhyan rakho	Ex 7
H	Exam-style twist (reciprocal / division-free)	$f$ aisa design karo ki update mein division na ho	Ex 8

Dhyan karo matrix guess ke dono signs, degenerate slope, dono limiting behaviours (fast converge, diverge), aur applied + exam flavours cover karta hai. Kuch bhi chhuta nahi.

---

Cell A — well-behaved simple root

Forecast: tum $2$ par start karte ho, true root $\approx 1.52$ hai. Kya $x_1$, $2$ se neeche land karega? Roughly kitna — ek chhota nudge ya bada jump?

Steps

1. Guess par evaluate karo: $f(2)=8-2-2=4$, $f^{\prime }(2)=12-1=11$. Yeh step kyun? Update ko current height aur current slope dono chahiye; hum inhe hamesha pehle compute karte hain.
2. Formula apply karo: $x_1=2-\frac{4}{11}=2-0.363636\dots =1.636364$. Yeh step kyun? Hum point $(2,4)$ se tangent par ride karte hain neeche jahan tak woh axis cross kare — woh crossing $x_1$ hai. Neeche figure mein lavender tangent dekho jo $x$-axis par drop ho raha hai.
3. Repeat karo: $f(1.636364)=4.379-1.636-2=0.743$ (3 dp tak), $f^{\prime }(1.636364)=3(2.6767)-1=7.030$. Phir $x_2=1.636364-\frac{0.743}{7.030}=1.530681$. Yeh step kyun? Nayi guess wapas feed karo — har pass nayi point par fresh tangent use karta hai.
4. Ek baar aur: $x_3=1.521452$, aur $x_4=1.521380$ — pehle hi 6 correct digits. Yeh step kyun? Digits ko double hote dekho per step (2 correct → 4 → 6): yeh quadratic-convergence signature hai jo Taylor series argument predict karta hai.

Verify: $x=1.521380$ ko $f$ mein plug karo: $1.521380^3-1.521380-2=3.52138-1.52138-2\approx 0.000$. Height essentially zero hai ✅. Aur $1.636364<2$ jaise forecast tha — pehla modest nudge, phir snap in ho jaata hai.

---

Cell B — doosre quadrant mein negative guess

Forecast: $-1$ par start karte hue, hum kis root ki taraf jaayenge — $+\sqrt{2}$ ya $-\sqrt{2}$? Sign dekho.

Steps

1. $f(-1)=1-2=-1$, $f^{\prime }(-1)=-2$. Yeh step kyun? Left branch par slope negative hai; woh negative sign exactly humein negative root ki taraf redirect karta hai.
2. $x_1=-1-\frac{-1}{-2}=-1-0.5=-1.5$. Yeh step kyun? Height over slope $\frac{-1}{-2}=+0.5$ hai; subtract karna humein left move karta hai, $-1.414$ ki taraf. Kabhi assume mat karo ki jump right jaata hai.
3. $x_2=\frac{1}{2}(-1.5+\frac{2}{-1.5})=\frac{1}{2}(-1.5-1.3333)=-1.416667$. Yeh step kyun? Same digits-doubling, bas zero ke across mirror hua.
4. $x_3=\frac{1}{2}(-1.416667-1.411765)=-1.414216$.

Verify: $(-1.414216{)}^2-2=2.000007-2\approx 0$ ✅. Hum $-\sqrt{2}$ par land kiye, na ki $+\sqrt{2}$ par — starting sign ne basin choose kiya. Ex 6 se compare karo.

---

Cell C — double root, convergence slow hokar linear ho jaati hai

Forecast: parent ne quadratic speed promise ki thi. Yahan digits double nahi honge. Guess: kya har step remaining error ko halve karta hai?

Steps

1. $f(2)=1$, $f^{\prime }(2)=2$. To $x_1=2-\frac{1}{2}=1.5$. Error gaya $1\to 0.5$. Yeh step kyun? Ordinary update — lekin dhyan karo error sirf halve hua.
2. $f(1.5)=0.25$, $f^{\prime }(1.5)=1$. $x_2=1.5-0.25=1.25$. Error $0.5\to 0.25$. Yeh step kyun? Phir exactly halve hua — yeh linear convergence hai, ek bit per step.
3. $f(1.25)=0.0625$, $f^{\prime }(1.25)=0.5$. $x_3=1.25-0.125=1.125$. Error $0.25\to 0.125$. Yeh step kyun? Pattern $x_{n+1}-1=\frac{1}{2}(x_n-1)$ hamesha hold karta hai — quadratic guarantee fail ho gayi kyunki usne simple root assume kiya tha ($f^{\prime }(\text{root})\ne 0$).

Verify: error sequence $1,0.5,0.25,0.125$ exactly $2^{-n}$ hai — ek geometric (linear) decay, parent ke mistake-callout #3 ko confirm karta hai ✅. Ex 1 se contrast karo jahan errors square hoti theen.

---

Cell D — degenerate step: slope exactly zero hai

Forecast: $x=0$ par tangent ki slope kya hai? Agar woh flat hai, to flat line $x$-axis ko kahan cross karti hai?

Steps

1. $f(0)=-1$, $f^{\prime }(0)=2\cdot 0=0$. Yeh step kyun? Humein divide karne se pehle $f^{\prime }$ check karna zaruri hai (parent mistake #2). Yahan woh exactly $0$ hai.
2. Update $x_1=0-\frac{-1}{0}$ undefined hai — division by zero. Geometrically $x=0$ par tangent horizontal hai (figure mein flat coral line dekho); ek horizontal line kabhi $x$-axis se nahi milti. Yeh step kyun? Yeh woh degenerate case hai jiske baare mein theory warn karti hai — recipe ka simply koi answer nahi hai.

Verify: $f^{\prime }(0)=0$ exactly hai, to $f(x_0)/f^{\prime }(x_0)$ ka zero denominator hai — koi finite $x_1$ exist nahi karta ✅. Fix: start perturb karo, e.g. $x_0=0.5$ deta hai $f^{\prime }=1\ne 0$ aur $1$ ki taraf theek se converge karta hai. (Check: $x_1=0.5-\frac{-0.75}{1}=1.25$, phir $x_2=1.025$, $1$ par close ho raha hai.)

---

Cell E — ek runaway start jo diverge karta hai

Forecast: bade $x$ ke liye, $\arctan x$ $\frac{\pi }{2}$ ki taraf flatten ho jaata hai — slope $\frac{1}{1+x^2}$ tiny ho jaata hai. Tiny slope + not-tiny height matlab ek huge jump. Guess: kya $x_0=2$ humein $0$ ke closer move karta hai ya hume aur door phek deta hai?

Steps

1. $f(2)=\arctan 2=1.107149$, $f^{\prime }(2)=\frac{1}{5}=0.2$. Yeh step kyun? Dhyan karo height ($1.11$) slope ($0.2$) ke relative badi hai — ek warning sign.
2. $x_1=2-\frac{1.107149}{0.2}=2-5.535744=-3.535744$. Yeh step kyun? Hum zero overshoot karte hain aur root se aur door land karte hain, doosri side par. Nearly-flat tangent (parent mistake #2) ne hume launch kiya.
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 $=+13.95$. Yeh step kyun? Distance har step par grow kar rahi hai ($2\to 3.54\to 13.95$) — textbook divergence.

Verify: $|x_1|=3.5357>|x_0|=2$, aur $|x_2|\approx 13.95>|x_1|$ ✅ — guesses run away kar rahe hain, confirm karta hai ki Newton-Raphson sirf local hai. Iske bajay $x_0=0.5$ se start karo (chhoti height, slope $0.8$) converge karta hai: $x_1=0.5-\frac{0.463648}{0.8}=-0.079560$, pehle hi $0$ ke kaafi close.

---

Cell F — do roots, start basin choose karta hai

Forecast: $x_0=4$ se start karo ($0.5$ ke right) — kya tum $3$ par pahunchoge ya $-2$ par? Phir $x_0=-1$ se start karo — same sawaal, opposite side.

Steps

1. Right start $x_0=4$: $f(4)=16-4-6=6$, $f^{\prime }(4)=7$. $x_1=4-\frac{6}{7}=3.142857$. Yeh step kyun? Hum vertex ke right hain, root $3$ ke owned branch par.
2. $f(3.142857)=9.8776-3.1429-6=0.7347$, $f^{\prime }(3.142857)=5.285714$. $x_2=3.142857-0.139002=3.003855$. Yeh step kyun? $3$ par homing — right basin. Figure dekho: vertex ke dashed line ke right sab kuch $3$ ki taraf flow karta hai.
3. Left start $x_0=-1$: $f(-1)=1+1-6=-4$, $f^{\prime }(-1)=-3$. $x_1=-1-\frac{-4}{-3}=-1-1.3333=-2.333333$. Yeh step kyun? Vertex ke left, to hum root $-2$ ke left basin mein gir jaate hain.
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: $4$ se: $x_2=3.003855$ $3$ ki taraf ja raha hai ✅. $-1$ se: $x_2=-2.019608$ $-2$ ki taraf ja raha hai ✅. Vertex $x=0.5$ ($f^{\prime }=0$) do basins of attraction ke beech ki boundary hai — Fixed-point iteration se links.

---

Cell G — ek real-world word problem (units dhyan rakho)

Forecast: zero interest ($x=1$) exactly $5000 deta hai (paanch deposits). Humein $6000 chahiye, to $r$ ek positive, smallish rate hai. Guess: kya $r$ 5% ke karib hai ya 15% ke?

Steps

1. $x_0=1.1$ se start karo (10% trial): $f(1.1)=1.4641+1.331+1.21+1.1-5=0.1051$, $f^{\prime }(1.1)=5.324+3.63+2.2+1=12.154$. Yeh step kyun? Ek sensible physical starting rate choose karo; height aur slope same units mein compute karo (pure numbers ab, kyunki $x$ ek ratio hai).
2. $x_1=1.1-\frac{0.1051}{12.154}=1.1-0.008648=1.091352$. Yeh step kyun? Ek Newton jump growth factor ko thoda lower karta hai — ek chhota correction kyunki hum close se start kiye.
3. $f(1.091352)=0.002$-ish, $f^{\prime }(1.091352)\approx 11.95$, deta hai $x_2=1.091342$. Yeh step kyun? Already converge ho gaya; zyada passes barely move karte hain — simple root par quadratic convergence.

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

---

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

Forecast: tumhe ek $f$ invent karna hai jiska root $1/a$ ho aur jiska update division cancel kar de. Kaun sa function $1/a$ ko root ki tarah rakhta hai lekin sirf $a$ ko multiplicatively involve karta hai?

Steps

1. Choose karo $f(x)=\frac{1}{x}-a$. Iska root wahan hai jahan $\frac{1}{x}=a$, yani $x=\frac{1}{a}$ — exactly target. Yeh step kyun? Hum $f$ design karte hain taaki iska zero woh number ho jo hum chahte hain.
2. Differentiate karo: $f^{\prime }(x)=-\frac{1}{x^2}$. Yeh step kyun? Update ko slope chahiye; dhyan karo ismein bhi $1/x^2$ hai — hum ab inhe cancel hone dene wale hain.
3. Update: $x_{n+1}=x_n-\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}=x_n+x_n^2\left(\frac{1}{x_n}-a\right)=x_n+x_n-a{x}_n^2=x_n(2-a{x}_n)$. Yeh step kyun? Do reciprocals algebraically cancel ho jaate hain, sirf multiplications aur ek subtraction reh jaate hain — koi division nahi. Yahi exam ka punchline hai.
4. $a=3$, $x_0=0.3$ ke saath iterate karo: $x_1=0.3(2-0.9)=0.3\cdot 1.1=0.33$; $x_2=0.33(2-0.99)=0.33\cdot 1.01=0.3333$; $x_3=0.3333(2-0.9999)=0.333333$. Yeh step kyun? Confirm karo ki yeh sirf multiplication use karke $1/3=0.\overline{3}$ par home karta hai.

Verify: $x_3=0.333333\approx \frac{1}{3}$; check karo $1/x_3=3.00000=a$ ✅. Har operation multiply ya subtract tha — division-free goal met ✅.

---

Recall Self-test: example ko cell se match karo

Kaun sa example linear (quadratic nahi) convergence dikhata hai, aur kyun? ::: Ex 3 — ek double root, jahan $f^{\prime }(\text{root})=0$ hai to Taylor guarantee fail ho jaati hai. Kis example mein koi valid $x_1$ hi nahi hai? ::: Ex 4 — $f^{\prime }(0)=0$ exactly hai, to hum zero se divide karte hain. Ex 6 mein $x$ ki kaun si value do basins of attraction ko separate karti hai? ::: Vertex $x=0.5$, jahan $f^{\prime }(x)=2x-1=0$. Ex 8 ka update division avoid kyun karta hai? ::: $f$ se $1/x$ aur $f^{\prime }$ se $1/x^2$ cancel ho jaate hain, $x_{n+1}=x_n(2-a{x}_n)$ reh jaata hai.

---

Connections

• Tangent line and linear approximation — yahan har step ek tangent hai, draw kiya aur axis par drop kiya.
• Taylor series — explain karta hai kyun Ex 1 digits double karta hai lekin Ex 3 nahi karta.
• Derivatives — definition and rules — har example mein $f^{\prime }$ supply kiya.
• Fixed-point iteration — Ex 6 mein basin idea fixed-point convergence picture hai.
• Bisection method — safe fallback jab Ex 4/Ex 5 misbehave karte hain (safeguarded Newton).
• Roots of polynomials — Ex 1, 6, 7 sab polynomial roots hain.

Concept Map