Newton-Raphson method for root finding: Visual walkthrough

4.1.33 · D2 · Maths › Calculus I — Limits & Derivatives › Newton-Raphson method for root finding

Step 1 — " $f (x) = 0$ " maangta kya hai aakhir?

KYA. Hamare paas ek curve hai. Flat ground ke upar iski height ko $f (x)$ kehte hain: ek horizontal position $x$ daalo, yeh height return karta hai. Kuch heights positive hain (curve ground ke upar), kuch negative (curve ground ke neeche). Root woh jagah hai jahan height bilkul zero hoti hai — jahan curve ground ko touch karta hai.

KYU. " $f (x) = 0$ solve karo" ka matlab itna hi hai: "dhundho kahan curve flat line ko cross karta hai". Ek straight line ke liye yeh instantly dikhta hai. $x^{5} - x - 1$ jaise wiggly curve ke liye, algebra us crossing ko isolate nahi kar sakta — isliye hume ek trick chahiye.

PICTURE. Ground horizontal $x$ -axis hai. Curve neeche dip karta hai aur upar uthta hai; red dot true crossing $r$ mark karta hai (" $r$ " sirf root ka naam hai).

Step 2 — Ek guess lagao, aur wahan do cheezein measure karo

KYA. Hum pehla guess lagate hain aur use $x_{0}$ kehte hain (chhota $0$ sirf matlab hai "guess number zero"). Us jagah hum $x$ -axis par khade hote hain, seedha upar dekhte hain, aur apne upar curve ki height measure karte hain: woh number hai $f (x_{0})$ .

KYU. Hum root par jump nahi kar sakte kyunki hum use dekh nahi sakte. Lekin ek foot ke neeche ground feel kar sakte hain: apne guess par hum local height measure kar sakte hain aur (agle step mein) local steepness. Yeh do local facts hi Newton-Raphson kabhi use karta hai.

PICTURE. $x_{0}$ par axis se curve ke point $(x_{0}, f (x_{0}))$ tak ek vertical plum segment — us segment ki length height $f (x_{0})$ hai.

Step 3 — Slope: curve yahan kitna steeply jhukta hai

KYA. Hamare point par slope ek single number hai jo batata hai curve wahan kitna tilted hai: "ek step right ke liye, kitne steps upar?" Hum ise $f^{'} (x_{0})$ likhte hain (padho "f-prime at $x_{0}$ "). Yeh exactly derivative hai.

YEH TOOL KYU. Hum slope kyun chahte hain, phir se height nahi? Kyunki slope hume predict karne deta hai: agar curve is exact tilt par chalte rahta, toh yeh ek straight line trace karta — aur ek straight line ground ko ek obvious, computable jagah hit karta hai. Height akeli batati hai hum kahan hain; slope batata hai kis taraf aur kitni tezi se chalna hai. Doosra sawaal jaanna zaroori hai, isliye derivative laate hain.

PICTURE. Hamare point par curve par ek tiny right triangle: ek chhota "run" (horizontal, teal) aur matching "rise" (vertical, orange). Slope $= \frac{rise}{run}$ .

Step 4 — Curve ki jagah tangent line lagao

KYA. Ab hum woh ek straight line draw karte hain jo hamare point par curve ko touch kare aur wahan uski slope share kare. Yeh tangent line hai. $x_{0}$ ke aas paas yeh curve se itni close hoti hai ki fark lagbhag dikh hi nahi pata.

KYU. Yahi poora engine hai (Tangent line and linear approximation): ek point ke paas, ek smooth curve apni tangent jaisi lagti hai. Curve mushkil hai (solve nahi ho sakti). Tangent easy hai (ek straight line — hum iski ground-crossing haath se nikal sakte hain). Toh hum easy stand-in solve karte hain aur umeed karte hain ki iski crossing real wali ke paas ho.

PICTURE. Same curve, same point, ab ek lambi straight orange tangent iske through. Notice karo kaise yeh $x_{0}$ par curve ko kiss karti hai aur sirf door jaane par alag hoti hai.

Step 5 — Easy sawaal: tangent ground ko kahan hit karta hai?

KYA. Hum tangent ka khud ka root chahte hain — woh $x$ jahan iski height $y$ zero ho. Us landing spot ko $x_{1}$ bolte hain (guess number one). Toh hum tangent equation mein $y = 0$ set karte hain aur $x_{1}$ padhte hain.

KYU. Hum curve ko zero set karke solve nahi kar sakte — yahi toh problem thi. Lekin tangent ek line hai, aur ek line zero ko exactly ek aisi jagah hit karti hai jo hum compute kar sakte hain. Woh crossing hamaara agla, behtar guess hai.

PICTURE. Tangent axis ke through slice karti hui; crossing point $x_{1}$ mark hai, $x_{0}$ se clearly true root $r$ ke zyada paas.

Step 6 — Jump ke liye solve karo, aur triangle se formula padhlo

KYA. Dono sides ko slope $f^{'} (x_{0})$ se divide karo horizontal jump isolate karne ke liye, phir $x_{1}$ akele laane ke liye dono sides mein $x_{0}$ jodo.

KYU. Hume recipe chahiye: "jahan main hoon, agle step kahan jaoon?" $x_{1}$ solve karna exactly yahi deliver karta hai, bina kisi naye assumption ke — sirf ek silent requirement ke: $f^{'} (x_{0})$ se divide karna tabhi legal hai jab $f^{'} (x_{0}) \neq = 0$ (Step 8 flat case handle karta hai).

PICTURE. Ek bada right triangle jiska vertical leg height $f (x_{0})$ hai aur horizontal leg jump $Δ = x_{1} - x_{0}$ hai. Tangent iski hypotenuse hai.

Formula Triangle se slope padhna — aur minus kahan paida hota hai

Tangent point $(x_{0}, f (x_{0}))$ se height $f (x_{0}) > 0$ par neeche jaati hai landing $(x_{1}, 0)$ tak ground par, aur left move karti hai ( $x_{1} < x_{0}$ , toh $Δ < 0$ ). Toh run $Δ$ par height $f (x_{0})$ girti hai: $f^{'} (x_{0}) = \frac{rise}{run} = \frac{0 - f ( x _{0} )}{x _{1} - x _{0}} = \frac{- f ( x _{0} )}{Δ} .$

numerator $0 - f (x_{0}) = - f (x_{0})$ ::: height zero tak giri, toh rise negative hai.
denominator $Δ$ ::: leftward run, khud negative. Do negatives $f^{'} (x_{0})$ ko positive banate hain — uphill slope se match karta hai. Ab isko $Δ$ ke liye solve karo: dono sides ko $Δ$ se multiply karo aur $f^{'} (x_{0})$ se divide karo, $Δ = - \frac{f ( x _{0} )}{f ^{'} ( x _{0} )} .$ Minus sign ek choice nahi — yeh seedha "rise $= 0 - f (x_{0})$ " se nikalta hai. Triangle ka horizontal leg literally $- \frac{f ( x _{0} )}{f ^{'} ( x _{0} )}$ hai, aur yeh negative hai (ek left step) exactly isliye kyunki curve ground ke upar tha.

Step 7 — Charon sign-combinations: $Δ$ kis taraf point karta hai?

KYA. Step 6 ne ek picture draw ki: curve ground ke upar ( $f (x_{0}) > 0$ ) aur uphill tangent ( $f^{'} (x_{0}) > 0$ ), deta hai leftward step ( $Δ < 0$ ). Lekin height positive ya negative ho sakti hai, aur slope uphill ya downhill — chaar combinations total, jaise formula ke behaviour ke "chaar quadrants". Single formula inhe automatically handle karta hai kyunki $f$ aur $f^{'}$ ke signs ratio ke andar hain.

KYU. Agar hum sirf ek panel dekhte, toh hum galti se maan lete "Newton hamesha left step karta hai" ya "height leg hamesha upar point karti hai". Dono galat hain. Step ki direction bilkul $Δ = - \frac{f ( x _{0} )}{f ^{'} ( x _{0} )},$ ke sign se decide hoti hai, toh charon padhte hain:

PICTURE. Chaar panels, table ki har row ke liye ek, har apni tangent ke saath aur apna arrow dikhata hai $Δ$ left ya right axis ki taraf point karta hua. Har arrow ko table se verify karo.

Step 8 — Repeat karo, aur step counter $n$ naam do

KYA. $x_{1}$ ko apna naya standing spot rename karo aur Steps 2–6 phir karo. "General guess" ke baare mein baat karne ke liye hum ek counting label $n$ introduce karte hain.

KYU. Ek tangent-jump close land karta hai, exact nahi — tangent ne curve chhod di jab hum door gaye. Toh hum naye spot par height aur slope re-measure karte hain aur ek fresh, shorter jump lete hain.

PICTURE. Ek frame mein do tangent jumps: ek lamba pehla hop $x_{0} \to x_{1}$ , phir bahut chhota doosra hop $x_{1} \to x_{2}$ jo almost $r$ par land karta hai. Ghatta hua gap hi point hai.

Step 9 — Kab rokein? Tolerance $ε$

KYA. Picture kabhi exactly $r$ par land nahi karti; har jump sirf gap shrink karta hai. Toh practically hum tab rukте hain jab guess "kaafi achha" ho. Hum ek tiny positive number $ε$ choose karte hain (Greek letter "epsilon", standard naam chhoti allowed error ke liye) aur rukте hain jab ya toh guess hilna band ho jaata hai ya height essentially zero ho.

KYU. Stopping rule ke bina, computer forever loop karta aise digits chase karte hue jo woh store bhi nahi kar sakta, aur haath se kaam mein kab quit karna hai pata hi nahi. Neeche do rules har step par measure ki do cheezein match karti hain (position aur height).

Step 10 — Degenerate cases: jab picture toot jaati hai

Newton-Raphson local hai. Chaar tarike geometry misfire karti hai — har ek real drawing hai, abstraction nahi.

(a) Flat spot, $f^{'} (x_{n}) = 0$ . Tangent horizontal hai. Ek horizontal line kabhi ground nahi hit karti (jab tak woh ground hi na ho). Zero-slope se divide karna woh algebra hai jo chilla raha hai ki jump infinite hai: $x_{n + 1}$ kahin door ud jaata hai.

(b) Near-flat, $f^{'} (x_{n}) \approx 0$ . Exactly zero nahi, lekin tiny. Tab $\frac{f}{f ^{'}}$ bahut bada hai — ek chhoti height tiny slope par — aur hum root se bahut door catapult ho jaate hain. Hamesha chhote slopes se bachao.

(c) Cycling. Kabhi kabhi jumps ek loop mein land karte hain. $f (x) = x^{3} - 2 x + 2$ ke liye $x_{0} = 0$ ke saath: tangent $0 \to 1$ bhejti hai, aur $1$ par tangent $1 \to 0$ — hamesha ke liye. Achha formula, basin of attraction ke bahar bura starting point.

(d) Double root. Agar curve sirf ground ko graze karta hai ( $f = 0$ aur $f^{'} = 0$ root par), toh slope exactly utna hi shrink hoti hai jitna height. Convergence quadratic se sirf linear tak sag jaata hai — errors square hone ki jagah half hoti hain.

PICTURE. Chaar mini-panels: horizontal tangent shoot out karti hui (orange), near-flat catapult (teal), $0 \to 1 \to 0$ cycle (plum arrows), aur ek grazing double root.

Ek-picture summary

Sab ek saath: curve, ek guess $x_{n}$ , iski height (vertical), iski tangent (straight stand-in), right triangle jiski legs height aur jump $Δ$ hain, aur landing $x_{n + 1}$ root $r$ ke paas. Triangle padho aur formula nikalta hai: horizontal leg $=$ $-$ height $\div$ slope, aur hum iske khilaf step karte hain — left ya right jaisa Step 7 ka sign table dictate kare, jab tak Step 9 ka tolerance $ε$ rukne ko nahi kehta.

Recall Poori walk ki Feynman retelling

Aap flat ground par khade ho aur kisi taraf door ground ek aisi pahadi se milti hai jiska base aap dekh nahi sakte. Aap woh meeting point chahte ho. Aap sirf do cheezein jaante ho jahan aap khade ho: pahadi seedhi upar aapke upar kitni unchi hai, aur woh aapke pair ke neeche kitni steeply tilt hai. Toh aap pretend karte ho ki pahadi us exact tilt par perfectly straight ramp mein chalti rahti hai, aur aap wahan chalte ho jahan woh imaginary ramp ground ko touch karta. Woh walk hai "height divided by steepness," aur aap pahadi ki taraf jaate ho, door nahi — yahi minus sign hai, jo triangle se seedha nikalta hai kyunki height zero ho gayi. Left ya right step karna sirf is par depend karta hai ki aap ground ke kis side par ho aur ramp kis taraf jhukti hai; formula ka minus aapke liye sort kar deta hai. Aap abhi wahan nahi ho, kyunki asli pahadi aapki ramp se curve ho gayi. Theek hai: naye spot par khado (ise step $n + 1$ kaho), naya height aur tilt feel karo, naya ramp khicho, phir chalo. Har walk jo bacha hai use roughly uske square tak kaat deta hai — teen achhe digits chhe ban jaate hain — provided aap kaafi paas se shuru kiye aur pahadi smooth hai. Aap rukте ho jab walk us se chhoti ho jaaye jitna aapne pehle se choose kiya ( $ε$ ). Trip karne ke sirf do tarike: aapka pair flat jagah par land kare (koi ramp ground tak nahi pahunchti), ya aapne itna bura shuru kiya ki aap loop mein hopping karte rahte ho.

Connections

Tangent line and linear approximation — Steps 3–6 hain hi linear approximation, reuse hua.
Taylor series — leftover error kyun squares hoti hai (Step 8).
Derivatives — definition and rules — Step 3 ka slope $f^{'}$ supply karta hai.
Fixed-point iteration — Step 8 ka repeat loop disguise mein.
Bisection method — safe fallback jab Step 10 misfire kare.
Roots of polynomials — jahan yeh sab use hota hai.