Implementing root-finding from scratch — Newton-Raphson, bisection: Worked examples

5.4.25 · D3 · Coding › Scientific Computing (Python) › Implementing root-finding from scratch — Newton-Raphson, bis

Yeh page self-contained hai: iske liye jo bhi chahiye — guard tests, tangent formula, error argument — sab kuch yahan full mein restate kiya gaya hai pehle use karne se. Jab koi word naya ho, hum use zero se rebuild karte hain.

Scenario matrix

Root-finding mein kuch situations hain jo alag behave karti hain. Yahan poori list hai. Har row ka "Example" column us section ka naam hai jo usse work karta hai (labels C1…C10 neeche headings se exactly match karte hain).

#	Case class	Kya special hai	Kaun sa method	Example
C1	Clean bracket, opposite signs	textbook bisection	bisection	C1
C2	Sign convention: kaun sa half survive karta hai	quadrant/sign bookkeeping	bisection	C2
C3	Endpoint exactly ek root hai	degenerate: $f (a) = 0$	bisection	C3
C4	Newton, good start, $f^{'} \neq = 0$	quadratic bliss	Newton	C4
C5	Newton ek flat spot ke paas ( $f^{'} \to 0$ )	huge/undefined step	Newton	C5
C6	Newton jo diverge / cycle karta hai	limiting bad behaviour	Newton	C6
C7	Double root ( $f = f^{'} = 0$ saath mein)	convergence slow hokar linear	Newton	C7
C8	Real-world word problem	units, modelling	either	C8
C9	Exam twist: derivative nahi diya	robust engine choose karo	bisection	C9
C10	Invalid bracket ( $f (a) f (b) > 0$ ), aur midpoint root hit kare	error case + early exit	bisection	C10

Hum teeno se milenge, plus pure error cases bhi.

C1 — Clean bracket, textbook bisection

Worked example

$f (x) = x^{2} - 5$ ke root ke roop mein $5$ nikalo $[2, 3]$ par. Forecast: andaaza lagao ki $5$ kahan baitha hai, aur guess karo kitni halvings tumhe 3 decimals ke liye chahiye. (Answer $2.236$ ke paas; roughly $lo g_{2} (1/0.001) \approx 10$ steps.)

Step 1 — Bracket check karo. $f (2) = 4 - 5 = - 1$ , $f (3) = 9 - 5 = 4$ . Yeh step kyun? Bisection tabhi legal hai jab $f (a) f (b) < 0$ ; yahan $(- 1) (4) = - 4 < 0$ ✓ — ek sign change, toh Intermediate Value Theorem se ek root andar trapped hai.

Step 2 — Midpoint. $m = \frac{2 + 3}{2} = 2.5$ , $f (2.5) = 6.25 - 5 = 1.25 > 0$ . Kyun? Midpoint uncertainty ko exactly half mein split karta hai — yahi ek move hai jo guarantee karta hai ki har step ek bit buy kare (uncertainty ki ek halving, jaise upar define kiya).

Step 3 — Woh half rakhho jo abhi bhi sign change kare. $f (2) = - 1 < 0$ aur $f (2.5) = + 1.25 > 0$ : sign flip left half mein hai. $[2, 2.5]$ rakhho. Kyun? Root wahan rehta hai jahan endpoints abhi bhi sign mein disagree karte hain.

Step 4 — Repeat karo. $m = 2.25$ , $f = 0.0625 > 0$ → $[2, 2.25]$ rakhho. $m = 2.125$ , $f = - 0.484 < 0$ → $[2.125, 2.25]$ rakhho. Kyun har repeat help karta hai? Har step ek endpoint ko midpoint se replace karta hai, toh interval width $1 \to 0.5 \to 0.25 \to 0.125 = \frac{1}{2 ^{n}}$ jaati hai — har baar exactly half, har step ek bit of precision add karke (linear convergence with ratio $\frac{1}{2}$ ). Figure

ise coloured bars ki stack ke roop mein dikhata hai: top bar

[2, 3]

hai; neeche har bar ek sub-interval hai jo uske upar wale ke andar baitha hai aur half length ka hai (

[2, 2.5]

, phir

[2, 2.25]

, phir

[2.125, 2.25]

), aur har bar

x = 5 \approx 2.236

par red dashed vertical line ko straddle karta hai — toh tum dekh sakte ho ki root trapped rehta hai jabki bars usp par close in karte hain.

Verify karo: true root $5 = 2.2360679 \dots$ hai; 4 steps ke baad hamara bracket $[2.125, 2.25]$ indeed ise contain karta hai, width $0.125 = \frac{1}{2 ^{3}}$ . ✓

C2 — Sign bookkeeping: kaun sa half survive karta hai

Worked example

$f (x) = x^{3} - x - 2$ on $[1, 2]$ : stored sign ko carefully track karo. Forecast: cubics wiggle kar sakte hain. Kya yeh bracket usi root ki taraf point karta rehega? (Haan — yahan sirf ek real root hai, $1.52$ ke paas.)

Step 1 — Bracket. $f (1) = 1 - 1 - 2 = - 2$ , $f (2) = 8 - 2 - 2 = 4$ . Product $- 8 < 0$ ✓. $f_{a} = - 2$ store karo (negative).

Step 2 — Midpoint. $m = 1.5$ , $f (1.5) = 3.375 - 1.5 - 2 = - 0.125$ . $f_{a}$ store karke $f_{m}$ se compare kyun karte hain? Stored $f_{a} = - 2$ ko $f_{m} = - 0.125$ ke against compare karo: same sign (dono negative), toh crossing left half mein nahi hai. $f_{a}$ store karna (har loop mein $f (a)$ recompute karne ki bajaye) function calls bhi bachata hai.

"Mixing conventions" concretely kya hota hai. Do consistent tarike hain decide karne ke liye ki kaun sa half rakhen: (A) "woh half rakho jinke endpoints mein sign change ho" — yaani left half $[a, m]$ rakho jab $f_{a} \cdot f_{m} < 0$ ; ya (B) mirror test "right half $[m, b]$ rakho jab $f_{m} \cdot f_{b} < 0$ ". Har ek apne aap mein correct hai. Bug yeh hai ki decide karne ke liye test (A) use karo lekin endpoint update karo jaise test (B) use kiya ho — e.g. conclude karo "sign change left mein hai" phir bhi $a$ ko $m$ tak upar move karo $b$ ko $m$ tak neeche move karne ki bajaye. Woh single mismatch us half ko rakhta hai jo root contain nahi karta, aur agla iteration silently galat jagah converge karta hai (ya raise hit karta hai jab false bracket apna sign change kho deta hai). Safe rule: ek stored value ( $f_{a}$ ) choose karo, use $f_{m}$ se compare karo, aur matching-sign endpoint update karo — ek convention, start to finish.

Step 3 — Left endpoint upar move karo. Kyunki $f_{a} \cdot f_{m} > 0$ , root $[1.5, 2]$ mein hai. $a = 1.5, f_{a} = - 0.125$ set karo. Kyun? Hum us endpoint ko update karte hain jiska sign match kiya ( $a$ , kyunki $f_{a}$ aur $f_{m}$ agree karte hain), ek valid bracket preserve karte hue.

Step 4 — Continue karo. $m = 1.75$ , $f = 5.359 - 1.75 - 2 = 1.609 > 0$ → ab $f_{a} = - 0.125$ aur $f_{m} = + 1.609$ disagree karte hain → $[1.5, 1.75]$ rakhho.

Verify karo: $x^{3} - x - 2$ ka real root $1.52137 \dots$ hai, jo $[1.5, 1.75]$ mein hai ✓. Stored-sign convention ne kabhi use nahi khoya.

C3 — Degenerate: ek endpoint already root hai

Worked example

$f (x) = x^{2} - 4$ on $[2, 5]$ — kisine tumhe ek lucky endpoint diya. Forecast: $f (2) = 0$ . Kya standard bracket test chalega bhi? (Nahi! $f (2) f (5) = 0$ , $< 0$ nahi.)

Step 1 — Guard tests (yahan full mein state kiye gaye hain). Koi bhi correct bisection teen guards se open hota hai, is order mein:

fa, fb = f(a), f(b)
if fa == 0: return a       # endpoint already an exact root
if fb == 0: return b       # other endpoint already an exact root
if fa * fb > 0:            # STRICTLY > 0, not >= 0
    raise ValueError("f(a) and f(b) must have opposite signs")

Reject test > 0 kyun hai na ki >= 0? Zero ke barabar product ka matlab hai ki ek endpoint ek root hai — yeh koi error nahi, yeh ek jeet hai. Toh > 0 sirf truly no-sign-change case ko reject karta hai, jabki uske upar ke do == 0 checks pehle exact-root endpoints ko pakad lete hain. Yeh exact order kyun? == 0 checks product test se pehle aane chahiye, taaki zero endpoint return ho sake na ki fall through kare.

Step 2 — Humari inputs trace karo. $f (2) = 0$ , toh pehla guard if fa == 0 fire karta hai aur a return karta hai. Order kyun matter karta hai: agar hum fa*fb tak pahunchte, toh $0 \cdot 21 = 0$ milta, jo > 0 nahi hai, toh koi error raise nahi hota — lekin loop phir run karta aur exact zero discard kar sakta tha. fa == 0 pehle check karna hi humein bachata hai.

Step 3 — Return karo. Hum $x^{*} = 2$ immediately return karte hain, zero iterations.

Verify karo: $f (2) = 2^{2} - 4 = 0$ ✓ exact. $[2, 5]$ par endpoint check skip karna evaluations waste karta ya root kho deta.

C4 — Newton, good start, quadratic bliss

Worked example Newton on

$f (x) = x^{2} - 2$ , $f^{'} (x) = 2 x$ , $x_{0} = 1.5$ se. Forecast: har step ke baad correct digits count karo — kya woh roughly double hote hain? (Haan: yahi quadratic convergence ki pehchaan hai.)

Step 1 — Update (yahan derive kiya gaya, borrowed nahi). $x_{n}$ par $f$ ki tangent line $y = f (x_{n}) + f^{'} (x_{n}) (x - x_{n})$ hai. $y = 0$ set karke $x$ ke liye solve karne par Newton step milta hai $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$ — us tangent ka x-intercept. Is $f$ ke liye yeh Babylonian rule $x_{n + 1} = \frac{x _{n}}{2} + \frac{1}{x _{n}}$ mein simplify hota hai. Yeh simplification kyun? $x - \frac{x ^{2} - 2}{2 x} = x - \frac{x}{2} + \frac{1}{x}$ .

Step 2 — Iterate karo. Figure

blue parabola

f (x) = x^{2} - 2

dikhata hai, aur har guess par ek yellow straight tangent line; agla guess padhne ke liye har tangent ko neeche follow karo jahan woh horizontal axis cross karta hai (ek red dot).

$x_{1} = \frac{1.5}{2} + \frac{1}{1.5} = 0.75 + 0.6667 = 1.41667$ (3 good digits).
$x_{2} = \frac{1.41667}{2} + \frac{1}{1.41667} = 0.708333 + 0.705882 = 1.414216$ (6 digits).

Step 3 — Itni tezi kyun (second derivative explicitly ke saath). Error $e_{n} = x_{n} - x^{*}$ likho jahan $x^{*} = 2$ . Convergence rate second derivative $f^{''}$ par depend karta hai, jo measure karta hai ki slope khud kaise change hoti hai — curve ka bending. Yahan $f (x) = x^{2} - 2$ , toh $f^{'} (x) = 2 x$ aur $f^{''} (x) = 2$ (ek constant: parabola har jagah same tarah bend karta hai). Standard error bound hai $e_{n + 1} \approx \frac{f ^{''} ( x ^{*} )}{2 f ^{'} ( x ^{*} )} e_{n}^{2} .$ $f^{''} (x^{*}) = 2$ aur $f^{'} (x^{*}) = 22$ plug in karo: constant hai $\frac{2}{2 \cdot 2 2} = \frac{1}{2 2} \approx 0.354$ . Kyunki $f^{'} (x^{*}) = 22 \neq = 0$ , yeh bound hold karta hai aur error $e_{n}^{2}$ ke proportional hai — har step mein square hota hai, toh correct digits ki sankhya roughly double hoti hai.

Verify karo: $2 = 1.41421356 \dots$ ; $x_{2} = 1.414216$ 5 places tak match karta hai ✓. Digit count $3 \to 6$ gaya: doubling confirmed. Constant check: $e_{1} = 1.5 - 2 \approx 0.0858$ , $e_{2} = x_{2} - 2 \approx 0.0000024$ , aur $e_{2} / e_{1}^{2} \approx 0.33$ — predicted $0.354$ ke paas. ✓

C5 — Newton ek flat spot ke paas ( $f^{'} \to 0$ )

Worked example Newton on

$f (x) = x^{2} - 2$ lekin $x_{0} = 0.001$ se start (slope ki almost flat bottom-ish region). Forecast: slope $f^{'} (0.001) = 0.002$ tiny hai. Kya pehla step chhota hoga ya gigantic? (Gigantic — near-zero slope se divide karna tumhe dur faar le jaata hai.)

Step 1 — Step compute karo. $f (0.001) = 0.00 1^{2} - 2 = - 1.999999$ , $f^{'} (0.001) = 2 (0.001) = 0.002$ . Step $= \frac{- 1.999999}{0.002} = - 999.9995$ . Itna bada kyun? Tangent almost horizontal hai, toh uska x-intercept bahut dur hai — ek horizontal line axis ko infinity par milti hai.

Step 2 — Jump karo. $x_{1} = 0.001 - (- 999.9995) = 1000.0005$ . Dekho

: blue curve near-zero-slope point par ek shallow yellow tangent carry karta hai, aur ek red arrow agla guess axis ke kaafi right mein flung dikhata hai.

Step 3 — Recovery. $x_{1} = 1000$ se slope ab huge aur honest hai, toh Newton eventually wapas aata hai — lekin usne iterations waste kiye. Agar $x_{0}$ exactly $0$ hota, toh $f^{'} (0) = 0$ aur code ZeroDivisionError raise karta — ek flat tangent kahin nahi point karta.

Verify karo: step magnitude $∣ - 999.9995 ∣ = 999.9995$ ✓, aur $x_{1} = 1000.0005$ ✓. Lesson: ek flat spot ke paas step size blow up hoti hai, C4 mein tiny confident steps ke bilkul opposite.

C6 — Newton jo diverge / cycle karta hai (limiting disaster)

Worked example Newton on

$f (x) = x^{1/3}$ (cube root), root $x^{*} = 0$ ke alawa kahin se bhi start karo. Forecast: ek hi root $0$ hai. Kya Newton uski taraf chalega ya door? (Door — yeh geometrically diverge karta hai.)

Step 1 — Derivative. $f (x) = x^{1/3}$ , toh $f^{'} (x) = \frac{1}{3} x^{- 2/3}$ . $f^{'}$ kyun check karein? Newton isse divide karta hai; yahan yeh galat direction mein grow karta hai.

Step 2 — Update simplify hota hai. $x_{n + 1} = x_{n} - \frac{x _{n}^{1/3}}{\frac{1}{3} x _{n}^{- 2/3}} = x_{n} - 3 x_{n}^{1/3} x_{n}^{2/3} = x_{n} - 3 x_{n} = - 2 x_{n}$ . Yeh catastrophic kyun hai? Har step $x$ ko $- 2$ se multiply karta hai: sign flip karta hai aur root se distance double karta hai.

Step 3 — $x_{0} = 1$ se iterate karo. $x_{1} = - 2$ , $x_{2} = 4$ , $x_{3} = - 8$ , … Dekho

: guess dots (red) axis par baithte hain aur yellow curved arrows har guess ko next tak hop karte hain —

1 \to - 2 \to 4 \to - 8

— outward left, right, farther left, kabhi

0

par green root ki taraf nahi.

Verify karo: $x_{n + 1} = - 2 x_{n}$ se $∣ x_{n} ∣ = 2^{n}$ milta hai, jo $\to \infty$ ✓. $[- 1, 1]$ par bisection ( $f (- 1) = - 1 < 0$ , $f (1) = 1 > 0$ ke saath) safely $0$ ko trap karta — eternal robustness-vs-speed trade-off.

C7 — Double root: quadratic convergence break down ho jaati hai

Worked example Newton on

$f (x) = (x - 3)^{2}$ , multiplicity two ka root, $x_{0} = 5$ se. Forecast: $x^{*} = 3$ par, $f (3) = 0$ aur $f^{'} (3) = 0$ dono. Kya Newton abhi bhi digits double karta hai? (Nahi — double root ise linear tak demote karta hai, sirf error half karte hue.)

Step 1 — Update. $f (x) = (x - 3)^{2}$ , $f^{'} (x) = 2 (x - 3)$ . Toh $x_{n + 1} = x_{n} - \frac{( x _{n} - 3 ) ^{2}}{2 ( x _{n} - 3 )} = x_{n} - \frac{x _{n} - 3}{2} = \frac{x _{n} + 3}{2}$ . Yeh kyun matter karta hai? Error obey karta hai $e_{n + 1} = \frac{1}{2} e_{n}$ — linear, quadratic nahi.

Step 2 — Iterate karo. $x_{0} = 5$ (error 2) → $x_{1} = 4$ (error 1) → $x_{2} = 3.5$ (error 0.5) → $x_{3} = 3.25$ (error 0.25). Har step error ko exactly half karta hai, bisection ki tarah.

Step 3 — Slowdown kyun? C4 se quadratic bound $e_{n + 1} \approx \frac{f ^{''} ( x ^{*} )}{2 f ^{'} ( x ^{*} )} e_{n}^{2}$ assume karta hai ki $f^{'} (x^{*}) \neq = 0$ . Double root par $f^{'} (x^{*}) = 0$ , woh formula zero se divide karta hai aur invalid hai; true rate linear tak collapse ho jaati hai. (Multiplicity $m$ ke liye, $f$ aur uske pehle $m - 1$ derivatives wahan vanish karte hain, aur linear rate $1 - \frac{1}{m}$ ban jaati hai.)

Verify karo: $x_{1} = \frac{5 + 3}{2} = 4$ ✓, $x_{2} = \frac{4 + 3}{2} = 3.5$ ✓, $x_{3} = 3.25$ ✓. Errors $2, 1, 0.5, 0.25$ — ratio $\frac{1}{2}$ , linear ✓.

C8 — Real-world word problem (units matter karte hain)

Worked example Ek projectile ki height

$h (t) = 20 t - 4.9 t^{2}$ metres hai, $t$ seconds mein. Yeh ground se kab takkaraata hai ( $h = 0$ , launch ke alawa)? Forecast: launch $t = 0$ hai; landing kuch seconds baad honi chahiye (guess near $4$ s).

Step 1 — Root problem set up karo. Hum $h (t) = 0$ chahte hain, yaani $t (20 - 4.9 t) = 0$ . Factor kyun? Ek root trivial $t = 0$ hai; physical landing doosre factor ka root hai.

Step 2 — Landing bracket karo. $h (3) = 60 - 44.1 = 15.9 > 0$ , $h (5) = 100 - 122.5 = - 22.5 < 0$ : sign change on $[3, 5]$ ✓. Yahan bisection kyun? Ek clean bracket exist karta hai aur hum correct units ke saath guaranteed answer chahte hain.

Step 3 — Squeeze karo.

$m = 4$ : $h (4) = 80 - 78.4 = 1.6 > 0$ . $[4, 5]$ kyun rakhte hain? $h (4) > 0$ aur $h (5) < 0$ sign mein abhi bhi disagree karte hain, toh crossing right half mein hai.
$m = 4.5$ : $h (4.5) = 90 - 99.225 = - 9.225 < 0$ . $[4, 4.5]$ kyun rakhte hain? Ab $h (4) > 0$ aur $h (4.5) < 0$ disagree karte hain, toh root left half mein hai. $t^{*} = 20/4.9$ ki taraf converge ho raha hai.

Verify karo (algebra + units): $20 - 4.9 t = 0$ solve karne par $t^{*} = 20/4.9 = 4.0816 \dots$ seconds milta hai, jo $[4, 4.5]$ mein hai ✓. Units: $[m\cdots^{- 1}] / [m\cdots^{- 2}] = [s]$ ✓ — answer ek time hai, jaisa hona chahiye.

C9 — Exam twist: derivative supply nahi kiya gaya

Worked example "

$f (x) = e^{x} + x$ ka root $[- 1, 0]$ par nikalo. Tumhe $f^{'}$ nahi diya gaya." Kaun sa engine, aur kyun? Forecast: Newton ko $f^{'}$ chahiye; kya tum differentiate kiye bina kaam kar sakte ho? (Haan — bisection use karo, ya derivative-free Secant method.)

Step 1 — Constraint padho. No $f^{'}$ ⇒ Newton off the table hai jab tak tum haath se differentiate nahi karte. Kyun? Newton update $f^{'} (x_{n})$ se divide karta hai; iske bina divide karne ke liye kuch nahi hai.

Step 2 — Bracket karo. $f (- 1) = e^{- 1} - 1 = 0.3679 - 1 = - 0.6321 < 0$ , $f (0) = 1 + 0 = 1 > 0$ : sign change ✓. Yeh kaafi kyun hai? Bisection ko sirf $f$ values aur ek sign flip chahiye — koi calculus nahi.

Step 3 — Squeeze karo. $m = - 0.5$ : $f = e^{- 0.5} - 0.5 = 0.6065 - 0.5 = 0.1065 > 0$ → $[- 1, - 0.5]$ rakhho. $m = - 0.75$ : $f = 0.4724 - 0.75 = - 0.2776 < 0$ → $[- 0.75, - 0.5]$ rakhho. $- 0.567$ ke paas converge ho raha hai.

Verify karo: $e^{x} + x = 0$ ka true root $x^{*} = - 0.567143 \dots$ hai (negative Omega constant), jo $[- 0.75, - 0.5]$ mein hai ✓. Exam-safe choice: bisection, kyunki isme zero derivatives chahiye.

C10 — Invalid bracket, aur ek midpoint jo root par land kare

Worked example Bisection loop ke do edge behaviours: (a)

$f (a) f (b) > 0$ "bracket" ke roop mein pass in kiya gaya, aur (b) ek midpoint jo exactly ek root hai, jiske liye early exit if fm == 0: return m chahiye. Forecast: har case mein code ko kya karna chahiye? (a) error ke saath refuse karo; (b) instantly stop karo aur midpoint return karo.

Step 1 — Invalid bracket (a). $f (x) = x^{2} - 2$ on $[2, 3]$ lo. $f (2) = 2 > 0$ , $f (3) = 7 > 0$ : product $14 > 0$ . Refuse kyun? Sign change nahi hona matlab Intermediate Value Theorem andar root ki koi guarantee nahi deta — guard if fa*fb > 0: raise ValueError (C3 mein full mein stated) fire karta hai. Continue karna meaningless hoga: bisection ek aisi crossing invent nahi kar sakta jo bracketed nahi hai.

Step 2 — Midpoint root hit kare (b). $f (x) = x^{2} - 4$ on $[1, 3]$ lo. $f (1) = - 3 < 0$ , $f (3) = 5 > 0$ : valid bracket. Midpoint hai $m = \frac{1 + 3}{2} = 2$ , aur $f (2) = 0$ exactly. Early exit kyun? Loop body guard carry karta hai if fm == 0 or (b - a)/2 < tol: return m. fm == 0 test ke bina, code fa*fm compare karta — lekin $f_{a} \cdot 0 = 0$ na <0 hai na handled hai, toh woh else branch mein fall karta aur exact root se door walk karta. Early exit is lucky bullseye ko catch karta hai.

Step 3 — Contrast karo. Case (a) koi jawab nahi deta (ek error — input illegal tha); case (b) ek step mein exact answer deta hai. Dono ko do tiny guards handle karte hain; yeh example precisely dikhata hai ki har ek kab fire karta hai.

Verify karo: (a) $f (2) f (3) = 2 \cdot 7 = 14 > 0$ ✓ toh raise trigger hota hai. (b) $f (2) = 2^{2} - 4 = 0$ ✓ exact, toh if fm == 0: return 2 fire karta hai — aur $[1, 3]$ par $x^{2} - 4$ ka true root indeed $2$ hai. ✓

Recall Matrix par quick self-test

Kaun sa cell Newton ki quadratic convergence ko perfect start ke bawajood break karta hai? ::: C7 — ek double root, kyunki $f^{'} (x^{*}) = 0$ ise linear tak demote karta hai. $x_{0} = 0.001$ (C5) se start karna Newton ko itna dur kyun bhejta hai? ::: Slope $f^{'}$ near zero hai, toh tangent almost flat hai aur uska x-intercept enormous hai — ek giant step. $f (x) = x^{1/3}$ (C6) par, Newton update kya reduce hota hai? ::: $x_{n + 1} = - 2 x_{n}$ , har step mein distance double karke aur sign flip karke — pure divergence. Koi derivative na ho (C9), do safe options batao. ::: Bisection ya secant method — dono ko sirf $f$ values chahiye. C10 mein bisection edge cases ko kaun se do guards handle karte hain? ::: if fa*fb > 0: raise (valid bracket nahi) aur if fm == 0: return m (midpoint exactly root hai). Bisection ke liye "ek bit of precision" ka matlab kya hai? ::: Uncertainty interval ki ek halving — ek single yes/no answer ki bare mein kaun sa half root hide karta hai. Multiplicity $m$ ke root ke liye, wahan kaun se derivatives vanish karte hain? ::: $f = f^{'} = \dots = f^{(m - 1)} = 0$ lekin $f^{(m)} \neq = 0$ .

Yeh bhi dekho parent note, Fixed-point iteration iteration ka ek aur view ke liye, scipy.optimize production tooling ke liye, aur Floating point arithmetic is liye ki f(m)==0 almost kabhi exactly kyun nahi hota.