Secant method: Question bank — Hinglish deep dive

4.8.9 · D5 · Maths › Numerical Methods › Secant method

Ground rules — is page mein use hone wale symbols aur assumptions

Traps se pehle, har cheez pin down karo taaki koi bhi symbol "free mein" na aaye.

Definition The players

$f$ — woh function jiska root hum dhundh rahe hain. Assumptions: $f$ root ke paas continuous hai, wahan twice differentiable hai (neeche ke Taylor argument ke liye zaroori hai), aur root simple hai (yani $f (r) = 0$ lekin $f^{'} (r) \neq = 0$ ). Is page pe har failure mode ek aisa case hai jahan in assumptions mein se koi ek tod diya gaya hai.
$r$ — true root, exact $x$ jahan $f (r) = 0$ . Yeh woh target hai jahan hum kabhi poori tarah land nahi karte.
$x_{n}$ — humara $n$ -th guess. $x_{0}, x_{1}$ do starting guesses hain.
$e_{n} = x_{n} - r$ — error: $n$ -th guess true root se kitna dur hai. Chhota $∣ e_{n} ∣$ = accha guess.
$Δ x = x_{n} - x_{n - 1}$ (do latest guesses ke beech ka "run") aur $Δ f = f (x_{n}) - f (x_{n - 1})$ (unki heights ke beech ka "rise"). Chord slope hai $Δ f /Δ x$ .
$K, C$ — simple constants (fixed numbers jo badhte nahi hain), "roughly is size ka" kehne ke liye use hote hain neeche ke error laws mein.

Golden ratio kyun? — error recurrence derive karna

Intuition

$e_{n + 1} \approx K e_{n} e_{n - 1}$ kahan se aata hai Secant step ko errors ke terms mein likho. $f (x) \approx f^{'} (r) e + \frac{1}{2} f^{''} (r) e^{2}$ use karte hue (root $r$ ke around $f$ ki do-term Taylor expansion, valid hai kyunki humne $f$ twice differentiable assume ki thi), $x_{n - 1}, x_{n}$ ko boxed form mein daalo aur simplify karo. Errors mein jo bhi linear hai woh cancel ho jaata hai kyunki chord $f$ ke straight part ko exactly fit karta hai; jo bachta hai woh hai curvature term: $e_{n + 1} \approx \frac{f ^{''} ( r )}{2 f ^{'} ( r )} e_{n} e_{n - 1} = K e_{n} e_{n - 1}, K = \frac{f ^{''} ( r )}{2 f ^{'} ( r )} .$ Padho aise: naya error purane dono errors ka product hai, ek curvature constant $K$ se scale hota hua. Do chhote errors milke ek bahut chhota error banate hain — isliye Secant fast hai. (Dhyan do $K$ ke liye chahiye $f^{'} (r) \neq = 0$ , yani simple root — ek aur reason kyun humne yeh assume kiya.)

True or false — justify karo

Secant method ko derivative

f^{'} (x)

chahiye

False. Yeh slope ko chord slope

\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

se approximate karta hai, isliye yeh kabhi

f^{'} (x)

ko touch nahi karta — yahi Newton-Raphson method par iski poori selling point hai.

Secant method hamesha root ko Bisection method ki tarah bracketed rakhta hai

False. Yeh ek open method hai: naya iterate ek chord ka

x

-intercept hai aur

[x_{n - 1}, x_{n}]

ke bahar land kar sakta hai. Koi sign change enforce nahi hoti, isliye koi bracket guarantee nahi hoti.

Agar

x_{0}

aur

x_{1}

root ke aas-paas hain (opposite signs of

f

), toh convergence guaranteed hai

False. Straddling pehle step mein help karta hai lekin preserve nahi hota. Kyunki hum hamesha do latest points rakhte hain, dono ek hi side pe drift kar sakte hain aur method diverge ho sakta hai.

Do starting points sirf safety ke liye chahiye; technically ek bhi kaam kar sakta

False. Do points isliye chahiye taaki ek straight line define ho sake. Ek point aur bina derivative ke tum koi chord nahi kheench sakte — method undefined hai.

Secant method Newton ki tarah quadratically converge karta hai

False. Iski order golden ratio

p = \frac{1 + 5}{2} \approx 1.618

hai — superlinear hai lekin Newton ke

p = 2

se strictly neeche.

Per function evaluation, Secant Newton ko beat kar sakta hai

True. Humare cost model ke under (ek unit per

f

f^{'}

), Newton har step mein

2

units kharcha karta hai; Secant purani value reuse karta hai aur ek unit kharcha karta hai. Do Secant steps mein

1.61 8^{2} \approx 2.6 > 2

milta hai, isliye jab

f^{'}

costly ho toh Secant often jeetta hai.

"Weighted average" form

\frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )}

boxed form se alag method hai

False. Yeh same iteration hai, algebraically rearranged. Boxed subtraction form bas numerically safer hai (less catastrophic cancellation) code mein.

Regula Falsi (False Position) aur Secant same intercept formula use karte hain, isliye woh same behave karte hain

False. Same formula, different bookkeeping. Regula Falsi ek bracketing endpoint rakhta hai (bounded, lekin sirf linear); Secant do sabse recent points rakhta hai (order 1.618, no bracket).

Spot the error

x_{n + 1} = x_{n} - f (x_{n}) \frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

Fraction ulta hai. Hum

f

ko slope se divide karte hain, isliye hona chahiye

\frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )}

(

Δ x

over

Δ f

, yani run over rise). Jaise likha hai yeh slope se multiply karta hai aur blow up ho jaata hai.

x_{n + 1}

compute karne ke baad,

x_{n - 1}

x_{n + 1}

se replace karo aur

x_{n}

ko rakhho."

Window slide galat hai. Naya pair do latest points hona chahiye:

x_{n}

ban jaata hai

x_{n - 1}

aur

x_{n + 1}

ban jaata hai

x_{n}

. Purane

x_{n}

x_{n + 1}

ke saath rakhne se superlinear order khatam ho jaata hai.

"Main har iteration mein

f (x_{n - 1})

aur

f (x_{n})

fresh recompute karta hoon."

Wasteful aur self-defeating hai. Efficiency advantage aata hai already-computed

f (x_{n})

ko next step ke

f (x_{n - 1})

ke roop mein reuse karne se. Recompute karna one-evaluation-per-step benefit ko uda deta hai.

"Denominator

f (x_{n}) - f (x_{n - 1})

kabhi zero nahi ho sakta, isliye guarding ki zaroorat nahi."

Dangerous hai. Flat region ke paas — ya convergence ke waqt jab dono

f

values tiny aur almost equal hon — denominator

Δ f \to 0

ho jaata hai aur

x_{n + 1}

explode kar deta hai. Hamesha tolerance aur iteration cap se guard karo.

"Kyunki Secant Newton ko approximate karta hai, iski error bhi

e_{n + 1} \approx C e_{n}^{2}

satisfy karti hai."

Nahi. Secant ki error recurrence hai

e_{n + 1} \approx K e_{n} e_{n - 1}

(yeh do points use karta hai, isliye dono purani errors aati hain), jo

p = 1.618

deta hai,

2

nahi. Mixed

e_{n} e_{n - 1}

product exactly wahi reason hai kyun order linear aur quadratic ke beech hai.

x_{n + 1}

ek weighted average hai, isliye yeh hamesha

x_{n - 1}

aur

x_{n}

ke beech hota hai."

Galat conclusion. "Weights"

f (x_{n})

aur

f (x_{n - 1})

same sign ke ho sakte hain, ek weight effectively negative ban jaata hai — isliye yeh convex average nahi hai aur

x_{n + 1}

interval ke bahar ja sakta hai. Sirf tab jab signs differ karein yeh genuinely convex blend hota hai.

Why questions

Secant method ko do starting guesses kyun chahiye jab Newton ko sirf ek chahiye?

Newton ek point ko exact slope

f^{'} (x_{n})

ke saath pair karta hai; Secant ke paas koi slope nahi hai, isliye woh chord define karne ke liye doosra point substitute karta hai. Yeh derivative ko extra evaluation ke badle mein trade karta hai.

Order exactly golden ratio kyun hai koi arbitrary number nahi?

Error recurrence

e_{n + 1} \approx K e_{n} e_{n - 1}

force karta hai

p = 1 + \frac{1}{p}

, yani

p^{2} - p - 1 = 0

, jiska positive root

\frac{1 + 5}{2}

hai. Do-point product structure mein baked in hai — upar derivation section dekho.

Secant woh bracketing kyun kho deta hai jo Regula Falsi (False Position) rakhta hai?

Secant hamesha sign ki parwah kiye bina oldest point discard karta hai; Regula Falsi deliberately us endpoint ko discard karta hai jiska sign

f (x_{n + 1})

ke sign se same ho, sign change preserve karta hua. Kya rakhna hai yahi woh choice hai jo sab kuch alag karta hai.

Secant sometimes lower order hone ke bawajood Newton se zyada efficient kyun ho sakta hai?

Efficiency measure hoti hai per cost unit (ek

f

ya ek

f^{'}

) se, per step se nahi. Secant ka cost hai ek unit per step;

1.61 8^{2} \approx 2.6 > 2

do steps mein matlab hai yeh Newton ko outrun kar sakta hai jab

f^{'}

expensive ya unavailable ho.

Boxed form cancellation kyun avoid karta hai jo "weighted average" form mein hoti hai?

Boxed form correction

f (x_{n}) \cdot Δ x /Δ f

compute karta hai aur ise

x_{n}

se subtract karta hai; convergence ke paas yeh correction tiny hoti hai aur

x_{n}

intact rehta hai. Average form do large nearly-equal products

x_{n - 1} f (x_{n})

aur

x_{n} f (x_{n - 1})

subtract karta hai, significant digits kho deta hai.

x_{n - 1} \to x_{n}

karne se Secant formula Newton's kyun ban jaata hai?

Chord slope

Δ f /Δ x = \frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}

exactly woh difference quotient hai jiska limit

f^{'} (x_{n})

define karta hai. Jab do points merge hote hain, chord tangent ban jaata hai — Finite differences dekho.

Edge cases

Agar

f (x_{0}) = f (x_{1})

exactly ho toh kya hoga?

Chord horizontal hai (slope

0

, isliye

Δ f = 0

), aur iski koi

x

-intercept nahi hai — method pehle hi step mein break ho jaata hai. Tumhe alag starting values pick karni padenge.

Agar

x_{0} = x_{1}

ho toh?

Run

Δ x = x_{n} - x_{n - 1}

aur rise

Δ f = f (x_{n}) - f (x_{n - 1})

dono zero hain,

\frac{0}{0}

milta hai. Method undefined hai: chord banane ke liye do alag points zaroori hain.

Agar

f

ke do guesses ke beech local extremum (flat spot) ho toh?

Chord slope

Δ f /Δ x

tiny ho jaata hai, correction

f \cdot Δ x /Δ f

huge ho jaata hai, aur next iterate bahut door ja sakta hai — possibly kisi aur root pe ya divergence ki taraf. Isliye flat regions hazardous hote hain.

Agar function ke multiple roots hon aur starting pair do roots ke beech ho toh?

Chord ka intercept do heights par depend karta hai, is baat par nahi ki tum kaun sa root chahte ho; Secant us root ki taraf converge ho sakta hai jo tumne intend nahi kiya tha, ya oscillate kar sakta hai. Iski koi mechanism nahi hai jo kisi particular root ko prefer kare.

Double (repeated) root par convergence ka kya hoga?

Yahan

f^{'} (r) = 0

hai, isliye humari "simple root" assumption fail hoti hai aur constant

K = f^{''} (r) /2 f^{'} (r)

blow up ho jaata hai. Newton ki tarah, Secant bhi dramatically slow ho jaata hai; superlinear

1.618

order linear ki taraf degrade ho jaata hai aur bahut zyada iterations chahiye hote hain.

Agar do starting guesses root se bahut dur hon toh?

Taylor picture sirf

r

ke paas hold karta hai, isliye dur se chord ek poor model hai aur pehla intercept wildly overshoot kar sakta hai. Open method hone ke nate koi safety net nahi hai — yeh diverge ho sakta hai, unlike Bisection method jo hamesha progress karta hai.

Convergence ke bilkul waqt expected behaviour kya hai, aur yeh trap kyun hai?

Jaise-jaise

x_{n} \to r

f (x_{n})

aur

f (x_{n - 1})

dono

0

ki taraf shrink hote hain, isliye

Δ f = f (x_{n}) - f (x_{n - 1}) \to 0

. Method ki yahi success ise numerically fragile banati hai — step-size tolerance

∣ x_{n + 1} - x_{n} ∣

par ruko,

f = 0

ka hamesha peecha mat karo.

Recall One-line self-test

Har answer dhako, poora page upar se neeche run karo. Agar tum har cheez ko ek poore sentence mein justify kar sako — "haan"/"nahi" se nahi — toh concept tumhara hai. Weak spots almost hamesha Edge cases aur bracketing items mein cluster karte hain; wahan Order of convergence aur Regula Falsi (False Position) revisit karo.