Secant method: Exercises — Hinglish deep dive

4.8.9 · D4 · Maths › Numerical Methods › Secant method

Woh ek formula jis par hum poore waqt depend karte hain: $x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )} .$

Ise parent note ke [!mnemonic] ki tarah padho: "New = Now minus f-Now times Run-over-Rise."

Upar ki picture woh ek hi geometry hai jo tumhe chahiye: curve par do dots, unke through ek seedha ruler (secant line — do curve points ko jodne wali chord), aur woh jagah jahan woh ruler horizontal axis ko cross karta hai. Wahi crossing agla guess hai.

Level 1 — Recognition

L1.1

Secant iteration formula likhо aur ek sentence mein batao ki har step kaunsa geometric object banata hai.

Recall Solution

$x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )} .$ Har step do latest points $(x_{n - 1}, f (x_{n - 1}))$ aur $(x_{n}, f (x_{n}))$ ke through secant line (ek seedhi chord) banata hai, phir us line ka $x$ -intercept (jahan woh horizontal axis se milti hai) ko next guess $x_{n + 1}$ ke roop mein leta hai.

L1.2

True ya false, ek-line reasons ke saath: (a) Secant method ko $f^{'} (x)$ chahiye. (b) Isko do starting guesses chahiye. (c) Yeh hamesha root ko bracket mein rakhta hai jaise bisection.

Recall Solution

(a) False. Yeh derivative ki jagah chord slope $\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}}$ use karta hai — koi $f^{'}$ kabhi compute nahi hoti. (b) True. Do points chahiye hote hain ek line define karne ke liye. (c) False. Yeh ek open method hai: naya point $[x_{n - 1}, x_{n}]$ ke bahar jump kar sakta hai aur sign change preserve nahi hoti. Woh bracketing property Regula Falsi (False Position) ki hai.

L1.3

Newton ke step mein kaunsi quantity ko Secant method replace karta hai, aur kis cheez se?

Recall Solution

Newton exact slope $f^{'} (x_{n})$ use karta hai. Secant ise finite-difference slope se replace karta hai: $\frac{f ( x _{n} ) - f ( x _{n - 1} )}{x _{n} - x _{n - 1}},$ jo do known heights se bana $f^{'} (x_{n})$ ka finite-difference estimate hai. Exact-slope version ke liye Newton-Raphson method dekho.

Level 2 — Application

L2.1

$f (x) = x^{2} - 3$ , root $3 \approx 1.73205$ . $x_{0} = 1, x_{1} = 2$ lo. $x_{2}$ aur $x_{3}$ compute karo.

Recall Solution

$f (1) = 1 - 3 = - 2$ , $f (2) = 4 - 3 = 1$ . $x_{2} = 2 - 1 \cdot \frac{2 - 1}{1 - ( - 2 )} = 2 - \frac{1}{3} = 1.66667.$ Humne KYA kiya: do latest points ko boxed formula mein plug kiya. KYUN: unke through chord ka $x$ -intercept root ke liye humara best straight-line guess hai. KAISA DIKHTA HAI: ruler point $(2, 1)$ se axis ki taraf girta hai aur $x = 2$ ke baayein utar ta hai.

Ab $f (1.66667) = 1.6666 7^{2} - 3 = - 0.22222$ . $x_{3} = 1.66667 - (- 0.22222) \cdot \frac{1.66667 - 2}{- 0.22222 - 1} = 1.66667 - (- 0.22222) \cdot \frac{- 0.33333}{- 1.22222} .$ $= 1.66667 - (- 0.22222) (0.27273) = 1.66667 + 0.06061 = 1.72727.$ Toh $x_{2} = 1.66667$ , $x_{3} = 1.72727$ — pehle se teen sahi digits.

L2.2

$f (x) = x^{3} - x - 2$ , root $\approx 1.52138$ . $x_{0} = 1, x_{1} = 2$ lo. $x_{2}$ compute karo.

Recall Solution

$f (1) = 1 - 1 - 2 = - 2$ , $f (2) = 8 - 2 - 2 = 4$ . $x_{2} = 2 - 4 \cdot \frac{2 - 1}{4 - ( - 2 )} = 2 - 4 \cdot \frac{1}{6} = 2 - 0.66667 = 1.33333.$ Recipe se ek clean step — $x^{3} - x - 2$ ka koi derivative nahi chahiya.

L2.3

$f (x) = e^{x} - 3 x$ , root $\approx 0.61906$ . $x_{0} = 0, x_{1} = 1$ lo. $x_{2}$ compute karo.

Recall Solution

$f (0) = e^{0} - 0 = 1$ , $f (1) = e - 3 = 2.71828 - 3 = - 0.28172$ . $x_{2} = 1 - (- 0.28172) \cdot \frac{1 - 0}{- 0.28172 - 1} = 1 - (- 0.28172) \cdot \frac{1}{- 1.28172} .$ $= 1 - (- 0.28172) (- 0.78021) = 1 - 0.21980 = 0.78020.$ Method ko kabhi parwah nahi hui ki $e^{x} - 3 x$ ka koi closed-form root nahi hai.

Level 3 — Analysis

L3.1

Dikhao ki do formula forms agree karte hain. Shuru karo $x_{n + 1} = x_{n} - f (x_{n}) \frac{x _{n} - x _{n - 1}}{f ( x _{n} ) - f ( x _{n - 1} )},$ se aur weighted-average form derive karo: $x_{n + 1} = \frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )} .$

Recall Solution

Sab kuch common denominator $D = f (x_{n}) - f (x_{n - 1})$ par rakho: $x_{n + 1} = \frac{x _{n} D - f ( x _{n} ) ( x _{n} - x _{n - 1} )}{D} .$ Numerator expand karo: $x_{n} f (x_{n}) - x_{n} f (x_{n - 1}) - x_{n} f (x_{n}) + x_{n - 1} f (x_{n}) .$ $x_{n} f (x_{n})$ terms cancel ho jaate hain, bachta hai $x_{n - 1} f (x_{n}) - x_{n} f (x_{n - 1})$ . Isliye $x_{n + 1} = \frac{x _{n - 1} f ( x _{n} ) - x _{n} f ( x _{n - 1} )}{f ( x _{n} ) - f ( x _{n - 1} )} . ■$ KYUN boxed form prefer ki jaati hai: weighted form convergence ke paas do almost-equal large products subtract karta hai, precision lose hoti hai (cancellation). Boxed form sirf $x_{n}$ mein ek chhota correction add karta hai.

L3.2

Numerical blow-up. $f (x) = x^{2} - 2$ ke liye maano do consecutive iterates hain $x_{n - 1} = 1.41421$ aur $x_{n} = 1.41422$ . Numbers ke saath explain karo ki agla step dangerous kyun hai.

Recall Solution

$f (1.41421) = 1.4142 1^{2} - 2 = - 0.0000106$ , $f (1.41422) = 1.4142 2^{2} - 2 = + 0.0000177$ . Denominator $= f (x_{n}) - f (x_{n - 1}) = 0.0000177 - (- 0.0000106) = 0.0000283$ — bahut chhota. Near-zero denominator se divide karne par un almost-equal $f$ values ki koi bhi rounding error massively amplify ho jaati hai. Fix: tolerance ke saath guard rakho if abs(denominator) < eps: stop, saath mein ek max-iteration cap bhi.

L3.3

Open-method jump. $f (x) = x^{2} - 2$ ke liye $x_{0} = 0, x_{1} = 3$ ke saath $x_{2}$ compute karo aur check karo ki woh $[0, 3]$ ke andar rehta hai ya nahi. Comment karo.

Recall Solution

$f (0) = - 2$ , $f (3) = 7$ . $x_{2} = 3 - 7 \cdot \frac{3 - 0}{7 - ( - 2 )} = 3 - 7 \cdot \frac{3}{9} = 3 - 2.33333 = 0.66667.$ Yahan $x_{2} = 0.66667 \in [0, 3]$ , toh yeh step andar rehta hai. Lekin kisi cheez ne isse force nahi kiya: kyunki Secant signs ignore karta hai, ek buri tarah choose kiya hua pair ek $x_{2}$ produce kar sakta hai jo starting interval ke bahar ho — yeh ek open method ki pehchaan hai. Bracketing methods jaise Bisection method kabhi aisa nahi kar sakte.

Level 4 — Synthesis

L4.1

Convergence ka order derive karo. Error recurrence $e_{n + 1} \approx K e_{n} e_{n - 1}$ (jahan $e_{n} = x_{n} - r$ ) diye hue, dikhao ki order $p$ equation $p^{2} - p - 1 = 0$ solve karta hai aur $φ = \frac{1 + 5}{2}$ ke barabar hai.

Recall Solution

Order- $p$ convergence ki defining relation assume karo: $e_{n + 1} = A e_{n}^{p}$ kisi constant $A$ ke liye. Index ek neeche shift karo: $e_{n} = A e_{n - 1}^{p}$ , toh $e_{n - 1} = (e_{n} / A)^{1/ p}$ . Dono ko $e_{n + 1} = K e_{n} e_{n - 1}$ aur $e_{n + 1} = A e_{n}^{p}$ mein substitute karo: $A e_{n}^{p} = K e_{n} (e_{n} / A)^{1/ p} = K A^{- 1/ p} e_{n}^{1 + 1/ p} .$ Dono sides par $e_{n}$ ki powers match karo: $p = 1 + \frac{1}{p} ⟹ p^{2} = p + 1 ⟹ p^{2} - p - 1 = 0.$ Positive root: $p = \frac{1 + 5}{2} \approx 1.61803$ . $■$

L4.2

Evaluation ke hisaab se efficiency. Newton 2 evaluations/step karta hai aur order 2 hai. Secant 1 new evaluation/step karta hai aur order $1.618$ hai. Dikhao ki do Secant steps (Newton ke evaluation budget se match karte hue) ek Newton step se effective order mein aage nikalta hai.

Recall Solution

Effective order per pair of evaluations:

Newton: order $2$ ka ek step per 2 evaluations $\Rightarrow$ effective order $2$ .
Secant: do steps, har ek order $1.618$ , compose hokar $1.61 8^{2} = 2.618$ per 2 evaluations. Kyunki $2.618 > 2$ , Secant har function evaluation par zyada accuracy extract karta hai jab $f^{'}$ evaluation ka cost $f$ evaluation jaisa ho. Isliye Secant win karta hai jab derivative expensive ho.

L4.3

Ek hi problem par methods compare karo. $f (x) = x^{2} - 2$ , $x_{0} = 1, x_{1} = 2$ ke liye, pehle step par har method kaun si information use karta hai list karo: Secant, Newton, Regula Falsi, Bisection.

Recall Solution

Method	Step 1 par use ki gayi info	Bracket rakhta hai?
Secant	$f (1), f (2)$ — do heights	Nahi
Newton	$f (2), f^{'} (2) = 4$ — ek height + slope	Nahi
Regula Falsi	$f (1), f (2)$ + sign check	Haan (same-sign end replace karta hai)
Bisection	sirf $f (1), f (2)$ ke signs	Haan (interval half karta hai)
Secant aur Regula Falsi ek hi $x$ -intercept formula share karte hain; sirf bookkeeping alag hai (kya bracket enforce hota hai) — aur yehi reason hai ki ek order $1.618$ hai aur doosra sirf linear.

Level 5 — Mastery

L5.1

Ek tolerance tak full multi-step run. $f (x) = cos x - x = 0$ solve karo $x_{0} = 0.5, x_{1} = 1$ ke saath. Tab tak iterate karo jab tak $∣ x_{n + 1} - x_{n} ∣ < 1 0^{- 4}$ . Har iterate report karo. (Root $\approx 0.73909$ .)

Recall Solution

Saare angles radians mein. $f (0.5) = cos 0.5 - 0.5 = 0.87758 - 0.5 = 0.37758$ . $f (1) = cos 1 - 1 = 0.54030 - 1 = - 0.45970$ .

$x_{2}$ ke liye step: $x_{2} = 1 - (- 0.45970) \cdot \frac{1 - 0.5}{- 0.45970 - 0.37758} = 1 - (- 0.45970) \cdot \frac{0.5}{- 0.83728} .$ $= 1 - (- 0.45970) (- 0.59717) = 1 - 0.27452 = 0.72548.$

$x_{3}$ ke liye step: $f (0.72548) = cos (0.72548) - 0.72548 = 0.74829 - 0.72548 = 0.02281$ . $x_{3} = 0.72548 - 0.02281 \cdot \frac{0.72548 - 1}{0.02281 - ( - 0.45970 )} .$ $= 0.72548 - 0.02281 \cdot \frac{- 0.27452}{0.48251} = 0.72548 - 0.02281 (- 0.56895) = 0.72548 + 0.01298 = 0.73846.$

$x_{4}$ ke liye step: $f (0.73846) = cos (0.73846) - 0.73846 = 0.73951 - 0.73846 = 0.00105$ . $x_{4} = 0.73846 - 0.00105 \cdot \frac{0.73846 - 0.72548}{0.00105 - 0.02281} .$ $= 0.73846 - 0.00105 \cdot \frac{0.01298}{- 0.02176} = 0.73846 - 0.00105 (- 0.59651) = 0.73846 + 0.00063 = 0.73909.$

Ab $∣ x_{4} - x_{3} ∣ = ∣0.73909 - 0.73846∣ = 0.00063 > 1 0^{- 4}$ , toh ek aur: $f (0.73909) = cos (0.73909) - 0.73909 = 0.73909 - 0.73909 \approx 0.0000005$ . Agla iterate $1 0^{- 4}$ se kaafi kam move karta hai, toh hum $r \approx 0.73909$ ke saath ruk jaate hain.

Iterate list: $0.5, 1, 0.72548, 0.73846, 0.73909$ .

L5.2

Ek stopping rule design karo. Ek robust Secant routine ka pseudocode likho jo (i) iterations cap kare, (ii) denominator guard kare, (iii) step size par stop kare. Har guard explain karo.

Recall Solution

secant(f, x0, x1, tol, maxit):
    f0 = f(x0);  f1 = f(x1)
    for k in 1..maxit:
        denom = f1 - f0
        if abs(denom) < 1e-14:        # guard (ii): flat chord → blow-up
            stop "denominator too small"
        x2 = x1 - f1*(x1 - x0)/denom  # boxed form (least cancellation)
        if abs(x2 - x1) < tol:        # guard (iii): step size small enough
            return x2
        x0, f0 = x1, f1               # window slide karo...
        x1, f1 = x2, f(x2)            # ...f1 reuse karo, ek naya eval per step
    stop "no convergence in maxit"    # guard (i): iteration cap

Har guard kyun: (i) infinite loops rokta hai jab method diverge/oscillate kare; (ii) L3.2 se near-zero denominator ko inf produce karne se pehle pakad leta hai; (iii) step-size test sasta hai aur directly measure karta hai ki kitna progress baaki hai. Note karo loop mein ek naya $f$ -evaluation — yahi L4.2 ka efficiency edge hai.

L5.3

Convergence-rate sanity check. Parent ka Worked Example 1 ( $f = x^{2} - 2$ ) ki run use karte hue, errors the $0.081 \to 0.014 \to 0.00042$ . Verify karo ki yeh linear ki jagah superlinear (order $\approx 1.618$ ) behaviour ke consistent hain.

Recall Solution

Linear convergence ke liye, $e_{n + 1} / e_{n}$ roughly constant hota. Yahan: $\frac{0.014}{0.081} = 0.173, \frac{0.00042}{0.014} = 0.030.$ Ratio girta hai ( $0.173 \to 0.030$ ), matlab remove ki gayi error ka fraction badhta rehta hai — yeh superlinear hai, linear nahi. Order estimate ka cross-check teen errors se: $p \approx \frac{l n ( e _{n + 1} / e _{n} )}{l n ( e _{n} / e _{n - 1} )} = \frac{l n ( 0.030 )}{l n ( 0.173 )} = \frac{- 3.507}{- 1.755} \approx 1.998.$ Sirf teen noisy points aur rounded errors ke saath yeh $2$ ke paas land karta hai; ek lamba clean run $φ \approx 1.618$ par settle hota hai. Key qualitative fact — ratios shrink rahe hain, digits multiply ho rahe hain — confirm ho gaya.

Recall Ek-line self-audit

Kya tum memory se (a) boxed formula likh sakte ho, (b) bata sakte ho ki derivative kyun nahi chahiye, (c) order $φ \approx 1.618$ ka naam le sakte ho, aur (d) L5.2 ke teen guards list kar sakte ho? Agar chaaron ka jawab haan hai, toh Secant method tumhara D4 level par apna hai.