Cubic spline interpolation — natural, clamped: Exercises

4.8.12 · D4 · Maths › Numerical Methods › Cubic spline interpolation — natural, clamped

Poore note mein, hum parent se liya hua notation reuse karte hain:

$h_{i} = x_{i + 1} - x_{i}$ interval $i$ ki width hai (do neighbouring $x$ -values ke beech ka gap).
$M_{i} = S^{''} (x_{i})$ moment hai — node $x_{i}$ par spline ki second derivative (yaani bendiness).
$i = 1, \dots, n - 1$ ke liye interior recurrence yeh hai: $h_{i - 1} M_{i - 1} + 2 (h_{i - 1} + h_{i}) M_{i} + h_{i} M_{i + 1} = 6 (\frac{y _{i + 1} - y _{i}}{h _{i}} - \frac{y _{i} - y _{i - 1}}{h _{i - 1}}) .$

Hum do piece formulas yahan dobara print kar rahe hain jo baar baar use hoti hain, taaki kuch bhi assume na karna pade:

Saral shabdon mein bolo ki ek cubic spline $S (x)$ ke liye neeche diye gaye mein se har ek kya equal hota hai: (a) $S_{i} (x_{i})$ , (b) interior node $x_{i}$ par $S^{'}$ ki continuity, (c) natural boundary condition, (d) clamped boundary condition.

Recall Solution 1.1

(a) $S_{i} (x_{i}) = y_{i}$ : $[x_{i}, x_{i + 1}]$ par ka piece left data point se hoke guzarna chahiye — yeh interpolation hai. (b) $S_{i - 1}^{'} (x_{i}) = S_{i}^{'} (x_{i})$ : left piece se node $x_{i}$ mein aane wala slope, right piece par jaane wale slope ke barabar hona chahiye — koi kink nahi. (c) Natural: $M_{0} = 0$ aur $M_{n} = 0$ — donon ends par zero curvature (ruler relaxed/straight hai wahan). (d) Clamped: $S^{'} (x_{0}) = f_{0}^{'}$ aur $S^{'} (x_{n}) = f_{n}^{'}$ — end slopes given numbers par pin kiye gaye hain.

Problem 1.2 (L1)

$x_{0} = 0, x_{1} = 2, x_{2} = 5, x_{3} = 6$ par data ke liye, saare widths $h_{0}, h_{1}, h_{2}$ calculate karo. Kitne unknown moments $M_{i}$ hain, aur recurrence kitne interior equations deta hai?

Recall Solution 1.2

$h_{0} = x_{1} - x_{0} = 2$ , $h_{1} = x_{2} - x_{1} = 3$ , $h_{2} = x_{3} - x_{2} = 1$ . $n + 1 = 4$ moments hain: $M_{0}, M_{1}, M_{2}, M_{3}$ . Interior recurrence $i = 1, \dots, n - 1 = 1, 2$ par chalti hai, jo 2 interior equations deti hai. Hum 2 short hain (4 unknowns, 2 equations), isliye hume 2 boundary conditions chahiye — exactly natural ya clamped.

Level 2 — Application

Problem 2.1 (L2)

Data $(0, 1), (1, 3), (2, 2)$ equal spacing $h = 1$ ke saath. Natural spline banao: $M_{0}, M_{1}, M_{2}$ nikalo.

Recall Solution 2.1

Natural $\Rightarrow M_{0} = 0, M_{2} = 0$ . Sirf $M_{1}$ unknown hai. $i = 1$ par interior recurrence, $h_{0} = h_{1} = 1$ ke saath: $h_{0} M_{0} + 2 (h_{0} + h_{1}) M_{1} + h_{1} M_{2} = 6 (\frac{y _{2} - y _{1}}{h _{1}} - \frac{y _{1} - y _{0}}{h _{0}}) .$ $0 + 4 M_{1} + 0 = 6 ((2 - 3) - (3 - 1)) = 6 (- 1 - 2) = - 18.$ Toh $4 M_{1} = - 18 \Rightarrow M_{1} = - \frac{9}{2} = - 4.5$ . Result: $M_{0} = 0, M_{1} = - 4.5, M_{2} = 0$ . Negative middle moment ka matlab hai ki spline peak par concave-down hai, jaise ek hump ke liye expect hota hai.

Problem 2.2 (L2)

Problem 2.1 ke moments use karke, $[0, 1]$ par cubic $S_{0} (x)$ likho aur verify karo ki $S_{0} (0) = 1$ aur $S_{0} (1) = 3$ .

Recall Solution 2.2

Piece formula use karo $i = 0$ , $h_{0} = 1$ , $x_{0} = 0$ , $x_{1} = 1$ , $y_{0} = 1$ , $y_{1} = 3$ , $M_{0} = 0$ , $M_{1} = - 4.5$ ke saath:

S_{0} (x) = 0 \cdot \frac{( 1 - x ) ^{3}}{6} + (- 4.5) \frac{x ^{3}}{6} + (\frac{1}{1} - 0) (1 - x) + (\frac{3}{1} - \frac{( - 4.5 ) ( 1 )}{6}) x .

Simplify karo: $- 4.5/6 = - 0.75$ , aur $3 - (- 0.75) = 3.75$ . Toh $S_{0} (x) = - 0.75 x^{3} + (1 - x) + 3.75 x = - 0.75 x^{3} + 2.75 x + 1.$ Check $S_{0} (0) = 1$ ✓. $S_{0} (1) = - 0.75 + 2.75 + 1 = 3$ ✓. Aur $S_{0}^{''} (x) = - 4.5 x$ , toh $S_{0}^{''} (0) = 0$ ✓ (natural).

Level 3 — Analysis

Problem 3.1 (L3)

Usi data $(0, 1), (1, 3), (2, 2)$ ke liye, $f_{0}^{'} = 1$ aur $f_{2}^{'} = - 1$ ke saath clamped spline banao. $M_{0}, M_{1}, M_{2}$ nikalo, phir $M_{0}$ ko natural case se compare karo aur fark explain karo.

Recall Solution 3.1

Boundary equations (parent se, $h_{0} = h_{1} = 1$ ): Left: $2 h_{0} M_{0} + h_{0} M_{1} = 6 (\frac{y _{1} - y _{0}}{h _{0}} - f_{0}^{'}) = 6 ((3 - 1) - 1) = 6 (1) = 6$ , yaani $2 M_{0} + M_{1} = 6$ . Right: $h_{1} M_{1} + 2 h_{1} M_{2} = 6 (f_{2}^{'} - \frac{y _{2} - y _{1}}{h _{1}}) = 6 (- 1 - (2 - 3)) = 6 (- 1 + 1) = 0$ , yaani $M_{1} + 2 M_{2} = 0$ . Interior ( $i = 1$ ): $M_{0} + 4 M_{1} + M_{2} = - 18$ (RHS Problem 2.1 se). Solve karo: Left se, $M_{0} = (6 - M_{1}) /2$ ; Right se, $M_{2} = - M_{1} /2$ . Interior mein substitute karo: $\frac{6 - M _{1}}{2} + 4 M_{1} - \frac{M _{1}}{2} = - 18.$ 2 se multiply karo: $(6 - M_{1}) + 8 M_{1} - M_{1} = - 36 \Rightarrow 6 + 6 M_{1} = - 36 \Rightarrow M_{1} = - 7.$ Phir $M_{0} = (6 - (- 7)) /2 = \frac{13}{2} = 6.5$ aur $M_{2} = - (- 7) /2 = 3.5$ . Result: $M_{0} = 6.5, M_{1} = - 7, M_{2} = 3.5$ . Comparison: Natural ne $M_{0} = 0$ diya (zero end curvature). Clamped $M_{0} = 6.5$ force karta hai: spline ko left end par bend karna padta hai taaki imposed $f_{0}^{'} = 1$ tak slope aa sake. Curvature ek slope pin karne ki keemat hai.

Problem 3.2 (L3)

Explain karo ki spline coefficient matrix diagonally dominant kyun hai aur yeh kyun guarantee karta hai ki moments uniquely determined hain. Tridiagonal structure ka reference do.

Recall Solution 3.2

Har interior row hai $h_{i - 1} M_{i - 1} + 2 (h_{i - 1} + h_{i}) M_{i} + h_{i} M_{i + 1} = \dots$ Diagonal entry hai $2 (h_{i - 1} + h_{i})$ . Do off-diagonal entries hain $h_{i - 1}$ aur $h_{i}$ , dono positive. Unka sum hai $h_{i - 1} + h_{i}$ , jo diagonal $2 (h_{i - 1} + h_{i})$ se strictly less hai. Toh $∣ diagonal ∣ > \sum ∣ off-diagonals ∣$ har row mein — yahi strict diagonal dominance ki definition hai. Ek strictly diagonally dominant tridiagonal matrix non-singular hoti hai, isliye system $A M = r$ ka ek unique solution hota hai. Practically, Tridiagonal systems & Thomas algorithm ise $O (n)$ mein bina pivoting ke solve kar leta hai — dominance elimination ko numerically stable rakhti hai.

Level 4 — Synthesis

Problem 4.1 (L4)

Data $(0, 0), (1, 1), (2, 0), (3, 1)$ , equal spacing $h = 1$ . Natural spline banao: $M_{1}, M_{2}$ ke liye $2 \times 2$ system set up karo, solve karo, aur saare chaar moments state karo.

Recall Solution 4.1

Yahan $n = 3$ , nodes $x_{0}, \dots, x_{3}$ . Natural $\Rightarrow M_{0} = 0, M_{3} = 0$ . $i = 1$ aur $i = 2$ par interior recurrence, saare $h = 1$ toh diagonal $= 2 (1 + 1) = 4$ , off-diagonals $= 1$ : $i = 1$ : $M_{0} + 4 M_{1} + M_{2} = 6 ((y_{2} - y_{1}) - (y_{1} - y_{0})) = 6 ((0 - 1) - (1 - 0)) = 6 (- 2) = - 12.$ $M_{0} = 0$ ke saath: $4 M_{1} + M_{2} = - 12$ . $i = 2$ : $M_{1} + 4 M_{2} + M_{3} = 6 ((y_{3} - y_{2}) - (y_{2} - y_{1})) = 6 ((1 - 0) - (0 - 1)) = 6 (2) = 12.$ $M_{3} = 0$ ke saath: $M_{1} + 4 M_{2} = 12$ . Solve karo ${4 M_{1} + M_{2} = - 12 M_{1} + 4 M_{2} = 12$ . Pehle se, $M_{2} = - 12 - 4 M_{1}$ . Substitute karo: $M_{1} + 4 (- 12 - 4 M_{1}) = 12 \Rightarrow M_{1} - 48 - 16 M_{1} = 12 \Rightarrow - 15 M_{1} = 60 \Rightarrow M_{1} = - 4.$ Phir $M_{2} = - 12 - 4 (- 4) = - 12 + 16 = 4.$ Result: $M_{0} = 0, M_{1} = - 4, M_{2} = 4, M_{3} = 0.$ Antisymmetric data $(0, 1, 0, 1)$ antisymmetric moments $(0, - 4, 4, 0)$ produce karta hai — ek clean sanity check.

Problem 4.2 (L4)

4.1 ke moments use karke, $[1, 2]$ par $S_{1} (x)$ likho aur node $x = 2$ par spline ka slope dono left piece $S_{1}$ aur right piece $S_{2}$ se evaluate karo taaki wahan slope continuity confirm ho sake.

Recall Solution 4.2

Left piece $S_{1}$ on $[1, 2]$ : $h = 1$ , $x_{1} = 1, x_{2} = 2$ , $y_{1} = 1, y_{2} = 0$ , $M_{1} = - 4, M_{2} = 4$ . $S_{1} (x) = - 4 \frac{( 2 - x ) ^{3}}{6} + 4 \frac{( x - 1 ) ^{3}}{6} + (1 - \frac{- 4}{6}) (2 - x) + (0 - \frac{4}{6}) (x - 1) .$ Boxed formula se slope: $S_{1}^{'} (x) = 4 \frac{( 2 - x ) ^{2}}{2} + 4 \frac{( x - 1 ) ^{2}}{2} - (1 + \frac{2}{3}) + (0 - \frac{2}{3}) .$ $x = 2$ par: $(2 - x)^{2} = 0$ , $(x - 1)^{2} = 1$ , toh $S_{1}^{'} (2) = 0 + 2 - \frac{5}{3} - \frac{2}{3} = 2 - \frac{7}{3} = - \frac{1}{3} .$ Right piece $S_{2}$ on $[2, 3]$ : $x_{2} = 2, x_{3} = 3$ , $y_{2} = 0, y_{3} = 1$ , $M_{2} = 4, M_{3} = 0$ . $S_{2}^{'} (x) = - 4 \frac{( 3 - x ) ^{2}}{2} + 0 - (0 - \frac{4}{6}) + (1 - 0) .$ $x = 2$ par: $(3 - x)^{2} = 1$ , toh $S_{2}^{'} (2) = - 2 + \frac{2}{3} + 1 = - 1 + \frac{2}{3} = - \frac{1}{3} .$ Dono $- \frac{1}{3}$ dete hain ✓ — $x = 2$ par slope continuous hai, jo poori construction confirm karta hai. Assembled curve neeche dekho.

Figure — Cubic spline interpolation — natural, clamped

Level 5 — Mastery

Problem 5.1 (L5)

Degenerate case. Sirf do points diye gaye hain: $(0, 0), (2, 8)$ , aur aap ek natural cubic spline banate ho. Aap actually kaun sa curve paate ho, aur "cubic spline" yahan kyun collapse ho jaata hai? Ek clamped spline se contrast karo jisme $f_{0}^{'} = 0, f_{1}^{'} = 12$ ho.

Recall Solution 5.1

$n = 1$ ke saath ek single interval hai aur sirf moments $M_{0}, M_{1}$ hain. Natural $M_{0} = 0$ aur $M_{1} = 0$ set karta hai — aur koi interior nodes nahi hain, toh recurrence koi equations hi nahi deti. Dono moments already fix hain. Kyunki $S^{''}$ linear hai aur dono ends par vanish karta hai, $S^{''} \equiv 0$ on $[0, 2]$ . Ek function jiska zero second derivative ho woh ek straight line hai. $(0, 0), (2, 8)$ interpolate karo: $S (x) = 4 x .$ Toh natural spline linear interpolant mein degenerate ho jaata hai — koi cubic curvature survive nahi karti. Clamped: ab $M_{0}, M_{1}$ boundary equations ke through pin hote hain. Left: $2 h_{0} M_{0} + h_{0} M_{1} = 6 (\frac{y _{1} - y _{0}}{h _{0}} - f_{0}^{'})$ with $h_{0} = 2$ : $4 M_{0} + 2 M_{1} = 6 (\frac{8}{2} - 0) = 24$ . Right: $h_{0} M_{0} + 2 h_{0} M_{1} = 6 (f_{1}^{'} - \frac{y _{1} - y _{0}}{h _{0}})$ : $2 M_{0} + 4 M_{1} = 6 (12 - 4) = 48$ . Solve karo ${4 M_{0} + 2 M_{1} = 24, 2 M_{0} + 4 M_{1} = 48}$ : pehle ko 2 se multiply karo: $8 M_{0} + 4 M_{1} = 48$ ; doosre ko subtract karo: $6 M_{0} = 0 \Rightarrow M_{0} = 0$ , phir $2 M_{1} = 24 - 0 \Rightarrow M_{1} = 12$ . Toh $M_{0} = 0, M_{1} = 12$ : clamped spline ek genuine cubic hai (use right end par slope $12$ hit karne ke liye curve karna padta hai). Takeaway: sirf do points ke saath, natural boring hai (ek line) lekin clamped phir bhi real cubic information carry karta hai.

Problem 5.2 (L5)

Non-uniform spacing. Data $(0, 0), (1, 0), (3, 0)$ (teen collinear-at-height-zero points, lekin unequal gaps $h_{0} = 1, h_{1} = 2$ ). Natural spline banao. Pehle answer predict karo, phir recurrence se verify karo, aur explain karo ki yeh collinear data ke through splines ke baare mein kya sikhata hai.

Recall Solution 5.2

Prediction: saare $y = 0$ hain, toh flat line $S (x) = 0$ interpolate karti hai aur $C^{2}$ hai zero curvature ke saath — natural conditions satisfy hoti hain. Toh hum expect karte hain ki saare moments zero honge. Verify karo: Natural $\Rightarrow M_{0} = M_{2} = 0$ . $i = 1$ par interior, $h_{0} = 1, h_{1} = 2$ ke saath: $h_{0} M_{0} + 2 (h_{0} + h_{1}) M_{1} + h_{1} M_{2} = 6 (\frac{y _{2} - y _{1}}{h _{1}} - \frac{y _{1} - y _{0}}{h _{0}}) .$ RHS $= 6 (\frac{0 - 0}{2} - \frac{0 - 0}{1}) = 0$ . LHS $= 0 + 2 (3) M_{1} + 0 = 6 M_{1}$ . Toh $6 M_{1} = 0 \Rightarrow M_{1} = 0$ . Saare moments zero $\Rightarrow S (x) \equiv 0$ , unequal spacing ke bawajood prediction se match karta hai. Lesson: ek natural spline ek straight line ko exactly reproduce karta hai (yahan zero line). Isliye splines Runge wobble se nahi suffer karte — woh curvature sirf wahan add karte hain jahan data demand karta hai, aur tridiagonal structure non-uniform $h_{i}$ ko automatically handle karta hai. (Yeh B-splines se connect hota hai, jahan flat-data reproduction property built-in hoti hai.)

Recall Final self-check — answers cover karo

Problem 2.1 mein, $M_{1}$ kya tha? ::: $- 4.5$ .
Problem 3.1 (clamped) mein, $M_{0}$ kya tha? ::: $6.5$ .
Problem 4.1 mein, chaar moments kya the? ::: $0, - 4, 4, 0$ .
Kaun sa slope value 4.2 mein $x = 2$ par continuity confirm karta tha? ::: $- \frac{1}{3}$ dono pieces se.
Do-point natural spline kisme collapse hua (5.1)? ::: Straight line $S (x) = 4 x$ mein.

Exercises — Cubic spline interpolation — natural, clamped

Level 1 — Recognition

Problem 1.1 (L1)