Exercises — Implementing numerical integration from scratch — trapezoidal, Simpson's

Simpson's stencil har pair ke liye 1-4-1 hota hai. Teen pairs ko six slices ke across chain karo.

• Endpoints ($x_0,x_6$) ek-ek parabola se belong karte hain → weight $1$.
• Odd-index nodes ($x_1,x_3,x_5$) pair-centres hain (woh peak jahan parabola bend karti hai) → weight $4$.
• Even interior nodes ($x_2,x_4$) do adjacent parabolas ke beech share hote hain → weight $2$.

Toh pattern hai $[1,4,2,4,2,4,1].$ Check karo: weights ka sum $1+4+2+4+2+4+1=18=3n$ hai, aur $\frac{h}{3}\cdot 3n\cdot (\text{avg }f)$ constant $f$ ke liye $(b-a)\cdot (\text{avg }f)$ recover karta hai. ✓

Points $=n+1$, isliye $n=9-1=8$. Phir $h=\frac{b-a}{n}=\frac{4-0}{8}=0.5$. Simpson ko $n$ even chahiye; $8$ even hai ✓ — allowed hai.

$h=\frac{2-0}{4}=0.5$. Nodes aur heights: $x=0,0.5,1,1.5,2\Rightarrow f=1,1.25,2,3.25,5.$ $T_4=\frac{h}{2}[f(x_0)+f(x_4)+2(f(x_1)+f(x_2)+f(x_3)\left)\right]$ $=\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25[6+2(6.5)]=0.25\cdot 19=\mathrm{4.75.}$ Exact value: ${\int }_0^2(x^2+1)dx=\frac{8}{3}+2=4.6\overline{6}$. Trapezoid overshoot karta hai (curve convex hai, seedhi lids uske upar baith jaati hain) approximately $\approx 0.083$ se.

Same nodes/heights jaise L2.1 mein hain. Weights $[1,4,2,4,1]$: $S_4=\frac{h}{3}[1+4(1.25)+2(2)+4(3.25)+5]=\frac{0.5}{3}[1+5+4+13+5]=\frac{0.5}{3}\cdot 28=4.6\overline{6}.$ Yeh exact $14/3$ ke exactly barabar hai, kyunki $x^2+1$ degree $2$ ka hai aur Simpson saari parabolas ko exactly integrate karta hai.

Prediction: $E(h)\propto h^2$. $h\to h/2$ half karne par $E\to (1/2{)}^{2}E=E/4$ milta hai. Toh error $4$ ke factor se shrink hona chahiye.

Check. $h=(b-a)/n$ ke saath:

• $n=2$ ($h=0.5$), nodes $0,0.5,1$, $f=1,1.648721,2.718282$: $T_2=\frac{0.5}{2}[1+2.718282+2(1.648721)]=0.25\cdot 7.015724=\mathrm{1.753931.}$ Error $E_2=1.753931-1.718282=0.035649$.
• $n=4$ ($h=0.25$), nodes $0,0.25,0.5,0.75,1$, $f=1,1.284025,1.648721,2.117000,2.718282$: $T_4=\frac{0.25}{2}[1+2.718282+2(1.284025+1.648721+2.117000)]=0.125\cdot 14.817774=\mathrm{1.852222.}$ Ruko — bracket carefully recompute karo: $1+2.718282=3.718282$; interior sum $=1.284025+1.648721+2.117000=5.049746$, doubled $=10.099492$; total $=13.817774$; $\times 0.125=1.727222$. Error $E_4=1.727222-1.718282=0.008940$.

Ratio $E_2/E_4=0.035649/0.008940\approx 3.99\approx 4$. ✓ $O(h^2)$ prediction hold karti hai.

Kyun: Ek cubic par parabola fit karne ki error leftover cubic term se aati hai. Symmetric interval $[-h,h]$ par cubic error term $x$ ka odd function hota hai, aur symmetric limits par ek odd function ka integral $0$ hota hai. Simpson nodes $-h,0,h$ symmetric hain, isliye odd leftover exactly zero integrate ho jaata hai — cubic term error mein kuch contribute nahi karta. Yahi woh "free" extra degree hai.

Confirm: true value ${\int }_{-1}^1{x}^{3}dx=0$ (odd function). Simpson ke saath $h=1$, $f(-1)=-1,f(0)=0,f(1)=1$: $S=\frac{1}{3}[(-1)+4(0)+1]=\frac{1}{3}\cdot 0=0.$ Exact, jaisa promise tha. ✓

Pehle do trapezoid values nikalo ($h=1$ phir $h=0.5$).

• $T_2$ ($n=2$, nodes $0,1,2$, $f=1,2,5$): $\frac{1}{2}[1+5+2(2)]=\frac{1}{2}\cdot 10=5.$
• $T_4$ ($n=4$) $=4.75$ L2.1 se.

Richardson combine karo: $S=\frac{4(4.75)-5}{3}=\frac{19-5}{3}=\frac{14}{3}=4.6\overline{6}.$ Yeh $S_4=4.6\overline{6}$ se exactly match karta hai. Kyun kaam karta hai: trapezoid error $C{h}^2+O(h^4)$ hai. $h$ half karne par leading term $\frac{1}{4}$ se scale hota hai; weights $4$ aur $-1$ over $3$ precisely isliye choose kiye gaye hain taaki $h^2$ terms cancel ho jaayein, ek aisi method chhod ke jo $O(h^4)$ tak accurate hai — yahi Simpson hai. Yeh Richardson extrapolation ka ek step hai; ise iterate karne par Romberg integration milti hai.

$h=0.2$, nodes $x_i=0,0.2,0.4,0.6,0.8,1.0$, heights $f=1,1.221403,1.491825,1.822119,2.225541,\mathrm{2.718282.}$ Simpson on $[0,0.8]$ (nodes $x_0..x_4$, weights $1,4,2,4,1$): $S=\frac{0.2}{3}[1+4(1.221403)+2(1.491825)+4(1.822119)+2.225541]$ Bracket $=1+4.885612+2.983650+7.288476+2.225541=18.383279$; $\times \frac{0.2}{3}=\mathrm{1.225552.}$ Trapezoid on $[0.8,1.0]$ (nodes $x_4,x_5$): $T=\frac{0.2}{2}[2.225541+2.718282]=0.1\cdot 4.943823=\mathrm{0.494382.}$ Total $=1.225552+0.494382=1.719934$. Error $\approx 0.00165$ — pure trapezoid se kaafi better, dangling slice ko honestly handle karta hai.

Trapezoid exact: koi bhi straight line, jaise $f(x)=3x+1$ on $[0,4]$. Trapezoid construction se degree-1 polynomials ke liye exact hota hai (seedha lid function hi hota hai). Exact value ${\int }_0^4(3x+1)dx=\frac{3}{2}(16)+4=28$; $n=1$ ke saath, $T_1=\frac{4}{2}[f(0)+f(4)]=2[1+13]=28$. ✓ Simpson bhi $28$ deta hai lekin tumhe kuch extra nahi milta — capture karne ke liye koi curvature hi nahi hai.

Error orders fail: $f(x)=\sqrt{x}$ lo $[0,1]$ par. Iska derivative $x=0$ par blow up karta hai, isliye $O(h^2)$/$O(h^4)$ ke peeche wala Taylor-series argument (jisme bounded higher derivatives ki zaroorat hoti hai) break down ho jaata hai. Observed convergence $h^2$ se slower hoti hai. Lesson: advertised error orders assume karte hain ki $f$ kaafi smooth hai (trapezoid ke liye bounded 2nd derivative, Simpson ke liye 4th). Kinks, cusps, aur singularities guarantee void kar dete hain — wahan scipy.integrate.quad ki adaptive machinery use karo.

Correct (weights $1,4,1$): $\frac{1}{3}[0+4(1)+4]=\frac{8}{3}=2.6667$. Exact ✓. Buggy (weights $4,4,4$): $\frac{1}{3}[4(0)+4(1)+4(4)]=\frac{1}{3}[0+4+16]=\frac{20}{3}=6.6667$. Bug se error $=6.6667-2.6667=4.0000$ — bahut hi galat answer ($2.5\times$ zyada bada). Sirf $x_2=2$ par extra weight (jahan $f=4$) ne akele $\frac{1}{3}\cdot 3\cdot 4=4$ add kar diya. Isliye endpoint weighting #1 silent bug hai: yeh crash nahi karta, bas quietly garbage return karta hai.

$S_4$ ($h=0.25$, nodes $0,0.25,0.5,0.75,1$, $f=0,0.00390625,0.0625,0.31640625,1$): $S_4=\frac{0.25}{3}[0+4(0.00390625)+2(0.0625)+4(0.31640625)+1]$ Bracket $=0+0.015625+0.125+1.265625+1=2.40625$; $\times \frac{0.25}{3}=\mathrm{0.200521.}$ Error $E_4=0.200521-0.2=0.000521$ — chhota lekin nonzero (degree $4$, Simpson ke exact degree $3$ se zyada hai). $S_8$ ($h=0.125$): weighted sum compute karne par $S_8=0.20003255$ milta hai, error $E_8\approx 0.00003255$. Ratio $E_4/E_8\approx 0.000521/0.00003255\approx 16.0=2^4$. ✓ $O(h^4)$ order confirm ho gaya.

Slice width hai ==$h=(b-a)/n$. Simpson's per-pair stencil hai 1-4-1, scale hota hai $h/3$== se. Trapezoid error order ::: $O(h^2)$ Simpson error order ::: $O(h^4)$ Highest exact degree — trapezoid / Simpson ::: 1 / 3 $n$ subintervals ke liye points ::: $n+1$ Richardson combo jo trapezoids se Simpson deta hai ::: $S=(4T(h/2)-T(h))/3$