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

Simpson's stencil is 1-4-1 per pair. Chain three pairs across six slices.

• Endpoints ($x_0,x_6$) belong to one parabola each → weight $1$.
• Odd-index nodes ($x_1,x_3,x_5$) are pair-centres (the peak the parabola bends to) → weight $4$.
• Even interior nodes ($x_2,x_4$) are shared between two adjacent parabolas → weight $2$.

So the pattern is $[1,4,2,4,2,4,1].$ Check: the weights sum to $1+4+2+4+2+4+1=18=3n$, and $\frac{h}{3}\cdot 3n\cdot (\text{avg }f)$ recovers $(b-a)\cdot (\text{avg }f)$ for a constant $f$. ✓

Points $=n+1$, so $n=9-1=8$. Then $h=\frac{b-a}{n}=\frac{4-0}{8}=0.5$. Simpson needs $n$ even; $8$ is even ✓ — allowed.

$h=\frac{2-0}{4}=0.5$. Nodes and 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 overshoots (curve is convex, straight lids sit above it) by $\approx 0.083$.

Same nodes/heights as L2.1. Weights $[1,4,2,4,1]$: $S_4=\frac{h}{3}[1+4(1.25)+2(2)+4(3.25)+5]=\frac{0.5}{3}[1+5+4+13+5]=\frac{0.5}{3}\cdot 28=4.6\overline{6}.$ This equals the exact $14/3$ exactly, because $x^2+1$ is degree $2$ and Simpson integrates all parabolas exactly.

Prediction: $E(h)\propto h^2$. Halving $h\to h/2$ gives $E\to (1/2{)}^{2}E=E/4$. So error should shrink by a factor of $4$.

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

• $n=2$ ($h=0.5$), 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 $0,0.25,0.5,0.75,1$, $f=1,1.284025,1.648721,2.117000,2.718282$: $T_4=\frac{0.25}{2}[1+2.718282+2(1.284025+1.648721+2.117000)]=0.125\cdot 14.817774=\mathrm{1.852222.}$ Wait—recompute the bracket carefully: $1+2.718282=3.718282$; 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 $E_2/E_4=0.035649/0.008940\approx 3.99\approx 4$. ✓ The $O(h^2)$ prediction holds.

Why: The error of fitting a parabola to a cubic comes from the leftover cubic term. On the symmetric interval $[-h,h]$ the cubic error term is an odd function of $x$, and the integral of an odd function over symmetric limits is $0$. The Simpson nodes $-h,0,h$ are symmetric, so the odd leftover integrates to exactly zero — the cubic term contributes nothing to the error. That is the "free" extra degree.

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

First get the two trapezoid values ($h=1$ then $h=0.5$).

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

Richardson combine: $S=\frac{4(4.75)-5}{3}=\frac{19-5}{3}=\frac{14}{3}=4.6\overline{6}.$ This matches $S_4=4.6\overline{6}$ exactly. Why it works: trapezoid error is $C{h}^2+O(h^4)$. Halving $h$ scales the leading term by $\frac{1}{4}$; the weights $4$ and $-1$ over $3$ are chosen precisely so the $h^2$ terms cancel, leaving a method accurate to $O(h^4)$ — which is Simpson. This is one step of Richardson extrapolation; iterating it gives Romberg integration.

$h=0.2$, 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 $\approx 0.00165$ — much better than pure trapezoid, honestly handling the dangling slice.

Trapezoid exact: any straight line, e.g. $f(x)=3x+1$ on $[0,4]$. Trapezoid is exact for degree-1 polynomials by construction (a straight lid is the function). Exact value ${\int }_0^4(3x+1)dx=\frac{3}{2}(16)+4=28$; with $n=1$, $T_1=\frac{4}{2}[f(0)+f(4)]=2[1+13]=28$. ✓ Simpson also gives $28$ but buys you nothing extra — there is no curvature to capture.

Error orders fail: take $f(x)=\sqrt{x}$ on $[0,1]$. Its derivative blows up at $x=0$, so the Taylor-series argument behind $O(h^2)$/$O(h^4)$ (which needs bounded higher derivatives) breaks down. The observed convergence is slower than $h^2$. Lesson: the advertised error orders assume $f$ is smooth enough (bounded 2nd derivative for trapezoid, 4th for Simpson). Kinks, cusps, and singularities void the guarantee — reach for scipy.integrate.quad's adaptive machinery there.

Correct (weights $1,4,1$): $\frac{1}{3}[0+4(1)+4]=\frac{8}{3}=2.6667$. Exact ✓. Buggy (weights $4,4,4$): $\frac{1}{3}[4(0)+4(1)+4(4)]=\frac{1}{3}[0+4+16]=\frac{20}{3}=6.6667$. Error from the bug $=6.6667-2.6667=4.0000$ — a wildly wrong answer ($2.5\times$ too big). The extra weight on $x_2=2$ (where $f=4$) alone added $\frac{1}{3}\cdot 3\cdot 4=4$. This is why endpoint weighting is the #1 silent bug: it doesn't crash, it just quietly returns garbage.

$S_4$ ($h=0.25$, 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 $E_4=0.200521-0.2=0.000521$ — small but nonzero (degree $4$ exceeds Simpson's exact degree $3$). $S_8$ ($h=0.125$): computing the weighted sum gives $S_8=0.20003255$, error $E_8\approx 0.00003255$. Ratio $E_4/E_8\approx 0.000521/0.00003255\approx 16.0=2^4$. ✓ The $O(h^4)$ order is confirmed.

The slice width is ==$h=(b-a)/n$. Simpson's per-pair stencil is 1-4-1, scaled by $h/3$==. Trapezoid error order ::: $O(h^2)$ Simpson error order ::: $O(h^4)$ Highest exact degree — trapezoid / Simpson ::: 1 / 3 Points for $n$ subintervals ::: $n+1$ Richardson combo giving Simpson from trapezoids ::: $S=(4T(h/2)-T(h))/3$