Numerical-method comparison

Every numerical-methods course teaches its techniques one at a time, but every exam question asks the harder question first: which one do I use here? A table of data and a request for $f(1.4)$ is not an instruction to run Newton's forward formula — it is an invitation to notice that $1.4$ sits near the start of the table. An integral with an odd number of panels is not a trap for Simpson's rule — it is a signal that trapezoidal is the only legal option. This node steps back from Newton's forward/backward and Stirling's interpolation, the trapezoidal and Simpson's $1/3$ rules, and the bisection, regula falsi and Newton-Raphson root-finders, and asks what single diagnostic — position in the table, parity of subintervals, availability of a bracket — decides between them. Getting this triage right, fast, is worth more marks than any individual formula.

The core insight

Each family of numerical methods offers a menu of formulae that agree in principle but differ in the diagnostic that should choose between them: for interpolation it is where $x$ sits relative to the tabulated points (minimize $|p|$ ), for integration it is the parity of the subinterval count, and for root-finding it is the trade-off between guaranteed convergence (bracketing methods) and speed (Newton-Raphson).

The core idea

Interpolation: read the table, not habit. Newton's forward, Newton's backward and Stirling's (central) formulae are all Taylor-like series in a scaled variable $p$ , and all three converge to the same underlying polynomial through the tabulated points. What differs is only which reference point and which set of differences the series is built from:

p=\frac{x-x_{\text{ref}}}{h}

where: $x$ is the point at which $f$ is required, $x_{\text{ref}}$ is $x_0$ (forward), $x_n$ (backward) or the nearest interior point (Stirling's), and $h$ is the uniform table spacing. The series truncates fastest, and error stays smallest, when $|p|$ is small — so the one rule that governs formula choice is: pick whichever reference point puts $x$ closest, and hence $|p|$ smallest.

Integration: parity decides, then accuracy. The trapezoidal rule joins consecutive points with straight lines (exact only for linear $f$ ) and works for any number of subintervals $n$ . Simpson's $1/3$ rule fits a parabola across every pair of panels (exact for cubics) and is markedly more accurate for the same $h$ — but it needs $n$ even, since panels are consumed two at a time. If $n$ is odd, Simpson's $1/3$ is simply not available and trapezoidal is the fallback.

Root-finding: bracket safety versus tangent speed. Bisection and regula falsi both start from a bracket $f(a)f(b)<0$ guaranteeing a root inside, and both are guaranteed to converge; bisection halves the bracket every step regardless of $f$ 's shape (predictable, linear convergence), while regula falsi uses the chord through $(a,f(a))$ and $(b,f(b))$ to typically converge faster, though it can stall on one side for strongly curved $f$ . Newton-Raphson abandons the bracket entirely, needing only one starting guess and $f'(x)\ne 0$ , and converges quadratically near a simple root — but has no safety net, so a poor guess or a near-zero derivative can send it diverging instead of converging. A rough speed ranking for a well-behaved problem:

\text{Bisection} \;<\; \text{Regula falsi} \;<\; \text{Newton-Raphson}

Figure

A number line showing a tabulated interval with the forward zone near the start where p is small and positive, the backward zone near the end where p is small and negative, and the central zone in the middle where Stirlings formula applies

Derivation AWhere Simpson's 1/3 rule comes from and why it needs an even n

The goal is to show that fitting a parabola through each consecutive triple of tabulated points, rather than a straight line through each consecutive pair, is what forces Simpson's $1/3$ rule to consume subintervals two at a time and hence demand an even total count $n$ , while simultaneously explaining why the resulting rule is exact for a cubic integrand and not just a quadratic one.

Step 1. Take three equally spaced points $x_0, x_1=x_0+h, x_2=x_0+2h$ with values $f_0,f_1,f_2$ , and fit the unique parabola $y=a+bx+cx^2$ through them (three points fix three coefficients). Integrating that parabola exactly from $x_0$ to $x_2$ and simplifying the algebra in terms of $f_0,f_1,f_2$ and $h$ gives the compact result $\int_{x_0}^{x_2} y\,dx = \frac{h}{3}(f_0+4f_1+f_2)$ — this is Simpson's $1/3$ rule applied to one panel-pair.

Step 2. Notice the rule was built from one triple of points spanning two subintervals. To cover a whole interval $[x_0,x_n]$ this way, the $n$ subintervals must be consumed exactly two at a time, with no leftover single subinterval — which is only possible if $n$ is even. If $n$ is odd, one subinterval is left stranded with no partner to form a parabola, and the composite rule cannot be built.

Step 3. Chain the one-panel-pair result across the whole table: summing $\frac{h}{3}(f_0+4f_1+f_2)$ , $\frac{h}{3}(f_2+4f_3+f_4)$ , and so on, and collecting repeated endpoints gives the composite formula used in practice,

\int_{x_0}^{x_n} f(x)\,dx \approx \frac{h}{3}\Big[f_0+f_n+4(f_1+f_3+\cdots)+2(f_2+f_4+\cdots)\Big]

where: $f_0,f_n$ are the two outer endpoints, the odd-indexed $f_i$ carry weight $4$ (they are each the middle of their own parabola), and the even-indexed interior $f_i$ carry weight $2$ (each is shared between two adjacent parabolas).

Step 4. Because a parabola has three free coefficients, it can be forced to match not only $f_0,f_1,f_2$ but, by symmetry of the integral of a cubic about the panel's centre, it integrates any cubic exactly too — one degree of exactness higher than the trapezoidal rule's straight-line fit, which is exact only for linear $f$ . This is the algebraic reason Simpson's $1/3$ rule is generally more accurate than trapezoidal for the same $h$ , and why it strictly requires an even $n$ .

Worked example

Estimate $\displaystyle\int_1^2 \frac{1}{x}\,dx$ (true value $\ln 2 \approx 0.6931$ ) with both rules, using $4$ subintervals ( $h=0.25$ ), and confirm the accuracy gap the derivation predicts.

Step 1. Tabulate $f(x)=1/x$ at $x=1,1.25,1.5,1.75,2$ : $f_0=1.0000$ , $f_1=0.8000$ , $f_2=0.6667$ , $f_3=0.5714$ , $f_4=0.5000$ .

Step 2. Trapezoidal rule: $\int \approx \frac{0.25}{2}\big[(1.0000+0.5000)+2(0.8000+0.6667+0.5714)\big] = 0.125(1.5000+4.0762)=0.69702$ . Error $=|0.69702-0.69315|=0.00387$ .

Step 3. Simpson's $1/3$ rule ( $n=4$ , even — the formula from Step 3 above applies): $\int \approx \frac{0.25}{3}\big[f_0+f_4+4(f_1+f_3)+2f_2\big] = \frac{0.25}{3}(1.5000+5.4856+1.3334)=0.69325$ . Error $=|0.69325-0.69315|=0.00010$ . Simpson's rule is roughly $38$ times more accurate than trapezoidal here — the parabola-versus-line argument from the derivation, made concrete on real numbers. Answer: trapezoidal gives $0.69702$ (error $0.00387$ ); Simpson's $1/3$ gives $0.69325$ (error $0.00010$ ).

Competitive stretch

Same table, formula choice for interpolation: to estimate $f(1.4)$ , use $x_0=1$ so $p=(1.4-1)/1=0.4$ (small — Newton forward is correct). To estimate $f(4.6)$ , use $x_n=5$ so $p=(4.6-5)/1=-0.4$ (small — Newton backward is correct). Trying backward for $x=1.4$ gives $p=(1.4-5)/1=-3.6$ , and forward for $x=4.6$ gives $p=3.6$ — both far too large, confirming that the near-reference formula was the right pick in each case.

Common traps

Choosing an interpolation formula from habit instead of checking $|p|$ for each candidate reference point — always identify where $x$ sits before picking forward, backward or central.
Assuming Simpson's $1/3$ rule is always available — it needs $n$ even, because it consumes subintervals in matched pairs (Step 2 of the derivation); odd $n$ forces trapezoidal instead.
Assuming Newton-Raphson is the universally "best" root-finder — it is fastest when it works, but a bad initial guess or $f'(x)\approx 0$ can make it diverge, whereas bisection and regula falsi are guaranteed to converge once a valid bracket is found.
Treating convergence speed as the only criterion for root-finders — bisection's slow-but-predictable behaviour is exactly what you want when robustness, not speed, is the priority.

Ask the tutor: If an exam question gives you a table or a function and doesn't say which method to use, describe it to the AI tutor and ask it to walk through the diagnostic (position, parity, or bracket) that picks the right one.

Why not just always use Stirling's central formula since it uses the most information around x

Stirling's formula is genuinely the most accurate choice when $x$ sits near the middle of the table, because it draws on differences from both sides symmetrically. But it needs an interior reference point with roughly equal data on both sides. If $x$ is near either end of the table, there simply are not enough points on the far side to build the central differences the formula needs, so forward or backward becomes the only workable choice even though Stirling's would, in principle, be more accurate if the data existed.

If Simpson's 1/3 rule is more accurate, why does the syllabus bother teaching the trapezoidal rule at all

Because Simpson's $1/3$ rule has a hard requirement — an even number of subintervals — that real data does not always satisfy. If you are handed a table with an odd number of intervals, or if the composite Simpson's formula genuinely does not apply, trapezoidal is not a "worse alternative" but the only legal option left. It is also simpler to compute and easier to reason about, which matters when you need a quick order-of-magnitude estimate rather than the best possible accuracy.

How do I decide between bisection, regula falsi and Newton-Raphson on an exam without a computer

Look at what you are given. If the question hands you a bracket $[a,b]$ with $f(a)f(b)<0$ and no derivative, bisection or regula falsi are the natural choices, and regula falsi is usually faster unless the function is strongly curved on one side. If instead you are given a single starting guess and can differentiate $f$ easily, Newton-Raphson is intended, and you should check that your guess is reasonably close to the root so the quadratic convergence and the non-vanishing-derivative condition both hold.

Numerical-method comparison

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads