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The absolute error is the absolute value of the approximation error, $|x_{\mathrm{approx}}-x|$, measured in the same units as the numerical reference $x$. For example, approximating $1/3$ by $0.33$ gives absolute error $|0.33-1/3|=1/300$.

Continue the initial value problem $y^{\prime }=y$, $y(0)=1$, to $t=4$ using the Euler-Heun embedded pair. Choose absolute and relative error tolerances $\mathrm{atol}=10^{-3}$ and $\mathrm{rtol}=0$, so the scalar scale is $s=10^{-3}$ throughout. This intentionally absolute-only test makes the local and accumulated errors directly comparable. Start with $h=0.2$, accept the Heun's method endpoint precisely when $E=|d|/s\le 1$, and after every trial use the capped adaptive step-size controllerFor $E=0$ use the growth factor two. Cap every trial at $4-t_n$; on rejection retain the old time and state. These choices specify an original computation, rather than a claim about the defaults of a particular software solver.

The initial trial has $E=20$ and is rejected. Retrying from $(0,1)$ with $h=0.9(0.2)/\sqrt{20}\approx 0.0402492236$ gives $E=0.81$ and is accepted. The run takes 315 accepted steps and one rejected trial. Increasing exponential growth makes the local error estimator $|d|=|y_n|h^2/2$ grow at a fixed $h$, so subsequent steps shrink roughly like $|y_n{|}^{-1/2}$. The final step is additionally shortened to land at $t=4$.

The right panel plots errors against the accepted-step number, all divided by the same $s$. The actual Heun local defect uses the restarted exact endpoint $y_n{e}^{h_n}$, while actual global error uses $e^{t_{n+1}}$. All accepted $E$ are at most one, and the actual Heun local defects here are even smaller, at most about $0.01143s$. Nevertheless the endpoint is $y_N\approx 54.5833912205$, versus $e^4\approx 54.5981500331$, with absolute error about $0.0147588=14.7588s$. Passing local tests therefore does not certify the original trajectory's global error.

This failure of a global guarantee has an exact explanation for this example. Put $R(h)=1+h+h^2/2$ and $g_n=e^{t_n}-y_n$. ThenEach new local defect is added to the previous error amplified by $e^{h_n}$. Iterating this identity, rather than merely inspecting $E$, accounts for accumulation. The variable-step global ODE error bound describes the corresponding mechanism for more general ordinary differential equations.

Repeating with $\mathrm{atol}=10^{-2}$ gives 98 accepted steps, one rejection and endpoint absolute error about $0.153252$; with $\mathrm{atol}=10^{-4}$ it gives 1001 accepted steps, two rejections and endpoint absolute error about $0.00145642$. Accuracy improves on this smooth problem, but the endpoint error remains about fifteen times the requested local scale. A different initial value problem can amplify or damp defects differently.

Consider the nonstiff scalar initial value problem $y^{\prime }=-y$, $y(0)=1$, on $0\le t\le 8$. Its exact solution is the exponential decay $y(t)=e^{-t}$. Use step-doubling error estimation with the explicit Euler method: from the same numerical starting value $y_n$, form one coarse step $C_n$ and two half steps giving $F_n$:Accept $y_{n+1}=F_n$ when the absolute and relative error tolerances test $E_n=|{\hat{e}}_n|/s_n\le 1$ passes, with $s_n=10^{-5}+10^{-3}\max (|y_n|,|F_n|)$. There is no Richardson extrapolation in the accepted value. Both approximations have order of a numerical method one, while this local error estimator has step exponent $r=2$. Consequently the adaptive step-size controller uses a square root, not a cube root:Start with $h_0=1$. A rejected trial retains both $t_n$ and $y_n$, caps $q_n$ at one, and retries. A zero estimate uses $q_n=5$. Before every trial, cap $h_n$ at $8-t_n$. Stop with a failure if floating-point arithmetic gives $t_n+h_n=t_n$, or after 100000 trials. These rules specify the example completely; they are not a prescription for every ordinary differential equation solver.

The first two trials, $h=1$ and $h=0.2$, have $E\approx 247.525$ and $E\approx 9.90099$ and are rejected. The next trial has $h\approx 0.0572049$ and $E\approx 0.81$ and is accepted. For positive decreasing numerical values, $s_n=10^{-5}+10^{-3}{y}_n$, so the local error estimator givesWhen the relative term dominates, this suggests nearly constant accepted steps, approximately $2(0.9)\sqrt{10^{-3}}$. When $y_n$ falls below the crossover size $10^{-5}/10^{-3}=0.01$, the absolute term increasingly permits longer steps. The sharp final decrease in $h$ only lands on $t=8$; it does not indicate a sudden loss of smoothness.

The actual signed local defect from the same numerical input isHere $O$ is Big O notation, obtained from Taylor's theorem for the exponential function. Thus ${\hat{e}}_n$ agrees with the leading term as $h_n\to 0$, but is not identical to ${\ell }_n$. At $h=1$, $y_n=1$, the estimate is $0.25$ while the actual local defect is $e^{-1}-1/4\approx 0.117879$. Their closeness improves for the smaller accepted steps. Had we added ${\hat{e}}_n$ using Richardson extrapolation, the accepted formula would instead be $y_n(1-h_n+h_n^2/2)$, of order of a numerical method two; the same raw difference would still scale as $h_n^2$ and would not estimate that improved value's leading $h_n^3$ defect.

The global error $g_n=y_n-e^{-t_n}$ obeys an exact recurrence relation involving the actual local defects:Each previous global error contracts, yet the local defects keep adding. In the plotted run, 101 steps are accepted and two trials are rejected. Every accepted $E_n$ is at most approximately $0.81$, while the largest $|g_{n+1}|/s_n$ is approximately $38.6927$. At $t=8$, the numerical value is approximately $0.00027184218$, compared with the exact $e^{-8}\approx 0.00033546263$, giving absolute error approximately $6.36204\times 10^{-5}$. The right-hand plot deliberately scales actual global error by the same $s_n$ used for local acceptance; it is not a new global error tolerance test.

Tightening both absolute and relative error tolerances by a factor ten gives 315 accepted steps and endpoint absolute error approximately $2.14922\times 10^{-5}$; tightening both by a factor one hundred gives 991 accepted steps and endpoint absolute error approximately $6.94171\times 10^{-6}$. This is a convergence comparison for this particular initial value problem. The endpoint error decreases by about a factor three per tenfold tightening, rather than by ten: a first-order Euler method accumulates defects across many steps, while its local error estimator varies quadratically with $h$. A variable-step global ODE error bound needs propagation information as well as accepted local tests.

The absolute stability boundary remains a separate constraint. On the Dahlquist test equation $y^{\prime }=-\lambda y$ with $\lambda >0$, the accepted two-half-step formula has stability function $R(-\lambda h)=(1-\lambda h/2{)}^2$. It strictly damps perturbations only for $0<\lambda h<4$. Its local error estimator is ${\lambda }^2{h}^2{y}_n/4$, which can pass an absolute-tolerance test at tiny $y_n$ even outside this damping interval. In a stiff two-mode linear system, a fast mode can therefore constrain the step long after its visible amplitude has become small. An adaptive step-size controller does not turn an explicit Euler method into an A-stable method. Official solver documentation likewise distinguishes explicit Runge-Kutta methods for nonstiff problems from implicit Runge-Kutta methods and backward differentiation formulas for stiff equations.^[ref]

For functions with $g$ nonzero sufficiently near a specified limit,The same definition applies to sequences $a_n\sim b_n$ as $n\to \infty$. Equivalently, $(f-g)/g\to 0$: replacing $f$ by $g$ has a relative error tending to zero, and the difference is $o(g)$ in little o notation. The limit and reference scale are essential parts of the statement.

For example, $n^2+n\sim n^2$ because $(n^2+n)/n^2=1+1/n\to 1$. Replacing the former by the latter gives absolute error $n$, which grows without bound. Thus asymptotic equivalence controls a ratio, rather than requiring the absolute error to vanish. This distinction is crucial when approximating a rapidly growing factorial with Stirling's formula.

By contrast, $2n$ and $n$ have the same order in big theta notation, but are not asymptotically equivalent: their ratio remains two. For positive functions, asymptotic equivalence gives both-sided constant bounds eventually, whereas constant bounds alone do not recover the leading coefficient.

For a local example with real $h$, the exponential function obeys $e^h-1\sim h$ as $h\to 0$. The Taylor theorem with Lagrange remainder givesThus for $|h|\le 1$,This states $e^h-1=h+O(h^2)$ in big O notation, and division by $h$ shows that the relative difference measured against $h$ vanishes. The diagram uses positive $h$; the bound is valid on both sides of zero. At $h=0.1$, the absolute error is approximately $0.00517092$ and the relative difference against $h$ is approximately $0.0517092$.

Even restricting to increasing sequences in $[0,1]$ with limit of a sequence one gives no common stopping index. For any positive integer $K$, defineThis is an increasing bounded sequence converging to one, but its absolute error is one through index $K$. Since $K$ can exceed any proposed index, the bounded monotone sequence theorem alone cannot say when the absolute error will fall below $1/2$. A zero difference between two successive terms is also insufficient: the initial plateau eventually ends.

There is no universal eventual rate either. Let $(r_n)$ be any proposed positive nonincreasing sequence tending to zero, with $r_n\le 1$. SetThen $(a_n)$ is increasing, lies in $[0,1]$, and converges to one, but for every sufficiently large $n$ its absolute error satisfies $1-a_n=\sqrt{r_n}>r_n$. Indeed, $\sqrt{r_n}/r_n=1/\sqrt{r_n}\to \infty$, so no fixed constant multiple of $r_n$ bounds the absolute error eventually. If a proposed rate is not nonincreasing, replace it, after a finite initial segment if necessary, by the decreasing envelope $s_n=\sup \{r_k:k\ge n\}$; this also tends to zero, and the same construction defeats the original rate. This is an original construction, rather than a quantitative claim attributed to the qualitative theorem.

For a concrete slow example, $a_n=1-1/\log (n+2)$ increases to one, where $\log$ is the natural logarithm. Its absolute error is $1/\log (n+2)$, so making it less than $\epsilon$ requires $n+2>\exp (1/\epsilon )$, using the exponential function. At $\epsilon =0.1$, this first happens at $n=22025$. The reciprocal and geometric examples in increasing and decreasing approximations to a common limit meet the same tolerance at $n=11$ and $n=4$, respectively. All three obey the same bounded monotone sequence theorem.

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

Apply the midpoint quadrature rule on each of $n$ equal panels of a partition of an interval, with $h=(b-a)/n$:Each quadrature node is a panel midpoint and each quadrature weight is $h$. For $f\in C^2[a,b]$, put $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$. Summing the panel quadrature errors givesfor some $\xi \in (a,b)$. Each panel contributes at most $K{h}^3/24$; there are $n=(b-a)/h$ panels, explaining the composite midpoint rule error of order $O(h^2)$ in big O notation. On a nonuniform partition of an interval with panel widths $h_j$ and largest width $H$, the corresponding bound is $K{\sum }_j{h}_j^3/24\le K(b-a)H^2/24$.

For $f(x)=x^2$ on $[0,1]$, summing the sampled polynomial exactly yieldsDoubling $n$ divides the absolute error by exactly four. For a general $C^2$ function, halving $h$ divides the displayed error bound by four, but the actual absolute error need not follow that ratio: cancellation can make it smaller or zero. Continuous functions still give convergence as the largest panel width tends to zero, by uniform continuity; without the bounded second derivative, the second-order rate is not guaranteed. Noise sensitivity of a quadrature rule explains a separate limitation when samples are inexact.

Composite Simpson's rule applies Simpson's rule to consecutive pairs of equal subintervals. Choose a positive even integer $n\ge 2$, put $h=(b-a)/n$, and take quadrature nodes $x_j=a+jh$ for $j=0,\dots ,n$. There are $n$ subintervals, $n/2$ two-subinterval panels, and $n+1$ quadrature nodes. ThenThe second sum is empty when $n=2$, recovering one panel. The quadrature weights, before multiplication by $h/3$, are $1,4,2,4,\dots ,2,4,1$. Every interior even-indexed quadrature node is an endpoint of two panels, so its two endpoint contributions add to two; every odd-indexed quadrature node is a midpoint with weight four. An odd $n$ cannot be covered by these pairs. The displayed formula requires equal spacing.^[ref]

For the exponential function on the interval $[0,1]$, the exact integral is $I=e-1$. With $n=2$, the formula gives $S_2\approx 1.718861151876593$, an overestimate of about $5.793234175\times 10^{-4}$. With $n=128$, it gives $S_{128}\approx 1.718281828494607$, an overestimate of about $3.556140064\times 10^{-11}$. The figure shows the intermediate refinements and compares every computed absolute error with the error bound for composite Simpson's rule.

For this exponential function, a geometric series also gives a compact independent formula. With $q=e^h$ and $h=1/n$, summing the $n/2$ panel contributions yieldsA Taylor series at $h=0$ givesThe Big O notation describes the remaining terms as $h$ tends to zero. Consequently $(S_n-I)/(S_{2n}-I)$ tends to sixteen: the computed ratios increase from about $15.65169$ to $15.99965$. Fourth-order truncation error is also obtained by Richardson extrapolation of the composite trapezoidal rule.^[ref]

Obtaining consistent computational results from the same input data, code and stated computational conditions. Exact bitwise agreement is stronger than agreement to a justified numerical or statistical tolerance.

An error bound is an established upper bound on a specified approximation error, on a stated domain and under stated assumptions. A numerical comparison at selected points can test an approximation but does not by itself establish a bound throughout that domain. An absolute error bound controls $|\tilde{f}-f|$; a relative error bound additionally specifies the reference scale.

Suppose $f$ has a fourth derivative that is a continuous function on the interval $[a,b]$, and $|f^{(4)}(x)|\le M$ there. With $I={\int }_a^{b}f(x)dx$, the signed quadrature error here is $I-S_n$. Then, for some point $\xi \in (a,b)$,In particular, a strictly positive fourth derivative makes $I-S_n<0$: composite Simpson's rule overestimates the integral.^[ref]

The bound's scale follows by adding the panel errors. A panel of width $2h$ has signed quadrature error $-h^5{f}^{(4)}({\xi }_j)/90$ and absolute error at most $M{h}^5/90$. There are $(b-a)/(2h)$ panels, so the triangle inequality gives $M(b-a)h^4/180$. The intermediate value theorem for the continuous fourth derivative replaces the average of the panel values $f^{(4)}({\xi }_j)$ by a single value $f^{(4)}(\xi )$, giving the signed formula.

The underlying single-panel calculation can be seen through the Peano kernel theorem. On the interval $[0,2]$, define the raw Peano kernel $K(t)=L[(x-t{)}_{+}^3]$, where $L$ is integral minus Simpson's rule and $(x-t{)}_{+}=\max (x-t,0)$ is the positive part. Then $K(t)=-t^3(4-3t)/12$ for $0\le t\le 1$ and $K(t)=-(2-t{)}^3(3t-2)/12$ for $1\le t\le 2$. It is strictly negative inside the interval and has integral $-1/15$. ThusThe uniform norm $\Vert f^{(4)}{\Vert }_{\infty }$ bounds the fourth derivative. The polynomial $f(x)=x^4/24$ attains equality. Scaling to endpoint spacing $h$ multiplies the coefficient by $h^5$, giving the sharp single-panel bound $h^5\Vert f^{(4)}{\Vert }_{\infty }/90$. The normalized Peano kernel is $K/3!$, so the external factorial must be omitted when using that normalization.

For the exponential function on $[0,1]$, $f^{(4)}(x)=e^x$, so $M=e$ and the bound is $e/(180n^4)$. Its leading actual absolute error is $(e-1)/(180n^4)$, smaller than the bound because a uniform upper bound on the fourth derivative does not use its variation across the interval. The theorem bounds truncation error with exact function evaluations; sample errors and roundoff error are additional contributions, as described by noise sensitivity of a quadrature rule.

For an increasing sequence converging to $L$, its absolute error is $e_n=L-a_n\ge 0$ and is nonincreasing. For a decreasing sequence, the absolute error is $e_n=a_n-L\ge 0$ and is also nonincreasing. This tells us that later terms cannot be worse, but does not supply a computable stopping index for a chosen tolerance.

Suppose an increasing lower approximation $a_n$ and a decreasing upper approximation $b_n$ enclose the same limit of a sequence $L$. Their observable gap $w_n=b_n-a_n$ givesIf $w_n\to 0$, the squeeze theorem certifies a common limit of a sequence; the midpoint has a certified absolute error at most half the gap. This is the principle behind the bisection method. In the symmetric example $a_n=1-1/n$, $b_n=1+1/n$, the gap is $2/n$, each endpoint's exact absolute error is $1/n$, and the midpoint happens to equal one exactly. The bound is valid even when such symmetry is absent.

A fixed upper bound by itself supplies less. If $a_n\uparrow L$ and $a_n\le M$ for all $n$, then $L\le M$ and $0\le L-a_n\le M-a_n$, but the last gap need not vanish. Taking $a_n=1-1/n$ and $M=2$ makes $M-a_n\to 1$. A shrinking bracket or an explicit remainder formula adds information that qualitative convergence does not provide.

For any finite set of quadrature nodes $x_1,\dots ,x_m$ in $[a,b]$, $a<b$, consider the polynomialEvery sample is zero, so every quadrature rule at these quadrature nodes gives $Q(p)=0$. Yet $p$ is nonnegative and strictly positive away from its finitely many polynomial roots; its continuity makes the integral positive. Consequently the smooth functions $0$ and $g(x)=p(x)/A$ have identical samples but definite integrals $0$ and $1$. Multiplying $g$ by any positive constant makes the difference arbitrarily large. This is an information limit, even with exact arithmetic; merely assuming smoothness without quantitative bounds does not remove it. The related degree ceiling for quadrature exactness uses the same squared nodal polynomial to limit polynomial exactness.

For example, the endpoint quadrature nodes $0,1$ give $p(x)=x^2(1-x{)}^2$, $A=1/30$, and $g(x)=30x^2(1-x{)}^2$. Both the zero function and $g$ produce zero under the one-panel trapezoidal rule, although $I(g)=1$. More generally, any estimate based only on their shared samples has absolute error at least $1/2$ for one of these two integrands, by the triangle inequality.

The obstruction also applies to a deterministic adaptive procedure that terminates after finitely many evaluations on the zero function: form $p$ using every quadrature node it visited. On $g$, the same zero responses lead to the same choices and answer. Certified quadrature error control therefore needs additional information, such as a known bound on relevant derivatives, rather than agreement among sampled approximations alone.

For the two sequences in bounded monotone sequence theorem,Their directions and bounds are explicit. One is a succession of lower approximations and the other a succession of upper approximations to the same limit of a sequence. To obtain absolute error strictly below $\epsilon >0$, it suffices to choose an integer $N>1/\epsilon$, and then every $n\ge N$ works. For $\epsilon =0.01$, the first such index is $101$; at $100$ the absolute error equals the tolerance.

A geometric sequence offers a faster example. The recurrence relation $c_1=1/2$, $c_{n+1}=(c_n+1)/2$ gives $c_n=1-2^{-n}$. Thus $(c_n)$ is an increasing bounded sequence with limit of a sequence one and absolute error $2^{-n}$. For $\epsilon =0.01$, $n=7$ already works, since $2^{-7}=1/128<0.01$, whereas $2^{-6}=1/64>0.01$. The bounded monotone sequence theorem guarantees both examples converge; the formulas explain their different speeds.

Interval bisection solves a target equation $f(x)=t$ by repeatedly halving a closed interval known to contain a solution. It is the bisection method applied to $g(x)=f(x)-t$, so the target becomes a zero of a function. This reduction from an equation to root finding is described in Demanet's notes.^[ref] Here the emphasis is on a continuous strictly increasing function: the target identifies one input, making interval bisection a way to evaluate its inverse function without an explicit inverse formula.

Let $f:[a_0,b_0]\to \mathbb{R}$ be a continuous function, with $a_0<b_0$ and $f(a_0)\le t\le f(b_0)$. The intermediate value theorem guarantees existence; a strictly increasing function also guarantees uniqueness. Check endpoint equality first. Otherwise, for the current root bracket $[a_k,b_k]$, evaluate its midpoint $m_k=(a_k+b_k)/2$. If $f(m_k)<t$, retain $[m_k,b_k]$; if $f(m_k)>t$, retain $[a_k,m_k]$; if $f(m_k)=t$, return that exact solution. These comparisons preserve the target bracket. General bisection method needs continuity and opposite signs of $g$ at the endpoints, without requiring monotonicity or a unique zero of a function; see limitations of bisection.

An original dimensionless polynomial example isHere $f(0)=0<10<30=f(3)$ and the derivative $f^{\prime }(x)=3x^2+1>0$, so there is exactly one solution. In fact $x_{\ast }=2$; knowing this lets us check the computed absolute error independently of the guarantee. The first four halvings give the following exact rational numbers:

The beauty of interval bisection is that the enclosure itself supplies the certificate: one comparison removes half the possible inputs, regardless of the steepness of the function graph. In exact arithmetic, after $k$ completed halvings without exact termination, the width and the bisection error bound for the new midpoint areThe reported midpoint after update four is $m_4=63/32$, rather than the last tested point $m_3=33/16$. Its certified absolute error is at most $3/32$, and its actual absolute error is $1/32$. For a requested position tolerance $10^{-3}$, the bisection stopping criterion requires eleven halvings: ten give $3/2048>10^{-3}$, while eleven giveThe actual absolute error is $1/4096$. The new midpoint need not be evaluated to obtain this certificate.

The signed output residual is $r=f(m)-t$; the position approximation error is $m-x_{\ast }$. They answer different questions: how close is the output to the target, and how close is the input to the solution? Driscoll and Braun explain why a small residual needs information about the derivative to imply small position error.^[ref] For this particular polynomial, direct factorization givesThus at $m_4=63/32$ the residual is $-13121/32768\approx -0.400421$, despite position absolute error $1/32=0.03125$. If a differentiable function satisfies $0<\ell \le f^{\prime }(x)\le L$ throughout the bracket, the mean value theorem gives the rigorous conversionOur example permits $\ell =1$ and $L=28$ on $[0,3]$. Strict increase alone supplies no positive lower derivative bound: $f(x)=x^3$ is strictly increasing but $f^{\prime }(0)=0$. A small residual alone is therefore not a universal position certificate; even the target equation $10^{-6}(x-2)=0$ has residual $10^{-6}$ at $x=3$, a full unit from its solution.

The guarantees require continuity and correct evaluations, not just narrowing endpoints. For the strictly increasing function $h(x)=x$ when $x<0$ and $h(x)=x+1$ when $x\ge 0$, the target $1/2$ lies between $h(-1)$ and $h(1)$ but is never attained: a jump discontinuity defeats the existence argument. With floating-point arithmetic or an approximate evaluation, an incorrect comparison can discard the true solution. A certified interval enclosure of $f(m)$ wholly below or above $t$ certifies the corresponding decision; an enclosure containing $t$ is inconclusive, rather than proof of equality. If rounding makes the midpoint equal an endpoint, report the precision limit; see bisection stopping criterion for the rounded position bound. Both figures are exact mathematical examples, not measured data.

An Lp norm measures the size of a measurable function by averaging a power of its absolute value with respect to a specified measure. For $1\le p<\infty$,when the integral is finite. The exponent $p$ controls how strongly large values influence the result. The Lp space identifies functions equal almost everywhere, so this becomes a norm on the corresponding equivalence classes. For $p=\infty$, use the essential supremum of $|f|$, as defined by the L-infinity norm. ^[ref].

For a two-coordinate vector $(3,4)$, treat the coordinates as a function on two inputs with counting measure. Then $\Vert (3,4){\Vert }_1=3+4=7$, $\Vert (3,4){\Vert }_2=\sqrt{9+16}=5$, and $\Vert (3,4){\Vert }_{\infty }=4$. The middle case is the Euclidean norm; the first adds the coordinate magnitudes, and the last takes their largest value. These are different answers to different questions about size, all satisfying the triangle inequality.

If instead each input has probability $1/2$, the same values give $\Vert f{\Vert }_1=3.5$ and $\Vert f{\Vert }_2=\sqrt{12.5}$, while $\Vert f{\Vert }_{\infty }=4$ remains unchanged. Thus the measure is not an optional label. On a probability space, the finite Lp norms are nondecreasing in $p$ whenever they are finite, by Hölder's inequality; the unnormalized two-coordinate example above decreases instead. Statements comparing different exponents must specify their measure convention.

Why are these sizes useful? If $e$ is an error function, its L1 norm measures aggregate absolute error, its L2 norm measures the square root of aggregate quadratic loss, and its L-infinity norm bounds error almost everywhere. Concentrating a large error on a set of small positive measure can give a modest finite Lp norm but a large L-infinity norm. The three-point example and diagram in L-infinity norm show this explicitly, including why a null set behaves differently from a rare event.

Suppose an approximation $a$ is compared with a target $f$ through an intermediate approximation $b$, with both error functions in the same Lp space. The identity $f-a=(f-b)+(b-a)$ and the Minkowski inequality combine two error budgets without any independence assumption. The L1 norm controls aggregate absolute error, the L2 norm controls the square root of aggregate quadratic loss, and the L-infinity norm controls the largest error almost everywhere. For equally weighted samples with the probability measure $1/n$ on each coordinate, the L2 norm is the root mean square error; the inequality still applies when this normalization is used consistently.

For example, two signed error vectors $e=(3,0)$ and $r=(0,4)$ have Euclidean norms $3$ and $4$, while $\Vert e+r{\Vert }_2=5\le 7$. Reinforcing errors $e=(3,0)$ and $r=(4,0)$ attain the bound $7$; opposing errors $e=(3,0)$ and $r=(-3,0)$ cancel completely. The Minkowski inequality always allows reinforcement. A smaller combined budget requires additional information about the errors, not just their separate norms.

Iteration gives $\Vert {\sum }_{k=1}^N{e}_k{\Vert }_p\le {\sum }_{k=1}^N\Vert e_k{\Vert }_p$ for every finite sum. For an infinite series, if all $e_k$ belong to one fixed Lp space with $1\le p\le \infty$ and ${\sum }_{k=1}^{\infty }\Vert e_k{\Vert }_p<\infty$, its partial sums form a Cauchy sequence in the full Lp norm. Completeness of this Banach space then gives norm convergence to an element of the same Lp space. A finite sum bound alone does not establish convergence, and completeness is an additional ingredient beyond the Minkowski inequality. ^[ref]

If $I_1\supseteq I_2\supseteq \cdots$ are nonempty closed intervals with finite endpoints, then at least one real number belongs to every member. If their lengths tend to zero, exactly one real number belongs to every member. The result turns an unlimited sequence of consistent enclosures into an existence guarantee.

Write $I_n=[a_n,b_n]$. Nesting makes the left endpoints nondecreasing and the right endpoints nonincreasing. Their bounds give finite supremum and infimum, withIndeed, every $b_n$ is an upper bound for all the $a_j$, so the least-upper-bound property gives ${\sup }_j{a}_j\le b_n$ for every $n$. This proves the displayed interval is nonempty. If $b_n-a_n\to 0$, its two endpoints must coincide. This is the connection to the completeness of the real numbers discussed in Günther Hörmann and David Langer, Einführung in die Analysis, §§3.15–3.16: nested intervals and order completeness.

The distinction between existence and uniqueness is useful in computation. If an unknown point $x_{\ast }$ lies in every $[a_n,b_n]$, the midpoint $m_n=(a_n+b_n)/2$ satisfiesThus the enclosure itself certifies an absolute error without knowing the answer. The bisection error bound is the special case where each update halves the width.

Closedness and a finite initial enclosure matter. The open intervals $(0,1/n)$ have no common point, while the closed but unbounded intervals $[n,\infty )$ also have empty intersection. A picture of a few intervals cannot certify the intersection of an infinite family; the hypotheses of the nested interval theorem do that work.

Numerical integration approximates a definite integral using finitely many sampled function values. For an integrand $f$ on a finite interval $[a,b]$, $a<b$, a quadrature rule has the formThe quadrature nodes $x_j$ tell us where to evaluate $f$; the quadrature weights $w_j$ turn those values into an approximation to the accumulated signed area. The exact definite integral depends on the integrand throughout the interval, whereas the quadrature rule sees only these samples. Integrating a chosen interpolant exactly still generally approximates the original integrand's integral.^[ref]

The fundamental theorem of calculus evaluates $I(f)$ as $F(b)-F(a)$ when an antiderivative $F$ is available. Numerical integration is useful when no convenient antiderivative is known, or each evaluation of the integrand requires an expensive computation. The task then becomes choosing informative quadrature nodes and controlling the quadrature error with as few evaluations as possible. Fixed data also permit a quadrature rule, but refinement cannot supply new information between observations.^[ref]

For an exact comparison, take the polynomial $f(x)=x^2$ on $[0,1]$. Its antiderivative $x^3/3$ gives $I(f)=1/3$. On a partition of an interval into $n\ge 1$ equal panels, the composite midpoint quadrature rule uses $n$ midpoint samples, each with quadrature weight $1/n$; the composite trapezoidal rule uses $n+1$ endpoint samples, with outer quadrature weights $1/(2n)$ and interior quadrature weights $1/n$. The existing derivations under these two quadrature rules giveIn particular, $M_1=1/4$ and $T_1=1/2$ already bracket $1/3$; two panels give $M_2=5/16$ and $T_2=3/8$. The convex function $x^2$ explains the opposite signs. A midpoint rectangle can cross the curve within its panel, so its total signed area, rather than pointwise ordering, determines the underestimate.

Define the signed approximation error as $E_n=Q_n(f)-I(f)$. In this example the absolute errors are $1/(12n^2)$ and $1/(6n^2)$: doubling $n$ reduces each by exactly four. A plot with both axes on a logarithmic scale therefore has slope $-2$, expressing order two with respect to panel refinement. This is an exact calculation for this polynomial, including $n=1$, rather than an observed rate extrapolated from a few computed answers.

For a general integrand, the composite midpoint quadrature rule and composite trapezoidal rule give second-order quadrature error bounds when $f$ has two continuous derivatives on $[a,b]$ and a known bound on $|f^{\prime \prime }|$. These hypotheses describe behavior between samples. A formal error formula containing an unknown derivative is not a numerical accuracy certificate or a justified sample count. Finite samples do not certify quadrature accuracy makes the missing information explicit. Errors in the evaluations and roundoff error must also be considered; noise sensitivity of a quadrature rule treats the former.

More elaborate choices exploit other information: Gaussian quadrature obtains high polynomial exactness from carefully chosen quadrature nodes, and Simpson's rule uses a quadratic interpolant. Adaptive numerical integration compares approximations and concentrates samples where they disagree, often saving evaluations. Such a difference is an error estimate, not a certified bound without further assumptions: both approximations can miss the same feature.^[ref]

For $f(x)=x^{1/3}$, the required derivatives areAt $a=8$ these give $f(8)=2$, $f^{\prime }(8)=1/12$, and $f^{\prime \prime }(8)=-1/144$. Setting $x-a=1$, the three-term Taylor polynomial yieldsBy the Taylor theorem with Lagrange remainder, the signed error isIt is positive, and ${\xi }^{-8/3}<8^{-8/3}=1/256$. ConsequentlyThe last comparison is exact because $5\cdot 4000=20000<20736$. Thus the estimate is an underestimate with the required rigorous absolute error bound.