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In exact arithmetic, after $k$ completed halvings the retained root bracket has width $W_k=b_k-a_k$ and midpoint $m_k$, whereThe absolute value bound holds for every zero of a function $x_{\ast }$ in the current root bracket, including the one approached by continued bisection method updates. A zero of a function discarded with an earlier half need not satisfy later bounds. Before update $k\ge 1$, the evaluated candidate is $m_{k-1}$, whose corresponding guarantee is $(b-a)/2^k$. After update $k$, the new midpoint $m_k$ has the smaller guarantee $(b-a)/2^{k+1}$, even without evaluating $f(m_k)$.

The guaranteed error halves at every step; the actual absolute value error need not decrease at every step or have successive ratio $1/2$. In the square root example it increases from $|m_6-\sqrt{2}|\approx 0.000151062$ to $|m_7-\sqrt{2}|\approx 0.003755188$, while both errors remain below their guarantees. An exact midpoint zero of a function instead gives zero error and immediate termination. For this example $\sqrt{2}$ is an irrational number, so no finite exact halving can produce it as a midpoint.

For a brief proof outline, each retained half has half the previous width and contains a zero of a function by the intermediate value theorem. Every point in the retained root bracket is at most half its width from its midpoint. If the process does not stop at a zero of a function, the nested interval theorem gives one common point as the widths tend to zero; continuity and the opposite endpoint signs force its function value to be zero. Uniqueness in the initial root bracket is unnecessary.

The bisection method finds a zero of a function by repeatedly halving a root bracket. Let $f$ be a real-valued continuous function on the closed interval $[a,b]$, with $a<b$. First check for endpoint zeros of a function; otherwise require strictly opposite signs at the endpoints. Evaluate the midpoint and retain the half with strictly opposite endpoint signs, stopping immediately if the midpoint is a zero of a function. The intermediate value theorem ensures that every retained root bracket still contains a zero of a function. This is the construction in MIT's root-finding notes.^[ref]

For a worked square root calculation, use the polynomial $f(x)=x^2-2$ on $[1,2]$. Write $[a_k,b_k]$ for the root bracket after $k$ halvings, with $[a_0,b_0]=[1,2]$. The candidate evaluated during update $k$ is the old midpoint $m_{k-1}$:The initial midpoint is $m_0=1.5$. After four halvings the new midpoint is $m_4=1.40625$, and the bisection error bound is $|m_4-\sqrt{2}|\le 0.03125$. Its actual error is approximately $0.00796356237310$. The last tested candidate $m_3=1.4375$ and the reported midpoint $m_4$ are different points.

The useful feature of the bisection method is that the retained width halves on every update, independently of the shape of the function graph. The bisection error bound therefore supplies an accuracy guarantee before the zero of a function is known. No derivative is needed, and a valid initial root bracket gives convergence without a close initial guess. This makes the bisection method a dependable fallback for the faster Newton root-finding iteration.

For a requested absolute tolerance $\epsilon >0$, stop when the retained root bracket has width at most $2\epsilon$, and report its midpoint. The bisection error bound then guarantees $|m_k-x_{\ast }|\le \epsilon$ in exact arithmetic. Using the ceiling function and the base-two logarithm, the required number of halvings isThe zero allows stopping before any update when the initial root bracket is already sufficiently narrow. For the polynomial $f(x)=x^2-2$ on $[1,2]$ with $\epsilon =10^{-3}$, $k_{\min }=\lceil {\log }_{2}500\rceil =9$. After nine halvings,Thus the reported midpoint meets the tolerance; its actual error from the square root $\sqrt{2}$ is approximately $0.000825500126905$. After eight halvings the bound is $1/512>10^{-3}$, so eight do not give this worst-case guarantee.

Cache the endpoint values. Each update of the bisection method needs just one new function evaluation, at the old midpoint. With no early zero of a function, $k$ updates therefore cost $k+2$ function evaluations, including the two initial endpoints. Returning the new midpoint requires no additional function evaluation; requesting its value would add one. The nine-update example needs eleven evaluations for the guarantee.

A small function value $|f(m_k)|$ alone does not guarantee a small position error: rescaling $f$ changes those values without changing its zeros of a function. In a practical implementation, use an explicit width test and an iteration limit. With floating-point arithmetic, also stop if the computed midpoint equals either endpoint: the root bracket can no longer shrink, so repeating the update would loop forever. Report this precision limit separately if the requested tolerance remains unmet. Reliable sign evaluations are needed to preserve the root bracket; compare signs directly rather than multiplying endpoint values, which can suffer overflow or underflow in floating-point arithmetic.

For a rounded midpoint $m$ still inside a valid root bracket, the geometric error bound is $\max \{m-a_k,b_k-m\}$; it need not equal $W_k/2$ because of roundoff error. A rigorous computed enclosure also needs conservative arithmetic for that bound and reliable endpoint signs. The exact-arithmetic bisection error bound should not be claimed unchanged after precision prevents further halving.

For practical tolerance conventions, SciPy's bisect combines absolute and relative tolerances through $|x-x_0|\le \mathrm{xtol}+\mathrm{rtol}|x_0|$, and provides a maximum iteration count and convergence status. These differ from the purely absolute width test above.^[ref]

A bracketing method finds a zero of a function by maintaining a root bracket and shrinking it as new values of the function become available. The bisection method always tries the midpoint; its special attraction is a predictable bisection error bound without any derivative evaluations.

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

Every Cauchy sequence of real numbers converges to a real number. Explicitly, the conditionimplies the existence of $L\in \mathbb{R}$ with $x_n\to L$. The hypothesis compares approximations with one another, without requiring a candidate answer. This makes completeness of the real numbers an existence principle: sufficiently consistent approximations cannot approach a missing destination. ^[ref].

The equivalent least-upper-bound property says that every nonempty set $A\subseteq \mathbb{R}$ bounded above has a supremum $s\in \mathbb{R}$. Precisely, every $x\in A$ satisfies $x\le s$, and for every $\epsilon >0$ some $x\in A$ satisfies $s-\epsilon <x$. The second condition excludes any smaller upper bound. No largest element of $A$ is required.

The rational numbers lack this property. The set $A=\{q\in \mathbb{Q}:q\ge 0,q^2<2\}$ contains one and has upper bound two, but has no supremum in $\mathbb{Q}$. A proposed rational supremum $s$ would have to satisfy $s^2=2$: if $s^2<2$, a sufficiently small rational increase still lies in $A$; if $s^2>2$, a sufficiently small rational decrease remains an upper bound. The irrationality of the square root of two excludes equality. In $\mathbb{R}$, its supremum is $\sqrt{2}$. ^[ref].

Construct this square root by the bisection method, starting with rational bounds $[l_0,u_0]=[1,2]$. At each step square the midpoint and retain the half whose endpoints bracket two after squaring. The first four resulting intervals are $[1,3/2]$, $[5/4,3/2]$, $[11/8,3/2]$, and $[11/8,23/16]$. After $k$ steps, $u_k-l_k=2^{-k}$ and $l_k^2<2<u_k^2$. Both endpoint sequences are Cauchy sequences because every later endpoint remains in the current interval. Completeness of the real numbers gives a common limit of a sequence $L$; continuity of squaring gives $L^2=2$. The midpoint approximates $L$ with error at most $2^{-(k+1)}$. Thus exact rational arithmetic produces certified accuracy while completeness of the real numbers supplies the exact answer.

Richard Dedekind approached completeness of the real numbers by filling gaps in the rational numbers with Dedekind cuts. In the preface to his 1872 Continuity and Irrational Numbers, he traced the problem to teaching differential calculus in 1858 and wanting arithmetic foundations for limits. His construction makes a missing boundary, such as $\sqrt{2}$, into a number defined by the separation it produces among rational numbers. ^[ref].

A credible interval is an interval $C(d)$ containing a prescribed amount $1-\alpha$ of posterior probability for an unknown scalar statistical parameter $\Theta$. For a continuous posterior distribution with posterior density $p(\theta \mid d)$, its endpoints satisfyThe conditioning includes the chosen statistical model and prior distribution. For a discrete posterior distribution, exact equality may be impossible because of its probability masses; an interval can instead be required to contain at least the requested posterior probability. An interval compresses the full posterior distribution into a range whose uncertainty can be communicated in the parameter's own units. See MIT 18.05, Class 16: probability intervals.

For the eight-head, two-tail example in Bayesian statistics, the posterior density is $2860{\theta }^9(1-\theta {)}^3$ on $[0,1]$. Its cumulative distribution function, obtained by integration of this polynomial, isSolving $F(a)=0.025$ and $F(b)=0.975$ by the bisection method givesThese are numerical approximations to the exact quantiles defined by those equations; the endpoints were computed from the cumulative distribution function, rather than read off a plot. The resulting equal-tailed credible interval leaves 2.5% posterior probability on each side. Its shaded area is 95% of the total area under the posterior density.

Credible intervals are not unique. The one-sided credible interval $[0,F^{-1}(0.95)]\approx [0,0.887334219]$ also has 95% posterior probability, but leaves all 5% outside it on the right. A highest density region instead selects points above a density threshold; it need not be an equal-tailed credible interval, and when the posterior density has separated peaks it need not be one interval at all. Thus a reported credible interval should specify its construction.

A 95% credible interval makes a statement about the statistical parameter conditional on these observed data and the Bayesian model. A 95% confidence interval instead refers to the coverage probability of a procedure under repeated sampling at a fixed parameter value. These are different conditioning statements: 95% posterior probability does not by itself imply 95% coverage probability for every fixed parameter. The confidence interval article develops that sampling interpretation.

Consider the population statistical model $F_{\epsilon ,x}=(1-\epsilon )N(0,1)+\epsilon {\delta }_x$. The following are deterministic statistical functional calculations, not observed contamination experiments. Compare the actual location bias with the linear approximation $\epsilon \mathrm{IF}(x;T,F)$. The mean bias is exactly $\epsilon x$. For positive $x$ exceeding the shifted median and $0\le \epsilon <1/2$, the generalized-inverse median bias is exactlyThis formula applies to both $x=1$ and $x=8$ throughout the plotted interval $0\le \epsilon \le 0.1$. If the contaminating atom lies inside the quantile jump containing probability $1/2$, the generalized-inverse median instead equals that atom.

For the Huber location estimator with $s=1$ and $k=1.345$, solve its actual population equation, rather than substituting the influence function:Writing $a=t-k$ and $b=t+k$, integration of the normal distribution giveswhere $\varphi$ is the probability density function of $N(0,1)$. Its derivative is $-[\Phi (t+k)-\Phi (t-k)]<0$, so the mixture score is strictly decreasing for $\epsilon <1$. For $\epsilon <1/2$, its limits at negative and positive infinity have opposite signs even with arbitrary distant point contamination. The displayed examples use the unique root, evaluated by the bisection method to numerical precision.

At $x=8$ and $\epsilon =0.01$, the exact mean, median, and Huber location estimator biases are $0.080000$, $0.012660$, and $0.016541$; the influence function predictions are $0.080000$, $0.012533$, and $0.016375$. At $\epsilon =0.05$, they become $0.400000$, $0.066012$, and $0.086241$, versus predictions $0.400000$, $0.062666$, and $0.081875$. The bounded alternatives control the effect of the distant atom, but their actual response curves bend away from the tangent.

The nearby atom $x=1$ gives a useful comparison: at $\epsilon =0.05$, the Huber location estimator bias is $0.060237$, slightly below its linear approximation $0.060874$, whereas the median bias is still $0.066012$. The direction and size of approximation error depend on the contaminating location. Even accurate local influence calculations must specify $F$, the scale, the cutoff, and the actual contamination fraction.

At a fixed temperature, the frequency maximizing $B_{\nu }$ and the wavelength maximizing $B_{\lambda }$ in Planck's law are different ways of describing the same black-body radiation. They are maxima of different densities. The inverse relation between frequency and wavelength does not preserve those maxima because its interval-conversion factor varies with wavelength.

For a short extremum calculation, substitute $x=h\nu /(k_{B}T)$ into $B_{\nu }$, leaving a positive constant times $x^3/(e^x-1)$. Substitute $x=hc/(\lambda k_{B}T)$ into $B_{\lambda }$, leaving a positive constant times $x^5/(e^x-1)$. Setting each derivative to zero givesSelect the positive roots: $x_{\nu }=2.821439372\dots$ and $x_{\lambda }=4.965114232\dots$. Although $x=0$ solves the rearranged equations, it is an endpoint, not either peak. For $m=3$ or $5$, the function $m(1-e^{-x})-x$ initially increases, its derivative $m{e}^{-x}-1$ crosses zero once, and it eventually becomes negative. Hence it has exactly one positive root. Each density tends to zero at both ends, so this root gives its unique maximum. The bisection method brackets these roots in $(1,3)$ and $(1,5)$.

Consequently Wien's displacement law givesThese constants agree with NIST's tabulation. ^[ref] The wavelength peaks at $3000$, $4500$, and $6000\mathrm{K}$ are $965.924$, $643.949$, and $482.962\mathrm{nm}$. Doubling temperature halves this peak wavelength.

At $6000\mathrm{K}$, the frequency peak is ${\nu }_{\max }\approx 352.736\mathrm{THz}$, whose converted wavelength is $c/{\nu }_{\max }\approx 0.849907\mu \mathrm{m}$. The $B_{\lambda }$ peak is instead $0.482962\mu \mathrm{m}$, corresponding to about $620.737\mathrm{THz}$. These numbers are compatible: equal-width frequency bins and equal-width wavelength bins partition the same radiation differently. Specifying which density was maximized is essential when using a peak to infer temperature.

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

If $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

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.

Interval root exclusion certifies that a function has no zero of a function on $X$ when a valid interval enclosure $F(X)$ excludes zero. For $g(x)=x(1-x)-1/3$ on $[0,1]$, the interval square range yields $G(X)=[-1/3,-1/12]$, which excludes zero. Direct interval arithmetic yields $[-1/3,2/3]$ and is inconclusive, even though the function has no zero of a function there.

Knowing only that a calculated interval enclosure contains zero does not certify a zero of a function: it may contain unattainable values. Even the positive constant function $f(x)=1$ has a valid, loose interval enclosure $[-1,2]$. A separate existence argument, such as reliable opposite endpoint signs together with continuity in the intermediate value theorem, is required; see root bracket and bisection method. Interval subdivision can resolve some inconclusive pieces, but remaining pieces must be reported as unresolved unless an existence argument succeeds. This makes interval root exclusion useful for systematically discarding regions in a search.

// Target: analysis.bigb

Several zeros of a function can lie in one initial root bracket: the bisection method seeks one, rather than enumerating them. For instance, the polynomial $(x-1)(x-2)(x-3)$ has opposite endpoint signs on $[0,4]$, but its first midpoint evaluation returns only the polynomial root $2$. The initial zero of a function need not be unique for the bisection error bound to apply to a retained zero of a function.

A zero of a function without a sign change cannot be located by the strict sign-change test alone. The polynomial $(x-1{)}^2$ touches zero at $1$ but is positive on either side. On $[0,3]$ both endpoint values are positive: although the closed interval contains a polynomial root, it fails the usual starting test for the bisection method. An endpoint or evaluated midpoint zero of a function can still be recognized directly.

Opposite signs certify a root bracket only with the required continuity. For $f(x)=1/x$, the values at $-1$ and $1$ have opposite signs, but $f$ is undefined at the midpoint $0$ and has no zero of a function. Hence $[-1,1]$ is invalid input for the bisection method; the intermediate value theorem does not apply across the gap.

A sequence of nested intervals keeps every new enclosure inside the previous one:Each step adds a restriction without discarding an earlier restriction. This is useful when an exact real number is described by increasingly precise bounds rather than by a finite decimal expansion. The bisection method, for instance, produces nested intervals that keep enclosing a zero of a function.

For a simple example, the closed intervals $I_n=[1-1/n,1+1/n]$ are nested and all contain $1$. Their lengths $2/n$ tend to zero, so the nested interval theorem leaves exactly that point. The family $J_n=[1-1/n,2+1/n]$ is also nested, but its intersection is the whole closed interval $[1,2]$. Nesting alone does not force uniqueness.

The open intervals $K_n=(0,1/n)$ illustrate a different possibility. Every finite collection has a nonempty intersection, but the intersection of all of them is empty: any positive candidate eventually exceeds $1/n$, and zero is excluded from every member. Each finite row in the diagram contains points; the full infinite family need not.