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Consider the original worked example $f(x)=x^3+x-1$ on $[0,1]$. First, existence: a polynomial function is a continuous function, and $f(0)=-1<0<1=f(1)$. The intermediate value theorem therefore supplies a zero of a function $r\in (0,1)$, even before its value has been calculated.

Second, uniqueness is a separate argument. The derivative is $f^{\prime }(x)=3x^2+1>0$. The derivative criterion for monotonicity, obtained from the mean value theorem, makes $f$ a strictly increasing function. Two distinct zeros of a function would contradict that strict increase. Thus this root bracket contains exactly one zero of a function; indeed the same derivative argument proves uniqueness on the entire real line.

Third, localization uses the existing bisection method. Set $[a_0,b_0]=[0,1]$; after $k$ completed halvings, retain $[a_k,b_k]$ with opposite endpoint signs. Each update tests the old midpoint $m_{k-1}=(a_{k-1}+b_{k-1})/2$. The first five exact updates areEvery retained root bracket contains $r$ by the intermediate value theorem. After ten halvings, exact arithmetic givesHence $0.681640625<r<0.6826171875$. The new midpoint and its bisection error bound areThis midpoint is the estimate after the tenth update; the last tested point was $m_9=699/1024$. In general the retained width is $2^{-k}$, so the bisection error bound is $2^{-(k+1)}$. Nine halvings already suffice for an error at most $10^{-3}$, since $2^{-10}=0.0009765625<10^{-3}$. This guarantee comes from the width of the root bracket, independently of any displayed decimal approximation to the unknown zero of a function. Endpoint signs must be established reliably if evaluations use floating-point arithmetic.

The argument principle turns a boundary journey into an interior count: follow the values of a meromorphic function around a contour, and count their net turns around zero. Every enclosed zero of a function contributes its multiplicity of a zero; every enclosed pole subtracts its order. There is no need to find each zero of a function separately.

Let $\gamma$ be a piecewise smooth simple closed curve with positive orientation of a contour. Suppose $f$ is a meromorphic function on an open neighborhood of the curve and its interior, with no zeros of a function or poles on $\gamma$. ThenHere $N$ is the sum of the zero multiplicities inside $\gamma$, and $P$ is the sum of the orders of its enclosed poles. The image contour $f\circ \gamma$ may cross itself or revisit a point. Its signed winding number, rather than the number of visually bounded regions, is the quantity that counts.^[ref]

For the polynomial $f(z)=z^3$ and the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, the image is $e^{3it}$. It goes around zero three times although its drawn outline is just one circle. The interior contains one distinct zero of a function, at zero, with multiplicity of a zero three: the count is three. Reversing the orientation of a contour reverses the winding number and the sign of the contour integral.

An argument must be followed continuously along the journey. Subtracting the principal arguments at the endpoints loses whole turns: a closed image contour has the same endpoint value even when its winding number is nonzero. Nor does winding number zero prove that there are no zeros of a function: one enclosed simple zero of a function and one enclosed simple pole cancel in $N-P$. Known poles must be added back to recover $N$.

The mechanism is short. Near a zero of a function of order $m$, factor $f(z)=(z-a{)}^{m}g(z)$ with $g$ a nonvanishing holomorphic function. Its logarithmic derivative is $f^{\prime }/f=m/(z-a)+g^{\prime }/g$, so its residue is $m$. A pole of order $m$ gives residue $-m$. The residue theorem adds those contributions. Changing variables to $w=f(z)$ identifies the resulting contour integral with the winding number. This is a local algebraic count joined to a global geometric count.^[ref]

The usefulness goes beyond explicit polynomials. Rouché's theorem keeps a count unchanged when a boundary perturbation cannot push the image contour through zero. The Nyquist stability criterion applies the same counting idea to poles governing a system under feedback control.^[ref]

Consider the rational functionInside the unit circle are a double zero of a function at $1/4$ and a simple pole at $-1/2$. The simple zero of a function at $2$ is outside. Thus $N=2$, $P=1$, and the winding number is one. The factorization checks the count exactly; the plot illustrates how the boundary records it.

For the concentric circle $|z|=r$, the winding number is zero for $0<r<1/4$, two for $1/4<r<1/2$, one for $1/2<r<2$, and two for $r>2$. At $r=1/4$ or $r=2$, the boundary contains a zero of a function; at $r=1/2$, it contains a pole. The theorem does not apply at those three radii. Moving a boundary changes the count only when it crosses an object being counted; homotopy invariance of winding number describes the unchanged count between crossings.

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.

The intermediate value theorem says that a real-valued continuous function cannot skip a value between two of its outputs. Let $a<b$ and let $f:[a,b]\to \mathbb{R}$ be a continuous function on the entire closed interval $[a,b]$. ThenThe word "between" includes both endpoint values: choose $c=a$ when $y=f(a)$, or $c=b$ when $y=f(b)$. If $y$ is strictly between them, every such $c$ lies in the open interval $(a,b)$. Continuity on $[a,b]$ means continuity at every interior point, together with the endpoint one-sided limits ${\lim }_{x\to a^{+}}f(x)=f(a)$ and ${\lim }_{x\to b^{-}}f(x)=f(b)$. No derivative is required.^[ref]

For finding a zero of a function, the useful special case isStrictly opposite endpoint signs certify a root bracket, because zero is strictly between $f(a)$ and $f(b)$. If an endpoint value is already zero, that endpoint supplies a zero of a function directly; an interior zero of a function need not exist. For example, $f(x)=x$ on $[0,1]$ has its only zero of a function at the left endpoint. The intermediate value theorem guarantees existence, but neither uniqueness nor a necessary sign-change test. Its numerical usefulness is that the bisection method can turn existence into successively tighter root brackets.

Three explicit functions separate the hypotheses from the conclusion. These are original mathematical illustrations, not measured data.

The polynomial function $h(x)=x^3-x=x(x-1)(x+1)$ is a continuous function on $[-3/2,3/2]$, with $h(-3/2)=-15/8$ and $h(3/2)=15/8$. The intermediate value theorem guarantees at least one zero of a function in the interior, while the polynomial factorization displays three: $-1,0,1$. Thus strict opposite signs do not establish uniqueness.

The everywhere-defined functionhas $j(-1)=-1$ and $j(1)=1$, yet no zero of a function. Its jump discontinuity at zero is exactly where the continuity hypothesis fails. A drawn vertical connector would be misleading: the function graph contains neither the intermediate heights nor a zero of a function at that jump. Even continuity only on the interior is insufficient: setting $q(x)=-1$ for $0\le x<1$ and $q(1)=1$ gives opposite endpoint signs on $[0,1]$ with no zero of a function, because the left one-sided limit at $1$ is not $q(1)$.

The polynomial function $t(x)=x^2$ on $[-1,1]$ is a continuous function, and $t(-1)=t(1)=1$, yet $t(0)=0$. Its root of a polynomial at zero has multiplicity of a root two. The function graph touches its horizontal tangent line there and stays positive on both sides: existence of a zero of a function does not require a sign change. More generally, near a real root of a polynomial $r$, a polynomial factorization $p(x)=(x-r{)}^{2m}u(x)$ with $m\ge 1$ and $u(r)\ne 0$ has the same sign on both sides sufficiently close to $r$, since the continuous function $u$ retains its sign. The intermediate value theorem does not rule out values outside the range between endpoint outputs.^[ref]

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

Let $R>0$ and let $f$ be a holomorphic function on an open neighbourhood of the closed disk $\{z:|z|\le R\}$. Assume $f(0)\ne 0$ and that $f$ has no zeros of a function on the boundary circle $|z|=R$. List its interior zeros of a function as $a_1,\dots ,a_N$, repeating each according to its multiplicity of a zero. With the natural logarithm, Jensen's formula statesThe empty sum is zero. This boundary-zero-free version is the holomorphic special case of the general identity in the source.^[ref]

Every interior zero of a function makes a positive contribution; a zero nearer the centre contributes more. Consequently boundary growth relative to $|f(0)|$ limits how many zeros of a function can lie well inside the disk. The Jensen zero-count bound makes this quantitative, while the integrated zero count retains the individual radial weights. The boundary statistic here is the circle log-mean of a holomorphic function; exponentiating gives a circle geometric mean of a holomorphic function.

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.

At a point of a smooth algebraic curve, its local ring is a discrete valuation ring. A local parameter is a generator $t$ of its maximal ideal, equivalently an element of valuation one. It measures the order of a zero of a function or pole by expressing functions as powers of $t$ times units. Over the complex numbers its analytic counterpart is a local coordinate vanishing simply at the point. For an elliptic curve in a Weierstrass equation of an elliptic curve, $t=-x/y$ is the standard parameter at the identity.

Let $F$ be twice continuously differentiable on an interval around a zero of a function $t_{\ast }$, with $|F^{\prime }(t)|\ge m>0$ and $|F^{\prime \prime }(t)|\le M$ there. Choose a closed interval $[t_{\ast }-\rho ,t_{\ast }+\rho ]$ inside that interval with $M\rho /(2m)<1$. Starting inside it makes the Newton root-finding iteration well defined, keeps it inside, and gives the displayed quadratic convergence bound for $e_r=t_r-t_{\ast }$. The nonzero derivative and sufficiently close start are essential hypotheses. The scalar estimate follows from the Taylor remainder argument in MIT's root-finding notes.^[ref]

For a short proof outline, apply Taylor's theorem between $t_r$ and $t_{\ast }$; subtract the defining tangent line equation. The linear error cancels, leaving $e_{r+1}=F^{\prime \prime }({\xi }_r)e_r^2/(2F^{\prime }(t_r))$. The displayed bound shrinks the error and preserves the chosen interval. If $F^{\prime \prime }(t_{\ast })\ne 0$, the error ratio tends to $|F^{\prime \prime }(t_{\ast })/(2F^{\prime }(t_{\ast }))|$; if it vanishes, convergence can be faster.

Fix a holomorphic function on a neighbourhood of a closed disk, with $f(0)\ne 0$. On an interval of positive radii whose circles contain no zeros of a function, differentiating Jensen's formula with respect to $x=\log R$ givesEach interior zero of a function contributes slope one for each unit of its multiplicity of a zero. Between consecutive zero moduli, $L_f(e^x)$ is an affine function; passing the modulus of a zero of multiplicity of a zero $m$ increases the adjacent slopes by $m$. The finite sum $\log |f(0)|+{\sum }_j\max (0,x-\log |a_j|)$ defines a continuous convex function of $x$ whose values agree with the circle log-mean of a holomorphic function at every boundary-zero-free radius. This describes the adjacent line segments without asserting a boundary-singularity extension of the integral.

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.

The Newton root-finding iteration seeks a zero of a function $F$ by repeatedly solving its tangent line approximation. At a current guess $t_r$ with derivative $F^{\prime }(t_r)\ne 0$, setGeometrically, follow the tangent line to the horizontal axis, then return to the function graph for the next tangent line. Close to a suitable zero of a function, the error becomes dramatically smaller.^[ref]

For a concrete square root calculation, take the polynomial $F(t)=t^2-2$ and start at $t_0=1$. The Newton root-finding iteration becomes $t_{r+1}=(t_r+2/t_r)/2$, givingThe errors use the unrounded iterates. After the first step, each iterate overestimates $\sqrt{2}$, and the exact identityexhibits quadratic convergence: once close, the number of correct decimal digits roughly doubles per step. This speed makes the Newton root-finding iteration useful when accurate derivatives are available.

Global convergence requires additional care. For the polynomial $F(x)=x^3-2x+2$, starting at $x_0=0$ gives $F(0)=2$, $F^{\prime }(0)=-2$, hence $x_1=1$. Then $F(1)=1$, $F^{\prime }(1)=1$ gives $x_2=0$. The sequence repeats $0,1,0,1,\dots$ forever, although a real polynomial root lies in $(-2,-1)$ by the intermediate value theorem. Neither point in the cycle is a zero of a function, and neither has zero derivative. A maintained sign-changing interval and fallback bisection method can protect a root solver from unsuitable Newton steps.

Numerical analysis designs algorithms that approximate mathematical answers while controlling cost and error. Its central question is how much of a computed answer can be trusted. Finding a zero of a function, evaluating an integral, solving a linear system, and evolving an ordinary differential equation all demand choices about accuracy, work, and the information available. MIT's numerical-methods course explicitly treats performance and accuracy as joint concerns.^[ref]

Three questions organize error analysis. Truncation error measures the approximation introduced by an algorithm even in exact arithmetic; roundoff error comes from finite precision; the condition number measures sensitivity inherent in the problem. Numerical stability asks how the chosen algorithm propagates perturbations. These distinctions suggest different remedies: refine an approximation, change precision or arithmetic, or improve the accuracy of the input.

For example, the square root $s(a)=\sqrt{a}$ has relative condition number $|a{s}^{\prime }(a)/s(a)|=1/2$ for $a>0$: a small relative change in $a$ produces about half that relative change in $s(a)$. In contrast, $g(a)=1/(1-a)$ has relative condition number $|a/(1-a)|$. Changing $a=0.99$ to $0.9901$ changes $g(a)=100$ to about $101.010101$: an input change of about $0.0101\%$ produces an output change of about $1.0101\%$. This sensitivity exists even with exact arithmetic. MIT's conditioning lecture separates the condition number from roundoff error in precisely this way.^[ref]

The Newton root-finding iteration replaces the graph by its tangent line at $x_0$ and uses that line's zero:Comparing with the printed reciprocal-ratio form givesThis step requires $f^{\prime }(x_0)\ne 0$. If $f(x_0)=0$, the iteration has already reached a zero of a function and the printed reciprocal-ratio expression need not be evaluated.