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On $I=[1,7/5]$, $f^{\prime }<0$ and $f^{\prime \prime }<0$, since $x^2<2$. Hence $|f^{\prime }|$ increases and has minimum $|f^{\prime }(1)|=14$. Alsowith equality at $x=1$. Therefore

There is a unique root $r$ in $I$: $f(1)=2>0$, $f(7/5)=-5.678464<0$, and the function is strictly decreasing. Moreover $f(6/5)=-1.382016<0$, so $r<6/5=x_0$.

It remains to justify that the iteration stays where the error estimate applies. Because $f$ is a concave function, its graph lies below each tangent line. At any $x\in [r,6/5]$, the tangent has nonpositive value $f(x)$ at $x$, negative slope, and nonnegative value at $r$. Its zero $N(x)=x-f(x)/f^{\prime }(x)$ therefore satisfiesBy mathematical induction, the Newton root-finding iteration stays in $[r,6/5]$ and decreases to some limit $L\ge r$. Continuity in $x_{k+1}-x_k=-f(x_k)/f^{\prime }(x_k)$ implies $f(L)=0$, so uniqueness gives $L=r$.

Now the supplied error bound applies at every step:This proves at least quadratic convergence, with an invariant interval and a nonvanishing derivative explicitly established.

Rolle's theorem is the equal-endpoint case of the mean value theorem. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, there exists $c\in (a,b)$ with $f^{\prime }(c)=0$. Geometrically, a tangent line somewhere in the interior is horizontal, like the secant line joining the endpoints. ^[ref]

For the polynomial $f(x)=x(x-2)$ on $[0,2]$, both endpoint values are zero and the derivative is $f^{\prime }(x)=2x-2$. Thus $c=1$, where $f(1)=-1$, and the horizontal tangent line is $y=-1$. The horizontal secant line is $y=0$: equal endpoint values do not force the interior tangent line to pass through those endpoints.

The absolute value function $f(x)=|x|$ on $[-1,1]$ exposes the importance of differentiability. It is continuous and $f(-1)=f(1)=1$, so the secant line has slope zero. However,and the derivative does not exist at zero because the one-sided derivatives disagree. Every existing derivative in the interior is nonzero. The corner fails a hypothesis of both Rolle's theorem and the mean value theorem, so neither conclusion is guaranteed.

For a short proof, the extreme value theorem supplies a greatest and a least value on $[a,b]$. If $f$ is a constant function, every interior point works. Otherwise at least one of these values differs from the common endpoint value, so it is attained at an interior extremum. At an interior local maximum or local minimum, differentiability forces the derivative to be zero: the difference quotients from the two sides have opposite weak signs and the same limit.

This turns repeated values into information about derivatives. Two distinct roots of a polynomial force a root of its derivative between them; iterating this idea gives the Rolle root count with multiplicities. More generally, a differentiable function with nonzero derivative throughout a real interval cannot repeat a value there, hence is injective. The beauty is that knowing only two equal values guarantees an exact local feature somewhere between them.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

A secant line is a straight line passing through two distinct points of a curve. For a graph $y=f(x)$ at $x=a$ and $x=a+h$, its slope is the difference quotient $(f(a+h)-f(a))/h$. As the second point approaches the first, a limiting secant line can give the tangent line.

In Cartesian coordinates, the slope of a nonvertical straight line is its vertical change divided by its horizontal change. The graph $y=mx+c$ has slope $m$; a vertical straight line has no finite slope. For a tangent line to a differentiable function, the slope is the derivative.

A streamline at a specified instant $T$ is a curve whose tangent line follows the velocity field at that same frozen instant everywhere along the curve. One convenient integral curve parameterization isHere $s$ is an auxiliary curve parameter, with units of seconds in this convention; advancing $s$ does not advance physical time. Any regular reparameterization preserves the tangent geometry. Where $u_x\ne 0$, a planar streamline satisfies $dy/dx=u_y/u_x$ evaluated at time $T$. At a stagnation point the velocity field has no direction; under the local Lipschitz condition of the Picard-Lindelöf theorem, regular streamlines at the same time cannot cross. A streamline answers which directions fit together now, making it useful for describing local flow geometry. ^[ref]

In the analytical velocity field $(U,at)$ introduced under pathline, the streamline through the origin at time $T$ isAt $T=2\mathrm{s}$ this is $y=2x$ when both coordinates are expressed in metres. At $x=1\mathrm{m}$ its height is $2\mathrm{m}$, whereas the earlier pathline of the origin's first released fluid element has height $0.5\mathrm{m}$. Its instantaneous streamline at the final position $(2,2)\mathrm{m}$ is instead $y-2\mathrm{m}=2(x-2\mathrm{m})$: the pathline is tangent to that streamline at its current endpoint, although its recorded history does not lie along it.

The green streamline uses the velocity field at $T=2\mathrm{s}$ throughout; the orange pathline uses times $0\le t\le 2\mathrm{s}$; the purple streakline uses different release times at the single observation time $T=2\mathrm{s}$. The grey arrows all show the instantaneous velocity direction at $T$. These are exact analytical curves with chosen units, not measured dye shapes.

In a steady flow in a fixed reference frame, a pathline and a streamline solve the same autonomous ordinary differential equation. Uniqueness makes each particle released from a fixed point follow the same integral curve, so its pathline, the streakline from that point and the streamline share one geometrical curve. Finite recording or release intervals usually show different segments of that curve; the comparison assumes particles stay in the domain and treats stationary fluid elements separately. Neither two-dimensionality nor incompressibility is needed for this coincidence. Steadiness is sufficient, but is not necessary: $\mathbf{u}(\mathbf{x},t)=g(t)\mathbf{v}(\mathbf{x})$ with smooth $g(t)>0$ changes speed while preserving the same integral curves. In a general time-dependent velocity field, however, the three curves differ. Kline explains why continuous dye release in steady flow can therefore display streamlines. ^[ref]

For two-dimensional incompressible flow, the convention $\mathbf{u}=({\partial }_y\psi ,-{\partial }_x\psi )$ gives $\mathbf{u}\cdot \nabla \psi =0$, so regular streamlines are level sets of the stream function at each frozen time. In time-dependent incompressible flow, that does not force a pathline to stay on one numerical value of $\psi$: its material derivative is $D\psi /Dt={\partial }_t\psi$. The distinction is the difference between a geometric snapshot and a parcel's history.

The tangent line to a differentiable function at $a$ is its affine first-order approximation $f(a)+f^{\prime }(a)(x-a)$. For a differentiable concave function, it is an upper bound throughout a convex interval: the secant-slope inequality and passage to the derivative give $f(x)\le f(a)+f^{\prime }(a)(x-a)$. This upper bound constructs valid envelopes in adaptive rejection sampling.

For a projective plane curve $C=V(F)$ and a nonsingular projective point $p$, the tangent is the line $F_x(p)x+F_y(p)y+F_z(p)z=0$. The coefficients are not all zero, and the Euler theorem for homogeneous functions makes this line pass through $p$. On an affine chart this is precisely the first-order part of a local defining equation, so the definition agrees with the usual tangent line and is invariant under projective transformations.

For a nonzero real constant $a$, the tangent line to $y=e^{ax}$ at $x_{\ast }$ passes through the origin precisely when $e^{a{x}_{\ast }}-x_{\ast }a{e}^{a{x}_{\ast }}=0$. Thus $x_{\ast }=1/a$, $y_{\ast }=e$, and its slope is $k=ae$. This gives the unique such tangent line. If $a=0$, the graph is the horizontal line $y=1$ and no tangent line through the origin exists. The equality between the derivative and the slope from the origin, $f^{\prime }(x_{\ast })=f(x_{\ast })/x_{\ast }$, is useful for other differentiable functions too.

At a point $T$ on a circle with centre $O$, the tangent line is perpendicular to the radius $OT$. For example, differentiating $|\mathbf{r}(s)-O{|}^2=R^2$ along a parametric curve on the circle gives $(\mathbf{r}-O)\cdot {\mathbf{r}}^{\prime }=0$, showing that the radial vector and tangent direction are orthogonal vectors. Conversely, the perpendicular straight line through $T$ has only that point in common with the circle: for a nonzero displacement $v$ along it, $|T+v-O{|}^2=R^2+|v{|}^2>R^2$.

A tangent vector at a point $p$ of a smooth manifold without boundary records an instantaneous velocity through $p$. Its base point is part of the data; measuring its length additionally requires a Riemannian metric. Intrinsically, a smooth curve $\gamma$ with $\gamma (0)=p$ defines its tangent vector by how every local smooth function changes along it:This is a derivation at a point. Two smooth curves through $p$ represent the same tangent vector if these derivatives agree for every such smooth function; equivalently, their coordinate velocity vectors agree in one manifold chart. The chain rule makes that agreement independent of the chosen manifold chart. ^[ref]

For an embedded submanifold of Euclidean space, the intrinsic tangent vector is represented by the ordinary derivative ${\gamma }^{\prime }(0)$, with its base point $p$ retained. A drawn arrow starting at $p$ represents that tangent vector; its endpoint $p+{\gamma }^{\prime }(0)$ need not lie on the embedded submanifold. Nor is a tangent vector required to be a unit tangent vector: a stationary smooth curve gives the zero vector, which belongs to the tangent space but has no direction to normalize.

Here are explicit calculations. In ${\mathbb{R}}^2$, $\gamma (t)=(t,t^2)$ and $\eta (t)=(t,2t^2)$ have different routes but the same tangent vector $(1,0)$ at $(0,0)$. For $f(x,y)=3x+y$, both give $v(f)=3$. For a circle of radius $R$, $c(t)=(R\cos (\omega t),R\sin (\omega t))$ has tangent vector $(0,R\omega )$ at $(R,0)$: changing $\omega$ changes the speed or reverses travel without changing the circle.

A reparametrization $\beta (s)=\gamma (h(s))$ with $h(0)=0$ gives, by the chain rule, ${\beta }^{\prime }(0)=h^{\prime }(0){\gamma }^{\prime }(0)$. A regular change of curve parameter has $h^{\prime }(0)\ne 0$. The singular choice $h(s)=s^3$ instead givesBoth describe the same parabola, whose tangent line at the origin is still the horizontal axis. The zero velocity diagnoses this singular change of curve parameter, whose inverse is not differentiable at zero; it does not show that the geometric tangent line is absent. The regular curve $\gamma$ supplies its nonzero direction.

The diagram is an analytical illustration of these exact parametric curves, with equal parameter increments marked on each route. It separates geometry from traversal: the tangent line depends on the route, while a velocity vector also depends on how the curve parameter advances.

A Taylor polynomial replaces a function near a chosen point by a polynomial with matching derivatives. If $f$ has derivatives through order $n$ at $a$, defineThe factorial denominator makes $T_{n,a}^{(j)}(a)=f^{(j)}(a)$ for $0\le j\le n$. This determines a unique polynomial of degree of a polynomial at most $n$. Order zero matches the value, order one also matches the tangent line, and order two also incorporates the second derivative. Matching more local information is useful because polynomials are easy to evaluate, differentiate, and integrate. The distinction between the finite Taylor polynomial and the infinite Taylor series is developed in OpenStax, Taylor and Maclaurin series.

For a computed example, let $f(x)=\sqrt{1+x}$ and expand at zero. Its first three derivatives giveAt $x=0.1$, the linear approximation is $1.05$ and the quadratic approximation is $1.04875$, compared with $\sqrt{1.1}\approx 1.048808848$. The quadratic correction captures the downward bending of the square root and reduces the absolute difference from about $0.00119115$ to $0.0000588482$.

The Taylor remainder turns this convenient approximation into a controlled calculation. On $[0,0.1]$, $f^{\prime \prime \prime }(x)=3/[8(1+x{)}^{5/2}]$ is positive and at most $3/8$. The Taylor theorem with Lagrange remainder therefore givesConsequently $1.04875<\sqrt{1.1}\le 1.0488125$ follows without using the displayed decimal evaluation as evidence. The Taylor polynomial and the Taylor remainder play different roles: the former supplies an answer that is easy to compute, while the latter specifies what accuracy is justified. Farther from the expansion point, the same low-degree polynomial need not remain accurate.

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.

A tangent line parallel to the vertical coordinate axis. For a regular level set $\Phi (x,y)=C$, the conditions $\partial \Phi /\partial y=0$ and $\partial \Phi /\partial x\ne 0$ make the tangent line vertical: the relation $d\Phi =0$ forces $dx=0$. The implicit function theorem then expresses $x$ locally as a differentiable function of $y$, although a graph $y(x)$ has no finite derivative there. The curve $x=y^2$ at the origin is a simple example.