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Polynomial interpolation makes Simpson's rule exact for every polynomial of polynomial degree at most two. To understand the extra polynomial degree, centre the coordinate at the midpoint: $y=x-m$, with endpoints $-h,h$. For any cubic polynomial $p$, let $p_2$ be its interpolating quadratic polynomial. The polynomial $p-p_2$ vanishes at all three quadrature nodes, sowhere $A$ is the coefficient of $x^3$ in $p$. As a function of $y$, this is an odd function, whose integral over the symmetric interval $[-h,h]$ is zero. Hence integrating $p_2$ also integrates $p$ exactly. This is the symmetry mechanism in the nodal-polynomial criterion for quadrature exactness.

In the worked example, the polynomial difference factors asIts definite integrals on $[-1,1/2]$ and $[1/2,2]$ are respectively $81/64$ and $-81/64$. Their cancellation concerns signed integrals, rather than pointwise accuracy of polynomial interpolation.

The degree of precision of a quadrature rule stops at three: on the same interval, the monomial $x^4$ gives $S(x^4)=69/8$, whereas ${\int }_{-1}^2{x}^{4}dx=33/5$. Their difference, exact integral minus approximation, is $-81/40$.

A cubic polynomial in one variable has polynomial degree three. Over a splitting field, its factorization of polynomials contains three polynomial factors of polynomial degree one when multiplicity of a root is counted.

For a degree-$n$ polynomial $p(X)=a_n{\prod }_{i=1}^n(X-r_i)$ over a field, with $a_n\ne 0$ and roots in an algebraic closure, its discriminant isIt is a polynomial with integer coefficients in the coefficients of $p$, and lies in the original coefficient field. Its vanishing is equivalent to a repeated root of a polynomial in an algebraic closure, even when the coefficient field itself contains no roots. This detects separability without solving the polynomial equation.^[ref]

For a quadratic polynomial, the expression is the quadratic discriminant $b^2-4ac$. For the cubic polynomial $X^3+bX+c$, it is $-4b^3-27c^2$. Thus $X^3-2$ has discriminant $-108$ over $\mathbb{Q}$, but after reduction to characteristic three its discriminant is zero and the resulting polynomial has a repeated root in its splitting field. Discriminants also indicate exceptional parameter values where distinct polynomial roots collide.

The first Lyapunov coefficient is the cubic radial coefficient in the center-manifold normal form of a smooth Hopf bifurcation, after an appropriate normalization. In $\dot{r}=\beta r+l_1{r}^3+\cdots$, its nonzero sign distinguishes stable and unstable bifurcating small cycles relative to the eigenvalue-crossing direction. If $l_1=0$, higher-order terms are required and the generic cubic Hopf theorem does not apply.

Let $A$ be real symmetric, let $b$ be a unit eigenvector satisfying $Ab=\lambda b$, and use scalar step sizes ${\alpha }_n$. The gradient of the displayed cubic polynomial function isOn the eigenline, $\nabla h(tb)=3\lambda t^{2}b$, so a gradient descent update remains on that same line and obeys the scalar recurrence in the title. Induction proves invariance for every iterate starting on the line, including an iterate that reaches zero. This is a statement about the geometry of the updates, not a convergence theorem: when $\lambda \ne 0$, the restriction $h(tb)=\lambda t^3$ is unbounded below in one direction.

A cubic polynomial over a field is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a linear polynomial factor, and conversely a root supplies one by the factor theorem.

The rational root theorem, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm 1,\pm 2,\pm 4$. Direct evaluation givesNone vanishes. Hence there is no rational root, provingThe same no-root argument alone would not decide a quartic polynomial, which could factor as two quadratic polynomials; its sufficiency here uses degree three.

Put $k=\beta /\gamma$. The nullclines of the space-clamped FitzHugh-Nagumo model areThe first nullcline is a downward-leading cubic polynomial, shifted vertically by $I$; the second nullcline is a line through the origin with slope $k>0$. Their intersections satisfyWriting $M=(1-\alpha +{\alpha }^2)/3$, differentiation givesIf $k>M$, $h$ increases strictly, so there is exactly one equilibrium. If $k=M$, $h$ is still strictly increasing, although its derivative vanishes at one point, so the equilibrium is again unique. If $0<k<M$, defineHere $h$ has a local maximum at $u_{-}$ and a local minimum at $u_{+}$; thereforeAt either endpoint there are two distinct equilibria, one a double root. These are the saddle-node bifurcation boundaries omitted by the generic one-or-three wording. At $k=M$ and $I=h((1+\alpha )/3)$ there is one triple root. This is the Cubic FitzHugh-Nagumo nullcline criterion.

There is a consistency issue in the printed model. If $w\equiv w_0$ is imposed while retaining its evolution equation, then $0=\beta v-\gamma w_0$ forces $v\equiv \gamma w_0/\beta$. Thus the full two-equation system has no nonconstant travelling wave with exactly constant $w$. Its constant solution exists only when $I=w_0-f(\gamma w_0/\beta )$. The intended wave calculation must treat the recovery variable as externally frozen, leaving the scalar reaction-diffusion equationFor this frozen-recovery reduction, letThe cubic polynomial $f$ has a local minimum at $a_{-}$ and a local maximum at $a_{+}$. Hence the conditiongives three distinct real roots $r_1<r_2<r_3$ of $g$, withThe outer roots are stable for the scalar spatially uniform dynamics and the middle root is unstable. Seek a travelling wave $v(x,t)=V(z)$, where $z=x-ct$. The travelling-wave reduction of a reaction-diffusion system givesLet $\delta =r_3-r_1$ and chooseIt has $V(-\infty )=r_3$, $V(+\infty )=r_1$ andSubstitution leavesConsequently the frozen-recovery cubic front isReflection $x\mapsto -x$ reverses the front and its speed. A nonzero speed requires $r_2\ne (r_1+r_3)/2$; equality gives a stationary front and is the Maxwell balance for a scalar reaction-diffusion front. Since the roots sum to $1+\alpha$, this stationary value is $I=w_0-f((1+\alpha )/3)$. For the simple admissible choice $I=w_0$, the roots are $0,\alpha ,1$ andThis particular front moves when $\alpha \ne 1/2$. At $\alpha =1/2$, other currents inside the displayed open interval give moving fronts.

Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_{\infty }(V)$; the two slower gates have the same time constant, ${\tau }_n(V)={\tau }_h(V)=\tau (V)$; and their limiting open fractions have a voltage-independent sum, $n_{\infty }(V)+h_{\infty }(V)=K$. The constant-sum assumption is essential: equality of two time constants by itself does not preserve a moving voltage-dependent sum.

For $q=n+h$, addition of the two gate ordinary differential equations givesThe specified initial condition is $q(0)=K$. Uniqueness for this first-order linear differential equation gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two ordinary differential equations areThese assumptions give a planar reduction of the Hodgkin-Huxley model. The FitzHugh-Nagumo model keeps its fast activation and slow recovery variable geometry, approximating the voltage response by a cubic polynomial and the recovery response by a linear function after nondimensionalization. The polynomial equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.

Take $\gamma >{\gamma }_c$ and let $\epsilon \to 0^{+}$ with the other parameters fixed. The nullclines are the cubic polynomial graph $w=f(v)$ and the line $w=\gamma v$. In the phase plane, $v$ increases below the cubic polynomial graph and decreases above it; $w$ increases below the line and decreases above it. The fast ordinary differential equation, in time $s=t/\epsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.

At $w=0$, the fast roots are $0,a,1$. Their linear stability follows from $f^{\prime }(0)=-a<0$, $f^{\prime }(a)=a(1-a)>0$ and $f^{\prime }(1)=a-1<0$. Thus a fixed $0<v^{\ast }<a$ falls towards the resting branch, whereas a fixed $a<v^{\ast }<1$ rises towards $v=1$. If $v^{\ast }>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading singular perturbation threshold; finite-$\epsilon$ corrections matter for initial values approaching $a$ with $\epsilon$.

LetThese are the roots of $f^{\prime }(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the critical manifold are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.

For this cubic polynomial, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R{)}^2(v-v_J)$, comparison of the quadratic coefficients givesThus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the phase plane and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.

The four-stage description concerns the singular limit. Close to a simple fold point of a critical manifold, balancing the local quadratic activation term with the slow drift gives $\Delta v=O({\epsilon }^{1/3})$ and $\Delta t=O({\epsilon }^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.

Move the constant term to the left and factor the cubic polynomial:Setting each factor to zero gives all three polynomial roots:

Write the remaining quadratic polynomial as $x^2+bx+c$. Comparing coefficients in $(x-2)(x^2+bx+c)$ with the given cubic polynomial gives $b-2=-1$ and $-2c=-2$, hence $b=c=1$. The coefficient of $x$ is then $c-2b=-1$, as required. Thus

Let $p(x)=x^3-5x-1$. Its values at useful points areSince a polynomial is a continuous function, the intermediate value theorem gives one polynomial root in each of the disjoint intervals $(-3,-2)$, $(-1,0)$ and $(2,3)$. A nonzero cubic polynomial has at most three distinct polynomial roots. Hence it hasThese three distinct polynomial roots already exhaust its degree of a polynomial, so each has multiplicity of a root one.

The remainder in part (b) involves an unspecified $\xi =\xi (x)$. Its existence for every $x$ does not ensure a differentiable selection. Even if such a selection existed and $f$ had an additional derivative, the product rule and chain rule would giveThe first term cannot be discarded or bounded from the given information. Neither $f^{(n+2)}$ nor ${\xi }^{\prime }$ is supplied by the original theorem.

For an explicit demonstration, interpolate the cubic polynomial $f(x)=x^3$ at zero and one. Then $p_1(x)=x$, andDifferentiating correctly givesTreating the remainder coefficient $x+1$ as a constant would omit $x(x-1)$ and give a wrong answer at every interior point. An existential intermediate point cannot be frozen when differentiating. Part (i) instead applies Rolle's theorem directly to the polynomial interpolation error.

A Newton-Cotes quadrature integrates a Lagrange interpolation polynomial at equally spaced quadrature nodes. For the Newton-Cotes closed quadrature, the endpoints are included: write $x_j=a+jh$, $h=(b-a)/n$, and useThese quadrature weights depend on the quadrature nodes, not on the sampled function. Open Newton-Cotes quadrature uses equally spaced interior quadrature nodes instead.

For Simpson's rule, use $n=2$ and set $x=a+ht$, so $0\le t\le 2$. The Lagrange cardinal polynomials becomeIntegrating these quadratic polynomials gives $1/3$, $4/3$, $1/3$, respectively. The change of variable contributes $h$, soIts degree of precision of a quadrature rule is three: it integrates every quadratic polynomial by construction, and also the cubic nodal polynomial $(x-a)(x-a-h)(x-b)$, whose integral vanishes by odd symmetry about $a+h$. Any cubic polynomial differs from its quadratic polynomial interpolant by a constant multiple of this nodal polynomial.

At $q=3$,The $y$ factor is now a cubic polynomial. Simpson's rule still integrates it exactly because its degree of precision of a quadrature rule is three. Exactness of tensor-product quadrature depends on the polynomial degree in each variable, rather than their sum.

Write $P(N)=f(N,0)$. Its three positive zeros of a function are $1/2,1,2$, and expansion and differentiation giveThus its local minimum and maximum occur respectively atwith valuesThe function graph approaches the value $1$ as $N\downarrow 0$, decreases through its first zero to the minimum, increases through its second zero to the maximum, and then decreases through $2$ to $-\infty$. This fixes the shape and all axis crossings of the cubic polynomial.

Changing $a$ translates the function graph vertically without moving either stationary point. For $a=1/2$ both extrema are above the horizontal axis: their heights are approximately $0.421109$ and $0.764076$. For $a=-1/2$ both are below it, at approximately $-0.578891$ and $-0.235924$. The function graph for $a=1/2$ therefore has only its final positive-axis crossing, whereas the function graph for $a=-1/2$ has only its first positive-axis crossing. The original plots mark the extrema and also show the equilibrium point branches used below.

With three distinct positive roots of an equation, the negative-leading cubic polynomial has the sign patternEquivalently its derivative is negative at the first root of an equation, positive at the middle root of an equation, and negative at the final root of an equation. The phase line arrows therefore point toward the two outer equilibrium points and away from the middle one:The middle equilibrium point separates the two basin of attraction in the bistability mathematical model parameter interval.

Use mathematical induction on the iteration. Initially $x_0=b=1\cdot b$. If $x_n={\lambda }_{n}b$, then $A{x}_n={\lambda }_{n}b/2$, $x_n^{T}b={\lambda }_n$, and $x_n^{T}A{x}_n={\lambda }_n^2/2$. ThereforeA gradient descent step with any scalar step length ${\alpha }_n$ consequently has the formThus every iterate remains on the same eigenline, withThis proves that gradient descent preserves an eigenline for a cubic product. The statement is invariance, not convergence to a minimum: along the negative direction of this line the cubic polynomial objective is unbounded below.

Simpson's rule approximates a definite integral by integrating the quadratic polynomial obtained by polynomial interpolation at two endpoints and their midpoint. For an interval $[a,b]$ with $a<b$, write $m=(a+b)/2$ and $h=(b-a)/2$. The three quadrature nodes are $a,m,b$, and the quadrature rule isThus its quadrature weights are $h/3,4h/3,h/3$. Their sum is $2h=b-a$, as required for exactness on constant functions. This is the three-node Newton-Cotes closed quadrature; its degree of precision of a quadrature rule is three, even though the interpolating polynomial has polynomial degree at most two.^[ref]

For an original worked example, take the cubic polynomial $f(x)=x^3-2x+4$ on the interval $[-1,2]$. Here $m=1/2$ and $h=3/2$, so the function evaluations giveSimpson's rule givesAn antiderivative supplies an independent exact definite integral:The quadratic polynomial through the same quadrature nodes is $p_2(x)=\frac{3}{2}{x}^2-\frac{1}{2}x+3$. It differs from $f$ between the quadrature nodes, but its definite integral agrees exactly. The picture shows why: the signed polynomial interpolation error has two lobes with opposite integrals.

Three function evaluations therefore capture more information about an integral than about the whole function. When an antiderivative is unavailable, composite Simpson's rule repeats this inexpensive local construction over a partition of an interval. For a sufficiently smooth function, halving the node spacing reduces its truncation error by approximately a factor of sixteen.