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An amplification root is a root of a polynomial governing a geometric mode $y_n={\xi }^n$ of a numerical recurrence on the Dahlquist test equation. Its modulus measures damping or growth per step. A repeated root produces additional modes such as $n{\xi }^n$; therefore a unit-circle root can be permitted by a bounded-root condition only when it is simple. The principal root approximates $e^z$ near zero, while additional roots can describe parasitic amplification roots.

Every square matrix satisfies its own characteristic polynomial. More explicitly, for an $n$-by-$n$ matrix $A$ over a commutative ring with identity, formThenHere $0$ is the zero matrix, and the constant term multiplies the identity matrix. This is a polynomial evaluated at a matrix; it is stronger than saying that each eigenvalue is a root of a polynomial. Neither diagonalizability nor distinct eigenvalues is required.^[ref]

The practical payoff is that arbitrarily high matrix powers can be reduced to a combination of $I,A,\dots ,A^{n-1}$. These powers arise when repeatedly applying a linear map, solving a linear recurrence relation, or evolving a finite state space model. The characteristic polynomial turns that repeated action into a fixed-order scalar recurrence. The minimal polynomial may reduce the number of required powers further.

For the homogeneous linear differential equation $P({\partial }_x)u=0$, a characteristic root is a root of a polynomial $P(r)=0$. A simple root supplies an exponential solution $e^{rx}$; a root of multiplicity $m$ supplies $x^j{e}^{rx}$ for $0\le j<m$. The sign of the real part of $r$ determines spatial growth or decay.

For the quadratic polynomial $p(x)=2x^2-6x+5$, the discriminant is $(-6{)}^2-4(2)(5)=-4$, so its roots of a polynomial are the complex conjugatesTheir sum is $3=-(-6)/2$, and their product is $(9+1)/4=5/2$, matching the constant coefficient divided by the leading coefficient. More revealingly, $r_{+}{r}_{-}=r_{+}\overline{r_{+}}=|r_{+}{|}^2$: the product is the square of a complex modulus. Coefficients in the real numbers therefore need not imply polynomial roots in the real numbers.

For any polynomial with coefficients in the real numbers, $p(\overline{z})=\overline{p(z)}$. Thus a nonreal root of a polynomial brings its complex conjugate along with the same multiplicity of a root. This follows by applying complex conjugation to the polynomial expression and its factors. It explains why sums of complex conjugate terms can produce coefficients in the real numbers.

A root of a polynomial lying in the complex numbers. It may be real: a nonreal complex root has nonzero imaginary part.

For two polynomials over the complex numbers $p,q$ of equal positive degree of a polynomial $n$ with no common root of a polynomial, there are unique polynomials $u,v$ of degree of a polynomial less than $n$ such that $up+vq=1$. To prove the bound, let $V_m$ be the vector space of polynomials of degree of a polynomial at most $m$, including the zero polynomial. The linear map $(a,b)\mapsto ap+bq$ from $V_{n-1}\times V_{n-1}$ to $V_{2n-1}$ has zero kernel of a linear map: if $ap=-bq$, each root of a polynomial of $p$, counted with its multiplicity of a root, must occur in $b$ because it does not occur in $q$, forcing $b=0$ by the degree of a polynomial bound, and then $a=0$. Both vector spaces have dimension $2n$, so the rank-nullity theorem makes the linear map a bijection, and $1$ has a unique preimage.

A root of a polynomial is double when its multiplicity of a root is exactly two. Equivalently, $p(x)=(x-x_0{)}^{2}q(x)$ with $q(x_0)\ne 0$. A generic small perturbation splits it according to square-root splitting of a double polynomial root.

If $P=z^{d}p$ and $p$ is nonnegative on the unit circle, a root of a polynomial $\zeta =e^{i{\theta }_0}$ there has even multiplicity of a root. Indeed $p(e^{i\theta })$ is a real analytic nonnegative function, whose first nonzero term in its Taylor series at a zero must have even order. The map $\theta \mapsto e^{i\theta }-\zeta$ has a simple zero, and multiplication by $e^{-id\theta }$ is nonvanishing, so the order equals the polynomial multiplicity.

For a polynomial $f$ over a field, $f(a)=0$ exactly when $X-a$ divides $f(X)$. Polynomial division by the monic linear factor givesThe constant polynomial remainder is thus the value obtained by evaluation. For $p(X)=X^3-3X^2+4$, the root of a polynomial $2$ gives factor $X-2$ and polynomial quotient $X^2-X-2$. The latter also vanishes at $2$, explaining the repeated factor. This provides a practical way to move between evaluation and factorization.

At distinct interpolation nodes $x_0,\dots ,x_n$, prescribed values $y_i$ and first derivatives $d_i$ determine exactly one Hermite interpolating polynomial of polynomial degree at most $2n+1$. The homogeneous data force a double root of a polynomial at every interpolation node. Hence the kernel of a linear map given by these $2n+2$ conditions is zero: a nonzero polynomial with those double roots of a polynomial would have polynomial degree at least $2n+2$. Equal finite dimensions of a vector space then give existence as well as uniqueness. This interpolation polynomial is distinct from the Hermite polynomials forming a classical orthogonal polynomial family.

Consider a neutral monoprotic acid $\mathrm{HA}$ dissolved in water, with no added salts or other acids. Assume dilute ideal thermodynamic activities, fixed temperature, water thermodynamic activity one, and negligible self-ionization of water. Write $C_T$ for its analytical concentration, $c=C_T/c^{\circ }$ with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and $\alpha$ for its weak acid dissociation fraction. Conservation of acid units and electric charge givesHere ${\mathrm{A}}^{-}$ is the conjugate base, and ${\mathrm{H}}^{+}$ is conventional shorthand for hydrated hydrogen ions, represented by hydronium in this model. Substitution into the dimensionless acid dissociation constant givesSolving this quadratic equation and rejecting its negative root of a polynomial gives an exact result within the stated model:The second form avoids subtracting nearly equal numbers. Unlike the approximation $\alpha \approx \sqrt{K_a/c}$, it retains depletion of neutral weak acid and always satisfies $0<\alpha <1$ for positive $K_a,c$. The broader weak acid mass and charge balances restore self-ionization of water when that omitted contribution matters.

For an original worked example, stipulate $K_a=10^{-5}$ without identifying a real substance. At $C_T=10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$, the weak acid dissociation fraction is approximately $0.0311267$, or $3.11267\%$, and $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=3.11267\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. A hundredfold dilution to $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ raises the fraction to $0.270156$, or $27.0156\%$, but lowers $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to $2.70156\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$. A completely ionized strong acid at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ would have $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\approx 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$: less than the more concentrated weak acid in the first calculation, despite its greater converted fraction.

The directions of both changes follow without relying on the plot. Put $x=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=c\alpha$. Rearranging the chemical equilibrium relation givesThus smaller $c$ forces smaller $x$, which in turn makes $\alpha$ larger. The derivative proves that total hydronium concentration and converted fraction move in opposite directions during dilution. In this model, adding solvent also makes additional acid units ionize at fixed total acid amount of substance, even though their ions are spread over more volume. That distinction explains why “more ionized” alone does not predict greater electrical conductivity.

Do not extend the no-background model to unlimited dilution: its prediction $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\to 0$ omits water's own ions. The weak acid mass and charge balances include that background. Added salts, other acids or bases, and nonideal activity coefficients also change the relation between analytical concentration, converted fraction and hydronium concentration.

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]

For positive matter and vacuum fractions with ${\Omega }_m+{\Omega }_{\Lambda }=1$, redshift drift vanishes when $s^2={\Omega }_m{s}^3+{\Omega }_{\Lambda }$, $s=1+z$. Factoring out the trivial observer root $s=1$ leaves ${\Omega }_m{s}^2-{\Omega }_{\Lambda }s-{\Omega }_{\Lambda }=0$, giving the displayed positive root of a polynomial. It lies at a positive redshift exactly when ${\Omega }_m<2/3$. It is independent of the Hubble constant at fixed density fractions and is not the redshift of the acceleration-to-deceleration transition.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

Write a rational root of a polynomial as $\gamma =a/b$ in lowest terms with $b>0$. Substituting into the monic integer-coefficient polynomial and multiplying by $b^d$ givesEvery term other than $a^d$ is divisible by $b$, so $b\mid a^d$. But $\gcd (a,b)=1$ implies $\gcd (a^d,b)=1$, and hence $b=1$. ThusThis is the monic case of the rational root theorem. The monic polynomial assumption is essential because it removes any leading coefficient that could supply prime factors of the denominator.

For $p=3$, a root would have valuation zero because five is a unit. It would therefore reduce to a unit cube modulo nine. The only cube residue classes modulo nine are $0,1,-1$, none equal to five. Thus no root exists in ${\mathbb{Q}}_3$.

For $p=5$, the valuation equation would be $3v_5(x)=1$, impossible because the valuation of a nonzero element of ${\mathbb{Q}}_5$ is an integer. Thus no root exists in ${\mathbb{Q}}_5$.

For $p=13$, $7^3\equiv 5(\mathrm{mod}13)$ and the derivative $3\cdot 7^2\equiv 4\not\equiv 0(\mathrm{mod}13)$. Hensel's lemma lifts this simple root to a root in ${\mathbb{Z}}_{13}\subseteq {\mathbb{Q}}_{13}$. Thus a root exists in ${\mathbb{Q}}_{13}$.

Let $\alpha ,\beta ,\gamma$ be the nonzero reductions of $a,b,c$. The square set in ${\mathbb{F}}_p$ has $(p+1)/2$ members, including zero. Multiplication by $\alpha$, and separately the transformation $z\mapsto \gamma -\beta z$, preserve this cardinality. Hence the two sets in the hint have total cardinality $p+1>p$ and must intersect. This gives a solution $(x_0,y_0)$ modulo $p$.

The solution cannot have both coordinates zero since $\gamma \ne 0$. If $x_0\ne 0$, choose any integral lift of $y_0$ and apply Hensel's lemma to the polynomial $a{X}^2+b{y}_0^2-c$ in $X$: its derivative at the reduced root is $2\alpha x_0\ne 0$. If $y_0\ne 0$, use the other variable instead. The nonzero derivative uses $p\ne 2$. Thus a solution lifts to $\boxed{x,y\in {\mathbb{Z}}_p}$ as required.

Here $4A^3+27B^2=31$. A nonzero ordinate of a torsion point must have its square dividing 31, so $y=\pm 1$. The curve equation then becomes $x(x^2+1)=0$, whose only integer solution is $x=0$. No point with $y=0$ exists over the rationals: the monic cubic $x^3+x+1$ has no integer root, since its only possible integer roots are $\pm 1$ and neither works.

It remains to test $(0,\pm 1)$. Doubling either gives abscissa $(1/(\pm 2){)}^2=1/4$, which is not an integer. If the original point were torsion its double would satisfy the integrality part of the Nagell–Lutz theorem, a contradiction. ThereforeThe rational point $(0,1)$ consequently has infinite order, so its multiples give infinitely many distinct rational points.

Take $n\ge 1$, as intended by the stated normalization. The Chebyshev polynomial recurrence $T_{r+1}=2x{T}_r-T_{r-1}$, with $T_0=1$ and $T_1=x$, shows by mathematical induction that $T_n$ has polynomial degree $n$ and leading coefficient $2^{n-1}$. Consequently $f_n=2^{1-n}{T}_n$ is a monic polynomial. At the ordered Chebyshev extreme points $t_j=-\cos (j\pi /n)$,and $\Vert f_n{\Vert }_{\infty }=2^{1-n}$.

Suppose a monic polynomial $q$ of the same polynomial degree had $\Vert q{\Vert }_{\infty }<2^{1-n}$. The difference $f_n-q$ has degree at most $n-1$ because its leading coefficients cancel. Its signs at the $n+1$ ordered extreme points strictly alternate. The intermediate value theorem supplies a distinct root of a polynomial between each consecutive pair, giving at least $n$ roots. This contradicts its degree bound. Hence the monic Chebyshev extremal polynomial satisfiesAt $n=0$ the printed formula instead gives $f_0=2$, which is not monic; the degree-zero minimizer is the constant one. This endpoint case requires the separate normalization $f_0=1$.

The differential operator is $(D-1{)}^2$, with $D=d/dt$. Because $e^t$ already occurs in the complementary function with a repeated root of a polynomial, an unmodified trial $C{e}^t$ gives zero. Put $f_p=e^{t}h(t)$. The product rule yieldsThus $h^{\prime \prime }=2$, and we may choose $h=t^2$, givingThe other terms arising on twice integrating $h^{\prime \prime }=2$ give $(C_1+C_{2}t)e^t$, exactly the complementary function, so they need not be included in the particular integral.