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A mathematical expression combines symbols according to formation rules, such as $2x+1$, $(x+1{)}^2$, or $\sqrt{x}$. With values for its variables and meanings for its algebraic operations, it can be evaluated. An equation relates two mathematical expressions; an expression alone need not assert anything.

For example, $2x+3$ evaluates to seven at $x=2$ and nine at $x=3$. Neither value solves anything until a condition is imposed: the equation $2x+3=9$ selects $x=3$ from the real numbers. This distinction is useful when moving from a formula to a question about which inputs satisfy it. Evaluation by replacing the variables with assigned values is described in College Algebra 2e, §1.1: Evaluating Algebraic Expressions.

Different written mathematical expressions can have the same value for every input in a common domain of a function. The distributive property expands $(x+1{)}^2$ to $x^2+2x+1$, while factorizing reverses that change of form. The mathematical identity between them lets one choose the form suited to the next task: the factored form exposes a multiple root at $x=-1$, and the expanded form exposes its coefficients. Simplifying should preserve values on the stated domain, rather than silently enlarging it.

If a sufficiently smooth function has the form $f(x)=(x-\alpha {)}^{m}q(x)$ with $m>1$ and $q(\alpha )\ne 0$, then the Newton root-finding iteration satisfiesThus the ratio of successive nonzero errors tends to $(m-1)/m$: the usual iteration has only linear convergence. Cancelling the common factor in $f/f^{\prime }$ produces a continuous extension at the multiple root, although the original quotient is undefined there. An implementation should stop upon finding an exact zero of a function.

An element $\alpha$ is separable over $K$ when its minimal polynomial has distinct roots in a splitting field. A polynomial has a repeated root exactly when it has a common root with its formal derivative. Hence
$\gcd (f,f^{\prime })=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the minimal polynomial of $\beta$. Repeatedly factor through the Frobenius untilwhere $g^{\prime }\ne 0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha ={\beta }^{p^h}$ has separable minimal polynomial $g$. PutThen $M/K$ is separable andEvery $\gamma \in L=K(\beta )$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity givesso $L/M$ is purely inseparable.

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts ${\beta }^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.

At $\kappa ={\kappa }_c=1/2$, the indicial equation has the multiple root $s=-1/2$. The general axial component isSince $B_{\varphi }=-2R{B}_z^{\prime }$, the azimuthal component isConsequentlyas $R\to \infty$; the same ratio is identically one when $D=0$. A magnetic field line has tangent parallel to $\mathbf{B}$, so its asymptotic angle $\vartheta$ with the $z$-axis satisfiesHence the field lines become helices making the constant anglewith the axis.

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Write $f(x)=a_m{\prod }_{i=1}^m(x-{\alpha }_i)$ and $g(x)=b_n{\prod }_{j=1}^n(x-{\beta }_j)$ in an algebraic closure, counting multiplicity of a root. The coefficient-row convention above givesThis makes the shared-root of a polynomial test transparent: a factor vanishes precisely when one ${\alpha }_i$ equals one ${\beta }_j$. Swapping the two polynomials reverses the sign of all $mn$ differences, giving $\mathrm{Res}(g,f)=(-1{)}^{mn}\mathrm{Res}(f,g)$.^[ref]

The value measures an algebraic product, not a normalized geometric distance. Replacing $f$ by $cf$ for nonzero $c\in K$ leaves its roots of a polynomial unchanged but multiplies the resultant by $c^n$. Therefore a small numerical resultant alone does not quantify how close two sets of roots of a polynomial are. Exact zero testing and numerical sensitivity are different questions.

A multiple root of a polynomial is also a root of a polynomial of its formal derivative. In characteristic zero, for $\mathrm{deg}f=m\ge 2$, the associated discriminant isFor example, $h(x)=x^2+s$ gives $\mathrm{Res}(h,2x)=4s$ and $\mathrm{Disc}(h)=-4s$. The two roots of a polynomial merge precisely at $s=0$.^[ref]

Exact reconstruction does not promise that small errors in power sums produce proportionally small errors in individual polynomial roots. For two roots, let $p_1=2$ and $p_2=2+\delta$. Newton's identities give $e_1=2$ and $e_2=1-\delta /2$, so the monic polynomial is $(z-1{)}^2-\delta /2$. Its polynomial roots are $1\pm \sqrt{\delta /2}$. A zero value of $\delta$ gives a multiple root; positive $\delta$ splits it into two real roots, and negative $\delta$ splits it into a complex-conjugate pair.

With $\delta =2{\epsilon }^2>0$, the second power sum changes by $2{\epsilon }^2$ while each polynomial root moves by $\epsilon$. The ratio $1/(2\epsilon )$ grows without bound as $\epsilon$ tends to zero. Thus coefficients can stay close while individual root locations are more sensitive. This is a reason to distinguish recovering coefficients by Newton's identities from subsequently approximating the polynomial roots.

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.