Sets, Relations and Functions

Functional equations

A functional equation is an equation whose unknown is a function, and the aim is to recover f from a relation that holds for every admissible value of the variable. Three techniques cover most problems at this level: direct substitution, elimination, and structural comparison.

In direct substitution, a shifted argument such as f(x+1) is handled by putting t=x+1, so that x=t-1. Rewriting the expression in t and then renaming t as x gives f directly. For instance, if f(x+2)=x^2, then f(t)=(t-2)^2, so f(x)=(x-2)^2.

When an equation links f(x) with f\left(\frac{1}{x}\right), replace x by \frac{1}{x} to obtain a second equation in the same two unknowns, and solve the pair like simultaneous linear equations. If the equation has the form af(x)+bf\left(\frac{1}{x}\right)=R(x), the determinant of the resulting system is a^2-b^2, so the method works only when a\neq \pm b. The solution typically contains terms such as \frac{1}{x} together with x.

Structural arguments are used for composed functions. If f is a polynomial and f\circ f is given, compare degrees using \deg(f\circ f)=(\deg f)^2, then compare coefficients of like powers of x to fix the remaining constants. Once f is determined, any requested derivative or value is computed directly.

Exponential-type equations such as f(x+y)=f(x)f(y), together with a given value at 1, force f(n)=c^n for natural numbers n, by induction. Sums of such values are finite geometric series, and \sum_{k=a}^{b}r^k=\frac{r^{b+1}-r^{a}}{r-1} for r\neq 1 can be used to solve for an unknown upper limit. When the arguments are linear in x, the sum is evaluated term by term after substitution, and a product such as 2^x f(x) simplifies to a power with a new base that again forms a geometric series.

For a quotient-type equation f(x-y)=\frac{f(x)}{f(y)}, putting y=1 gives f(x)=f(1)\,f(x-1), a recurrence whose solution is geometric. Setting x=y gives f(0)=1, which fixes the starting value of the recurrence.

When the composite g\circ f and the inner function f are known, determine g by writing t=f(x), expressing x in terms of t with a branch consistent with the domain, and substituting. For example, if g(f(x))=2x+1 and f(x)=x^2 for x\ge 0, put t=x^2 so that x=\sqrt{t}, giving g(t)=2\sqrt{t}+1. To evaluate g at a point without first finding the formula, solve f(x)= that point and use g(f(x)). If f(x)=x^2-1 and the point is 15, then x=4 and g(15)=(g\circ f)(4).

Rationalising a surd, \frac{1}{\sqrt{x+1}-\sqrt{x-1}}=\frac{\sqrt{x+1}+\sqrt{x-1}}{2}, is the first step whenever a function is defined through an integral that involves such differences, after which the power rule fixes the coefficient of each fractional power.

Identities for explicit formulas are verified by direct algebra. A ratio such as \frac{f(x)}{f(x+1)} for a rational rule can simplify to a negative multiple of f(x^2), so signs must be tracked with care. Expressions with surds are simplified by factoring, for example 1-x=(1-\sqrt{x})(1+\sqrt{x}), which makes cancellations visible; the total then often collapses to 0.

• Substitution — replace the variable by a new symbol and rewrite the equation, e.g. f(x+1) with t=x+1.
• Reciprocal pairing — use x\to\frac{1}{x} to create a second equation, e.g. solve af(x)+bf(1/x)=R(x) only when a^2\neq b^2.
• Geometric recurrence — from f(x)=c\,f(x-1) conclude f(n)=c^n.

Relations as sets of ordered pairs

A relation R from a set A to a set B is a subset of the Cartesian product A\times B, and xRy means that the ordered pair (x,y) belongs to R. A relation in A is a subset of A\times A. Every function is therefore a relation, but a relation is a function only when each element of its domain has exactly one image.

To list a relation defined by an equation, find all ordered pairs in the given sets that satisfy it. For a linear equation ax+by=c over positive integers, write x=\frac{c-by}{a} and keep only those values of y for which c-by is positive and divisible by a. Divisibility conditions, such as parity, often decide which values of y are admissible, and the remaining pairs are then counted directly.

A relation given by a quadratic expression is best studied after factoring. For example, x^2-5xy+4y^2=(x-y)(x-4y), so the relation is the union of the two sets of pairs satisfying x=y and x=4y. Reflexivity holds because (x-x)(x-4x)=0 for every x. Symmetry fails because a pair such as (4,1) can satisfy the relation while (1,4) does not, and transitivity fails for a similar reason.

The three standard properties are reflexivity ((a,a)\in R for all a), symmetry ((a,b)\in R\Rightarrow(b,a)\in R) and transitivity ((a,b),(b,c)\in R\Rightarrow(a,c)\in R). The inverse relation R^{-1}=\{(y,x):(x,y)\in R\} inherits all three properties. Reflexivity and symmetry transfer directly, and for transitivity, (x,y),(y,z)\in R^{-1} gives (y,x),(z,y)\in R, hence (z,x)\in R and so (x,z)\in R^{-1}.

Relations defined by powers, such as x^2=y^3 on \mathbb{N}, must be tested against each property separately. Reflexivity requires x^2=x^3, which holds only for x=1. Symmetry fails since 8^2=4^3 but 4^2\neq 8^3. Transitivity can fail even when two linked pairs exist, because the pair joining their outer elements need not satisfy the relation.

For a relation R from A to B, the domain is the set of first coordinates that occur, the range is the set of second coordinates that occur, and the codomain is B. The range is always a subset of the codomain and equals it only when every element of B is used. For a relation defined by 4x-3y=1 on a finite set, the admissible y lie in a single residue class modulo 4, so the domain is a proper subset of A.

Counting pairs in A\times B that satisfy a condition is done by fixing one coordinate at a time. For a condition such as |x|<y, the value y=0 contributes nothing because |x| is never negative. For a value of y exceeding the largest possible |x|, every element of the first set pairs with it.

A relation fails to be a function when some first coordinate has two different second coordinates, or when some element of the domain has none. For a relation of the form \{(xy,\,x-y)\}, the first coordinate 0 can be paired with every integer -y, which already rules out a function. Similarly, a value that can be produced by two different pairs with different second coordinates disqualifies a relation of the form \{(xy,\,x+y)\}.

Set operations laws

Set operations obey laws parallel to those of arithmetic, but cancellation does not carry over. Union and intersection are commutative and associative, and each distributes over the other: A\cup(B\cap C)=(A\cup B)\cap(A\cup C) and A\cap(B\cup C)=(A\cap B)\cup(A\cap C). De Morgan's laws, (A\cup B)'=A'\cap B' and (A\cap B)'=A'\cup B', convert complements of unions and intersections.

The difference A-B equals A\cap B'. The condition A-B=\varnothing says only that A\subseteq B. The sets are equal only if B\subseteq A as well, so an empty difference must never be read as equality.

Equality of sets can be tested through a single operation. A\cup B=A\cap B holds exactly when A=B, because every element of the union must lie in the intersection. Likewise the symmetric difference A\triangle B=(A-B)\cup(B-A) is empty exactly when A=B.

Cancellation fails for both operations: A\cap B=A\cap C does not force B=C, and A\cup B=A\cup C does not force B=C either. For intersections, take A=\varnothing with B and C two different singletons. For unions, take A=\{1\}, B=\{2\} and C=\{1,2\}. Any proposed cancellation must therefore be checked against a concrete counterexample.

Identities involving differences are verified by splitting A\cup B into the disjoint pieces A-B, B-A and A\cap B. In this decomposition, A\cup(B-A)=A\cup B holds, and (A\cup B)-(A-B)=B holds because only the elements of A-B are removed. A statement such as A=(A\cup B)\cup(A-B) is not an identity, since the right side is simply A\cup B, which equals A only when B\subseteq A.

The three pieces A-B, B-A and A\cap B together recover the whole union, so (A-B)\cup(B-A)\cup(A\cap B)=A\cup B for all sets. Subtracting this union from A\cup B leaves the empty set, which is a quick shortcut for expressions of this kind.

Complements interact with differences in a controlled way. Since A-B=A\cap B', we have A-B=B'-A', because B'-A'=B'\cap A. Replacing a set by its complement in only one place changes the result, however: A-B'=A\cap B, while A'-B=A'\cap B', and these are not equal in general.

• A-B=A\cap B' — the elements of A outside B; e.g. \{1,2,3\}-\{2\}=\{1,3\}.
• A\triangle B=(A-B)\cup(B-A) — empty only when A=B.
• Cancellation — A\cup B=A\cup C does not imply B=C.

Range of a function

The range (image) of f:A\to B is the set \{f(x):x\in A\}, which is a subset of the codomain B. To count the elements of the range, map each element of the domain and collect the distinct outputs, since repeated values are counted only once. For example, if each integer is sent to its largest prime factor, many inputs share the same output, so the range is just the set of distinct primes that occur.

Since \log_{10} is strictly increasing, the minimum of \log_{10}u occurs where u is least, so the extremum of the function is found from the extremum of the expression inside. Completing the square, x^2+2x+11=(x+1)^2+10, whose least value is 10 at x=-1. Hence the minimum of the logarithm is \log_{10}10=1.

For a linear fractional function f(x)=\frac{ax+b}{cx+d} with c\neq 0 and ad-bc\neq 0, the domain excludes x=-\frac{d}{c} and the range excludes y=\frac{a}{c}. Solving y=f(x) gives x=\frac{b-dy}{cy-a}, which is undefined exactly at y=\frac{a}{c}. For f to be onto, the codomain must be this reduced range.

Functions defined through |x| are handled by splitting the domain at the point where each absolute value changes sign, computing the range on each piece, and taking the union of the results. Each piece is usually a linear function whose range is an interval, and the union of these intervals may be a half-line, which is how a piecewise function gets its range.

For an expression formed by adding a constant to a bounded trigonometric term, the range follows from the bounds of that term alone. Since 0\le \sin^2 x\le 1 for every real x, the expression c+\sin^2 x takes exactly the values in [c,\,c+1], and both endpoints are attained.

Maxima of expressions with square roots are found by squaring. For \sqrt{u}+\sqrt{v} with u,v\ge 0, squaring gives u+v+2\sqrt{uv}, so the maximum depends on both u+v and uv. In a typical case with u=2-x and v=2+x, the domain is [-2,2] and the squared expression is 4+2\sqrt{4-x^2}, which is largest at x=0, giving the maximum \sqrt{8}.

• Range — the set of actual outputs, always a subset of the codomain; e.g. \{f(x)\}\subseteq B.
• Onto condition — for \frac{ax+b}{cx+d}, the codomain must exclude \frac{a}{c}.
• Bounded term — 0\le\sin^2 x\le 1 gives the range [c,c+1] for c+\sin^2 x.

Domain of functions

The domain of a real-valued function is the set of all real numbers x for which the formula gives a real value. To find it, examine each part of the formula, write down its restriction, and take the intersection of these restrictions. The main restrictions are: a denominator must be nonzero; the radicand of an even root must be nonnegative; and the argument of a logarithm \log_a u must be strictly positive, with a>0 and a\neq 1. Odd roots impose no restriction, so a cube root of a real expression is defined for every real value of that expression, including zero and negative values. Exponential functions a^{x} with a>0, a\neq 1 are defined for every real x, so 3^{x} has domain (-\infty,\infty).

For an even root, the radicand must be at least zero. In \sqrt{1-(x-1)^{2}}, the condition (x-1)^{2}\le 1 means |x-1|\le 1, so 0\le x\le 2. Since the radicand is zero at the endpoints and zero is allowed under a root, the endpoints are included and the domain is the closed interval [0,2]. When several roots appear in a sum, the domain is the intersection of the individual conditions. For \sqrt{2-x}+\sqrt{2+x} this gives -2\le x\le 2, a closed interval.

When the radicand sits in a denominator, the condition becomes strict: \frac{1}{\sqrt{u}} requires u>0, not u\ge 0. Thus \frac{1}{\sqrt{x-|x|}} has no domain at all, because x-|x| equals 0 for x\ge 0 and is negative for x<0. By contrast, \frac{1}{\sqrt{|x|}-x} has a nonempty domain, since only the points where the denominator vanishes, here x=0 and x=1, are removed, giving \mathbb{R}\setminus\{0,1\}.

Composition of functions

The composition f\circ g is defined by (f\circ g)(x)=f\left(g(x)\right), so the inner function g is applied first. Composition is in general not commutative, so f\circ g and g\circ f usually differ. For a repeated composition such as f\circ f\circ f(a), apply f three times, starting from a. A point a with f(a)=a is a fixed point, and then every further application of f returns a, so f\circ f\circ f(a)=a. In the same way, if f(0)=0, every iterated composition of f evaluated at 0 equals 0.

For linear functions f(x)=px+q and g(x)=rx+s, we get (f\circ g)(x)=prx+ps+q and (g\circ f)(x)=prx+rq+s. Hence f\circ g=g\circ f holds exactly when the constant terms agree, ps+q=rq+s. Equating the constant terms is the standard way to find an unknown coefficient. The same method applies to f(x)=ax-b and g(x)=cx+d. The condition f\circ g=g\circ f reduces to d(a-1)=b(1-c), a relation among the parameters.

When the outer function is unknown but a relation involving a composite is given, write the expression in terms of a single variable u equal to the inner function. This determines the outer function, which can then be evaluated at the required inner expression.

Subsets and power set

A set with n elements has 2^{n} subsets. Its power set P(A), the set of all subsets of A, therefore has 2^{n} elements. The empty set is a subset of every set, and every set is a subset of itself. For example, a set with 10 elements has 2^{10}=1024 subsets in its power set.

A proper subset is a subset different from the set itself. A set with n elements therefore has 2^{n}-1 proper subsets. Sets that are pairwise disjoint have a union whose cardinality is the sum of their sizes, so the number of elements must be found before the count of subsets is taken.