Functions, Relations & Functional Equations

Axiomatic Definition

A function $f:A\to B$ is a binary relation that associates each element $x\in A$ with exactly one element $f(x)\in B$. $A$ is the domain, $B$ is the codomain, and $f(A)=\{f(x):x\in A\}\subseteq B$ is the range.

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1. Classification of Mappings

📊Visual Mathematical Intuition

Function Mappings, Horizontal Line Test & Inverse Reflection

Bijective Condition:f \text{ Invertible} \iff f \text{ is One-to-One (Injective) & Onto (Surjective)}

Inverse Symmetry:$(x, y) \in f \iff (y, x) \in f^{-1} \text{ (Reflected across } y = x \text{)}$

Function mapping taxonomy (injective, surjective, bijective), the horizontal line test for invertibility, and reflection of $f(x)$ across $y = x$ yielding inverse $f^{-1}(x)$.

• Injective (One-to-One): $f(x_1)=f(x_2)⟹x_1=x_2$. (Geometrically: Any horizontal line intersects the graph at most once; derivative $f^{\prime }(x)>0$ or $f^{\prime }(x)<0$ strictly monotonic).
• Surjective (Onto): $\text{Range}(f)=\text{Codomain}(B)$.
• Bijective (Invertible): Both injective and surjective $⟺$ Inverse function $f^{-1}:B\to A$ exists uniquely.

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2. Parity, Periodicity & Transformations

Even and Odd Functions

• Even: $f(-x)=f(x)$ $⟹$ Symmetric about the $y$-axis; ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$.
• Odd: $f(-x)=-f(x)$ $⟹$ Symmetric about the origin (rotation by $\pi$); ${\int }_{-a}^{a}f(x)dx=0$.

Periodicity

A function $f(x)$ has fundamental period $T>0$ if $f(x+T)=f(x)$ for all $x\in \text{Dom}(f)$.

• If $f(x)$ has period $T$, then $f(ax+b)$ has period $\frac{T}{|a|}$.
• If $f(x+a)=-f(x)$, then $f(x+2a)=-f(x+a)=f(x)⟹T=2a$.
• If $f(x+a)=\frac{1}{f(x)}$, then $f(x+2a)=f(x)⟹T=2a$.

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3. Classical Functional Equations

Cauchy's Classical Functional Equations (for continuous $f$)

$$\begin{array}{rl}f(x+y)& =f(x)+f(y)⟹f(x)=kx\\ f(x+y)& =f(x)f(y)⟹f(x)=a^x=e^{kx}\\ f(xy)& =f(x)+f(y)⟹f(x)=k\ln x\\ f(xy)& =f(x)f(y)⟹f(x)=x^n\end{array}$$

Polynomial Identity

If $P(x)$ is a polynomial satisfying $P(x)\cdot P\left(\frac{1}{x}\right)=P(x)+P\left(\frac{1}{x}\right)$, then:

$$P(x)=1\pm x^n(n\in \mathbb{N})$$