Concept 09 / 10Sets, Relations and Functions
Even, Odd and Periodic Functions
Symmetry of even and odd functions and their algebra, the unique even-odd decomposition, the fundamental period and rules for f(ax + b), sums and products, cases where the LCM rule over-estimates the period, and periods forced by functional equations.
Symmetry and repetition are the two structural features that most often shorten JEE calculations, especially in definite integrals and in counting the roots of equations.
Even and odd functions. Let the domain be symmetric about (if is in it, so is ). The function is even if for all , and odd if for all . The graph of an even function is symmetric about the -axis; the graph of an odd function is symmetric about the origin (rotating it by leaves it unchanged). Examples of even functions: , , , . Odd: , , , , , and . If an odd function is defined at , then , so . The function is the only function that is both even and odd. Most functions, for example , are neither.
Algebra of parity, which works like the signs and under multiplication: even even even, odd odd even, and even odd odd. Sums of two even functions are even and sums of two odd functions are odd, but even odd is generally neither. For composition, if the inner function is even the composite is even; if both are odd the composite is odd. Differentiation switches parity (the derivative of an even function is odd, and of an odd function is even), and the integral of an odd function is even.
Every function on a symmetric domain splits uniquely into an even part and an odd part:
.
Replacing by confirms that the first term is even and the second is odd. For uniqueness, if , then is both even and odd, hence zero. For this decomposition gives and .
Periodic functions. is periodic if there is some with for all in the domain (and in the domain whenever is). The smallest such is the fundamental period, usually just called the period. Standard periods: , , , have period ; , , , , have period ; has period ; and have period . A constant function is periodic but has no fundamental period, because every positive works. The Dirichlet function (equal to on rationals and on irrationals) also has every positive rational as a period and so has no smallest one.
Rules. If has period , then has period : compressing the graph horizontally by a factor compresses the period by the same factor, while the shift has no effect. If and have periods and with a rational ratio, then and are periodic, and the LCM of and is a period, but not necessarily the smallest one. For fractions, . The LCM over-estimates when the combination has extra symmetry. In the diagram, and each have period , yet their sum repeats every , because and , so shifting by just swaps the two terms. Always test whether a smaller shift already works. If the ratio of the periods is irrational, as for or , the sum is not periodic. Also non-periodic: , and , since the input to is not advancing at a steady rate.
Key Takeaways:
- Even: , symmetric about the -axis. Odd: , symmetric about the origin, and if defined.
- Even odd odd; odd odd even; the derivative switches parity.
- , uniquely.
- Period of is .
- The LCM of the periods is a period of the sum, but check smaller candidates such as half the LCM.
- and have period ; has period .
- An irrational ratio of periods means no period at all.
- Pitfall: calling 'even plus odd, so neither, so not periodic'. Parity and periodicity are unrelated; it has period .
Brain-Teaser 1: Find the period of .
Solution: The LCM rule suggests , since and each have period . But simplify first: . The period of is , so the period of is . Rewriting the expression as a single trigonometric function exposes the true period, which the LCM rule over-estimated by a factor of .
Brain-Teaser 2: A function satisfies for all real . Show that is periodic and find a period.
Solution: Replace by : . Adding this to the original equation, , so . Shifting, for all . Applying this twice, . So is a period. A concrete example is , using with . The move is to generate a second copy of the equation by shifting, combine the two to eliminate a term, and iterate until returns to itself.
Brain-Teaser 3: is odd and periodic with period , and defined everywhere. Prove that , and deduce (assuming is integrable).
Solution: By periodicity, , and by oddness, . So , and . For the integral, periodicity means integrating over any interval of length gives the same value, so , which is because is odd. For example, vanishes at and has zero integral over , and is undefined at , which is consistent because the result requires to be defined there. Combining the two symmetries produces facts that neither symmetry gives on its own.
Practice this concept
Free · no account required · instant solutions