# Even Function , Odd Function , Periodic Function

If f : X → Y is a real valued function such that for all x ∈ D

⇒ − x ∈ D ( where D = domain of f ) and :

if f(− x) = f(x) for every x ∈ D then f is said to be an even function

and if f(− x) = − f(x) then f is said to be odd function.

The graph of an even function is symmetric about the y−axis (i.e. if (x , y) lies on the curve, then (− x , y) also lies on the curve) and that of an odd function is symmetric about the origin (i.e. if (x , y) lies on the curve , then (− x , − y) also lies on the curve).

### Notes:

⋄ Some times it is easy to prove that f(x) − f(−x) = 0 for even function and f(x) + f(−x) = 0 for odd functions .

⋄ A function can be even or odd or neither even nor odd.

⋄ Every function defined in symmetric interval D

( i.e. x∈ D ⇒ −x ∈ D) can be expressed as a sum of an even and an odd function.

$\large f(x) = (\frac{f(x) + f(-x)}{2}) + (\frac{f(x) – f(-x)}{2})$

Let     $\large h(x) = (\frac{f(x) + f(-x)}{2}) \; and \; g(x) = (\frac{f(x) – f(-x)}{2})$

It can now easily be shown that h(x) is even and g(x) is odd.

⋄ The first derivative of an even function is an odd function and vice versa. This is left as an exercise for you to prove.

⋄ If x = 0 ∈ domain of f, then for odd function f(x) which is continuous at x = 0 , f(0) = 0 i.e. if for a function, f(0) ≠ 0, then that function can not be odd.

It follows that for a differentiable even function f ‘(0) = 0 i.e if for a differentiable function f ‘(0) ≠ 0 then the function f cannot be even

Illustration : Which of the following functions is (are) even , odd or neither:

(i) f(x) = x2 sinx

(ii) $\large f(x) = \sqrt{1 + x + x^2} – \sqrt{1 – x + x^2}$

(iii) $\large f(x) = log(\frac{1-x}{1+x})$

(iv) $\large f(x) = log(x+ \sqrt{1 + x^2})$

(v) f(x) = sinx – cosx

(vi) $\large f(x) = \frac{e^x + e^{-x}}{2}$

Solution: (i) $\large f(-x) = (-x)^2 sin(-x)$

$\large f(-x) = – x^2 sinx$

= -f(x)

Hence , f(x) is odd .

(ii) $\large f(-x) = \sqrt{1 + (-x) + (-x)^2} – \sqrt{1 – (-x) + (-x)^2}$

$\large = \sqrt{1 – x + x^2} – \sqrt{1 + x + x^2}$

= – f(x)

Hence f(x) is odd.

(iii) $\large f(-x) = log(\frac{1-(-x)}{1 + (-x)})$

$\large = log(\frac{1+x}{1-x})$

= – f(x)

Hence f(x) is odd.

(iv) $\large f(-x) = log(-x + \sqrt{1 + (-x)^2})$

$\large f(-x) = log(\frac{(-x + \sqrt{1 + x^2}) (x + \sqrt{1 + x^2})}{(x + \sqrt{1 + x^2})})$

$\large = log(\frac{1}{x + \sqrt{1 + x^2}})$

= – f(x)

Hence f(x) is odd.

(v) f(-x) = sin(-x) – cos(-x) = -sinx – cos x

Hence f(x) is neither even nor odd.

$\large f(-x) = \frac{e^{-x} + e^{-(-x)}}{2}$

$\large = \frac{e^{-x} + e^x}{2}$

= f(x)

Hence f(x) is even

### Extension of Domain:

Let a function be defined on certain domain which is entirely non-negative (or non positive). The domain of f(x) can be extended to the

set X = {−x : x∈ domain of f(x)} in two ways:

(i) Even extension: The even extension is obtained by defining a new function f(−x) for x ∈ X , such that f(−x) = f(x).

(ii) Odd extension: The odd extension is obtained by defining a new function f(-x) for x∈X, such that f(−x) = − f(x)

Illustration : If $\displaystyle f(x) = \left\{\begin{array}{ll} x^3 + x^2 & \mathrm{for}\; 0 \le x \leq 2 \\ x+2 & \mathrm{for} \; 2 < x \le 4\end{array} \right.$

Then find the even and odd extension of f(x).

Solution: The even extension of f(x) is as follows:

$\displaystyle g(x) = \left\{\begin{array}{ll} -x + 2 & \mathrm{,}\; -4 \le x < -2 \\ -x^3 + x^2 & \mathrm{,} \; -2 < x \le 0 \end{array} \right.$

The odd extension of f(x) is as follows:

$\displaystyle h(x) = \left\{\begin{array}{ll} x – 2 & \mathrm{,}\; -4 \le x < -2 \\ x^3 – x^2 & \mathrm{,} \; -2 < x \le 0 \end{array} \right.$

### Periodic Function:

A function f: X →Y is said to be a periodic function if there exists a positive real number p such that :

f(x + p) = f(x) , for all x ∈ X .

The least of all such positive numbers p is called the principal period or simply period of f.

All periodic functions can be analyzed over an interval of one period within the domain as the same pattern shall be repetitive over the entire domain.

Example :

sinx , cosx, secx are periodic functions with period 2π

tanx, cotx are periodic with period π

Example :

f(x) = x − [x]

f(x) is periodic with period 1

(can you prove it mathematically?).

There are two types of questions asked in the examination. You may be asked to test for periodicity of the function or to find the period of the function.

In the former case you just need to show that f(x + T) = f(x) for same T ( > 0) independent of x whereas in the latter, you are required to find a least positive number T independent of x for which f(x + T)= f(x) is satisfied.

### The following points are to be remembered:

If f(x) is periodic with period p, then af(x) +b where a , b ∈ R (a ≠ 0) is also periodic with period p.

∎ If f(x) is periodic with period p , then f(ax + b) where a, b ∈ R ( a ≠0) is also period with period $\large \frac{p}{|a|}$ .

∎ Let f(x) has period p = m/n (m, n ∈ N and co-prime) and g(x) has period q = r/s (r, s ∈ N and co-prime) and let t be the LCM of p and q i.e. $\large t = \frac{LCM \; of (m , r)}{HCF \; of (r , s)}$

Then t shall be the period of f + g provided there does not exist a positive number k (< t) for which

f(k + x) + g(k + x) = f(x) + g(x) , else k will be the period.

The same rule is applicable for any other algebraic combination of f(x) and g(x)

### Remarks:

LCM of p and q always exist if p/q is a rational quantity. If p/q is irrational then algebraic combination of f and g is non-periodic.

∎ sinnx, cosnx, cosecnx and secnx have period 2π if n is odd and π if n is even.

∎ tannx and cotnx have period π whether n is odd or even.

∎ A constant function is periodic but does not have a well-defined period.

∎ If g is periodic then fog will always be a periodic function. Period of fog may or may not be the period of g.

∎ If f is periodic and g is strictly monotonic (other than linear) then fog is non-periodic.

Illustration : Find the periods (if periodic) of the following functions, where [.] denotes the greatest integer function.

(i) $\large f(x) = e^{ln(sinx)} + tan^3 x – cosec(3x-5)$

(ii) f(x)= x – [x – b], b ∈ R

(iii) $\large f(x) = \frac{|sinx + cosx|}{|sinx| + |cosx|}$

(iv) $\large f(x) = tan\frac{\pi}{2} [x]$

Solution: (i) $\large f(x) = e^{ln(sinx)} + tan^3 x – cosec(3x-5)$

Period of elnsinx = 2π , tan3x = π

cosec(3x-5) = 2π/3 ; period = 2π

(ii) f(x) = x –[x-b] = b+{x –b}

⇒  period = 1

(iii) $\large f(x) = \frac{|sinx + cosx|}{|sinx| + |cosx|}$

Since period of |sinx + cox| = π and   period of  |sinx| + | cosx|  is π/2.

Hence  f(x) is periodic  with π  as  its  period.

(iv) $\large f(x) = tan\frac{\pi}{2} [x]$

$\large tan\frac{\pi}{2} [x + T] = tan\frac{\pi}{2} [x]$

$\large \frac{\pi}{2} [x + T] = n \pi + \frac{\pi}{2} [x]$

Hence , Period = 2