Periodic Function

These are the function, whose value repeats after a fixed constant interval called period, and which makes a class of a widely used function.

        A function f of x, such that:

        f(T + x) = f(x) ∀ x ε domain of f.

The least positive real value of T for, which above relation is true, is called the fundamental period or just the period of the function.

        e.g. for f(x) = sin x ∀ x ε R.

We know that sin (2∏ + x) = sin x, ∀ x ε R

so f(x) = sin x is a periodic function with a period of 2∏ radians.

Rules for finding the period of the periodic functions

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

(ii) If f(x) is periodic with period, then f(ax + b), where a ε R -{0} and b ε R, is also periodic with period p/|a|.

(iii) let us suppose that f(x) is periodic with period p and g(x) is periodic with period q. Let r be the L.C.M. of p and q, if it exists.

(a) If f(x) and g(x) cannot be interchanged by adding a least positive number k, then r is the period of f(x) + g(x).

(b) If f(x) and g(x) can be interchanged by adding a least positive number k and if k < r, then k is the period of f(x) + g(x). Otherwise r is the period.

Illustration:  Find the period of the following functions

(i) f(x) = sinx + {x}   

(ii) f(x) = tan(x/3) + sin 2x.

(iii) f(x) = |sinx| + |cosx|      

(iv) f(x) = ((1+sin x)(1+sec x))/((1+cos x)(1+cosec x))

Solution:

(i) Here f(x) = sinx + {x}

Period of sinx is 2p and that of {x} is 1. But the L.C.M. of 2p and 1 does not exist. Hence sinx + {x} is not periodic.

(ii) Here f(x) = tanx/3 + sin2x. Here tan(x/3). Here tan(x/3) is periodic with period 3p and sin2x is periodic with period p. 

Hence f(x) will be periodic with period 3p.

(iii) Here f(x) = |sinx| + |cosx|

 

 

Now, |sinx| = √sin2x = √((1+cos2x)/2),              which is periodic with period ∏.

Similarly, |cosx| is periodic with period ∏.

Hence, according to rule of LCM, period of f(x) must be ∏.

But |sin((∏/2)+x)| = |cos x| and |cos((∏/2)+x)| = |sin x|.

Since ∏/2 < ∏, period of f(x) is ∏/2.

(iv)   f(x) =((1+sin x)(1+sec x))/((1+cos x)(1+cosec x))  =

               ((1+sin x)(1+cos x)sin x)/((1+cos x)(1+sin x)cos x) = tan x

Hence f(x) has period ∏.

  

Note:        For f(x) = |sin x| + |cos x|

                The period of both |sin x| and |cos x| is ∏

                But they are related with phase difference ∏/2  i.e.

                |sin x| = |cos (x + ∏/2)|

                |cos x| = |sin (x + ∏/2)|

So the period of the function f(x) is ∏/2.

 

Illustration:       Prove that the period of y = sin x is 2∏

Solution:

        Let T be the period of f(x) = sin x

i.e.    f(T + x) = f(x)

        =>  sin (T + x) = sin x

        => T + x = n∏ + (-1)nx, n ε 1                        .........(i)

Let    n = 0

        T + x = x

        => T = 0

But we want a positive real value for T.

Let  n = 1,

        T + x = ∏ - x

        => T - ∏ = 2x              (Is it possible? Think)

 

No it is not possible because LHS is constant and RHS is a continuous variable.

Now, Let n = 2 in equation (i)

        T + x = 2∏ + x

       => T = 2∏

Therefore period of y = sin x is 2∏

Note: If we cannot find T independent of x, then y = f(x) is not periodic.

Example:

Find the period of y = cos √x and y = x sin x if possible

Ans. These are non-periodic function

  

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