Question

8. This problem is about the definition of periodic function. We assume you already know intuitively what periodic means, and now we want a formal definition. For simplicity, we will restrict ourselves to functions with domain R. A naive (but incorrect) definition of periodic function with period T is

   f (x + T ) = f (x)

   Without accompanying words, this is not a good definition because it does not introduce the variables x and T and it does not explain their role. For which values of x and T does the above definition have to be valid?

   Here is an attempt at a definition, with various ways to complete it:Definition. Let f be a function with domain (−∞, ∞). We say that

   f is periodic when...
   (a) Foreveryx∈(−∞,∞)andforeveryT >0,f(x+T)=f(x).
   (b) For every x ∈ (−∞,∞) there exists T > 0 such that f(x+T) = f(x). (c) There exists T > 0 such that x ∈ (−∞, ∞) =⇒ f (x + T ) = f (x). (d) There exists T > 0 such that for every x ∈ (−∞,∞), f(x+T) = f(x). (e) For every T > 0 there exists x ∈ (−∞,∞) such that f(x+T) = f(x).

9. One or more of the above are valid ways to complete the definition of periodic function. Identify which ones are correct and which ones are wrong. For any property which is wrong, show it by giving an example of a function which satisfies the property but is not periodic, or an example of a function which is periodic but does not satisfy the property. It is okay to give your examples as equations or graphs.

Answer 1

The properties (c) and (d) are valid ways to complete the definition of a periodic function.
Indeed, a function f with domain   is periodic if there exists a T > 0 such that x implies that f(x+T)=f(x) or equivalently, there exists a T > 0 such that for every x in , f(x+T)=f(x).

The properties which are wrong or are not valid ways to complete the definition of a periodic function are (a),(b) and (e).

For (a), consider the function f with domain |R given by, f(x)=sin x for all x in |R.
Observe that, f(x+2)=f(x) for all x in |R. Hence, f is periodic with period 2, by definition (c) or (d).
However, choosing x=0 and T=/2, we observe that f(x+T)=f(/2)=sin(/2)=1 0=sin(0)=f(0)=f(x).
Hence, f is a periodic function which doesn't satisfy property (a). Hence, (a) is a wrong property.

For (b), consider the function f with domain |R given by, f(x)=[x] for all x in |R, where for any real number x, [x] is the
greatest integer less than or equal to x.
For any x in |R, consider any real number T such that 0 < T < [x]-x+1 ( say T=([x]-x+1)/2 )
Then, x < x+T < [x]+1. Hence, [x]   x < x+T < [x]+1
   Thus, [x+T]=[x], and hence, f(x+T)=f(x).
Thus, for every x in |R, there exists T > 0 such that f(x+T)=f(x).
Hence, f satisfies property (b).
However, f is not periodic as if it were so, then by definition (c) or (d), there exists T > 0 such that for all x in |R,
f(x+T)=f(x). There are two cases:
i. T 1 : In this case, [T] 1 and thus, f(100+T)=[100+T]=100+[T]   100=[100]=f(100) . This is a contradiction.
ii. 0<T<1 : In this case, 1>1-T>0 and thus, [1-T]=0. Hence,  f(1-T+T)=[1-T+T]=[1]=1   0=[1-T]=f(1-T). This is
again a contradiction to the fact that T is a period of f.
Hence, f is not periodic.
Thus, (b) is not a valid way of completing the definition of a periodic function.


For (e), consider the function f with domain |R given by, f(x)=x-[x] for every real number x.
Observe that, f(1+x)=1+x-[1+x]=1+x-(1+[x])=x-[x]=f(x) for any real number x.
Thus, by definition (c) or (d), f is a periodic function with period 1.
Let T=0.4 ; Obviously, T>0.
However, if there exists an x in |R such that f(x+T)=f(x), then x+0.4-[x+0.4] = x-[x] and thus, 0.4=[x+0.4]-[x].
This is a contradiction since the right hand side of the last equation is an integer while the left hand side is not.
Hence, for the particular value of T=0.4, there does not exists an x in |R such that f(x+T)=f(x).
Thus, the periodic function f does not satisfy the property (e).
Hence, (e) is a wrong property.