Question

Are the following functions periodic?
f(t) = Abs[t]^3 over -1/2 < t < 1/2
and f(t) = f(t+1).

Answer 1

I don't see how the first function is periodic, given that for the same value for f(t), a different distance between points on the graph arises as you move up or down it.

Answer 2

For the second graph, I'm not quite sure how to interpret it.

Answer 3

This is all pre-fourier analysis and I know that each function needs to be continuous and their derivatives sectionally continuous. However, that still doesn't tell me that the functions are periodic which I'm having trouble discerning.

Answer 4

yes those are dirichlets conditions but u dont need them to check periodicity

Answer 5

for the first function u can see it looks like

Answer 6

Yeah

Answer 7

so its not really periodic

Answer 8

I agree. The second function is though? It looks a lot like the definition of a function that repeats itself every period, i.e. f(t) = f(t + T)

Answer 9

yes its the basic definition
A function f is said to be periodic with period P, if
\[f(x+P)=f(x)\]

Answer 10

so the second function is periodic with period 1

Answer 11

Excellent. Thanks for clarifying it. Much appreciated

Answer 12

no problem :)