A Can You Prove the Series Is Periodic?

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The discussion revolves around proving that the series Σ (sin(2^n x) / 2^n) is periodic, specifically with a period of 2π. Participants express confusion about how to demonstrate periodicity for a series, noting that each term in the series has a period of 2π, which implies that the entire series should also be periodic if it converges. Suggestions include using the property of sine functions and examining the behavior of partial sums to establish the periodic nature formally. Ultimately, one participant successfully proves the periodicity using insights gained from the discussion, emphasizing the importance of understanding the properties of sine functions and series convergence. The conversation highlights the challenges in transitioning from informal reasoning to formal proof in mathematical contexts.
  • #31
DaveE said:
Anyway, now I'm confused. Can you explain, regardless of how many values T can take (because it's value wasn't specified), why it isn't iff in either interpretation? Is it the "if" part you don't like, or the "only if" part?
If you define T as the smallest value, then if T is "the" period, we know f(x)=f(x+T) for all x, but if some value T satisfies f(x)=f(x+T) for all x, that doesn't imply T is "the" period but rather just "a" period. So it comes down to how one defines T. I've always seen it defined as the smallest value that satisfies f(x)=f(x+T) for all x, but apparently, not everyone defines it that way.
 
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  • #32
vela said:
If you define T as the smallest value, then if T is "the" period, we know f(x)=f(x+T) for all x, but if some value T satisfies f(x)=f(x+T) for all x, that doesn't imply T is "the" period but rather just "a" period. So it comes down to how one defines T. I've always seen it defined as the smallest value that satisfies f(x)=f(x+T) for all x, but apparently, not everyone defines it that way.
In this case, can you prove that ##2\pi## is the fundamental period of the function in question? It's easy to see that it does not have period ##\pi##. But, it might be tricky to show that it can't be anything else less than ##2\pi##.
 
  • #33
PeroK said:
In this case, can you prove that ##2\pi## is the fundamental period of the function in question? It's easy to see that it does not have period ##\pi##. But, it might be tricky to show that it can't be anything else less than ##2\pi##.
I'd offer the physicist's proof: Look at the graph. ;)
 
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  • #34
PeroK said:
In this case, can you prove that ##2\pi## is the fundamental period of the function in question? It's easy to see that it does not have period ##\pi##. But, it might be tricky to show that it can't be anything else less than ##2\pi##.
I believe it's a theorem that either the fundamental period is ##2\pi/n## for some ##n##, or the function has arbitrarily small periods. The second one can't be true because the function is continuous and not constant (probably? Maybe this is a trick question haha). The first one doesn't feel impossible to manage but I agree isn't easy.

Edit to add: it is not constant, since ##f(0)=0## and ##f(\pi/2)=1##
 
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