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A comment on A Simple Calculus Exercise. Parent comment
floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2.
> And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))?
If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.
floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2.
> And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))?
If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.