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$f(x) = \sin x + \cos x$ defined for $\Bbb R^+$ cannot be a periodic function because domain of a periodic function should be unbounded.

This statement was given in my book as a question, "Why $f(x)$ is not periodic for $x \in \Bbb R^+$ and the reason given was that the domain of a periodic function should be unbounded.

I don't think this statement is true, because $\sin(x + 2\pi) = \sin x$ for all $x$. So it does not matter if negative $x$ are in the domain or not and same for cosine.

Is this statement true ? (I highly doubt that) If yes why ?

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    $\begingroup$ A good point: the definition of a function of period $P$ should be a function whose graph is invariant under translation of the $x$-axis by $P$: i.e., $f(x \pm P) = f(x)$. Just saying $f(x + P) = f(x)$ for $x$ in the domain of $f$ doesn't give this. The wikipedia article on this gets this wrong. What does your book give as the definition? $\endgroup$ – Rob Arthan Sep 9 '17 at 21:13
  • $\begingroup$ The set ${\mathbb R}^+$, whatever its definition, is unbounded. This means your source is sloppy. $\endgroup$ – Christian Blatter Sep 10 '17 at 9:33
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The usual definition of a periodic function presented in undergraduate calculus classes is something like

$f$ is periodic with period $T$ if for all $x \in \mathbb{R}$ we have $f(x+T) = f(x)$.

In particular, $f(x)$ must be defined for all real numbers. Hence every function that is periodic has, by definition, an unbounded domain. Implicit in the definition is that $f(x+kT) = f(x)$ for all integers $k$. As a pathological counterexample, consider the following pathological function:

For each $n\in\mathbb{N}$, choose (inductively) some $x_n \in (0,1)$ such that $x_i - x_j$ is irrational for all $i\ne j$. Define a function $f$ by setting $$ f(x_n + n + m) = 1, $$ where $m$ and $n$ range over the natural numbers. This function is pretty far from periodic, if you ask me (it doesn't really repeat itself in any meaningful way from one interval to the next), but if you drop the requirement that a periodic function be defined for all of $\mathbb{R}$, you could perhaps argue that $f$ is 1-periodic.


Another approach is to consider that a periodic function is really defined on an $n$-dimensional torus (a circle, in the case of functions that we want to think of as taking real inputs, but we can also talk about periodicity in $\mathbb{R}^n$). A periodic function on $\mathbb{R}$ then becomes some kind of lifting of a function on the circle via a covering map. Again, this would imply that the domain (thought of as a subset of $\mathbb{R}$) must be unbounded, though it needn't imply that the domain is all of $\mathbb{R}$.

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I'd say this is a matter of definition (i.e., of language).

Just because one definition doesn't apply doesn't mean we can have another one that is reasonable. For example.

f is periodic with period T if $\exists T > 0$ such that $\forall x > 0 $ we have $f(x+T)=f(x)$.

No reason this can't be done. A covering map can also be defined with the circle as the base space and $\mathbb{R}^+$ as the covering space.

Like Rob Arthan's comment above, I'd ask - "periodic under which definition?"

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