Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think I have answered the following puzzle:

Rotations

Under what conditions can you rotate the graph of a function about the origin, and still have the resulting graph being the graph of a function? If the graph of a function cannot be rotated about the origin without ceasing to be the graph of a function, might there be other points which could act as centre of rotation and preserve the property of being the graph of a function?

I think the answer is that the function has to be a line. Otherwise we can find an angle which lets us rotate the graph so that one $x$ maps to two $y$s.

Question: Is this correct? And if yes, how can I prove it?

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

For any set of points on the plane $\mathbb R^2$, there's at least one angle by which it can be rotated so that the result is not a graph of a function from $\mathbb R$ to $\mathbb R$.

Specifically, let $A$ and $B$ be any two distinct points in the set, and rotate the set so that the line $AB$ becomes vertical. Now the images of $A$ and $B$ share the same $x$ coordinate but have different $y$ coordinates, which contradicts the requirement that a function should map each $x$ value to a single $y$ value.

(If the set of points does not contain two distinct points, then it is either a singleton or empty; neither of these can be the graph of a function from $\mathbb R$ to $\mathbb R$ in any orientation.)


However, the question you quote doesn't explicitly say that the graph would have to remain a graph of a function under any rotation; instead, I'd read it as asking about whether, given a function $f: \mathbb R \to \mathbb R$ and an angle $\alpha$, the graph of $f$ rotated by $\alpha$ is a graph of some function $g: \mathbb R \to \mathbb R$.

This seems a more complicated question, although it's easy to find sufficient conditions: for example, $\alpha \equiv 0 \pmod{2\pi}$ obviously works for any $f$, and, as you note, for functions of the form $f(x) = ax + b$ any value of $\alpha$ except $\alpha \equiv \frac\pi 2 - \arctan a \pmod\pi$ will produce the graph of another function of the same form.

More generally, if $f$ is Lipschitz continuous with Lipschitz constant $K$, then its graph can be rotated by any angle of the form $\alpha + n\pi$, where $|\alpha| < \frac \pi 2 - \arctan K$ and $n$ is an integer, so that the result is still the graph of a function. In particular, the function $f(x) = \sin x$ mentioned by Henning is Lipschitz continuous with $K = 1$.

In fact, we can generalize this slightly:

Definition: We'll say a function $f: \mathbb R \to \mathbb R$ is $(J,K)$ skew Lipschitz continuous if there exist constants $J,K \in \mathbb R$ such that, for all distinct $x_1,x_2 \in \mathbb R$, $$J \le \frac{f(x_1)-f(x_2)}{x_1-x_2} \le K.$$ (In particular, any $K$-Lipschitz continuous function $f: \mathbb R \to \mathbb R$ is $(-K,K)$ skew Lipschitz continuous.)

Claim: Rotating the graph of a $(J,K)$ skew Lipschitz continuous function by an angle $\alpha + n\pi$, where $$-\frac\pi2-\eta < \alpha < \frac\pi2-\theta$$ $\eta=\arctan J$, $\theta=\arctan K$ and $n$ is an integer, yields the graph of a $(J',K')$ skew Lipschitz function, where $J' = \tan(\alpha+n\pi+\eta)$ and $K'=\tan(\alpha+n\pi+\theta)$.

(Proof left as an exercise.)

I believe this is getting close to a sufficient and necessary condition, at least for continuous and differentiable functions, although there remain some edge cases to be taken care of: for instance, rotating the graph of a $(J,K)$ skew Lipschitz function $f$ by exactly $\alpha = \frac\pi2-\arctan K$ may or may not yield the graph of a function, even if $K$ is the smallest upper skew Lipschitz constant for $f$; it depends on whether the equality $\frac{f(x_1)-f(x_2)}{x_1-x_2}=K$ is ever actually attained, or only approached from below.


The second part of the question is easy to answer, though: the center of rotation does not matter. To see why, note that rotation around any point $A$ can be decomposed into translation of $A$ to the origin, rotation around the origin and translation of the origin back to $A$. And since the graph of a function remains the graph of a function under translation, only the rotation matters.

share|improve this answer
add comment

Another interpretation of the question is whether there is some nontrivial angle the graph can be rotated and still stay a graph. This is somewhat more interesting than whether all angles work -- for example the graph of $\sin x$ can be rotated by up to $\pi/4$ and still stay a graph of something.

For the second part of the question note that rotation about a non-origin can always be decomposed into a rotation about the origin and a translation in the plane. Translations clearly don't change whether a set of points is the graph of some function.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.