Equation of a function after rotation of the plane

Question

I have a function $\mathbb{R}\to\mathbb{R}$, $y=f(x)$. I want to find its equation $y'=g(x')$ when the basis $(x,y)$ is rotated by $a$. $(x',y')= R_a[(x,y)]$, where $R_a$ is the rotation of angle $a$.

I assume that f is symétric (f(x)=f(-x)) and that $g$ is still a function after the rotation !

I thought naively that I could rotate by -a, compute f, and then rotate by a. But this is wrong when $f$ is linear and commutes with $R_a$ ! More precisely, I thought that $$g(x)= P_y \circ R_a (0,f \circ P_x \circ (R_{-a}[(x,0)]))$$, where $P_x$ and $P_y$ are the projectors $R^2 \rightarrow R$ on the two coordinates

Thank you for your help.

Perhaps you're looking for something related to the rotation matrix? http://math.stackexchange.com/questions/346672/2d-rotation-of-point-about-origin/346686#346686 — apnorton, Jul 29 '13 at 13:13
You are applying rotation matrix to each point in the plane around the origin. Symmetry about y-axis cannot be obtained by rotation. Anti-symmetry about origin is attainable by a rotation of $ \pi$. — Narasimham, Jan 07 '15 at 22:44

Eric Auld · Answer 1 · 2013-07-29T13:20:08.450

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I know of no answer in general unless the function has a sort of symmetry about the origin. For instance, on the real numbers the only non-identity rotation is multiplication by $-1$. We can say something about $g(-x)$ if $g$ is odd or even, for example.

edited Jul 29 '13 at 13:20

answered Jul 29 '13 at 13:14

Eric Auld

28,127

Oh, you should edit the question to mention that. – Eric Auld Jul 29 '13 at 13:33
I am talking about rotations of the PLANE, not the line ! – vv33d Jul 29 '13 at 13:51
But your function is defined on the line, not the plane. – Eric Auld Jul 29 '13 at 13:52
I rotate the coordinates (x,y) – vv33d Jul 29 '13 at 13:59

Equation of a function after rotation of the plane

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