1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Point-Rotational Symmetry

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Given f(x), find an expression to check whether f(x) has rotational symmetry about any arbitrary point (h, v).


    2. Relevant equations
    If [tex]f(x) = f(-x)[/tex] then the function is symmetrical about the y-axis.

    If [tex]f(x) = -f(-x)[/tex] then the function is point-rotational about the origin.


    3. The attempt at a solution
    I don't know how to deal with "general" functions, so I did this:

    Let [tex]f(x) = x^3 + v[/tex] because I know it has this symmetry.
    [tex]f(-x) = (-x)^3 + v[/tex]
    [tex]f(-x) = -x^3 + v[/tex]
    [tex]-f(-x) = x^3 - v[/tex]

    Then I found the difference between f(x) and -f(-x).

    [tex]f(x) - (-f(-x)) = (x^3 + v) - (x^3 - v)[/tex]
    [tex]f(x) + f(-x) = 2v[/tex]
    [tex]f(x) = -f(-x) + 2v[/tex]

    This gives me point-rotation about (0, v). But how do I do this for just a "general" function?

    Also, I can guess that point-rotation about (h, 0) will be something like [tex]f(x) = -f(-x+2h)[/tex], based on knowledge of transformations, but how do I show this as elegantly as above with (0, v)?

    From this, I can guess that if [tex]f(x) = -f(-x+2h)+2v[/tex] for a real point (h, v), then the function's graph is point-rotational about (h, v). Is this correct?

    Cheers,
    Unit

    p.s. I made this problem up for myself, if that's okay.
     
    Last edited: May 28, 2010
  2. jcsd
  3. May 29, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You don't mean "general function" you mean "general point".

    Given a point (u, v) in the xy-plane, you can "translate" (u, v) to (0, 0) by subtracting u from x and v from y. f(x) becomes f(x- u) and y= f(x) becomes y- v= f(x- u) or, finally, y= f(x- u)+ v. Now, what happens if you replace x with -x?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook