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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?


    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


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    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?
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