Point-Rotational Symmetry

  • Thread starter Unit
  • Start date
  • #1
182
0

Homework Statement


Given f(x), find an expression to check whether f(x) has rotational symmetry about any arbitrary point (h, v).


Homework 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.


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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
964
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?
 

Related Threads on Point-Rotational Symmetry

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
5
Views
990
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
3
Views
28K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
2
Views
1K
Top