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Graph reflections

  1. Jun 20, 2009 #1

    Mentallic

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    I know how to take the reflection of a graph in the [itex]y=x[/itex] line, or more formally, finding the inverse function. All I really do is switch the x and y variables in the function.

    e.g. [itex]y=x^2[/itex], [itex]x=y^2[/itex]

    I tried taking the same idea and extending it to a reflection in the y=mx line, m constant. But I encountered problems as such:

    Take the function [itex]y=(x+1)^2[/itex], reflect it in the line [itex]y=2x[/itex] or [itex]x=y/2[/itex]

    I tried using the same idea as before, so I substitute all x and y variables as such and this is the result:

    [tex]y=(x+1)^2 : 2x=(y/2+1)^2[/tex]

    But when I graph both functions, it doesn't look correct. The new 'reflected' function looks much too fat/shallow.
    Could someone please explain what I'm doing wrong. Where is my logic flawed here?
     
  2. jcsd
  3. Jun 20, 2009 #2

    tiny-tim

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    Hi Mentallic! :smile:

    Your first transformation was

    0 1
    1 0

    which is a reflection.

    Your second transformation is

    0 1/2
    2 0

    It leaves the line y = 2x invariant not because it is a reflection about that line, but because it is a reflection about y = x combined with a shear: :wink:

    0 1
    1 0

    and

    1/2 0
    0 2
     
  4. Jun 20, 2009 #3

    Mentallic

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    Oh no, matrices. My worst enemy! I tried to learn them off the Massechusetts (spelling) videos on youtube, but failed miserably after a few lectures.

    Is it possible to extend this explanation into another form other than matrices?
     
  5. Jun 21, 2009 #4

    tiny-tim

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    ah … this is your chance to get a better understanding of matrices. :wink:

    First, can you see that

    3 0
    0 3

    is an expansion (everything gets 3 times bigger)?

    Second,

    3 0
    0 1

    stretches in the x-direction only (leaving the y coordinates the same),

    and likewise

    1 0
    0 2

    stretches in the y-direction only (leaving the x coordinates the same).

    Finally,

    3 0
    0 2

    is a shear, which stretches 3 times in the x-direction but only 2 times in the y-direction, and

    0 3
    2 0

    is a reflection in the x = y line, combined with a shear.

    Does that make sense? :smile:
     
  6. Jun 21, 2009 #5

    Mentallic

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    Surprisingly, yes :bugeye:

    But this is how I learnt (if you can even call it that) matrices:

    Excuse me for not using latex, as I don't know how to create matrices...

    [a c] [e]
    [b d] [f]

    Is equivalent to: ax+cy=e and bx+dy=f

    Now, as for all the matrices you've shown, I don't know what they're supposed to mean when they don't have the second 'box' next to them.

    i.e.

    3 0
    0 3

    this is equivalent to 3x and 3y? It doesn't make sense to me when they're not expressed as functions.
     
  7. Jun 21, 2009 #6

    tiny-tim

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    I do know how to :approve:

    but it takes so long I can't be bothered! :rolleyes:
    Nooo, that should be:

    [a c] [x] = [e]
    [b d] [y] = [f]
    [3 0]
    [0 3]

    is (in algebra) a set of instructions,

    and (in geometry) a transformation,

    and it means that if you put a vector next to it:

    [3 0] [2]
    [0 3] [3]

    then it converts that vector to another vector:

    [3 0] [2] = [6]
    [0 3] [3] = [9]

    and similarly

    [0 3]
    [2 0]

    is the rule that converts as follows:

    [0 3] [2] = [9]
    [2 0] [3] = [4]

    So when you write a matrix on its own, it's a rule (like a computer program),

    and you can put any "input" vector next to it, and get an "output" vector. :smile:
     
  8. Jun 21, 2009 #7

    Mentallic

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    omgosh :yuck:

    Maybe I should go learn matrices for real this time... Be back in a bit after
    i've acquired more knowledge on the topic :smile:
     
  9. Jun 21, 2009 #8

    Mentallic

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    In one last attempt, I tried reflecting [itex]y=(x+1)^2[/itex] in the line [itex]y=4x[/itex]
    This case is slightly different since the line now intersects the parabola at one point (1,4) and when I tried the same thing I did previously:

    So I plotted [tex]4x=(\frac{y}{4}+1)^2[/tex] and this parabola intersected the the line at the same point, which is consistent with the features of other functions being reflected in the y=x line. I guess they don't look like the same because they have been sheared as you said tiny-tim :smile:

    Still, I need to learn these matrices as they keep popping up in the most awkward places and catch me off-guard.
     
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