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Conventionality when Graphing Transformations.

  1. Aug 6, 2012 #1
    If you think about graphing an equation like f(x) = x; you think about a line through the origin in two dimensional space, where the horizontal axis represents the domain and the vertical the image.
    How can you get the input and the output of a transformation in the same picture? In the previous case neither the input nor the output even exist in ℝ^2. If you take that to higher dimensions it becomes even weirder. How can a transformation from ℝ^n to ℝ^m be represented in ℝ^(n+m).
    I get how it works, we've been introduced to the concept in middle school. I'm just wondering if there is some elementary reason or if it's just convention. And also, who decided that the constant output of a function like g(x, y) = x - y = 0 should be omitted from its graph? Why is it that in this case we're allowed to map only all the possible inputs to the transformation and and leave out the fact that a third component z exists but is just a constant. By the f(x) = x graphing convention shouldn't the graph of x - y = 0 be a line through the origin in three dimensional space that stays in the xy-plane?
     
  2. jcsd
  3. Aug 7, 2012 #2

    Simon Bridge

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    I don't understand your question about general transformations. I think your language is ambiguous.

    In the last case
    That is exactly what it is - the trouble with graphing this the way you suggest is that we only have 2D paper so best practice is to choose to orient the image to maximize the information we can use. In this case, we usually choose an orientation looking along the z axis. (Technically there are an infinite possible extra dimensions ... you don't suggest drawing more than three I notice :) )

    So what you are talking about is a rule-of-thumb ... like rounding to 2dp or crossing your axis at (0,0,0).

    Note:
    if you wrote z=x-y, then the set x-y=0 is the intersection of z in the x-y plane. Does it still make sense to represent x-y=0 in 3D?
     
    Last edited: Aug 7, 2012
  4. Aug 8, 2012 #3

    Bacle2

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    You're right in that what you are representing is the collection of points given by

    {(x,f(x))} , or {(x,y),z:=f(x,y)} , etc -- meaning we cannot graphically-represent

    higher dimensions. Re g(x,y) , its graph, re the previous rule, is actually given by

    {(x,y,g(x,y)}={(x,y,0)} . You're right that sometimes assumptions are made (implicitly)

    and you need to guess from the context; the graph of x may mean the graph of f(x)=x,

    of from f(x,y)=x , or f(x,y,z)=x, etc. Usually , from the context one should be able to

    tell which one is being referred-to. Re your comment that neither the input nor the

    output exist in ℝ2, they actually do, and they are given by the subsets

    {(x,0)} and {(0,x)} ( or, more generally, for y=f(x), {(x,0) and (0,f(x)) ).

    In addition, if I understood you well, there is the

    issue of the choice of embedding: a figure that can "live" in a certain space, can also

    live in higher dimensions. You are right in the strict sense that it should be specified in

    advance what the ambient space is-- you can define functions going into circles,

    or any other topological space.
     
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