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Drawing circles with different distance functions

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    In ##\mathbb{R}^2##, draw a unit circle for taxicab distance ##(d_t)##, euclidean distance ##(d_e)##, and max distance ##(d_s)##.

    2. Relevant equations

    ##d_e = \sqrt{(x -x_1)-(y-y_1)}##
    ##d_s = \text{max}\{|x-x_1|,|y-y_1|\}##
    ##d_t=|x-x_1|+|y-y_1|##

    3. The attempt at a solution

    Euclidean distance is the most familiar one. It will just be our everyday unit circle.

    For taxicab, will the unit circle be a shape of a square?

    For max distance I am not so sure about to.
     
  2. jcsd
  3. Jan 29, 2014 #2

    HallsofIvy

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    Try a simple example like [itex](x_1, y_1)= (0, 0)[/itex] and the distance is 1.

    Then with the Euclidean distance you have [itex]\sqrt{x^2+ y^2}= 1[/itex] which is the same as [itex]x^2+ y^2= 1[/itex]. Yes, that is a circle with center at (0, 0) and radius 1..

    With the taxicab metric, that is |x|+ |y|= 1. If x and y are both positive it is x+ y= 1 which is the line segment from (1, 0) to (0, 1). If x is negative and y is positive, it is -x+ y= 1 which is the line segment from (0, 1) to (-1, 0). If x and y are both negative, it is -x- y= 1 which is the line segment from (-1, 0) to (0, -1). Finally, if x is positive and y is negative, it is x- y= 1 which is the line segment from (0, -1) to (1, 0). Yes, that is a square with its diagonals horizontal and vertical.

    With the max distance it is max(|x|,|y|)= 1. If x and y are both positive and x< y that is |y|= y= 1 which is the horizontal line segment from (0, 1) to (1, 1). If x and y are both positive and y< x, that is |x|= x= 1 which is that vertical line segment from (1, 0) to (1, 1). You check the other quadrants:

    If x is negative and y is positive and x< y .... If y< x ....
     
  4. Jan 29, 2014 #3
    Thanks for the help!
     
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