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In $\mathbb(R)^2$ sketch $B((1,2),3)$, the open ball of radius $3$ at the point $(1,2)$, with the following norms:

  1. the Euclidean norm $\parallel .\parallel_2$,
  2. the supremum norm $\parallel .\parallel_\infty$,
  3. the $1$-norm $\parallel .\parallel$,
  4. the norm given by $$\parallel x\parallel_e:=\sqrt{9x_1+ x_2^2},$$ for $x=(x_1,x_2) \in \mathbb{R}^2$,

For 1.) I think it is a circle with centre $(1,2)$ and radius of $3$ (as said in question).

For 2.) I think it is a square (using a dashed line as opposed to a solid line) with centre $(1,2)$ and where the square crosses the $x$-axis at points $(4,0)$ and $(-2,0)$ and crosses the $y$-axis at $(0,5)$ and $(0,-1)$.

I'm unsure for 3.) and 4.).

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Since you have "balls" for various norms in $\mathbb{R}^2$, how about drawing the boundaries (not necessarily "smooth" curves) of those balls by writing the equations for each norm equal 3? – hardmath Feb 22 '13 at 2:58
up vote 1 down vote accepted

You are correct, 1) is obvious a circle and 2) is a square. I copied the following images from WolframAlpha, each with reference to the page, but I will leave out case 1). I will furthermore use the derived formulas from bubba, so the credit belongs to him.

2) $\max\{|x-1|,|y-2|\} = 3$ Reference here. case 2)

3) $|x-1|+|y-2| = 3$ Reference here. case 3)

4) $ 9(x-1)^2+(y-2)^2=9$ Reference here. case 4)

Note: Please excuse my poor Gimp skills.

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$$\mathbf{x} \in B_2((1,2),3) \Longleftrightarrow \Vert \mathbf{x} - (1,2)\Vert_2 = 3 \Longleftrightarrow \sqrt{(x-1)^2 + (y-2)^2} = 3$$

$$\mathbf{x} \in B_\infty((1,2),3) \Longleftrightarrow \Vert \mathbf{x} - (1,2)\Vert_\infty = 3 \Longleftrightarrow \max\{\vert x-1\vert, \vert y-2\vert\} = 3$$

$$\mathbf{x} \in B_e((1,2),3) \Longleftrightarrow \Vert \mathbf{x} - (1,2)\Vert_e = 3 \Longleftrightarrow \sqrt{9(x-1)^2 + (y-2)^2} = 3$$

And so on. Figure out the shapes described by the equations on the right-hand sides.

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+1 as I used your derived formulas. – k1next Feb 22 '13 at 7:43

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