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Let's define a 2-vector: $$ v \in \mathbb{R}^2,\ \ \ v=[v_x,v_y] $$ We then have a 'maximum metric-based' 'distance + rotation' function $d_{max}(v)=[|v_x|+|v_y|,v_{\theta}]$ and a 'distance + rotation' function for the euclidean metric $d_{euclidean}(v)=[\sqrt{v_x^2+v_y^2},v_{\theta}]$. I want a function $f(s,\theta)$ ($s$ for scalar distance and $\theta$ for rotation) such that: $$ d_{euclidean}(v)=f(d_{max}(v)) \\ d_{max}(v)=f^{-1}(d_{euclidean}(v)) $$ Note that $f^{-1}(v)$ is just the inverse function to $f(v)$ such that $f^{-1}(f(v))=f(f^{-1}(v))=v$. My question is, what is this function $f(v)$ and what is the inverse function to this function $f^{-1}(v)$? (Define them in terms of $v_x$ and $v_y$).

($v_{\theta}$ is just the normal rotation component of $v$ (measured in Radians).)

(My math-speak may be a bit wonky and this question is lacking tags, so feel free to correct in a way that does not change the question and add appropriate tags)

Edit: the output of all functions mentioned are 2-vectors (if they were scalar, there would be multiple possible 'correct' outputs)

god of llamas
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  • Do you want a function $f(v)$ that takes as input $v$, or a function $f(d)$ that takes as input $d_\max(v)$? Your notation is inconsistent. Also note that in the latter case no such function is possible. –  May 08 '16 at 22:31
  • @Rahul Sorry for not specifying; the output of all of the functions mentioned are vector. – god of llamas May 08 '16 at 22:34
  • still, $f(v)$ means domain of $f$ is $\mathbb R^2$ but $f(d_{\mbox{max}})$ implies domain is (a subset) of set of metrics on $\mathbb R^2$ – ugur efem May 08 '16 at 22:36
  • @Rahul (I thought about the scalar possibility but now I see what you mean, and as a consequence I am forced to have the output be vector) – god of llamas May 08 '16 at 22:37
  • @ugurefem I'm not sure what you mean/what point you are trying to make – god of llamas May 08 '16 at 22:40
  • @ugurefem It is forced to be $\mathbb{R}^2$ because otherwise the function would not have an answer for certain inputs. – god of llamas May 08 '16 at 22:41
  • what is the domain of $f$? – ugur efem May 08 '16 at 22:42
  • but $f(d_{\mbox{max}}(v))$ means it is $\mathbb R$ – ugur efem May 08 '16 at 22:43
  • No such function ($f: \mathbb{R} \to \mathbb{R}$) can exist. If it did then $f(|x|+|y|) = \sqrt{x^2+y^2}$. Take $y=0$ to get $f(|x|) = \sqrt{x^2} = |x|$ but then $f(|x|+|y|) = |x| + |y| \not= \sqrt{x^2+y^2}$ when $xy\not = 0$. – Winther May 08 '16 at 22:45
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    @ugurefem I see your point now... I think I've found a 'solution' which is to add a "rotation" component to the output. I'll edit the question now. – god of llamas May 08 '16 at 22:46
  • now this makes sense, and @Winther already explained the answer – ugur efem May 08 '16 at 22:50
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    The reason a mapping from the max-norm to the euclidean norm does not exist is that vectors with the same max-norm does not neccessarily have the same euclidean norm. Say if $d_{\rm max} = c$ theen $|x| = \epsilon c$ and $|y| = c(1-\epsilon)$ for some $\epsilon \in [0,1]$. However $d_{\rm euclidean} = c\sqrt{2\epsilon^2 -2\epsilon + 1}$ which depends on $\epsilon$. – Winther May 08 '16 at 22:56
  • With your new formulation it is now possible as the euclidean/max distance + an angle is enough to uniquely define a point in $\mathbb{R}^2$. For example let $\theta$ be the standard polar angle then $$(v_x,v_y) = d_{\rm euclidean}(\cos\theta,\sin\theta) \implies d_{\rm max} = d_{\rm euclidean}(|\cos(\theta)| + |\sin(\theta)|)$$ and $d_{\rm euclidean} = \frac{d_{\rm max}}{|\cos(\theta)| + |\sin(\theta)|}$. Thus $f(d,\theta) = \left(\frac{d}{|\cos(\theta)| + |\sin(\theta)|},\theta\right)$ is such a function. – Winther May 08 '16 at 23:15

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