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I wish to find the inverse of $\dfrac{x}{\|x\|}$, where $x \in \mathbb{R}^2$

Let's do this.

Let $$y_1 = \dfrac{x_1}{\sqrt{x_1^2+x_2^2}}$$

$$y_2 = \dfrac{x_2}{\sqrt{x_1^2+x_2^2}}$$

Then $$y_1 = \dfrac{x_1}{\sqrt{x_1^2+x_2^2}} \Rightarrow \dfrac{x_2^2}{x_1^2} = \dfrac{1 - y_1^2}{y_1^2}$$

Similarly

$$y_2 = \dfrac{x_2}{\sqrt{x_1^2+x_2^2}} \Rightarrow \dfrac{x_2^2}{x_1^2} = \dfrac{y_2^2}{1-y_2^2}$$

It seems however we combine the above equations, the $x_1$ and the $x_2$ will both drop out. Leaving only the $y$s

How to proceed from here?

StubbornAtom
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Olórin
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  • You mean the inverse function, right? In that case, it clearly doesn't exist. –  Feb 07 '16 at 20:23
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    This function is not 1-1, so you are sunk. It maps all points to the unit circle, so it is no onto either. – ncmathsadist Feb 07 '16 at 20:35

2 Answers2

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The answer is quite easy if you think of what is going on with the transformation. We are taking a vector of $\mathbb{R}^2$ and assigning to it the unique unit vector which has the same direction and orientation.

So the inverse of each unit vector $u$ is the ray which goes from 0 to u (without including the 0).

Note however that there is not an inverse function, since the mapping is not one-to-one.

Jsevillamol
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  • That is the inverse image. There is no inverse function. – ncmathsadist Feb 07 '16 at 20:35
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    @ncmathsadist The OP did not specify otherwise. – Jsevillamol Feb 07 '16 at 20:43
  • @Jsevillamol I am a little confused, this question is inspired from http://math.stackexchange.com/questions/365920/prove-a-square-is-homeomorphic-to-a-circle there is a well known homeomorphism between the square and the disk, and people are telling me there is no inverse...how can that be? – Olórin Feb 07 '16 at 21:13
  • @MathNewb Let's see what's going on. You have defined the mapping over $\mathbb{R}^2$, which is the whole cartesian plane! The mapping is certainly an homeomorphism between the circumference of radii 1 and the frontier of the square $[-1,1]\times[-1,1]$. – Jsevillamol Feb 07 '16 at 21:28
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Let $x_1 = cos \theta$, $x_2 = sin \theta$, where $\theta = arctan(\frac{x_2}{x1})$, then $$y_1 = cos(arctan(\frac{x_2}{x_1})$$ $$y_2 = sin(arctan(\frac{x_2}{x_1})$$

runaround
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