1

Let $B_a$ be the open ball{$x: |x|^2<a$} in $\mathbb{R}^k, |x|^2 = \sum x_i^2$. What is the inverse of the function $x \mapsto \frac{ax}{\sqrt{a^2 - |x|^2}}$?

Here I want some justification to equate $|x|^2$ with $x^2$, I don't know to justify; but otherwise I don't know how to solve it.

WishingFish
  • 2,412
  • 5
    You can't equate $|x|^2$ with $x^2$ because that makes no sense; what is $x^2$ when $x$ is the vector $(1,3,2)$? What you should try instead is to give the function value a name, say $y = ax/\sqrt{a^2-|x|^2}$, observe that this is a vector $x$ times a scalar $c=a/\sqrt{a^2-|x|^2}$, so you can divide by $c$ and get $x = y/c$, and then try to express $c$ in terms of $y$ instead of $x$. –  Jun 27 '13 at 22:22
  • 3
    Note that the map leaves the rays emanating from $0$ invariant, so all you have to do is to invert the function of the radius $r \mapsto \frac{ar}{\sqrt{a^2 - r^2}}$. – Daniel Fischer Jun 27 '13 at 22:25
  • Hi @RahulNarain - I tried to express $c$ in terms of $y$ - mission impossible. Because $y$ is defined in terms of $c$. Could you illuminate a bit? Thank you. – WishingFish Jun 27 '13 at 23:00
  • Hi @DanielFischer - could you explain a little bit more details that why I can treat $|x|^2$ as $x^2$ by seeing $x$ as $r$? Thanks. – WishingFish Jun 27 '13 at 23:06
  • 1
    @user83036 You can write $x = r\cdot \xi$ where $\lVert\xi\rVert = 1$. Then you have $F(r\cdot\xi) = f(r)\cdot\xi$. – Daniel Fischer Jun 27 '13 at 23:12

1 Answers1

3

Try to think a minute about what this function does; as Rahul Narain stated, it multiplies a vector by a "scalar" which depends on the norm of the vector.

Second, remark the spherical invariance of this scalar; any vector with norm $r$ (origin to any point on the sphere of radius $r$) will have an associated scalar $\frac{a}{\sqrt{a^2 - r^2}}$.

Third, note that this scalar tends to $1$ when the radius tends to $0$, and that it tends to infinity when the radius tends to $a$. In other words, it "maps" the open $a$-ball to $\mathbb{R}^n$.

Now the only inverse you have to find is that of that last sentence... :) (Well, you do have to understand that mapping as well ;).

Ok, I gave you some time to think about it, now let's see the answer. As we said it's a mapping from $\mathbb{R}^n$ to the open $a$-ball, the question that drives us crazy is which one?

Take a look at the norm of the resulting vector. For any vector of $\mathbb{R}^n$ (this includes any vector in any open ball), we can write $\vec{x} = r\vec{u}$, where $\vec{u}$ is some unit vector, and $r$ is the norm of $\vec{x}$.

The norm of the "image" of $\vec{x}$ by our function (say $\vec{y}$) is $\frac{ar}{\sqrt{a^2 - r^2}}$. Because our application preserves the orientation, we can also write $\vec{y} = s\vec{u}$ (where $\vec{u}$ is the same than before), and we have $|y|=s=\frac{ar}{\sqrt{a^2 - r^2}}$.

If we simply inverse this last relation (see below) we obtain the inverse function, that is $ \frac{a s}{\sqrt{a^2 + s^2}} \vec{u} = \frac{a y}{\sqrt{a^2 + |y|^2}} $.

Details about the inversion: $$ s^2 (a^2 - r^2) = a^2r^2 \iff a^2s^2 = r^2(s^2+a^2) \iff r = \frac{as}{\sqrt{a^2+s^2}} $$

Jonathan H
  • 2,233
  • 16
  • 29
  • So you mean I have to find a map from $\mathbb{R}^n$ to $a$-ball rather than computation? – WishingFish Jun 27 '13 at 23:38
  • Oh, I guess so. Then I just have to justify that $f \circ f^{-1} = f^{-1} \circ f = Id.$ – WishingFish Jun 27 '13 at 23:39
  • 1
    Not a map, the inverse map. What this hint gives you is that the inverse function is also of the form "scalar times vector", and that the scalar is equivalent to $\frac{a}{r}$ when $r$ tends to infinity, and that it tends to 1 when r tends to 0. – Jonathan H Jun 27 '13 at 23:43
  • Thank you Sh3ljohn, I followed your proof but couldn't understand the last step. Could you tell me how did you magically cooked the inverse? What are you inversing when you say "If we simply inverse this"? I tried to inverse $s$ obtained $r = \frac{a}{\sqrt{1+s^2a^2}}$. Since we need to express $x$ in terms of $y$, and $x = ru$, so $x = \frac{a}{\sqrt{1+s^2a^2}}u$. – WishingFish Jun 28 '13 at 12:51
  • 1
    @user83036 Try to see if my last edit makes it clearer :) – Jonathan H Jun 28 '13 at 13:48
  • Ah sorry, it sudden come to my mind, so I start to draft on a piece of paper without finishing it... So my idea is like this: apparently, the mapping preserves direction. So if we see f(x) as $\Big|\frac{ax}{\sqrt{a^2 - |x|^2}}\Big| \vec{u}$, the the inverse function is some scalar times $\vec{u}$. And $\Big|\frac{ax}{\sqrt{a^2 - |x|^2}}\Big| \vec{u} = \frac{ar}{\sqrt{a^2 - r^2}} \vec{u}$, so $f^{-1}$ is just the reciprical of $\frac{ar}{\sqrt{a^2 - r^2}}$ – WishingFish Jun 28 '13 at 14:05
  • 1
    @user83036 I would say yes, but I'm not sure what you mean by "reciprocal"; don't forget that the scalar is NOT a constant, since it depends on the norm of the input vector. So the reciprocal is no just the denominator on the numerator here, that's why we do this calculation at the end, but I think you got it ;) – Jonathan H Jun 28 '13 at 14:16
  • Yeah, got it. I overlooked the fact that the scalar is not a constant, and tried to fuel my brain with Latte... But your explanation got me right out of there, thanks! – WishingFish Jun 28 '13 at 14:25
  • @user83036 You're welcome, don't forget to mark it as an answer if you find it satisfactory ;) – Jonathan H Jun 28 '13 at 14:28
  • Fantastic answer, thanks! – WishingFish Jun 28 '13 at 14:49