$f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right) $
How do I find inverse function? I have trouble here, because there are 2 variables and I am not sure how to do it?
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1Hint: what is $f(f(x, y))$? – Apple Jun 18 '16 at 17:34
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Your function $f$ is from $\mathbb{R}^2$ to $\mathbb{R}^2$, so it can be represented in a matrix we say for example $A$. Now you find $A^{-1}$ if it is exists – A s Jun 18 '16 at 17:35
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Do you have any reason to believe there is an inverse to this function? You are mapping every point in $R^2$ (except (0, 0)) to $R$. – user247327 Jun 18 '16 at 17:36
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2Only linear functions can be represented by matrices. This function is not linear. – user247327 Jun 18 '16 at 17:37
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Are you from Croatia? Or maybe Serbia? – Farewell Jun 18 '16 at 17:41
4 Answers
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The procedure is the same as always when you have to find an inverse:
Set $u = \frac{x}{x^2+y^2}, \quad v = \frac{y}{x^2+y^2}$ and solve for $x,y$.
You can easily see immediately that $u^2+v^2 = \frac{x^2+y^2}{(x^2+y^2)^2} = \frac{1}{x^2+y^2}$ and therefore $x^2+y^2 = \frac{1}{u^2+v^2}$.
From the equations above we then get $$x = u\cdot (x^2+y^2) = \frac{u}{u^2+v^2}$$
and similarly
$$y = v\cdot (x^2+y^2) = \frac{v}{u^2+v^2}$$
So $f$ is it's own inverse on $\mathbb R^2 \setminus \{0\}$.
In fact, this particular map is well known as circle inversion. It is basically mirroring the points on a mirror shaped like an unit circle.
flawr
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your function maps every point $\mathbf{p}\in\mathbb R \setminus \{0\}$ to $\mathbf{p} / (\|\mathbf{p} \|^2)$, the same point but divided by its norm squared. So the direction is unchanged but the size is inverted. To 'undo' this you simply apply the same function again.
Slugger
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Let $f(x,y)=(a,b)\implies f^{-1}(a,b)=(x,y)$ Therefore$$a={x\over x^2+y^2}\ ;\ b={y\over x^2+y^2}$$
$$a^2+b^2 = \frac{x^2+y^2}{(x^2+y^2)^2} = \frac{1}{x^2+y^2}$$ So $$x^2+y^2 = \frac{1}{a^2+b^2}$$.
Now check$$x = a (x^2+y^2) = \frac{a}{a^2+b^2}$$
Do exactly the same operations with $y$ and you are left with $$x={a\over a^2+b^2}\ ;\ y={b\over a^2+b^2}$$
As said before (and it fortunately turns out that) $f^{-1}(a,b)=\left({a\over a^2+b^2},{b\over a^2+b^2}\right)=f(a,b)$
Qwerty
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@AnaMatijanovic I have solved it entirely for you. Happy? :-) Please accept if you liked my answer. – Qwerty Jun 18 '16 at 17:59
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@user84413 In a previous comment (which the OP has deleted) she asked me to do the solving explicitly. So I did it. – Qwerty Jun 18 '16 at 18:14
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$$ \left(x_{inv},y_{inv}\right) \rightarrow \left(\frac{a^2 x}{x^{2}+y^{2} }, \frac{a^2y}{x^{2}+y^{2} } \right) $$
The inverse of right bracket is the original left bracket itself.
The two way mapping can be geometrically best illustrated /explained as an inversion geometrically. The red circle with radius $ a $ reflects outside this circle to inside and vice-versa. $$ r\rightarrow 1/r $$ if inverted with respect to unit circle and, $$ r\rightarrow a^2/r $$ if inverted with respect to a circle radius $a$.
Narasimham
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