My Math Forum inverse function of a multivariable function

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 February 23rd, 2013, 10:24 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 inverse function of a multivariable function could someone show me step by step how to find an inverse function of the following one... $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$
 February 23rd, 2013, 05:40 PM #2 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: Inverse function of a multivariable function There is no inverse. The function is not injective, e.g. $f(1,1)=f(2,2)$.
 February 23rd, 2013, 09:57 PM #3 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 Re: inverse function of a multivariable function well in my notes it actually is a bijective and cont fcn (thus a homeomorphism) between $f:S^{1}\to square$, they even give that inverse $f^{-1}(x,y)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x ^2+y^2}})$ but I have no clue how to find an inverse of a multivariable function
 February 24th, 2013, 08:17 AM #4 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: Inverse function of a multivariable function Then you should have stated that the domain of your function is $\mathbb{S}^1$ in the first place! If you don’t state your question clearly, how are people going to understand your problem, let alone help you? So, your function $f$ is from the circle $\mathbb{S}^1=\{(x,y)^2+y^2=1\}" /> to the square $\Sigma=\{(x,y)x|+|y|=1\}" />. Draw a straight line from the origin making an angle $\theta$ ($0\leq\theta<2\pi$) anticlockwise from the positive x-axis. This line intersects $\mathbb{S}^1$ and $\Sigma$ at $(\cos\theta,\sin\theta)$ and $\left(\frac{\cos\theta}{|\cos\theta|+|\sin\theta|} ,\frac{\sin\theta}{|\cos\theta|+|\sin\theta|}\righ t)$ respectively. What $f$ does is map the first point to the second point, so the inverse $f^{-1}$ maps the second point to the first point. It’s just a matter of expressing the coordinates of the first point in terms of coordinates of the second point, really. Let $x=\frac{\cos\theta}{|\cos\theta|+|\sin\theta|}$ and $y=\frac{\sin\theta}{|\cos\theta|+|\sin\theta|$. We want first to express $\cos\theta$ in terms of $x$ and $y$. Notice that $\sqrt{x^2+y^2}=\frac{1}{|\cos\theta|+|\sin\theta|}$. Hence $\cos\theta=\frac{\cos\theta}{|\cos\theta|+|\sin\th eta|}\,/\,\frac{1}{|\cos\theta|+|\sin\theta|}=\frac{x}{\sq rt{x^2+y^2}}$ Similarly $\sin\theta=\frac{\sin\theta}{|\cos\theta|+|\sin\th eta|}\,/\,\frac{1}{|\cos\theta|+|\sin\theta|}=\frac{y}{\sq rt{x^2+y^2}}$

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# finding inverse of function of multiple variables

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