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I have been all over the internet but only got surprised in how no one mentions what the general equation(formula) is and how to derive it, for a map $\mathbb{R}^n \rightarrow S^n$ \ $P$where $S^n=\{(x_0,...x_n) \in \mathbb{R}^n+1| x_0^2+...+x_n^2=1\}$ and $P=(1,0...0)$.

So it is a stereographic projection but instead of the map "input;point on sphere" "output;Some point on$\mathbb{R}^n$" I am trying to find the opposite. So for some point $y \in Y=\{(0,y_1...,y_n) \in \mathbb{R}^{n+1}\}$, I want to find the point $x=(x_0,...,x_n) \in S^n$\ $P$ that $x,y,P=(1,0...,0)$ are all collinear.

I don't know how to derive this, I don't even know what the equation should look like. Can someone help me please?? Thank you....

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  • $\begingroup$ I don't think you want an equation (which is statement that two things are equal) but a formula. $\endgroup$ – Cheerful Parsnip Oct 19 '15 at 21:17
  • $\begingroup$ Sorry about my English skills, sure, then yes I am looking for a formula or maybe in this case a mapping, $\endgroup$ – Melba1993 Oct 19 '15 at 21:18
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It is known the stereographic projection $\pi : S^n\smallsetminus N\longrightarrow \Bbb R^n$ is given by mapping $$(x_1,\ldots,x_{n+1})\mapsto \frac{1}{1-x_{n+1}}(x_1,\ldots,x_n)$$

If you haven't proven this, consider the line $\lambda (N-x)+x$ that passes through a point $x$ in the sphere and the north pole $N=(0,\ldots,0,1)$. The equation $\lambda(x-N)+N=(y_1,\ldots,y_n,0)$ gives that $\lambda(x_{n+1}-1)=-1$ so that $\lambda =\dfrac{1}{1-x_{n+1}}$ and the rest is $\lambda x_i=y_i$ for $i=1,\ldots,n$, which is the above.

To find its inverse, set $y_i = \frac{x_i}{1-x_{n+1}}$. Note that $y_1^2+\cdots+y_n^2=\dfrac{ x_1^2+\cdots+x_n^2}{(1-x_{n+1})^2}=\dfrac{ 1-x_{n+1}^2}{(1-x_{n+1})^2}=\dfrac{1+x_{n+1}}{1-x_{n+1}}$ so that $$x_{n+1}=\frac{y_1^2+\cdots+y_n^2-1}{y_1^2+\cdots+y_n^2+1}$$

This gives the remaining equations for the $x_i$ since $y_i(1-x_{n+1})=x_i$.

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  • $\begingroup$ Thanks for answering, but one thing; how can you have $x_{n+1}$ in your final formula when $x=(x_1,....,x_{n+1}$ is your output(unknown)? $\endgroup$ – Melba1993 Oct 19 '15 at 21:32
  • $\begingroup$ Oh hang on, so you do have an expression for $x_{n+1}$in terms of $y$... $\endgroup$ – Melba1993 Oct 19 '15 at 21:37
  • $\begingroup$ @Melba1993 Right, I am saying the map sends a point $(y_1,\ldots,y_n,0)$ to those $x_i$. $\endgroup$ – Pedro Tamaroff Oct 19 '15 at 21:37
  • $\begingroup$ Brillinat, I got it. Thanks a lot Pedro! $\endgroup$ – Melba1993 Oct 19 '15 at 22:05
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the answer $(1,0..,0)+c(-1,y_1,...,y_n)=(1-c,cy_1,..,cy_n)$

$(1-c)^2+c^2\mid Y\mid^2 =1$, $c=2/\mid Y\mid^2$

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  • $\begingroup$ Can you please elaborate on how that is obtained? $\endgroup$ – Melba1993 Oct 19 '15 at 21:16

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