I'm working on an example on where I am required to construct a map which sends a point on the unit circle (i.e. ${X_1}^2+{X_2}^2=1$) to a line $L$ and vice versa.

The example says: "To construct this map, project from the south pole onto the line $X_2=1$ as shown in the figure below. A simple computation gives $$x_1=4t/(4+t^2), ~ x_2=(4-t^2)/(4+t^2), ~ t=2x_1/(x_2+1)."$$

enter image description here

(The picture is taken from Koblitz, Neal "p-adic numbers, p-adic analysis, and zeta-functions", p. 112, which can be found online here: http://citeseerx.ist.psu.edu/viewdoc/download?doi= - I hope it's ok that I used it in this question.)

Now, as I understood it, there is the aim to have the following maps: $\varphi: B_1(0) \rightarrow L, x=(x_1,x_2) \mapsto t$ where then $t$ is defined as $t=2x_1/(x_2+1)$. Vice versa $ \psi: L \rightarrow B_1(0), t \mapsto x=(x_1,x_2)$ with $x_1, ~x_2$ being defined as follows: $x_1=4t/(4+t^2), ~ x_2=(4-t^2)/(4+t^2)$.

Moreover I think, that $t$ is supposed to denote the $x_1$-coordinate of the points on $L$. (EDIT: changed this as pointed out in the comments)

My questions:

  1. Are my ideas so far correct?
  2. I do not understand which "simple computation" yields these results. I tried to work with orthogonal projection and constructing the line equation, but it never resulted in the required results...

Any ideas would be very much appreciated!

EDIT: As pointed out, I say that any point $C$ on $L$ has coordinates $(t,1)$. Then by constructing the line $g$ through the south pole $(0,-1)$ and the point $X=(x_1,x_2)$, I come to the conclusion that if $C \in g$ it has the coordinates $(t, 1)=(2x_1/(x_2+1),1)$ which solves one direction :)

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    $\begingroup$ From the figure it seems that $t$ is the position along the line, so it is oriented in the $x_1$ direction. $\endgroup$ – N74 Oct 28 '17 at 9:14
  • $\begingroup$ Thank you, that makes sense - I changed it in my post. $\endgroup$ – SallyOwens Oct 28 '17 at 9:32

After the edit it should be straightforward how to proceed: $$t={2x_1 \over x_2+1}$$ so $$x_2={2x_1 \over t} -1.$$ Using the equation for the circle, $$x_1^2+\left ( {2x_1 \over t} -1 \right )^2=1$$ $${4x_1^2 \over t^2} - {4x_1 \over t} +1+x_1^2=1$$ $$4x_1^2 - 4x_1t +t^2x_1^2=0.$$ This equation has $x_1=0$ as solution, corresponding to the "south pole". The other solution is $$4x_1+t^2x_1=4t$$ so $$x_1={4t \over 4+t^2}.$$

I think you can find $x_2$ by yourself, now.


You can find the inverse map by solving the system $$t = {2x_1\over1+x_2} \\ x_1^2+x_2^2=0$$ as in N74’s answer, or perform a “simple” trigonometric computation. Considering the angle that the dotted line makes with the $y$-axis, we have $x_1 = \sin\left(2\arctan\frac t2\right)$ and $x_2=\cos\left(2\arctan\frac t2\right)$. It’s fairly easy to derive the identities $\sin(\arctan y)={y\over\sqrt{1+y^2}}$ and $\cos(\arctan y)={1\over\sqrt{1+y^2}}$ and using the identity for the sine of a double angle, we get $$\begin{align} x_1 &= 2 \sin\left(\arctan\frac t2\right)\cos\left(\arctan\frac t2\right) \\ &= 2{t/2 \over \sqrt{1+t^2/4}}{1 \over \sqrt{1+t^2/4}} \\ &= {t \over 1+t^2/4} \\ &= {4t \over 4+t^2}. \end{align}$$ A similar calculation using the double-angle cosine identity yields an expression for $x_2$ as a rational function of $t$.


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