Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need some help (hints or an answer) in finding the actual equation of this bijections from the reals $ \mathbb{R} $ to $ (0,1) $. We may assume that the radius of the circle is $ \dfrac{1}{2} $.

enter image description here

Also, is it possible to transform this geometric representation to an order-preserving isomorphism?

Thanks!

Also, I am aware there are much simpler isomorphisms between $ \mathbb{R} $ and $ (0,1) $.

share|improve this question
1  
See Stereographic projection at en.wikipedia.org/wiki/Stereographic_projection for inspiration. –  B. S. Jan 23 '13 at 6:59
1  
"(This is my first post!)" Then what's this? –  Rahul Jan 23 '13 at 7:45

2 Answers 2

up vote 2 down vote accepted

The point $x$ in $(0,1)$ is sent to the point $y(x)$ in $\mathbb R$ if there exists some angle $\theta$ in $(0,\pi)$ such that $\cos\theta=2x-1$ and $y=a\cot\theta$, where $a$ is the distance between the center of the circle and the line. Since $\sin\theta=\sqrt{1-\cos^2\theta}$, this yields $$ y(x)=\frac{a(2x-1)}{\sqrt{1-(2x-1)^2}}=\frac{a(2x-1)}{\sqrt{4x(1-x)}}. $$ For $y\mapsto x(y)$ sending $\mathbb R$ to $(0,1)$, use the inverse mapping $$ x(y)=\frac12\left(1+\frac{y}{\sqrt{y^2+a^2}}\right). $$ Both functions $x\mapsto y(x)$ and $y\mapsto x(y)$ are increasing diffeomorphisms.

share|improve this answer
    
Thank you sir! great answer! –  Cornelius Johnson Jan 23 '13 at 8:46

Just do the work.

Assume that the circle is $x^2+y^2 = 1$. Assume that the line is $y=-k$ for some positive real number $k$.

Then, the point $(x, k)$ will get mapped to the point $(\frac {x}{\sqrt{x^2+k^2}}, \frac {-k} { \sqrt{x^2+k^2}})$.

If you want to map it to $(0, 1)$ as opposed to $(-1, 1)$, then take the transformation $t \rightarrow \frac {t+1} {2} $. Alternatively, use another circle but that makes the derivations ugly.

share|improve this answer
    
Calvin I dont see how this represents a bijection from R to (0,1). Since (x,k) doesn't lie in R. I am probably just not seeing it, could you define an explicit function? –  Cornelius Johnson Jan 23 '13 at 7:15
    
the map $f(x) = \frac {x}{\sqrt{x^2+k^2}}$ is a map from $\mathbb{R}$ to $(-1, 1)$. That was implicit in your diagram. –  Calvin Lin Jan 23 '13 at 7:17
    
Oh ok thanks you so much. You are right. Sorry I forgot to mention the circle has radius 1/2, which implies the top line segment is actually (0,1) not (-1,1) –  Cornelius Johnson Jan 23 '13 at 7:23
1  
@CorneliusJohnson It's just much easier to do it on $(-1, 1)$, since that's the standard unit circle that we're used to. As I said, if you want to use $(0, 1)$, just do a further transformation. The direct approach will give you quite ugly algebra that takes away the intuition of what's happening. –  Calvin Lin Jan 23 '13 at 7:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.