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.

Circle $C_1$ is tangent to the curves $y=e^x$ and $y=-e^x$ and the line $x=0$, and for $n>1$ circle $C_n$ is tangent to both curves and to $C_{n-1}$, how can I find the radius of any circle $C_k$?

This is related to my previous question: summing series usings circles inside curves

share|improve this question
1  
let $r$ be the radius of $C_1$ and $(x,e^x)$ the point of tangency. then we have $(r-x)^2+e^{2x}=r^2$ and $e^{2x}/(r-x)=1$ –  yoyo Dec 26 '11 at 19:40

1 Answer 1

up vote 7 down vote accepted

The circle with centre $(a,0)$ and radius $r$ is tangent to $x=0$ if $r=|a|$. It is then tangent to $y=e^x$ and $y=-e^x$ if there is $x$ such that $e^{2x} + (a-x)^2 = a^2$ and $2 e^{2x} + 2(x-a) = 0$. Eliminating $a$, the equation for $x$ is $(1-2x) e^{2x} - x^2 = 0$. This has two real solutions, approximately $-.7478435353725761$ and $.4580318389459811$, corresponding to $a = -.5237489523800181$ and $2.957464268427510$ respectively. I don't think there are "closed-form" solutions, even with the use of the Lambert W function.

Once you have the solution for the first circle, similar methods will get further circles. Thus the circle with centre at $(a_2, 0)$ and radius $r_2$ is tangent to $y=e^x$ and $y=-e^x$ and $x = 2 a$ (and thus to the first circle) if $r_2 = |a_2 - 2 a|$ and $e^{2x} + (a_2 - x)^2 = r_2^2$ and $2 e^{2x} + 2 (x - a_2) = 0$ For the case $a = -.5237489523800181$, I get $a_2 = -1.308841381001648$.

Here's a picture.

enter image description here

share|improve this answer
    
pretty picture :) –  ae0709 Dec 27 '11 at 1:22

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.