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 have the following question:

Given a line $y=\theta(1-x)$ where $0<x<1$, $0<y<1$ and $0<\theta<1$, I have a collection of curves $$ y^K=1-(1-x)^K $$ parametrized by an integer $K>2$. For any $K$, the related curve intersects the line at only one point. For this intersection point let $$ A=y^K $$ On the other hand I have \begin{equation} \prod_{i=1}^K {y_i}=1-\prod_{i=1}^K(1-x_i) \end{equation} where $x_i$ and $y_i$ are some points on $y=\theta(1-x)$ and let $$ B=\prod_{i=1}^K {y_i} $$ Is it possible that $B<A$?

Thank you very much in advance for any constructive comment.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

Replace $1-x$ by $x$. Then you have the line $\ \ell\colon\ y=\theta x$ and for a given $K>2$ the curve $$\gamma: \quad x^K+y^K=1\ ,$$ where everything is restricted to the window $[0,1]^2$. The curve $\gamma$ intersects $\ell$ in a point $p:=(u,v)$. Obviously $v=\theta u$, and as $p\in\gamma$ we get the equation $$u^K(1+\theta^K)=1\ .$$ This implies $$A:=v^K=\theta^K u^K={\theta^K\over 1+\theta^K}\ .$$ On the other hand you are given $K$ points $(x_i,y_i)\in\ell$, chosen such that $$\prod_{i=1}^K x_i + \prod_{i=1}^K y_i=1\ .$$ As $y_i=\theta x_i$ for each $i$ we have $(\theta^{-K}+1)\prod_{i=1}^K y_i=1$ or $$B:=\prod_{i=1}^K y_i={1\over \theta^{-K}+1}={\theta^K\over 1+\theta^K}=A\ .$$

share|improve this answer
2  
Thank you very much for the solution.. –  Seyhmus Güngören Jul 27 '12 at 10:28
add comment

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.