A book I am using has a problem which includes two points on the graph of $y=\ln x$, $M_1(x_1, y_1)$ and $M_2(x_2, y_2)$ and identifies the middle of the chord $M_1 M_2$ between them as $I(x_e, y_0)$. What do you suppose is meant by $x_e$ in this case? Is it $x=e$ or $x$ such that $y=e$ or ? I could put the whole problem here, but I hope to solve it myself once I understand the notation.

Update: the whole problem (please don't solve it yet or I won't get to. I would like to understand the problem better. I've done some thinking and computing and have some good ideas, but I don't understand the problem fully. Thus, hints would be more appreciated than an answer.):

Klein & Reeb, Problem I.1.10

Soient $M_1(x_1,x_y)\,M_2(x_2,y_2)$ deux points du graphe de $y=\ln x$. Soit $I(x_e,y_0)$ le milieu de la corde $M_1 M_2$. D'aprés la concavité du graphe $y_0 < \ln x_0$; en déduire: $2 \sqrt{x_1 x_2} < x_1 + x_2$.

My translation:

$M_1(x_1,x_y)\,M_2(x_2,y_2)$ are two points on the graph of $y=\ln x$. $I(x_e,y_0)$ is the middle of the chord $M_1 M_2$. According to the concavity of the graph $y_0 < \ln x_0$; deduce $2 \sqrt{x_1 x_2} < x_1 + x_2$.

Update: given GPerez's comment below which agrees with a thought I had too about $x_e$, I'd say it's possible (but not yet conclusive in my mind) the book has a typo and $x_0$ was meant instead.

Update: my thought process on solving the problem was approximately thus: since $y=\ln x$ is concave, if $M_1$ is negative, then $0<x_1<1$ since $\ln x < 0$ for $0 < x < 1$. If $x_e$ wasn't a typo, I had supposed that, since it was the x-coordinate of the midpoint of $M_1 M_2$ it might mean that $M_2$ had to be located to the right of either where $y$ attains the value $e$ or where $x=e$, and knowing this would indicate that $M_2$ is to the right of that point since it is the right endpoint. From these facts, we would then deduce that $0<x_1<1$ and (for the case where we suppose $x_e$ means $x=e$:) $e<x_2<\cdot$ where $\cdot$ here is some value to the right of $x_2$ or (for the case where we suppose that $x_e$ means $y=e$:) $1<x_2<\cdot$ where $\cdot$ here is some value to the right of $x_2$. Then using the rules of logarithmic manipulation we could expand the LHS and RHS of the inequality given and compare term by term to prove it based on our knowledge of the range of possible values for $x_1$ and $x_2$ supposed above. That was my plan, but given the uncertainty of the notation and the incompletion, as of yet, of my computations, I've not concluded the solution.

  • $\begingroup$ Can you give more context? As stated, the question is unanswerable. Also note that your question asks about $I(x_e, y_0)$ while the title includes $l(x_e, y_0)$. Wich one is it? $\endgroup$ – AlexR May 4 '15 at 16:17
  • $\begingroup$ what is the title of this book? $\endgroup$ – Dr. Sonnhard Graubner May 4 '15 at 16:19
  • $\begingroup$ @ AlexR, I updated the title. $\endgroup$ – Joe May 4 '15 at 16:19
  • $\begingroup$ The book is by Klein & Reeb. M.P.C. $\endgroup$ – Joe May 4 '15 at 16:20
  • 1
    $\begingroup$ It seems that $x_e$ is to be $x_0$, since in the second sentence he uses $x_0$, which hadn't been spoken of before, unless it was supposed to be $x_e$. It also makes sense since it's paired with $y_0$. $\endgroup$ – GPerez May 4 '15 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.