1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find where one graph is bigger than the other

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data

    My question is part of a bigger one that I'm using in a graphical proof. I have two functions, tanh(ap) and 2a/(1+a^2). The a is the input and p is a constant. I'm trying to find the value for p which is the bridging point between the two functions having 1 intersection and 2 intersections. To do this, I said that we want to find the value for p which makes tanh(ap) greater than 2a(1+a^2) for every point on the interval (0,1) (because the second function's max is at 1, so the two intersections will be on the sides of a=1)

    So I have tanhap>2a/(1+a^2)>0, but I'm not sure how to go about solving it. I can't set them equal to each other at a=1 because tanha never reaches 1 so I get an undefined answer. Is there something I can do with limits?
  2. jcsd
  3. Oct 18, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Do I understand that your problem is to find $$p: \tanh(px) > \frac{2x}{1+x^2}: x\in (0,1)$$... what's wrong with putting LHS=RHS and finding the intersection in terms of p as a starting point?

    You can gain an understanding of what p does by plotting tanh(px), and 2x(1+x^2), on the same axis, in the interval, for several values of p using a math-script program like matlab, mathematica, or gnu/octave.

    But I think I see your problem ... if you had a hard interval like (0,1] then your requirements cannot be met since at x=1, RHS>LHS ... and this is unambiguous. However, you only need LHS>RHS for x values arbitrarily close to 1.

    Interestingly, the RHS has a turning point at x=1 ... the slope of the RHS is a maximum at the origin, so any value of p that makes the slope of the LHS the same at the origin will make the LHS > RHS for much of the interval. But I suspect that you can only make the other two intersection points arbitrarily close together as p -> infinity. I suppose a limit formulation would be to find p that satisfies the conditions for an interval (0,z] in the limit z -> 1. Trouble is, p(x) is assymtotic at x=1.
    Last edited: Oct 18, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook