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!

Finding a function that goes through given points

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a function of a decaying logarithmic trend that passes through a set of 3 points (at x=1,2,3) whose sum is S, where S>0.

    2. Relevant equations


    3. The attempt at a solution
    starting point: i assumed the equation had the form: $$y= -kln(x+1)+S$$ with the condition that must satisfy:

    $$S=Σ(y(x))$$ from 1→3

    but I don't know what to do next
     
  2. jcsd
  3. Oct 11, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    In your proposed generic form, I see no reason to choose x+1 rather than x+an unknown constant. And I don't understand the final S; did you mean this to be the same as the given S or something different?
    If we make it x+c as well as a final +d (say) then we have three unknowns (c, d, k) but effectively only one equation. So presumably the question expects a simpler generic form, but it is not clear what. If I had to guess, I would ditch d first.

    Edit:After a little background reading, seems like you should keep d and ditch c. Logarithmic decay appears to be defined as the functional inverse of exponential decay, i.e. x=Ae-ky.
     
  4. Oct 11, 2015 #3
    Oh, no actually not necessary, I was just trying to find a convenient way to set up the graph on my calculator; my proposed form would have essentially looked like the same graph as that of x=Ae^(-ky)

    Same S.


    I tried your suggested formula. Since the A was left as a free variable to be determined, i had two equations two unknowns and was able to solve. much thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted