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Finding maxima of a function

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    I have been trying to find the value of x that maximizes the function is section 2.

    z is a variable distributed using a standard normal distribution i.e. can vary between -∞ and ∞, but generally is between -4 and 4. m varies between 0 and 1. c varies the same way as z.

    x is always greater than c (so the function is always real).


    2. Relevant equations

    http://www4b.wolframalpha.com/Calculate/MSP/MSP6341gbhe843gi5189e300002d76b658d53h3hd0?MSPStoreType=image/gif&s=45&w=271.&h=47 [Broken].

    Alternate Wolfram-Alpha link:
    http://www.wolframalpha.com/input/?...qrt((x-c)/m))*(erf(z/sqrt(2))-erf(x/sqrt(2)))


    3. The attempt at a solution

    I basically attempted to differentiate it (which is fine), and it gives me a really complicated solution. I set this to zero (to find the turning point), and am having trouble solving that equation. I was able to find specific values of this maxima by setting the other variables: z, m, c to specific values. For example,

    http://www.wolframalpha.com/input/?...2))*(erf(0.5/sqrt(2))-erf(x/sqrt(2))),+maxima

    Sorry, this isn't exactly homework, but for a research project I'm working on in college. Any help/guidance will be greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 27, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If it is not homework, there is no guarantee that there is a nice, analytic solution.

    You can simplify the problem a bit: Ignore the constant prefactor of sqrt(pi/2), substitute x/sqrt(2) by another variable and do the same for z, c and m.

    This should lead to something like
    $$y'=\sqrt{\frac{x'-c'}{m'}} \left( erf(z') - erf(x') \right)$$

    A maximum of y is a maximum of y^2 as well, so you can square the whole expression and look for a maximum of this. Even if there is no analytic solution, it could be easier to analyze numerically.
     
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