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Rearrange for x. Is this even possible?

  1. Jul 24, 2013 #1
    Hi all,

    1. The problem statement, all variables and given/known data

    I'm working on a problem related to an MRI signal equation. Essentially I want to derive an equation that can tell me the point at which f(x) is maximum. I have differentiated with respect to x and double checked that there are no errors.

    But now I am stumped at how to rearrange for x. Or even if it is possible to rearrange for x. Here is the equation:

    2. Relevant equations

    0 = ax-1.5 + be-cx * (cx-0.5 + 0.5x-1.5)

    3. The attempt at a solution

    Is it possible to rearrange for x? If so what technique should be used? I am happy to have a go at doing the algebra myself but I'm completely stumped!

    Any help will be greatly appreciated :smile:
     
  2. jcsd
  3. Jul 24, 2013 #2

    haruspex

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    No, there's no algebraic way to rearrange this. You could simplify it greatly by multiplying through by x1.5. If c is small you could expand the exponential as a power series, multiply out, discard terms beyond x2, and solve the quadratic. You could then improve on that by iterative methods.
     
  4. Jul 24, 2013 #3

    HallsofIvy

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    As haruspex said, you can start by multiplying through by [itex]x^{1.5}[/itex] to get [itex]0= a+ be^{-cx}(cx+ 0.5)[/itex]. If you let y= bcx+ 0.5b, x= ((y/b)- 0.5)/c the equation becomes [itex]0= a+ ye^{y/bc}e^{0.5/c}[/itex]. Divide both sides by bc and you have [itex]0= a/bc+ (y/bc)e^{y/bc}e^{0.5/c}[/itex] which is the same as [itex](y/bc)e^{y/bc}= -ae^{-0.5/c}/bc[/itex].

    Let z= y/bc and it becomes [itex]ze^x= -ae^{-0.5/c}/bc[/itex]. The solution to that is [itex]z= y/bc= W(-ae^{-0.5/c}/bc)[/itex] so that [itex]y= bcx+ 0.5b=bcW(-ae^{-0.5/c}/bc)[/itex] and [itex]x= W(-ae^{-0.5/c}/bc)- 0.5/c[/itex].

    There "is no algebraic way to rearrange this" because W, the "Lambert W function", used above, is not an "elementary function". It is defined as the inverse function to [itex]f(x)= xe^x[/itex].
     
    Last edited by a moderator: Jul 24, 2013
  5. Jul 27, 2013 #4
    Alternate Solution

    Apart from using special functions you can use plotting techniques to obtain elements in the solution set.

    Solve the equation into two functions of x on either side of the equal sign.

    -ax^-1.5/ (cx^-.5+.5x^-1.5)= be^-cx

    This can be simplified to

    -a/(cx+.5)= be^-cx taking that x≠0 (which is a solution in itself).

    Plotting these two functions on the same axes will show the intersections of the graph i.e. where they are equal to each other.

    Like I said, this is just an alternate way to doing this problem without knowledge of special functions (but with knowledge of some language like maple or mathematica).
     
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