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Finding the Intersection Of 2 Equations (difficult)

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation:

    [tex]e^{2x}=3x^2[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I can find an approximate solution with a graphing calculator easily, but I am interested how you would find the exact solution.

    I can take the natural log of both sides and wind up with.

    [tex]2x=\ln{3x^2}[/tex]

    I'm not sure how I could proceed from here to get it into terms of x.

    Would perhaps another method be better?
     
  2. jcsd
  3. Aug 29, 2010 #2

    Mentallic

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    There is no analytic solution to this problem. You can solve it in terms of the lambert W function though.

    The lambert W function is as follows:
    if [tex]a=ue^u[/tex] then [tex]u=W(a)[/tex] where u is some function of x and a is a constant.

    [tex]3x^2=e^{2x}[/tex]

    [tex]3x^2e^{-2x}=1[/tex] divided through by e2x

    [tex]xe^{-x}=\frac{1}{\sqrt{3}}[/tex] divided through by 3 then square root of both sides

    [tex]-xe^{-x}=\frac{-1}{\sqrt{3}}[/tex] negative of both sides.

    So now we have it in the lambert W form and thus the solution is

    [tex]-x=W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

    [tex]x=-W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

    But again, we can't express the solution in terms of elementary functions so you will only get a good approximation at best for the solution.
     
  4. Aug 30, 2010 #3

    HallsofIvy

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    Mentallic, I'm not sure why you would say using the Lambert W function (which you did very nicely, by the way) would not be an "analytic" solution. Surely, it is as analytic as, say, using the function ln(x) to solve [itex]e^x= 2[/itex].
     
  5. Aug 30, 2010 #4

    Mentallic

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    Ahh I think I've mixed up the meaning of analytic with 'expressible in terms of elementary functions', my mistake :biggrin:
    Thanks!
     
  6. Aug 30, 2010 #5
    hmm. Interesting.

    I'll bet If you wanted to take this even further you could use the series expansion for the lambert w function.

    [tex]w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}[/tex]

    [tex]-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}[/tex]

    Otherwise a computer would be required to calculate that inverse function. This would allow you to compute the solution to any degree of accuracy. Right?
     
  7. Aug 30, 2010 #6

    Mentallic

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    I really need to learn about series expansions, they seem to exist in numerous forms for every possible function!

    You could even simplify the summation,

    [tex](-1)^{n-1}\left(\frac{-1}{\sqrt{3}}\right)^n=(-1)^{n-1}(-1)^n.3^{-n/2}=(-1)^{2n-1}.3^{-n/2}=-3^{-n/2}[/tex]

    therefore the sum is:

    [tex]
    -W\left(\frac{-1}{\sqrt{3}}\right)=\sum^{\infty}_{n=1}\frac{n^{n-2}}{3^{n/2}(n-1)!}
    [/tex]

    And yes, you can solve it to any degree of accuracy you wish :smile:
     
  8. Aug 31, 2010 #7

    ehild

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    x can be obtained to any degree of accuracy also with an iterative process

    [tex]x_{i+1}=-\sqrt{\frac{e^{2x_i}}{3}}[/tex]

    I got -0.3906....



    ehild
     
    Last edited: Aug 31, 2010
  9. Aug 31, 2010 #8
    It is better to say that the solution of the equation is negative real number [itex]x = -y, \; y > 0[/itex] and the iteration should be:

    [tex]
    y_{i + 1} = \frac{e^{-y_{i}}}{\sqrt{3}}
    [/tex]
     
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