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Math Help - Transcendental equations

  1. #1
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    Transcendental equations

    Could anyone solve this equation ?
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  2. #2
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    Actually, you have a system of four equations in u, v, n, and h. I notice, in particular, that the second and third equations have v both in and exponent and not in an exponent. There is no way to solve such an equation in terms of "elementary functions". I recommend a numerical solution.
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  3. #3
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    How to solve by numerical method ?
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  4. #4
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    Yeah, this one's hopeless when it comes to traditional system methods. Here's a suggestion for a numerical approach:

    Equations 2,3,4 are all of the form A(1-x)=Bx, for the variable x being n,m,h respectively, which reduces to x=A/(A+B). So, isolate these in terms of v and sub them in equation 1.

    A(1-n)=Bn ----- n=A/(A+B)
    C(1-m)=Dm ----- m=C/(C+D)
    E(1-h)=Fh ----- h=E/(E+F)
    where
    A=.01(v+10)/(e^{(v+10)/10-1)}
    B=.125e^{v/80}
    C=.1(v+25)/(e^{(v+25)/10-1)}
    D=4e^{v/18}
    E=.07e^{v/20}
    F=1/(e^{(v+30)/10)+1}
    f(v)=36n^4(v-12)+120m^3*h(v+115)+.3(v+10.598)

    You can now sub eqn 1 for n,m,h so you have an equation involving only v equaling zero. This is way too hairy to simplify and isolate v, but you can easily compute values for f(v) and approximate the zeroes of the function. On a graph, it looks like f(v) is an increasing function but at a decreasing rate, crossing the x-axis around x~1. Assuming this is the only solution, you should be able to write a small program to approximate it to any arbitrary level of accuracy. The closest I can get for you with my handheld calculator is x=.8607 plus or minus .00006.
    Last edited by Media_Man; April 12th 2009 at 12:35 PM.
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