Is that any relationship between a and a', b and b', c and c'?
I can short R1 = p/(10 power to [(a(dR-dx)^b + c)] and R2 = R1 = p/(10 power to [(a'(dR-dx)^b' + c')]
I don't thing you can use X^2 - sX + p = 0 (s=R1+R2, p = R1*R2)
What I mean maybe you can short R1+R2 and R1*R2, so when you put into f(dR) it will be easy to see.
I have not working on math for long long time and I m kind busy to now cos I taking 3 class with full time work. If I have and more idea I will let you know. I need abit more time
Quote:
Originally Posted by lengcallrobot
Thank for your quick reply Pokemon.
R1 = 10 power [(1/10)*(PdB - a(dR-dx)^b - c)]
R2 = 10 power [(1/10)*(PdB - a'(dR^ b') - c')]
where a different from a' (another variable independant from a, not a conjugate)
similarely for b and b', and c and c'.
f(dR) = R1*R2*(S^2) - (2^(2*Clim) - 1)*(1 + S*(R1 + R2)) = 0
Actually, I also think ask what you suggest by represent (R1+R2)=s and R1*R2 = p
than use the equation X^2 - sX + p = 0 to find R1 and R2 and then to find dR.
However, I have only one equation but 2 variables.
Thank you anyways for your effort.
I will try the find the way out today (by using approximation method some where)
and if I can, I will informe you as well.
Best regards
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