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# Help with Bessel Function

Posted 10 years ago
 I need help solving for a variable in modifed bessel function of the second kind and I have no idea. Can anyone help? Below is the equation and and what I've put in Mathematica. I need to solve for K. d = 0.000012401 a = 0.0382 solve[0.5 + ((2 [Sqrt]x)/0.000012401) (BesselK[1, (0.000012401/(2 [Sqrt]x))]/BesselK[0, (0.000012401/(2 [Sqrt]x))]) == 1/(4*0.0382), x]
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Posted 10 years ago
 kn is a modified Bessel function of the third kind, proportional to K{n+1/2). Also Sqrt is coded incorrectly. Here's what I get for numerical solution of your problem: In[14]:= d = 0.000012401; a = 0.0382; FindRoot[0.5 + ((2 Sqrt[x])/ 0.000012401) (BesselK[3/2, (0.000012401/(2 Sqrt[x]))]/ BesselK[1/2, (0.000012401/(2 Sqrt[x]))]) == 1/(4*0.0382), {x, .1}] Out[15]= {x -> 1.55154*10^-10 - 9.43222*10^-21 I} In[16]:= Chop[%] Out[16]= {x -> 1.55154*10^-10} 
Posted 10 years ago
 Hello D. Moore,here is your equation to start from: eq10 = 1/2 + ((2 Sqrt[k])/d) BesselK[1, (d/(2 Sqrt[k]))] / BesselK[0, (d/(2 Sqrt[k]))] == 1/(4 a);It can be rewritten by substituting (d/(2 Sqrt[k])) -> z : eq11 = 1/2 + (1/z) BesselK[1, z] /BesselK[0, z] == 1/(4 a);You want to solve it numerically using FindRoot, therefore you have to reformat the equation in a way that the right handside is equal to zero. And you replace the symbol a by its value: eq12 = eq11[[1]] - 1/(4 a) /. { a -> 0.0382}Now solve using FindRoot: sol1 = FindRoot[eq12, {z, 1.}]To get back to k, you enter sol1 in this equation: sol2 = k -> (d / (2 z))^2 /. sol1 /. d -> 0.000012401Finally you want to check that the result is correct: eq10 /. sol2 /. { a -> 0.0382, d -> 0.000012401}Regards,Michael
Posted 10 years ago
 Thanks so much