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# How to determine (m) in term of (n) of Nakagami function

Posted 11 years ago (2*(m)^m)/Gamma[m]*(1/2)^((2 m - 1)/n)*E^(-m (1/2)^(2/n)) = 0.778m>0, n: Natural Number {1,2,3,4,....}How to determine (m) in term of (n) of Nakagami function as mentioned above..However, I have tried to use Reduce but unfortuantly I did not get result!!Thanks,Yahia
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Posted 11 years ago
 You are right and have got the point!!. Thanks
Posted 11 years ago
 @Yahia, You may want to simplify the equation first. Usually wrapping a Log function is a good idea here on the both sides of this equation. However, Mathematica will not convert Log [a b]  to Log + Log [ b] since the latter contains more leaf count. We can use a repeated replace function to expand the logrithm function:In:= Log[(2*(m)^m)/Gamma[m]*(1/2)^((2 m - 1)/n)*E^(-m (1/2)^(2/n))] //. {Log[x_*y_] :> Log[x] + Log[y], Log[x_/y_] :> Log[x] - Log[y], Log[x_^y_] :> y*Log[x]}Out= -2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] - Log[Gamma[m]]Then you can numerically solve the Out with a given "n":In:= n = 1; FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}]Out= {m -> 1.01603} If you want to check the number of the real roots, you may plot this function. For some of values of "n" there are actually two real roots, which I found graphically. Finally, to test the behavior of the roots of this equation, you cannot find a better solution without using Manipulate function: Manipulate[  sol = FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] -       Log[Gamma[m]] == Log[0.778], {m, 1}];  sol2 = FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] -       Log[Gamma[m]] == Log[0.778], {m, 5}];  Plot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] -     Log[Gamma[m]] - Log[0.778], {m, 0.5, 5},    PlotRange -> {{0, 5}, {-0.5, 0.5}},    Epilog -> {PointSize[0.02], Point[{m, 0} /. sol[]],     Point[{m, 0} /. sol2[]]}] , {n, 1.2, 1.72}] Posted 11 years ago
 Yes Mr. Yang, you are right but I have tried to approximate to the relationship between m and n by a simple equation e.g.  m=nX +Y, according to the following results:m                n 0.78           20.84           30.88           40.91           50.93           60.95           70.97           80.98           90.99          10Is it possible to predict (X and Y)?...Thanks alot for your cooperationRegards,Yahia
Posted 11 years ago
 @Yahia, I do not see a simple relation between m and n. Maybe you can let 2^(-2/n) be 1 and (-1 + 2 m)/n be zero for large n:In:= Limit[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] - Log[Gamma[m]], n -> Infinity]Out= -m + Log + m Log[m] - Log[Gamma[m]]There seems to exist a limit for m, which is  In:= FindRoot[-m + Log + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}]Out= {m -> 1.10214}Graphically you can see In:= Table[{                  n,                    m /. FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}][]                },                 {n, 1, 80}];In:= ListPlot[%, PlotStyle -> {PointSize[0.02]}] I am assuming that the X and Y you are talking about is simply from the first order Talyer expansion of the fitted model. This is not a prediction, but simply find the derivative of the above curve.