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

Questions
1

Gamma.jpg

(2*(m)^m)/Gamma[m]*(1/2)^((2 m - 1)/n)*E^(-m (1/2)^(2/n)) = 0.778

m>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

2

@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 to Log + Log [ b] since the latter contains more leaf count. We can use a repeated replace function to expand the logrithm function:

In[6]:= 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[6]= -2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + m Log[m] - Log[Gamma[m]]

Then you can numerically solve the Out[6] with a given “n”:

In[19]:= n = 1; FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}]

Out[20]= {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[2] + m Log[m] - 
     Log[Gamma[m]] == Log[0.778], {m, 1}];
 sol2 = FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + m Log[m] - 
     Log[Gamma[m]] == Log[0.778], {m, 5}];
 Plot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + 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[[1]]], 
    Point[{m, 0} /. sol2[[1]]]}]
 , {n, 1.2, 1.72}]
![ScreenShot2013-08-13at6.46.11PM.png|408x301](upload://9YCWv1VpjIi8DTj5lBOFMmiA2zl.png)
3

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 2
0.84 3
0.88 4
0.91 5
0.93 6
0.95 7
0.97 8
0.98 9
0.99 10

Is it possible to predict (X and Y)?..Thanks alot for your cooperation

Regards,
Yahia

4

@Yahia,

I do not see a simple relation between m and n.

6771res.PNG
6771res.PNG872×522 7.73 KB

Maybe you can let 2^(-2/n) be 1 and (-1 + 2 m)/n be zero for large n:

In[14]:= Limit[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + m Log[m] - Log[Gamma[m]], n -> Infinity]
Out[14]= -m + Log[2] + m Log[m] - Log[Gamma[m]]

There seems to exist a limit for m, which is

In[15]:= FindRoot[-m + Log[2] + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}]
Out[15]= {m -> 1.10214}

Graphically you can see

In[16]:= Table[{
                  n,  
                  m /. FindRoot[-2^(-2/n) m + (1 - (-1 + 2 m)/n) Log[2] + m Log[m] - Log[Gamma[m]] == Log[0.778], {m, 1}][[1]]
                }, 
                {n, 1, 80}];
In[17]:= 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.

5

You are right and have got the point!!. Thanks