# Message Boards

0
|
9116 Views
|
8 Replies
|
1 Total Likes
View groups...
Share
GROUPS:

Posted 10 years ago
 Hi, I am trying to solve an equation that has a MarcumQ function in it. One of the MarcumQ properties is that Limit[MarcumQ[1, 3, \[Mu]], \[Mu] -> Infinity] = 1  This is obvious when the function is plotted while varying its arguments Plot[{MarcumQ[1, 1, \[Mu]], MarcumQ[1, 2, \[Mu]], MarcumQ[1, 3, \[Mu]], MarcumQ[1, 4, \[Mu]], MarcumQ[1, 5, \[Mu]]}, {\[Mu], 0, 8}]  However, the Mathematica Limit function seems to be unable to give an answer. I wonder, what the reason could be. Thanks in advance. Saf
8 Replies
Sort By:
Posted 10 years ago
Posted 10 years ago
 Hi, you are right about the zero in the plotted range. I should have started the plot from a real number, which is a little bit bigger than zero. It seems that didn't make any difference though. However, as seen in the example, I included in the my question, the value of the marcumQ function evaluates, indeed, to 1, when the value of the second argument goes to infinity. The book I am reading for the time being, confirms that as well.But, you have a point in presenting the mathematical definition of the marcumQ function. I guess, I have to ask a mathematician one day. Thanks for your answer.
Posted 10 years ago
 Sorry One of the MarcumQ properties is that Limit[MarcumQ[1, 3, \[Mu]],\[Mu] -> Infinity] = 1 the picture does not suggest thatand if one is willing to accept $8$ as a good approximation for $\infty$ than the graphical result agrees with the Fundamental theorem of calculus. If you had have the $\mu$ in the first argument of MarcumQ the limit would be meaningfuland now you have to ask yourself what $\lgroup\frac{x}{a}\rgroup^{\mu-1} I_{\mu-1}(a x)$ for $\mu\rightarrow\infty$ is. For this approximations may exist.
Posted 10 years ago
 In Connections between the Generalized MarcumQ-Function and a class of HypergeometricFunctions the equation ${Q_{m}(a,b)=1 - Q_{1-m}(a,b)}$ is found.
Posted 10 years ago
 Hi, I disagree with most of what you said. But thanks anyways for your participation.
Posted 10 years ago
 Actually, N[MarcumQ[1, 0, 1]] equals 0.606531. Also, MarcumQ[1, 1, 0] equals 1. So obviously, the condition for a or b to be 0 is relaxed.
Posted 10 years ago
 It appears that Mathematica does not know this information about MarcumQ. It also seems not to be able to compute a power series expansion about mu=0, oddly just returning the MarcumQ function it self: In[1]:= Series[MarcumQ[1, 3, \[Mu]], {\[Mu], 0, 2}] Out[1]= MarcumQ[1, 3, \[Mu]] This is so, even though Mathematica knows how to compute derivatives, e.g.: In[2]:= D[MarcumQ[1, 3, \[Mu]], {\[Mu], 3}] Out[2]= 3 E^(1/2 (-9 - \[Mu]^2)) \[Mu] Hypergeometric0F1Regularized[1, ( 9 \[Mu]^2)/4] - E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[1, ( 9 \[Mu]^2)/4] - 27/2 E^(1/2 (-9 - \[Mu]^2)) \[Mu] Hypergeometric0F1Regularized[2, ( 9 \[Mu]^2)/4] + 9 E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[2, ( 9 \[Mu]^2)/4] - 81/4 E^(1/2 (-9 - \[Mu]^2)) \[Mu]^3 Hypergeometric0F1Regularized[3, ( 9 \[Mu]^2)/4] 
Posted 10 years ago
 Hi, Well, it seems that the reason is the mathematical definition of the function itself. Despite the fact that this property is obvious graphically, Mathematica seems unable to resolve it analytically. I hope that the program-developers will add this to its properties in a future update.Thanks for your efforts and regards.