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Question about MarcumQ function

Saf Al

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

POSTED BY: Saf Al

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David Reiss

David Reiss, Scientific Arts

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]

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 BY: David Reiss

Udo Krause

Udo Krause, NOrganization

Posted 10 years ago

0 is not positive, the manual states next thing the limit $b\rightarrow\infty$ (or $\mu\rightarrow\infty$) requests the integral $\int_\infty^\infty ... dx$ why this should evaluate to 1?

POSTED BY: Udo Krause

Saf Al

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 BY: Saf Al

Saf Al

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.

POSTED BY: Saf Al

Saf Al

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 BY: Saf Al

Udo Krause

Udo Krause, NOrganization

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 that and 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 meaningful and 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.