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Question about MarcumQ function

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
8 Replies
POSTED BY: Udo Krause
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

Sorry

One of the MarcumQ properties is that Limit[MarcumQ[1, 3, \[Mu]],\[Mu] -> Infinity] = 1

the picture does not suggest that

MarcumQ limit

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

MarcumQ correct position limit

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.

POSTED BY: Udo Krause
POSTED BY: Udo Krause
Posted 10 years ago

Hi, I disagree with most of what you said. But thanks anyways for your participation.

POSTED BY: 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

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
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
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