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Question about MarcumQ function

Posted 11 years ago
POSTED BY: Saf Al
8 Replies

0 is not positive, the manual states

marcumQ

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
Posted 11 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 11 years ago

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

POSTED BY: Saf Al
Posted 11 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 11 years ago
POSTED BY: Saf Al
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