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A little mouse on the pampas
A little mouse on the pampas
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I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$ and $k \rightarrow \infty$

  f =(1 - \[Alpha]α) * Sum[(\[Alpha]*α*(\[Nu]ν - k))^\[Nu]^ν/Exp[Log[\[Nu]Exp[Log[ν!]]/ E^(\[Alpha]*α*(\[Nu]ν - k)), {\[Nu]ν, 0, k}]

It works when this is substituted...

 f/. {\[Alpha]α -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha]α -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$ and $k \rightarrow \infty$

  f =(1 - \[Alpha]) * Sum[(\[Alpha]*(\[Nu] - k))^\[Nu]/Exp[Log[\[Nu]!]]/ E^(\[Alpha]*(\[Nu] - k)), {\[Nu], 0, k}]

It works when this is substituted...

 f/. {\[Alpha] -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha] -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$ and $k \rightarrow \infty$

  f =(1 - α) * Sum[(α*(ν - k))^ν/Exp[Log[ν!]]/ E^(α*(ν - k)), {ν, 0, k}]

It works when this is substituted...

 f/. {α -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {α -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

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PiE
PiE
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I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$ and $k \rightarrow \infty$

  f =(1 - \[Alpha]) * Sum[(\[Alpha]*(\[Nu] - k))^\[Nu]/Exp[Log[\[Nu]!]]/ E^(\[Alpha]*(\[Nu] - k)), {\[Nu], 0, k}]

It works when this is substituted...

 f/. {\[Alpha] -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha] -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$

  f =(1 - \[Alpha]) * Sum[(\[Alpha]*(\[Nu] - k))^\[Nu]/Exp[Log[\[Nu]!]]/ E^(\[Alpha]*(\[Nu] - k)), {\[Nu], 0, k}]

It works when this is substituted...

 f/. {\[Alpha] -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha] -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$ and $k \rightarrow \infty$

  f =(1 - \[Alpha]) * Sum[(\[Alpha]*(\[Nu] - k))^\[Nu]/Exp[Log[\[Nu]!]]/ E^(\[Alpha]*(\[Nu] - k)), {\[Nu], 0, k}]

It works when this is substituted...

 f/. {\[Alpha] -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha] -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?

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PiE
PiE
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How to Calculate Factorial when Overflow/Underflow occurs

I'm trying to compute this function, which sums to 1 for $0 < \alpha < 1$

  f =(1 - \[Alpha]) * Sum[(\[Alpha]*(\[Nu] - k))^\[Nu]/Exp[Log[\[Nu]!]]/ E^(\[Alpha]*(\[Nu] - k)), {\[Nu], 0, k}]

It works when this is substituted...

 f/. {\[Alpha] -> 0.5, k -> 20} = 1.0

But when I try higher values of $\alpha$ (i.e., 0.95), then there appears to be factorial overflow/underflow. For example,

 f/. {\[Alpha] -> 0.95, k -> 40} = 1200.43

which is incorrect. The function should sum to 1.

How do you handle this using Mathematica?