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Topic: Re: Evaluating Exponential functions
Replies: 0

 leigh pascoe Posts: 30 Registered: 3/2/10
Re: Evaluating Exponential functions
Posted: Apr 6, 2012 5:57 AM

Oops, sorry. I see now that I forgot the underlines in the function
definition (despite puzzling over it for several hours!). Please
disregard the question.

LP

Le 05/04/2012 13:54, leigh pascoe a =C3=A9crit :

> Dear Experts,
>
> I am working with an age specific risk function
>
> inc1[x_, M_, \[Tau]_, \[Phi]_] :=
> M/\[Tau] \[Phi]^M E^(-(x/\[Tau])) (1 - E^(-(x/\[Tau])))^(M - 1)
>
> where x is the age and M, tau and Phi are constants. This function
> plots, can be integrated numerically and Manipulated easily in Mathematica.
>
> I am also interested in the slightly more complicated function
>
> inc3[x, M, \[Tau], \[Phi], \[Tau]2] := (
> E^(-(x/\[Tau])) (1 - E^(-(x/\[Tau])))^(-2 +
> M) (1 - E^(-(x/\[Tau]2))) (-1 + M) \[Phi]^M)/\[Tau] + (
> E^(-(x/\[Tau]2)) (1 - E^(-(x/\[Tau])))^(-1 + M) \[Phi]^M)/\[Tau]2
>
> This function has an additional parameter tau2. When tau2==tau the two
> functions should be identical. However the second function will not
> Plot, Integrate or be Manipulated in Mathematica. In fact it will not
> even evaluate for specific values of the parameters. e.g.
>
> In[10]:= inc1[10, 12, 8, .65]
> inc3[10, 12, 8, .65, 8]
>
> Out[10]= 0.0000596328
>
> Out[11]= inc3[10, 12, 8, 0.65, 8]
>
> The output for the two functions should be identical. What am I not
> understanding here? How can I define this function so that I can
> numerically integrate it and Plot its values over a range of x.
>
> Thanks for any suggestions.
>
> LP