Re: Evaluating Exponential functions

*To*: mathgroup at smc.vnet.net*Subject*: [mg125952] Re: Evaluating Exponential functions*From*: David Bailey <dave at removedbailey.co.uk>*Date*: Mon, 9 Apr 2012 05:33:57 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <jlmfak$mhr$1@smc.vnet.net>

On 06/04/2012 11:08, leigh pascoe wrote: > 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 > You have missed out the _ characters in the definition of inc3 - so it doesn't match anything useful. You need: inc3[x_, M_, \[Tau]_, \[Phi]_, \[Tau]2_] ..... David Bailey http://www.dbaileyconsultancy.co.uk