Re: use a function within a new function

*To*: mathgroup at smc.vnet.net*Subject*: [mg70951] Re: use a function within a new function*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Thu, 2 Nov 2006 06:48:38 -0500 (EST)*Organization*: The Open University, Milton Keynes, UK*References*: <ei9nuo$1ps$1@smc.vnet.net>

aitor69gonzalez at gmail.com wrote: > Hello, > > I have a function like: > phi(a,t) = sin(2*pi*(t-a)/T' + a/T) > where "T" is itself a function of "a" like here: > #### > Plot3D[Sin[2*\[Pi]*((t - a)/3 + a/(a + 1))], {t, 0, 9}, {a, 0, 9}, -----------------------------------------------^^^^^^^^^^^^^^^^^^^^ Note that here were are going from o to 9, > AxesLabel -> {"t", "a", "ex"}, Lighting -> False, > ViewPoint -> {0, 0, 100}, PlotPoints -> 40, Mesh -> False]; > #### > Therefore, I would like to define T separately. Something like here: > #### > T[a_] = a + 1; > phi[a_, t_] = Sin[2*\[Pi]*((t - a)/3 + a/T[a])]; > Plot3D[phi[a, t], {t, 0, 720}, {a, 0, 720}, AxesLabel -> {"t", "a", --------------------^^^^^^^^^^^^^^^^^^^^^^^^ whereas here we are going from 0 to 2*Pi. > "ex"}, > Lighting -> False, ViewPoint -> {0, 0, 100}, PlotPoints -> 40, > Mesh -> False] > #### > But as you can see, both plots do not look like the same. How can I > define a "T" as a function of a separately? > Thank you in advance. > > Aitor > The issue is that you uses different ranges for t and a in both expressions. The following expression will give you a graph identical to the first one: T[a_] = a + 1; phi[a_, t_] = Sin[2*Pi*((t - a)/3 + a/T[a])]; Plot3D[phi[a, t], {t, 0, 9}, {a, 0, 9}, AxesLabel -> {"t", "a", "ex"}, Lighting -> False, ViewPoint -> {0, 0, 100}, PlotPoints -> 40, Mesh -> False]; Regards, Jean-Marc