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Re: use a function within a new function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg70942] Re: use a function within a new function
*From*: "." <ggroup at sarj.ca>
*Date*: Thu, 2 Nov 2006 06:47:52 -0500 (EST)
*References*: <ei9nuo$1ps$1@smc.vnet.net>
It does work. The error is in your two plots; you have chosen
different ranges for your two parameters a & t. Try these two:
Plot3D[Sin[2*\[Pi]*((t - a)/3 + a/(a + 1))], {t, 0, 9}, {a, 0, 9},
AxesLabel -> {"t", "a", "ex"}, Lighting -> False,
ViewPoint -> {0, 0, 100}, PlotPoints -> 40, Mesh -> False];
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]
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},
> 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",
> "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
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