Re: Sin*Cos + Log
- To: mathgroup at smc.vnet.net
- Subject: [mg113315] Re: Sin*Cos + Log
- From: Norbert Marxer <marxer at mec.li>
- Date: Sat, 23 Oct 2010 07:06:13 -0400 (EDT)
- References: <i9m83i$jeu$1@smc.vnet.net>
On Oct 20, 10:09 am, Sam Takoy <sam.ta... at yahoo.com> wrote: > Dear Bob, > > Thanks. > > My hope is that I can define f without mentioning the arguments, as in > > f = Sin*Cos + Log > > It doesn't make much difference in this case, but in my case I have something > along the lines of > > f[t_, x_, y_, z_] = g[t, x, y, z]*h[t, x, y, z] + Derivative[0, 0, 1, 0][h][t, > x, y, z] etc > > and I'd like to say > > f = g*h + Derivative[0, 0, 1, 0][h] > > skipping the arguments. > > Is that possible? > > Thanks! > > Sam > > ________________________________ > From: Bob Hanlon <hanl... at cox.net> > To: Sam Takoy <sam.ta... at yahoo.com>; mathgr... at smc.vnet.net > Sent: Tue, October 19, 2010 7:20:45 AM > Subject: Re: Sin*Cos + Log > > f[x_?NumericQ] := Sin[x]*Cos[x] + Log[x] > > f /@ {x, 2, 2.} > > {f[x], Log[2] + Cos[2] Sin[2], 0.314746} > > x /: NumericQ[x] = True; > > f /@ {x, y} > > {Log[x] + Cos[x] Sin[x], f[y]} > > Bob Hanlon > > ---- Sam Takoy <sam.ta... at yahoo.com> wrote: > > ============= > Hi, > > I'm working on a project that involves manipulating lots of functions. > It would be much easier if I could manipulate functions without > evaluating them and then evaluate them at the end. To this end, is there > a way to endow > > f = Sin*Cos + Log > > with meaning and then somehow evaluate > > f[x]? > > Many thanks in advance, > > Sam Hello I think that you have to specify the functions which will use your arguments. Note that f = Sin*Cos + Log (see FullForm) also contains the function Plus, and I'm sure your argument(s) should not be used there. But you could use something like that: f := Sin[##] Cos[##] + Log[##] & and f[x] will give you Log[x] + Cos[x] Sin[x] Or f := g[##]*h[##] + Derivative[0, 0, 1, 0][h][##] & and f[t,x,y,z] will give you g[t, x, y, z]*h[t, x, y, z] + Derivative[0, 0, 1, 0][h][t, x, y, z] I hope this helps. Best Regards Norbert Marxer