Re: Putting an If in my function
- To: mathgroup at smc.vnet.net
- Subject: [mg101073] Re: Putting an If in my function
- From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
- Date: Tue, 23 Jun 2009 07:06:27 -0400 (EDT)
- References: <email@example.com>
Dear lobotomy, "Nomen est omen", the old Romans used to say. And they were often right. I'm not sure what you are trying to do here, with your function in an unneccessary, overly complex, pure function notation. Why not write: myFun[a_, g_, f_, x_, d_, w_] := d (1 + (E^g - E^w)/f)^(f x) + (a (-1 + (1 + (E^g - E^w)/f)^(f x)) f)/( E^g - E^w) which is much more readable (at least in Mathematica)? In:= Limit[myFun[a, g, f, x, d, w], g -> w] Out= d + a f x You use the undefined 'n' in d + a n. I guess this must be f x... You can add a definition for myFun for this special case like this (no If[ ] necessary): myFun[a_, g_, f_, x_, d_, w_] := d + a f x /; g == w In:= myFun[a, w, f, x, d, w] Out= d + a f x If you want to use an If in the definition this would be something like: myFun[a_, g_, f_, x_, d_, w_] := If[g===w, d (1 + (E^g - E^w)/f)^(f= x) + (a (-1 + (1 + (E^g - E^w)/f)^(f x)) f)/( E^g - E^w),d + a f x] Cheers -- Sjoerd On Jun 22, 10:21 am, Lobotomy <labb... at gmail.com> wrote: > Hi, this is my function > > #5 (1 + ((E^#2 - 1) - (E^#6 - > 1))/#3)^(#3*#4) + (#1*((1 + (((E^#2 - 1) - (E^#6 = - > 1))/#3))^(#3*#4) - > 1))/(((E^#2 - 1) - (E^#6 - 1))/#3) &[a, g, f, x, d, w] > > in the case when w==g the denominator equals zero. > > in this case i would like to rewrite the formula above to the much > simpler > d+a*n. How is this done? I've tried > > If[w == g, % = d + a*n], but this is not working. Another thing is > where to put the "If"