Re: Why the loop gain no result?

*To*: mathgroup at smc.vnet.net*Subject*: [mg126083] Re: Why the loop gain no result?*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Sun, 15 Apr 2012 03:19:22 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201204140710.DAA27683@smc.vnet.net>

>From the documentation: "Unless an explicit Return is used, the value returned by For is Null." Use Print For[a = 1, a < 5, a++, Print[Solve[a*x^2 - 2 x + 1 == 0, x]]] {{x->1},{x->1}} {{x -> 1/2 - I/2}, {x -> 1/2 + I/2}} {{x -> (1/3)*(1 - I*Sqrt[2])}, {x -> (1/3)*(1 + I*Sqrt[2])}} {{x -> (1/4)*(1 - I*Sqrt[3])}, {x -> (1/4)*(1 + I*Sqrt[3])}} Or accumulate the results sol1 = {}; For[a = 1, a < 5, a++, AppendTo[sol1, Solve[a*x^2 - 2 x + 1 == 0, x]]]; sol1 {{{x -> 1}, {x -> 1}}, {{x -> 1/2 - I/2}, {x -> 1/2 + I/2}}, {{x -> (1/3)*(1 - I*Sqrt[2])}, {x -> (1/3)*(1 + I*Sqrt[2])}}, {{x -> (1/4)*(1 - I*Sqrt[3])}, {x -> (1/4)*(1 + I*Sqrt[3])}}} Table is more straightforward. sol2 = Table[Solve[a*x^2 - 2 x + 1 == 0, x], {a, 4}] {{{x -> 1}, {x -> 1}}, {{x -> 1/2 - I/2}, {x -> 1/2 + I/2}}, {{x -> (1/3)*(1 - I*Sqrt[2])}, {x -> (1/3)*(1 + I*Sqrt[2])}}, {{x -> (1/4)*(1 - I*Sqrt[3])}, {x -> (1/4)*(1 + I*Sqrt[3])}}} Alternatively, Clear[a] f[a_] = Solve[a*x^2 - 2 x + 1 == 0, x] {{x -> (1 - Sqrt[1 - a])/a}, {x -> (1 + Sqrt[1 - a])/a}} sol1 == sol2 == f /@ Range[4] True Bob Hanlon On Sat, Apr 14, 2012 at 3:10 AM, Huafeng Cao <geotom.cao at gmail.com> wrote: > Dear all, > I use the' For' loop to solving an equation, such as : > For[i = 1, i < 5, i++, a = i; Solve[a*x^2 - 2 x + 1 == 0, = x]].The > 'a' in the body must be updated in every loop with the 'i'. But i > cann' t gain the result,Somewhere is disorder? I am confused. > Thinks in advanced! > > > > > > Geotom.Cao > GUCAS >

**References**:**Why the loop gain no result?***From:*Huafeng Cao <geotom.cao@gmail.com>