Re: FindRoot with a vector of unknowns

*To*: mathgroup at smc.vnet.net*Subject*: [mg125077] Re: FindRoot with a vector of unknowns*From*: Szabolcs Horvát <szhorvat at gmail.com>*Date*: Mon, 20 Feb 2012 02:53:00 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <jhqmed$f70$1@smc.vnet.net>

On 2/19/2012 1:29 PM, Sam Takoy wrote: > Hi, > > Is there an elegant way to implement what I am trying to do here, that > is solve for a vector of unknowns: > > FindRoot[x - {1, 2, 3} == {0, 0, 0}, {x, {1, 1, 1}}] > > I can do this writing a loop, but hoping for a "vectorized" solution. > The documentation explicitly mentions that FindRoot supports vector variables. So what's wrong here? You'll immediately see the problem if you evaluate the equation alone: In[6]:= x - {1, 2, 3} == {0, 0, 0} Out[6]= {-1 + x, -2 + x, -3 + x} == {0, 0, 0} Plus[] has attribute Listable which means that it will auto-thread over lists. ## What is the workaround? One way is defining a function to use in FindRoot: f[x : {__?NumericQ}] := x - {1, 2, 3} FindRoot[f[x], {x, {1, 1, 1}}] (* ==> {x -> {1., 2., 3.}} *) The special pattern used in the definition of f[] is essential: it prevents f[] from evaluating when symbols (not numbers) are passed to it. ## This is good when f[] is a big function, but are there easier workarounds for tiny functions this this one? I am not aware of any, but I'd very much like to see some! (Looking forward to the answers) -- Szabolcs Horvát Visit Mathematica.SE: http://mathematica.stackexchange.com/