Re: Re: naive question

*To*: mathgroup at smc.vnet.net*Subject*: [mg92881] Re: [mg92844] Re: [mg92781] naive question*From*: Francisco Gutierrez <fgutiers2002 at yahoo.com>*Date*: Thu, 16 Oct 2008 05:07:16 -0400 (EDT)*Reply-to*: fgutiers2002 at yahoo.com

Dear Friends: I also want to thank all the other responses that came after Bob' and Valerie's --which by the way reveal the very big flexibility of Mathematica. Performing an imaginary "union" operation, I counted at least five distinct, powerful solutions. Best, Francisco --- On Tue, 10/14/08, Francisco Gutierrez <fgutiers2002 at yahoo.com> wrote: From: Francisco Gutierrez <fgutiers2002 at yahoo.com> Subject: [mg92881] [mg92844] Re: [mg92781] naive question To: mathgroup at smc.vnet.net Date: Tuesday, October 14, 2008, 4:01 AM Many thanks to Bob and Valerie for their fast and ultraeffective responses. Best, Francisco --- On Mon, 10/13/08, Bob Hanlon <hanlonr at cox.net> wrote: From: Bob Hanlon <hanlonr at cox.net> Subject: [mg92881] [mg92844] Re: [mg92781] naive question To: fgutiers2002 at yahoo.com, mathgroup at smc.vnet.net Date: Monday, October 13, 2008, 6:26 AM expr = Fit[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, {1, x1, x2}, {x1, x2}]; Flatten[CoefficientList[expr, {x2, x1}]] // Most {1.55556,0.555556,0.444444} Most is used to drop the coefficient (0) of the (x1*x2) term Cases[expr, _?NumericQ, Infinity] {1.55556,0.555556,0.444444} Or using FindFit rather than Fit {a, b, c} /. FindFit[{{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, a + b*x1 + c*x2, {a, b, c}, {x1, x2}] {1.55556,0.555556,0.444444} Bob Hanlon ---- Francisco Gutierrez <fgutiers2002 at yahoo.com> wrote: ============= Dear Friends: I am using Fit within a function, and I want it to produce only a list of coefficients, without the variables. For example, if I plug into fit the following: Fit[{{1,2,3},{4,5,6},{7,8,9}},{1,x1,x2},{x1,x2}], Mathematica produces: 1.55556+0.555556 x1+0.444444 x2 I want to get only the following list: {1.55556,0.555556,0.444444} It seem awfully simple, but I haven't managed. How can I do it? Thanks Francisco -- Bob Hanlon