Re: Can this problem be solved in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg84461] Re: Can this problem be solved in Mathematica
- From: Bill Rowe <readnewsciv at sbcglobal.net>
- Date: Wed, 26 Dec 2007 05:07:01 -0500 (EST)
On 12/25/07 at 6:25 AM, msalazardc at yahoo.com (Manuel Salazar) wrote: >When b=0 y = 358.55x2 + 5811x + 17579 >when b=0.1 y = 1320.3 x^3 + 1558 x^2 + 5237.9 x + 15630 >when b=0.2 y = 3185.7 x^3 + 1607.2 x^2 + 5059.9 x + 13879 >when b=0.3 y = 5552.8 x^3 + 588.57 x^2 + 5277.7 x + 12290 >when b=0.4 y = 8299.1 x^3 - 1284 x^2 + 5821.4 x + 10840 You really haven't made it clear what you mean by solving this. I assume you want some function of b to give the coefficients in the equations. If so, yes this can be solved in Mathematica. However, there will not be an unique function that does this. To find a function, the first thing I would do after putting the coefficients in a more useful form is to plot them, i.e., coeff = {{0, 358.55, 5811, 17579}, {1320.3, 1558, 5237.9, 15630}, {3185.7, 1607.2, 5059.9, 13879}, {5552.8, 588.57, 5277.7, 12290}, {8299.1, -1284, 5821.4, 10840}}; GraphicsGrid[ Partition[ ListPlot[#, DataRange -> {0, .4}, Frame -> True, Axes -> None] & /@ Transpose@coeff, 2], ImageSize -> Large] The plots suggest the coefficients for x^3 and the constant in each equation are linear functions of b. So, In[28]:= a b + c /. FindFit[Transpose@{Range[0, .4, .1], Last /@ coeff}, a b + c, {a, c}, b] Out[28]= 17407.2-16818. b yields an equation for the constant term and In[29]:= a b + c /. FindFit[Transpose@{Range[0, .4, .1], First /@ coeff}, a b + c, {a, c}, b] Out[29]= 20830.7 b-494.56 gives an equation for the coefficients of x^3 The plots show the coefficients of x and X^2 are not linear functions of b. There is no immediately obvious choice for the functional form. So, lacking any additional information, I would use Interpolation for these coefficients. So for x^2 In[30]:= f = Interpolation[Transpose[{Range[0, .4, .1], coeff[[All, 2]]}]]; and for x In[31]:= g = Interpolation[Transpose[{Range[0, .4, .1], coeff[[All, 3]]}]]; Putting this altogether Table[xcube x^3 + f[b] x^2 + g[b] x + const // Expand, {b, 0, .4, .1}] // TableForm This table gives that do not quite match for either x^3 or the constant term. So, they must not be exactly linear despite the appearance of the plot. This can be corrected by using Interpolation or by using a different model in FindFit. -- To reply via email subtract one hundred and four