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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.
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