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## RE: solving equation system

 From: Hall, Benjamin Subject: RE: solving equation system Date: Fri, 22 Apr 2005 08:54:05 -0400

```I'm not sure how to change the output format.  One work-around is to do it
manually

sprintf('\n%15.5e',coeffs)

but maybe someone on the list has some additional suggestions?

-----Original Message-----
Sent: Thursday, April 21, 2005 6:29 AM
To: 'Hall, Benjamin'
Subject: AW: solving equation system

Thank you very much for your help!

I like this solution, because this way is very short!

I tried it and I got a new problem. My result is as follows:

coeffs =

1.0e+02 *

-0.00000
0.00004
-0.00504
0.29366
-2.69074

I don`t like that the program multyplies the complete vector with the value
1.0e+02 I would like the writing you have:

-1.0222e-05
3.5786e-03
-5.0372e-01
2.9366e+01
-2.6907e+02

Do you know how I can change this?

With best regards

Brigitte Diekhaus

-----Ursprüngliche Nachricht-----
Gesendet: Dienstag, 19. April 2005 20:06
Betreff: RE: solving equation system

It looks like you're trying to fit a 4th-order polynomial to the data

xx = [124.5; 111;    88.9;  75.5;  63  ];
yy = [ 29.3; 126.7; 236.4; 284.7; 315.5];

which you can do using polyfit

coeffs = polyfit( xx, yy, 4 )

coeffs =

-1.0222e-05
3.5786e-03
-5.0372e-01
2.9366e+01
-2.6907e+02

-----Original Message-----
Sent: Tuesday, April 19, 2005 1:52 PM
Subject: Re: solving equation system

Looks simple. Define a coefficient matrix and solve it :

octave:6>
a=[1,124.5,124.5^2,124.5^3,124.5^4;1,111,111^2,111^3,111^4;1,88.9,88.9^2,88.
9^3,88.9^4;1,75.5,75.5^2,75.5^3,75.5^4;1,63,63^2,63^3,63^4]
a =

1.0000e+00   1.2450e+02   1.5500e+04   1.9298e+06   2.4026e+08
1.0000e+00   1.1100e+02   1.2321e+04   1.3676e+06   1.5181e+08
1.0000e+00   8.8900e+01   7.9032e+03   7.0260e+05   6.2461e+07
1.0000e+00   7.5500e+01   5.7002e+03   4.3037e+05   3.2493e+07
1.0000e+00   6.3000e+01   3.9690e+03   2.5005e+05   1.5753e+07

octave:7> f=[29.3,126.7,236.4,284.7,315.5]'
f =

29.300
126.700
236.400
284.700
315.500

octave:8> a\f
ans =

-2.6907e+02
2.9366e+01
-5.0372e-01
3.5786e-03
-1.0222e-05

I hope that someone can point out a shorter way to define the coefficient
matrix since all its rows are of the form [1,x,x^2,x^3,x^4].

HTH

On Tuesday 19 April 2005 12:15, Brigitte Diekhaus wrote:
> hello,
>
> it`s my first time using octave and I want to solve the following equation
> system:
>
> 29.3 = A*124.5^4+B*124.5^3+C*124.5^2+124.5*D+F,
> 126.7 = A*111^4+B*111^3+C*111^2+D*111+F,
> 236.4 = A*88.9^4+B*88.9^3+C*88.9^2+D*88.9+F,
> 284.7 = A*75.5^4+B*75.5^3+C*75.5^2+D*75.5+F,
> 315.5 = A*63^4+B*63^3+C*63^2+D*63+F },
>
> I need to find the variables {A,B,C,D,F}
>
> to get a function like this:
>
> y(x)=a*x^4+b*x^3+c*x^2+d*x+e
>
> I Don`t know the syntax I have to use with octave to solve this equation
> system.
>
>
> Brigitte
>
>
>
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-------------------------------------------------------------
Octave is freely available under the terms of the GNU GPL.

Octave's home on the web:  http://www.octave.org
How to fund new projects:  http://www.octave.org/funding.html
Subscription information:  http://www.octave.org/archive.html
-------------------------------------------------------------

-------------------------------------------------------------
Octave is freely available under the terms of the GNU GPL.

Octave's home on the web:  http://www.octave.org
How to fund new projects:  http://www.octave.org/funding.html
Subscription information:  http://www.octave.org/archive.html
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