solving for coefficients in 4th order polynomial

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nctt
nctt 2024-1-14
编辑： Torsten 2024-1-18
so in this code i get
Empty sym: 0-by-1
for coefficients so it doesnt give the numerical values i tried double() command but i think i used it wrong can you help me with that?
clear all
clc
syms a4 a3 a2 a1 a0
x=[0:5];
y=[15,10,9,6,2,0];
eqn=a0+a1.*x+a2.*x.^2+a3.*x.^3+a4.*x.^4==y;
[a4,a3,a2,a1,a0]= solve(eqn)

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采纳的回答

Shivam
Shivam 2024-1-14
Ran in:
Hi,
From the provided information, I understand that you are trying to get the coefficients of the 4th-degree polynomial using six sets of x and y values.
You can observe that for the 4th-degree polynomial, you have 5 unknown coefficients (a0, a1, a2, a3, a4), and with 6 data points, it is not the correct approach to solve the equations simultaneously for each x and y pair.
You can use the least squares method, which finds the polynomial coefficients that minimize the sum of the squares of the differences between the observed values (y) and those predicted by the polynomial model.
You can follow the below workaround using 'polyfit' to find the coefficients of a 4th-degree polynomial:
x = [0:5];
y = [15, 10, 9, 6, 2, 0];
% Fit a 4th-degree polynomial to the data
p = polyfit(x, y, 4);
% Display the coefficients
a4 = p(1)
a4 = 0.1875
a3 = p(2)
a3 = -1.9398
a2 = p(3)
a2 = 6.2986
a1 = p(4)
a1 = -9.4272
a0 = p(5)
a0 = 14.9802
You can refer to the following documentation to know more about 'polyfit' function:
https://www.mathworks.com/help/releases/R2023b/matlab/ref/polyfit.html
I hope it helps.
Thanks
  2 个评论
Torsten
Torsten 2024-1-18
编辑：Torsten 2024-1-18
Ran in:
x=[0:5].';
y=[15,10,9,6,2,0].';
A = [ones(6,1),x,x.^2,x.^3,x.^4];
a = A\y;
a0 = a(1)
a0 = 14.9802
a1 = a(2)
a1 = -9.4272
a2 = a(3)
a2 = 6.2986
a3 = a(4)
a3 = -1.9398
a4 = a(5)
a4 = 0.1875

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更多回答（1 个）

Hassaan
Hassaan 2024-1-14
编辑：Hassaan 2024-1-14
Ran in:
@nctt
  1. Polyfit: Use polyfit to find the coefficients of the polynomial that best fits your data. The polyfit function finds the coefficients of a polynomial of a specified degree that fits the data in a least-squares sense.
  2. Degree of the Polynomial: In your case, you have 6 data points, so you can fit a polynomial of degree 5 (or less).
If you want to visualize this polynomial against your data, you can use the polyval function to evaluate the polynomial at points along the x-axis and plot it
clear all
clc
x = [0:5];
y = [15,10,9,6,2,0];
% Fit a 5th degree polynomial
p = polyfit(x, y, 5);
% The coefficients are in descending powers
a4 = p(1);
a3 = p(2);
a2 = p(3);
a1 = p(4);
a0 = p(5);
fprintf("%f %f %f %f %f",a0,a1,a2,a3,a4)
-11.750000 10.291667 -4.208333 0.708333 -0.041667
% Evaluate polynomial
x_fit = linspace(min(x), max(x), 100);
y_fit = polyval(p, x_fit);
% Plot
plot(x, y, 'o', x_fit, y_fit, '-');
legend('Data Points', 'Fitted Polynomial');
xlabel('x');
ylabel('y');
title('Polynomial Fit');
------------------------------------------------------------------------------------------------------------------------------------------------
If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.
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