# Given a set of date point x and y, estimate three polynomial functions (2nd degree, 3rd degree, 4th degree) that will best fit the data.

조회 수: 5 (최근 30일)
John Doe 2021년 11월 27일
댓글: Walter Roberson 2021년 11월 27일
Given a set of date point x and y, estimate three polynomial functions (2nd degree, 3rd degree, 4th degree) that will best fit the data. Compute new set of y data from the three functions and determine which of the three functions will give the least norm difference.
This is what I've came up with, but still not sure if it really satisfy the problem.
x=[0.9 1.5 3 4 6 8 9.5];
y=[0.9 1.5 2.5 5.1 4.5 4.9 6.3];
p=polyfit(x,y,3)
xp=0.9:0.1:9.5;
yp=polyval(p,xp);
plot(x,y,'o',xp,yp)
xlabel('x'); ylabel('y')

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### 답변 (1개)

Walter Roberson 2021년 11월 27일
The question is expecting you to also polyfit with degree 2 and 4, and to polyval with those, and to compute the norm() of the projected values against the actual values, and then to see whether degree 2, degree 3, or degree 4 gave the best fit.
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John Doe 2021년 11월 27일
편집: John Doe 2021년 11월 27일
Is it like this?
x = [0.9 1.5 3 4 6 8 9.5];
y = [0.9 1.5 2.5 5.1 4.5 4.9 6.3];
p = polyfit(x,y,2);
l = polyfit(x,y,3);
t = polyfit(x,y,4);
xp = 0.9:0.1:9.5;
yp1 = polyval(p,xp);
yp2 = polyval(l,xp);
yp3 = polyval(t,xp);
xlabel('x');
ylabel('y');
plot(x,y,'o',xp,yp1,yp2,yp3)
Walter Roberson 2021년 11월 27일
x = [0.9 1.5 3 4 6 8 9.5];
y = [0.9 1.5 2.5 5.1 4.5 4.9 6.3];
p = polyfit(x,y,2);
l = polyfit(x,y,3);
t = polyfit(x,y,4);
xp = 0.9:0.1:9.5;
yp1 = polyval(p,xp);
yp2 = polyval(l,xp);
yp3 = polyval(t,xp);
xlabel('x');
ylabel('y');
plot(x,y,'o',xp,yp1,xp,yp2,xp,yp3)

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