八度中具有数据的抛物线图

Question

I have been trying to fit a parabola to parts of data where y is positive. I am told, that P1(x1,y1) is the first data point, Pg(xg,yg) is the last, and that the top point is at x=(x1+xg)/2. I have written the following:

x=data(:,1);
y=data(:,2);
filter=y>0;
xp=x(filter);
yp=y(filter);
P1=[xp(1) yp(1)];
Pg=[xp(end) yp(end)];
xT=(xp(1)+xp(end))/2;
A=[1 xp(1) power(xp(1),2) %as the equation is y = a0 + x*a1 + x^2 * a2
   1 xp(end) power(xp(end),2)
   0 1 2*xT]; % as the top point satisfies dy/dx = a1 + 2*x*a2 = 0
b=[yg(1) yg(end) 0]'; %'
p=A\b;
x_fit=[xp(1):0.1:xp(end)]; 
y_fit=power(x_fit,2).*p(3)+x_fit.*p(2)+p(1);

figure 
plot(xp,yp,'*')
hold on 
plot(x_fit,y_fit,'r')

And then I get this parabola which is completely wrong. It doesn't fit the data at all! Can someone please tell me what's wrong with the code?

My parabola

Answer 1

Well, I think the primary problem is some mistake in your calculation. I think you should use three points on the parabola to obtain a system of linear equations. There is no need to calculate the derivative of your function as you do with

dy/dx = a1 + 2*x*a2 = 0

Instead of a point on the derivative you choose another point in your scatter plot, eg the maximum: PT = [xp_max yp_max]; and use it for your matrix A and b.

The equation dy/dx = a1 + 2*x*a2 = 0 does not fulfill the basic scheme of your system of linear equations: a0 + a1*x + a2*x^2 = y;

By the way: If you don't have to calculate your parabola necessarily in this way, you can maybe have a look at the Matlab/Octave-function polyfit() which calculates the least squares solution for your problem. This would result in a simple implementation:

p = polyfit(x, y, 2);
y2 = polyval(p, x);
figure(); plot(x, y, '*'); hold on;
plot(x, y2, 'or');