How do I get the minima of a parabola?
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I have a function that starts with 2 matrices (7x1 in this case)
tol=input('enter tolerance matrix: ');
pix=input('enter pixel matrix: ');
plot(tol,pix);
syms x
a=polyfit(tol,pix,3);
b=poly2sym(a);
c=ezplot(diff(b,x),[0 100]);
"c" generates a parabola. I need to find the exact minima. Mainly the x value of the minima. How do i get it.
Thanks for the help
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Star Strider
2014년 4월 17일
I don’t see why you need to use the Symbolic Math Toolbox for this. I suggest:
% Create data:
tol = linspace(1,10,7);
pix = polyval(poly([1 3 5]),tol) + 0.5.*(rand(1,7)-.5);
% Fit:
a = polyfit(tol,pix,3);
d1a = polyder(a); % First derivative
d2a = polyder(d1a); % Second derivative
ip = roots(d1a); % Inflection points
mm = polyval(d2a,ip); % Second derivative evealuated at inflection points
fprintf(1,'\n\tMinimum at %.3f\n', ip(mm>0))
fprintf(1,'\n\tMaximum at %.3f\n', ip(mm<0))
figure(1)
plot(tol, pix, '-b')
hold on
plot(ip, polyval(a,ip), '+r')
hold off
grid
If you get complex roots, this becomes more interesting to plot but the maths are the same.
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Star Strider
2014년 4월 17일
John — Noted. In his code, jchris14 mentions fitting a 3-degree polynomial with polyfit. Reading between the lines...
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Jos (10584)
2014년 4월 17일
Why not solve this analytically?
P = polyfit(x,y,2) % fit parabola
% vertex is a minimum when P(1) > 0, a maximum when P(1) < 0
% (no vertex for a straight line when P(1)=0 !)
h = -P(2) / (2*P(1)) % x value of the minima (maxima)
k = P(3)-((P(2)^2) / (4*P(1))) % y value of the minima (maxima)
% polyval(P,h) will give the same value!
%(h,k) is the vertex point
% y = (p(1)*(x-h)^2) + k % vertex notation
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Star Strider
2014년 4월 17일
In his code, jchris14 is using polyfit to fit a 3-degree polynomial, not actually a parabola. Otherwise I agree. He also wants to do this using MATLAB functions, and I’m not going to discourage that.
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