How do I get the minima of a parabola?

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

Answer 1

[Star Strider]

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.

[jchris14]

thanks for the help Star. I actually was seeking the wrong command. Rather than finding a minima, i just need to find the inflection point of the parabola. I used roots and voila... it works! thanks

[Star Strider]

My pleasure!

[John D'Errico]

The inflection point of a parabola?

Last time I checked, a parabola has no inflection point, since that is a point where the second derivative would pass through zero, so change sign. Since a parabola has constant second derivative, unless the parabola is degnerate, i.e., a straight line, it will never have a zero second derivative. And even then, it will have no inflection point.

Must be the new math.

[Star Strider]

John — Noted. In his code, jchris14 mentions fitting a 3-degree polynomial with polyfit. Reading between the lines...

[jchris14]

John- sorry, i meant turning point not inflection point. I dont really have any use for the inflection point of the 3-degree polynomial.

Answer 2

[Jos (10584)]

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

[Star Strider]

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.

[jchris14]

Jos- thanks for the tip. I was thinking about doing it this way but using the command "roots" simplified the whole sequence of commands and shortened the whole code to just 4 or 5 lines.