Taking second derivative and solving it for when it equals 0

Here is the code that I have:

clear all; close all
m = xlsread('1 MgO.xlsx','Sheet1','B1:J18');
temp = m(:,6);
MgO = m(:,3);
plot(temp, MgO,'*')
xlabel('Temperature (K)'); ylabel('Shear Modulus (GPa)'); title('MgO')
hold on

% Line of best fit
p = polyfit(temp,MgO,3);
y = polyval(p,temp);
plot(temp,y,'-')
SSE=sum((MgO-y).^2); SST=sum((MgO-mean(MgO)).^2); Rsquared=1-SSE/SST;
text(100,100,sprintf('y = %.3x^3 + %.3fx^2 + %.3fx + %.3f, R^2 = %.3f',p(1),p(2),p(3),p(4), Rsquared))

I'm trying to figure out how to get the second derivative and find at what temperature it equals 0. temp is my xaxis and MgO is my yaxis. I tried using the diff command but it didn't really work. Could someone help please?

Answer 1

[Star Strider]

I am not certain what you are doing. Taking the derivative of data will significantly amplify any noise that might be present, so in general, it is best to take the derivative of a smooth signal. Also, you are fitting a 3° polynomial, so the second derivative of that (regardless of the method you use) is giong to be a linear (1°) function.

If you want to take the derivative of filtered data (eliminating as much noise as possible first), use the gradient (link) function. Use it two times to get the second derivative.

If you want to take the derivative of your fitted function ‘p’, one option is to use the polyder (link) function, again twice. You can then easily solve for the x-intercept of the linear relation this produces.

Note that the second derivative is 0 at a Saddle point (link).

[Christina Kersten]

So what I'm trying to do is take the second derivative of the equation of the line of best fit and find where that equals to zero. So would I do polyder(polyder(p))?

[Star Strider]

Yes.

If you define it as:

d2p = polyder(polyder(p));

the x-intercept is:

xint = -d2p(1)/d2p(2);

[Christina Kersten]

Sorry, could you explain how dividing the first and second element of d2p gives you where d2p = 0?

[Star Strider]

It should be:

xint = -d2p(2)/d2p(1);

That was what I intended to write, and I thought that was what I wrote.

The derivation comes from:

y = d2p(1)*x + d2p(2)
0 = dp2(1)*xint + d2p(2)

with the x-intercept defined by the point where y=0.

Answer 2

[Walter Roberson]

Take the derivative

https://www.mathworks.com/help/matlab/ref/polyder.html

and take the roots()