Finding min point of a second derivative function
Afficher commentaires plus anciens

What I have done is:
f=@(x)2./sqrt(pi).*integral(@(t)exp(-t.^2),0,x);
fplot(f,[-5,5])
DELTA=0.01;
X=-5:DELTA:5;
Y=f(X);
DY_DX=diff(Y)./DETLA;
But it does not work, is there an easier way to the first/second derivative? (not symbolic)
Réponses (1)
Eduard Reitmann
le 3 Août 2018
You were almost there. Hope this helps. The zero in the differential is a bit crude (just to keep the vectors the same length), but a small enough step size should give you are very accurate answer.
f = @(x) (2./sqrt(pi)).*integral(@(t) exp(-t.^2),0,x);
dx = 0.01;
x = (-5:dx:5)';
y = arrayfun(f,x);
dydx = [0;diff(y)./dx];
d2ydx2 = [0;diff(dydx)./dx];
[dy2d2x_min,minpos] = min(d2ydx2);
x_min = x(minpos)
figure;
plot(x,[y dydx d2ydx2],x_min,dy2d2x_min,'*')
legend('erf(x)','erf''(x)','erf''''(x)')
Catégories
En savoir plus sur Mathematics dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!