How can I build power function through point?

1 visualizzazione (ultimi 30 giorni)
Please, help me! I have the coordinates of the curve y = [150 150 130 130 130 100 100 100 80 80 80 60], x = [52000 70000 74000 96000 100000 170000 220000 245000 530000 720000 810000 4900000]. This curve well described by power function. How can I build the same curve passing through the point A(68856, 210) and get it equation?
  6 Commenti
Irene Dzhavadova
Irene Dzhavadova il 16 Gen 2019
Torsten, thank you very much for the answer!
But I have an additional question. If I will have not only point A, but several points, how can I approximate these several points and a raised curve?
Torsten
Torsten il 16 Gen 2019
If points are added, you will have to fit your model again. As a result, you'll get different values for a, b and c.

Accedi per commentare.

Risposta accettata

Sayyed Ahmad
Sayyed Ahmad il 16 Gen 2019
I would do something like this:
x = [52000 70000 74000 96000 100000 170000 220000 245000 530000 720000 810000 4900000];
y = [150 150 130 130 130 100 100 100 80 80 80 60];
p=polyfit(log(x),sqrt(y),2)
x1=linspace(min(x),max(x),100);
y1=polyval(p,log(x1)).^2;
plot(x,y,'r*',x1,y1,'b-')
I hope that is waht you looking for.
Poly.JPG
  3 Commenti
Sayyed Ahmad
Sayyed Ahmad il 16 Gen 2019
Modificato: Sayyed Ahmad il 16 Gen 2019
you can shift it.
y(a)-(y1)=dy
y=f(x)+dy
but you could be only met one point. But If you do that, the deviation will increase. I don't want to recomend you this way.
I don't know what is the aim of this calculation. I will try to reduce the standard diviation with changing in the value n in polyfit.
p=polyfit(x,y,n)
n could be 1, 2, 3, 4, 5 and so on.
Try to change them to find out your acceptable p.
x = [52000 70000 74000 96000 100000 170000 220000 245000 530000 720000 810000 4900000];
y = [150 150 130 130 130 100 100 100 80 80 80 60];
for i=1:5
p=polyfit(log(x),sqrt(y),i)
x1=linspace(min(x),max(x),100);
y1=polyval(p,log(x1)).^2;
plot(x,y,'r*',x1,y1,'-')
hold on
end
Poly.JPG
Irene Dzhavadova
Irene Dzhavadova il 16 Gen 2019
Sayyed Ahmad, thanks! I'll try this.

Accedi per commentare.

Più risposte (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by