How to define a special function with some points

6 vues (au cours des 30 derniers jours)
Mojtaba Mohareri
Mojtaba Mohareri le 10 Oct 2020
I want to define a function like this
f(0.2)=1.42007;
f(0.4)=1.88124;
f(0.5)=2.12815;
f(0.6)=2.38676;
f(0.7)=2.65797;
f(0.8)=3.94289;
f(1)=3.55975;
to use these values in a For Loop. How can I define function f?
Thanks in advance.

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Ameer Hamza
Ameer Hamza le 10 Oct 2020
Modifié(e) : Ameer Hamza le 10 Oct 2020
You can use interp1()
x = [0.2 0.4 0.5 0.6 0.7 0.8 1];
y = [1.42007 1.88124 2.12815 2.38676 2.65797 3.94289 3.55975];
f = @(xq) interp1(x, y, xq);
Then you can also evaluate in for in-between points
>> f(0.2)
ans =
1.4201
>> f(0.3)
ans =
1.6507
>> f(0.55)
ans =
2.2575
  6 commentaires
Ameer Hamza
Ameer Hamza le 10 Oct 2020
It happens when the input xq goes beyond the range of values in x. In your case, if xq is less than 0.2 or higher than 1.0, interp1 will give NaN. To avoid this, use extrapolation.
clear all;
clc;
x = [0.2 0.4 0.5 0.6 0.7 0.8 1];
y = [1.42007 1.88124 2.12815 2.38676 2.65797 3.94289 3.55975];
f = @(xq) interp1(x, y, xq, 'linear', 'extrap');
x = 0.6;
h = 0.4;
D(1,1) = (f(x + h) -2*f(x)+ f(x - h))/(h^2)
for i=1:2
h = h/2;
D(i + 1,1) = (f(x + h) -2*f(x)+ f(x - h))/(h^2);
for j=1:i
D(i + 1,j + 1) = (4^j*D(i + 1,j) - D(i,j))/(4^j - 1)
end
end
Mojtaba Mohareri
Mojtaba Mohareri le 10 Oct 2020
Modifié(e) : Mojtaba Mohareri le 10 Oct 2020
I understood. It works properly. Thank you so much for your consideration.

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