How to define a special function with some points

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.

Answer 1

[Ameer Hamza]

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

[Mojtaba Mohareri]

Thank you very much.

This is my code

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);
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

But the output is as follows

D =

NaN 0 0

26.2652 NaN 0

1.2600 -7.0751 NaN

Could you please tell me what the "NaN" is?

[Mojtaba Mohareri]

When I compute (f(x + h) -2*f(x)+ f(x - h))/(h^2) in Command Window separately, it equels to 1.2600, but actually D(1,1)=(f(x + h) -2*f(x)+ f(x - h))/(h^2) which is NaN.

[Walter Roberson]

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;
f(x+h)

ans = 3.5598

f(x)

ans = 2.3868

x-h

ans = 0.2000

(x-h) - X(1)

ans = -5.5511e-17

f(x-h)

ans = 1.4201

D(1,1) = (f(x + h) -2*f(x)+ f(x - h))/(h^2)

D = 1.2894

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

D = 2×1

1.2894 26.2652

●

D = 2×2

1.2894 0 26.2652 34.5905

●

D = 3×2

1.2894 0 26.2652 34.5905 1.2600 0

●

D = 3×2

1.2894 0 26.2652 34.5905 1.2600 -7.0751

●

D = 3×3

1.2894 0 0 26.2652 34.5905 0 1.2600 -7.0751 -9.8528

●

[Walter Roberson]

If you look closely at this, you will see that 0.6 - 0.4 is not equal to 0.2 and instead is slightly less. Because of that, interp1() considered you to be doing extrapolation and the default extrapolation is NaN.

[Ameer Hamza]

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]

I understood. It works properly. Thank you so much for your consideration.