How to define a special function with some points

2020 10 10
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更新时间：2020 10 10
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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 2020-10-10
编辑：Ameer Hamza 2020-10-10

0 个投票

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 个评论
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Mojtaba Mohareri
Mojtaba Mohareri 2020-10-10
编辑：Mojtaba Mohareri 2020-10-10
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
Mojtaba Mohareri 2020-10-10
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
Walter Roberson 2020-10-10
Ran in:
Ran in:
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
Walter Roberson 2020-10-10
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
Ameer Hamza 2020-10-10
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 2020-10-10
编辑：Mojtaba Mohareri 2020-10-10
I understood. It works properly. Thank you so much for your consideration.

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