I'm calculating using the order 3 polynomial interpolation method. But why does an array appear?

%polinomial
disp('polinomial interpolation')
clear
clc

x=1.3;
x1=0.0;
x2=0.5;
x3=1.0;
x4=1.5;
y1=log(x1);
y2=log(x2);
y3=log(x3);
y4=log(x4);
m3=log(x);

% dif 1
xa=(y2-y1)/(x2-x1);
xb=(y3-y2)/(x3-x2);
xc=(y4-y3)/(x4-x3);

% dif 2
xd=(xb-xa)/(x3-x1);
xe=(xc-xb)/(x4-x2);

% dif 3
xf=(xe-xd)/(x4-x1);

% f(1.3)
f(x)= (x1)+(xa)*(x-x1)+(xd)*(x-x1)*(x-x2)+(xf)*(x-x1)*(x-x2)*(x-x3);
fprintf ('%f\n',(f(x)))
et=(m3-f(x))/m3*100;

it says that theres an error in line 30 and i dont know how to fix it. please help

Answer 1

[Dyuman Joshi]

The syntax for defining an anonymouts function is incorrect in your code.

Also, you are trying to take logarithm of 0, which does not lead to good things, see below -

clear
clc

%polinomial
disp('polinomial interpolation')

polinomial interpolation

x=1.3;
x1=0.0;
x2=0.5;
x3=1.0;
x4=1.5;
y1=log(x1)

y1 = -Inf

y2=log(x2);
y3=log(x3);
y4=log(x4);
m3=log(x);

% dif 1
xa=(y2-y1)/(x2-x1);
xb=(y3-y2)/(x3-x2);
xc=(y4-y3)/(x4-x3);

% dif 2
xd=(xb-xa)/(x3-x1);
xe=(xc-xb)/(x4-x2);

% dif 3
xf=(xe-xd)/(x4-x1);

% f(1.3)

%% Corrected Syntax
f = @(x) (x1)+(xa)*(x-x1)+(xd)*(x-x1)*(x-x2)+(xf)*(x-x1)*(x-x2)*(x-x3);
fprintf('%f\n',f(x))

NaN

et=(m3-f(x))/m3*100

et = NaN

Answer 2

[Steven Lord]

When you receive warning and/or error messages please show the full and exact text of those messages (all the text displayed in orange and/or red in the Command Window.) The exact text of the message may be useful in determining what's going on and how to avoid the warning and/or error.

In this case I suspect I know the cause of the error.

f(x)= (x1)+(xa)*(x-x1)+(xd)*(x-x1)*(x-x2)+(xf)*(x-x1)*(x-x2)*(x-x3);

You can't assign to element 1.3 of an array. If you want f to be a function you can call with a specific value of x as input, define it as an anonymous function.

f = @(x) (x1)+(xa)*(x-x1)+(xd)*(x-x1)*(x-x2)+(xf)*(x-x1)*(x-x2)*(x-x3);

But you should use array multiplication, not matrix multiplication here so your function is vectorized, can be called with a vector or array as input not just a scalar.