Help vectors size doesn't match...why?

This is my code to evaluate Lagrange where x_in are the given interpolation points and x_out is suppose to be f(x_in), but it keeps saying x_in and x_out don;t have the same vector length

x = -1:50;
y = 1.0 ./ (1+9*x.^2);
x_in=linspace(-1,1,100);
x_out=(1.0) ./ (1+9*(x_in).^2);
y_out=make_Ln(x,x_in,x_out);
plot(x,y,':',x_in,y_out,'*');
title('Lagrange Interpolation:x_in=linspace(-1,1,100)')
legend('y','Interpolation');
figure;

[Zoltán Csáti]

Show your make_Ln function.

[Guillaume]

And also give us the exact error message you get, particularly the bit that shows you the failing instruction.

[cakey]

Here is my function call:

if true
  % code
endfunction y=lagrange(x,x_in,x_out)

n=size(x_in,2); L=ones(n,size(x,2)); if (size(x_in,2)~=size(x_out,2)) fprintf(1,'x_in and x_out must have same elements'); y=NaN; else for i=1:n for j=1:n if (i~=j) L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j)); end end end y=0; for i=1:n y=y+x_out(i)*L(i,:); end end

if true
  % code
end

[cakey]

if true
  % code
endfunction y=lagrange(x,x_in,x_out)

n=size(x_in,2); L=ones(n,size(x,2)); if (size(x_in,2)~=size(x_out,2)) fprintf(1,'x_in and x_out must have same elements'); y=NaN; else for i=1:n for j=1:n if (i~=j) L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j)); end end end y=0; for i=1:n y=y+x_out(i)*L(i,:); end end

[cakey]

Error using plot Vectors must be the same lengths.

[cakey]

Sorry it is this one:

y_out=make_Ln(x,x_in,x_out)
n=size(x_in,2); 
L=ones(n,size(x,2)); 
if (size(x_in,2)~=size(x_out,2)) 
    fprintf(1,'x_in and x_out must have same elements'); 
    y=NaN; 
else 
    for i=1:n 
        for j=1:n 
            if (i~=j) 
                L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j)); 
            end 
        end 
    end 
    y=0; 
    for i=1:n 
        y_out=y_out+x_out(i)*L(i,:); 
    end 
end

Answer 1

[Julia]

Hi,

x_in and x_out have the same length (100).

x and y are shorter (52), but I don't know if this bothers your function make_Ln().

I suppose y_out and x_in do not match, but without knowing your function make_Ln() I cannot say more.

To plot x_in and x_out is no problem

plot(x,y,':',x_in,x_out,'*')

[cakey]

Here it is

   y=lagrange(x,x_in,x_out)
  n=size(x_in,2); L=ones(n,size(x,2)); if (size(x_in,2)~=size(x_out,2)) fprintf(1,'x_in and x_out must have same elements'); y=NaN; else for i=1:n for j=1:n if (i~=j) L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j)); end end end y=0; for i=1:n y=y+x_out(i)*L(i,:); end end

[Julia]

Your problem is the definition of L. In the last loop you create a vector y with length 52 (but you need 100). This is because of the second dimension of L (which is only 52).

plot(x,y,':',x_in(1:52),y_out,'*');

works. But I don't think you can use that plot ... I don't get much information out of it.

[cakey]

Oh I am sorry I forgot I changed the names here is how it is supposed to look

y_out=make_Ln(x,x_in,x_out)
n=size(x_in,2); L=ones(n,size(x,2)); if (size(x_in,2)~=size(x_out,2)) fprintf(1,'x_in and x_out must have same elements'); y=NaN; else for i=1:n for j=1:n if (i~=j) L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j)); end end end y=0; for i=1:n y_out=y_out+x_out(i)*L(i,:); end end

[cakey]

Julia, how are you able to tell the forst lengths were 100 and that one was 50? How can I see this?

[Julia]

I guessed that you changed the name, otherwise it would give you a completely different error ;)

How to find the length/size of a matrix or vector:

>> x=-1:50;
>> length(x)

ans =

      52

>> [m,n]=size(x)

m =

       1

n =

      52

[Patrik Ek]

Cakey i think that you should take a look at the debugger. You can easily set breakpoints in the code and check dimensions, values, etc. It is also possible to set a breakpoint if error. Both under the editor menu under breakpoints, or simply by typing dbstop error. This can be of help in the future and also to make it easier to find solutions of problems.

[cakey]

Thank you!