how to write this summation?

how to write this summation in matlab

Edit

if k=1,..n, and stored at each iteration of k ,i.e

I tried to use the solution in symbolic

clear;clc
h=0.1; x=0:h:2;n=10;m=3;
w = @(i,j,k) (-1)^(i+j)*cos((j*pi/h)*(x-x(k)));
syms I J
J=1:m;
for k=1:n
f(k)=symsum(symsum(w(I,J,k), J, 1, I), I, 1, m);
end

but the Error

Error using mupadmex
Error in MuPAD command: A square matrix is expected.
[(Dom::Matrix(Dom::ExpressionField()))::exp]

Error in sym/privBinaryOp (line 1693)
            Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});

Error in sym/mpower (line 189)
        B = privBinaryOp(A, p, 'symobj::mpower');

Error in @(i,j,k)(-1)^(i+j)*cos((j*pi/h)*(x-x(k)))

Moreover, I used a alternate codes with loop condition but i don't know if it's right

clear;clc
h=0.1; x=0:h:2;n=10;m=3;
w = @(i,j,k) (-1)^(i+j)*cos((j*pi/h)*(x-x(k)));
    for k=1:n;
    for i = 1:m;
    for j = 1:i;
    f(k)= sum(sum(w(i,j,k) ) );
    end
    end
    end

[Guillaume]

Is xk a constant or are you missing a k index in your summation?

As it is your expression can be reduced to a single sum from i = 1:2:m of cos(i*pi/h*(x-xk) as all the other terms cancel out.

Answer 1

[Guillaume]

I don't think your equation is right. The way to implement that summation properly would be:

 [ii, jj] = ndgrid(1:m);
 hh = pi*(x-xk)/h;
 result = sum(sum(triu((-1).^(ii+jj)  .* cos(jj*hh))))

However, because most of the terms cancel out, it could be simplified to:

hh = pi*(x-xk)/h;
result = sum(cos((1:2:m)*hh))

[javad ebrahimi]

thanks for your answere;I learned a lot from this answer

[work wolf]

Thank you @ Guillaume, what about if k=1,..n, and stored at each iteration of k ,i.e

Please, see above, i edited my Question.

[Guillaume]

It's not difficult to calculate fk(x) for each x and k, but for a given k, your summation can be greatly simplified, to the point that it looks like something is missing.

For a given k, look at all the terms generated when j = 1, which happens for all i from 1 to m. Let's call hh = pi*(x-xk)/h, which depends neither on j or i. The terms are:

j = 1; i= 1:m
(-1)^2*cos(hh) + (-1)^3*cos(hh) + (-1)^4*cos(hh) + ... + (-1)^(m+1)*cos(hh)

It's basically:

cos(hh)-cos(hh)+cos(hh)+ ... + (-1)^(m+1)*cos(hh)

If m is even that sum reduces to 0, otherwise it reduces to cos(hh). The same happens will all the other j values, the sum reduces either to cos(j*hh) or 0.

Therefore:

fk(x) = sum(cos(j*pi/h*(x-xk))), for j = 2:2:m, if m is even
fk(x) = sum(cos(j*pi/h*(x-xk))), for h = 1:2:m, if m is odd

[work wolf]

Really, many thanks. Please, can you show code to comply with the equation above without reduce terms in symbolic and for loop way?

Answer 2

[javad ebrahimi]

Hi work wolf this code can help you

clc
clear
p=0;
h=2;
x=2;
xk=3;
for i=1:10
    for j=1:10
       p=p+(-1)^(i+j)*cos((j*pi/h)*(x-xk));
    end
end

[Guillaume]

Your second for loop should go from 1 to i to comply with the equation:

for i = 1:10
   for  j = 1:i
      ...

[work wolf]

Thanks.