Summation with resultant vector's index as upper limit

Hello,

I'm trying to write this summation in Matlab

I couldn't find any example of this anywhere, maybe someone can help me with the syntax here.

Answer 1

[Torsten]

https://de.mathworks.com/help/matlab/ref/conv.html#bucr92l-2

[Guillaume]

Or to put it clearly, your formula is the convolution of the 4 element vectors a and b, divided by 2

[Kristoffer Lindvall]

Great! Thanks, I will use that. Also, how would you compute the sum if there was only a, not a and b.The convolution command takes two vectors.

[Torsten]

Then set b = ones(size(a),1)

[Kristoffer Lindvall]

It seems the conv function does kind of what I want. Conv gives back a vector length 2*n-1, where n = length(b) , but the sum I have c has length n.

This is the whole sum I am trying to compute. I know this can be done with for loops but from what I heard summations are more efficient. The sum command is not so intuitive imo.

[Torsten]

Use a for-loop for such exotic sums.

[Guillaume]

It seems the conv function does kind of what I want.

It does exactly what you want. By default it returns the full convolution. It sounds like you want the 'same' option:

c = conv(a, b, 'same') / 2;

[Kristoffer Lindvall]

Yeah, I tried that but I got a different result.

The full conv gives c_full = [0.5 2 2], and same conv gives c_same=[2 2]. The answer I need is c = [0.5 2]. Atleast for the first sum in the formula.

a = (1:2);
b = (1:2);
c1 = conv(a,b)/2;

c2 = conv(a,b'same')/2;

This is the best I could do for the entire formula

n = length(a);
c = zeros(1,n);
for k=1:n
    c(k) = sum(a(1:k).*b(k:-1:1)/2)+sum(a(2:n-k).*b(2+k:n-k)/2)+sum(a(2+k:n-k).*b(2:n-k)/2);
end

[Bruno Luong]

How hard to extract k+1 first elements of conv full result?