OK, this is true ?multiplying scalar 'k' by matrix 'v'

such as

Z = zeros(size(W1));
  kernel = [Z, W1, Z; W3, Z, W4; Z, W2, Z];
  filtered = conv2(vk, kernel, 'same');
  vk = (1-w)*vk + filtered./W;

where 'k' is iteration
      'v' is matrix

[Jan]

How could we decide if "it is true"? It depends, on what you are exactly searching for.I do not see "k" in the code, so what does it mean, that it is a "iteration"? I cannot find "V" also.

[Mary Jon]

vk that not 'function' or variables;

'k' is iteration counter, 'v' is matrix

Answer 1

[Image Analyst]

I don't see v * k anywhere in your code, so I guess the answer is no.

[Mary Jon]

I get this code from you(matlab answers) when asked you to write my equation ,where I told you 'k' iteration counter,'v' matrix!

yor answer was,

 Z = zeros(size(W1));
    kernel = [Z, W1, Z; W3, Z, W4; Z, W2, Z];
    filtered = conv2(vk, kernel, 'same');
    vk = (1-w)*vk + filtered./W;

why you write 'v' and 'k'as (vk)?