Formula for norm(x)

Hi everyone, please describe me the formula for norm (x) where x is a complex vector?

[James Tursa]

Cross posted here:

https://stackoverflow.com/questions/73807235/nomalization-of-complex-number-in-c

Answer 1

[James Tursa]

This question appears to be a follow-up to this post:

https://www.mathworks.com/matlabcentral/answers/1809695-what-exactly-norm-x-function-do?s_tid=srchtitle

You gave your attempt in that post, but it didn't do the sum properly. Here is a corrected version of your code with some edits. I would recommend to use double for the sum and normalized value:

double norm = 0.0;
for(int i=0;i < data_size; i++){
    // real_number and img_number are the vectors
    norm = norm + real_number[i]*real_number[i] + img_number[i]*img_number[i];  
}
norm = sqrt(norm);

If there is the possibility that the int*int result could overflow (I don't know the range of your numbers), then cast those numbers to doubles first. E.g.,

    norm = norm + (double)real_number[i]*(double)real_number[i] + (double)img_number[i]*(double)img_number[i];  

[Manu Chaudhary]

Thank you Sir. I will change everything to double for avoid any unnecessary error. Thank you for great help.

Answer 2

[Matt J]

Assuming you're only interested in p=2, if you already have code that computes norm(x) for real x, you can extend it to complex x via,

norm(x) = norm( [norm(real(x)), norm(imag(x)) ] )

which is easily verified below,

x=complex(rand(5,1), rand(5,1));

norm(x)

ans = 1.5694

norm( [norm(real(x)), norm(imag(x)) ] )

ans = 1.5694

Alternatively, if you have an implementation of abs, you could do norm(abs(x),p) which will work for any p-norm

p=3;
norm(x,p)

ans = 1.2935

norm(abs(x),p)

ans = 1.2935