# Formula for norm(x)

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Manu Chaudhary 2022 年 9 月 22 日

Hi everyone, please describe me the formula for norm (x) where x is a complex vector?
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James Tursa 2022 年 9 月 22 日

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### 採用された回答

James Tursa 2022 年 9 月 22 日

This question appears to be a follow-up to this post:
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];
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Manu Chaudhary 2022 年 9 月 22 日
Thank you Sir. I will change everything to double for avoid any unnecessary error. Thank you for great help.

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### その他の回答 (1 件)

Matt J 2022 年 9 月 22 日

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

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