Talk:Cohn's theorem

Missing complex conjugate sign in the conjugate reciprocal formula

Latest comment: 3 years ago1 comment1 person in discussion

If $f (z) = a 0 + \dots + a n z n$ is a polynomial, then its conjugate reciprocal is given by reversing and conjugating the coefficients, i.e., $f * (z) = a n + \dots + a 0 z n$ .

The formula for conjugate reciprocal (when $z \neq 0$ ) should be $f * (z) = z n f (1/ z)$ , i.e., there should be a complex conjugate sign on both the input and output of $f$ .

Steps:

f (z) = a 0 + \dots + a n z n

⇒

f (1/ z) = a 0 + \dots + a n z - n

⇒

f (1/ z) = a 0 + \dots + a n z - n

⇒

f (1/ z) = a 0 + \dots + a n z - n

⇒

z n f (1/ z) = a 0 z n + \dots + a n =: f * (z)

.

The article was missing a complex conjugate on the input of $f$ , so I added it in. This matches the formula used in the Lehmer–Schur algorithm article.

Arthur Cohn made the same mistake all throughout his 1921 PhD dissertation 😄. [1] (See e.g. equation (1) on page 112.)

[1] https://gdz.sub.uni-goettingen.de/id/PPN314393552

X-Fi6 (talk) 01:45, 4 December 2022 (UTC)Reply

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Talk:Cohn's theorem

Missing complex conjugate sign in the conjugate reciprocal formula