8
$\begingroup$

I'm interested in looking at the details of Gauss' method of determining the sign of the Gauss sum in his "Summatio Quarumdam Serierum Singularium", and I was wondering if anyone knew if there was an English or French translation available at all? I've tried searching on the internet and haven't found anything. Also, are there any modern presentations of the Gauss' method that you would recommend?

Thanks for your help.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ D M Bressoud, On the value of Gaussian sums, J Number Theory 13 (1981) 88–94, MR 82b:10046, "gives a new proof, closely paralleling that of Gauss...." $\endgroup$
    – Gerry Myerson
    Commented Nov 17, 2010 at 4:58
  • $\begingroup$ Is this part of his Disquisitiones arithmeticae? $\endgroup$
    – Sylvain JULIEN
    Commented Oct 12, 2023 at 19:18

3 Answers 3

Reset to default
9
$\begingroup$

"The determination of Gauss sums" by Berndt and Evans (Bull. Amer. Math. Soc., Vol. 5, Number 2 (1981), 107-129.) contains an exposition of the original proof due to Gauss. It also includes a short historical outline of various classical and modern proofs due to Dirichlet, Cauchy and others.

Improve this answer
$\endgroup$
0
3
$\begingroup$

there is a German translation: pages 463-495 in: http://www.amazon.co.uk/Untersuchungen-hohere-arithmetik-Chelsea-Publishing/dp/0821842137

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ (English + French)/2 = German? $\endgroup$
    – Adam Hughes
    Commented Dec 4, 2010 at 1:04
  • 5
    $\begingroup$ actually, i would say that English $-$ French $\to$ German. $\endgroup$
    – Suvrit
    Commented Dec 4, 2010 at 1:52
  • $\begingroup$ Unfortunately I only speak a tiny bit of German (I've just taken a very basic introductory course), but thanks for pointing out that there is a German translation. $\endgroup$
    – Lea M
    Commented Dec 7, 2010 at 2:07
2
$\begingroup$

An English translation of Baumgart's exposition of Gauss's fourth proof can be found here.

Improve this answer
$\endgroup$
1
  • $\begingroup$ That looks great-thanks so much for providing the link. $\endgroup$
    – Lea M
    Commented Dec 7, 2010 at 2:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .