8
$\begingroup$

According to the MacTutor essay "D'Arcy Thompson on Greek irrationals" (which I take to be a version of Thompson's original essay whose only liberty with the original text is giving English translations where Thompson gives Greek words and phrases), Proclus, following Adrastus, asserted that $2+8+50+288+\dots$ (the sum of twice the squares of the "side numbers" $1,2,5,12,\dots$) equals $1+9+49+289+\dots$ (the sum of the squares of the "diagonal numbers" $1,3,7,17,\dots$). Here is the relevant excerpt:

"The table of side and diagonal numbers has many other properties. For instance, as Proclus tells us, the sum of the squares of two adjacent diagonals = twice the sum of the squares on the two corresponding sides: e.g. $3^2+7^2=2(2^2+5^2)$. And, in Chapter xxiii he shows, following Adrastus, that the sum of the squares of 'all' the diagonals is equal to twice the sum of the squares of 'all' the sides."

I haven't been able to find the text that Thompson is referring to here. In any case, I'm curious whether the original text seems to be an early attempt at manipulating divergent series.

Interestingly, under modern approaches to regularizing divergent series, the equality asserted by Proclus fails: the two divergent sums differ by $\frac12$.

[Note: This question was originally posted at MathOverflow, but it seems more appropriate here.]

$\endgroup$
5
  • 1
    $\begingroup$ It is hard to imagine that there were something similar to the modern concept of "sum of a divergent series"... $\endgroup$ Commented Apr 1, 2022 at 16:45
  • $\begingroup$ IMO we have to read them as infinite sums (whose "law" is quite obvious). $\endgroup$ Commented Apr 1, 2022 at 16:46
  • 1
    $\begingroup$ According to the masters thesis An Introduction to Pythagorean Arithmetic, Proclus addresses side and diagonal numbers in Commentary on Plato's Republic - specifically commentary on the eighth book. However, as Mauro has commented, it is very unlikely that anything involving infinity would have been entertained by the Greeks. $\endgroup$
    – nwr
    Commented Apr 1, 2022 at 17:31
  • 1
    $\begingroup$ @Mauro ALLEGRANZA: You are quite right, it was anachronistic of me to use the term "series" in this context. Still, it is not obvious to me what kind of sum is meant or what kind of laws such sums are assumed to satisfy. $\endgroup$ Commented Apr 1, 2022 at 19:53
  • $\begingroup$ I still haven't been able to find the text that Thompson is referring to. Can anyone steer me toward an English translation? $\endgroup$ Commented Apr 9, 2022 at 20:18

1 Answer 1

7
$\begingroup$

As nwr commented above, the original text by Proclus is his commentary on Plato's Republic, and Cambridge University Press is in the process of publishing an English translation. Unfortunately, I think the relevant section is in Volume 2, which has not been published yet. I don't read Greek, but maybe someone who does can find the right section in the Greek text.

However, it may not be necessary to dig up the exact section from Proclus. D'Arcy Thompson cites an earlier paper by A. E. Taylor, Forms and Numbers: A Study in Platonic Metaphysics. Taylor notes that the convergents of the continued fraction of $\sqrt{2}$ are $${1\over 1}, {3\over 2}, {7 \over 5}, {17 \over 12}, {41 \over 29}, {99 \over 70}, \ldots.$$ The numerators are the "diagonal numbers" and the denominators are the "side numbers." I would guess that the claim that "the sum of the squares of 'all' the diagonals is equal to twice the sum of the squares of 'all' the sides" is probably just the claim (in modern language) that the above sequence of fractions converges to $\sqrt{2}$. In particular, it seems unlikely to me that divergent series are being considered here.

$\endgroup$
1
  • 1
    $\begingroup$ Good observation! Though I don't know, now that you've said it, I suspect you're correct. Language/terminology drifts... :) $\endgroup$ Commented Apr 1, 2022 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.