Jost Burgi and Logarithms

Just found an article on the net (part of the LOCOMAT project) by Denis Roegel about Jost Burgi's "Progress Tabulen” of 1620. This is the publication that many people use to justify Burgi as an independent inventor of logarithms. In fact, he had made his method known to Tycho Brahe and others prior to the actual publication of Napier's Canon of Logarithms.
Roegel suggest that, in fact, what Burgi has done is :

There is no doubt that Bürgi came very close to the notion of logarithm, but if we
call logarithms a correspondence between two infinite continua representing real numbers, such that multiplication becomes addition, then Bürgi obviously only partly attained it, whereas Napier undoubtedly hit it.47 We can of course understand why this happened. It is in particular clear that Napier’s kinematic approach based on the theory of proportions gave him naturally a firm base in two infinite continua.48 We can also observe that Napier went beyond his needs. He needed only to define logarithms of sines, but his definition goes beyond. Bürgi, on the other hand, basically only needed to define the multiplication of numbers between 1 and 10 (or 108 and 109) and he therefore introduced a restricted way to do these computations, which at the same time prevented him from defining the general logarithm function which he didn’t really need.

A fair comparison between Napier and Bürgi requires a clear definition of the notion of logarithms, but it also makes it necessary to distinguish the abstract notion of logarithm from the knowledge of logarithmic computation. It is by confusing these two notions that many authors were led to attribute the discovery of logarithms to Bürgi or some of his forerunners, when they have in fact only produced what are, admittedly, tables of logarithms.

The date on the article is Jan 11, 2011, only a few days before the date (Jan 31) on which Burgi died in 1632. For those who haven't heard of Burgi, he is considered one of the great mechanics of the period, and created beautiful Armillaries and mechanized globes that are still works of both engineering and artistic beauty. He communicated with Brahe and Kepler, and was urged to publish his "Tablulen" by both.
From Wikipedia:

The most significant artifacts designed and built by Burgi surviving in museums are:

* Several mechanized celestial globes (now in Paris, Zuerich (Schweizerisches Landesmuseum), Stuttgart (Wurttembergisches Landesmuseum) and Kassel (Orangerie,2x,1580-1595) )
* Several clocks in Kassel, Dresden (Mathematisch Physikalischer Salon) and Vienna (Kunsthistorisches Museum)
* Sextants made for Keppler (at the National Technical Museum in Prag)
* The Mond-Anomalien-Uhr (a mechanical model of the irregularities of the motion of the Moon around the Earth)

Roegel doesn't deny the creative greatness of Burgi, but suggest it is greatness of a 2nd order, or as he says in his paper:

That being said, our purpose was in no way to diminish Bürgi’s contributions. Instead, we are a great admirer of his technical and mathematical contributions. But we have felt that justice had not been given to the Progress Tabulen and that wishful thinking by Swiss landsmen had somewhat distorted Bürgi’s work, and that it had again to be put straight.

Although we do not consider that Bürgi discovered or invented logarithms, we think
it is still appropriate to quote the words of Cajori in 1915: “The facts as they are known to-day assign to Napier the glory of a star of the first magnitude as the inventor of logarithms who first gave them to the world, and to Bürgi the glory of a star of lesser magnitude, but shining with an independent light.” [from Florian Cajori. Algebra in Napier’s day and alleged prior inventions of logarithms. In Knott [92], pages 93–109.]

A Nice, but Tough, Problem

Click on images to see higher resolution

This one is for my old teaching friend, Al Harmon, who has asked and answered problems for me and from me for many years... This is a variation of a problem that was on the "Wild About Math" website, adjusted a little to make it more compatible to students at HS level..

If you take a number (let's use 200 for an example) and think of all the ways you could write it as a sum or positive integers... (100+100, or 2 + 105 + 93 or... 1+1+1+.....+1) these are called the partitions of the number. Now if you take all the integers in the partition and multiply them together, which one would give the largest product? Obviously 1+1+1...+1 will not amount to a very large product.... 100* 100 is lots larger, and 66* 67 * 67 is even larger...almost 3 (10 ⁵),... which might make you think... aha...more small numbers is better... so 2¹⁰⁰ will be great! ...Well it is better, its about 1.126 (10³⁰ ) which is way larger than the others... but can we make it bigger???... What if we experiment a little with other values we try five, and find 5 ⁴⁰ = about 9 (10 ²⁷) .... big, but not as big as 2¹⁰⁰ ...

well, what about 4 ⁵⁰... wait, that is exactly the same as 2¹⁰⁰ by the exponent laws... so using all twos or all fours gives the same result... and fives yeilds less

Hmmmm..... We can't use all threes, but just for the heck of it, we try using two threes in place of three of the twos... what would 2⁹⁷ (3²) give us...... aha... more.... just a little more, but more... about 1.46 (10³⁰)... would more threes help??? would lots of threes help??? We could try using all threes and as many twos as would be left over at the end... 66 threes and 1 two would give us 6.18 (10³¹) more than any of the others. But Is THAT the largest... how can we convince ourselves that using threes is better than anything else?

Side trip to remember some ideas about logarithms.... remember that if a>b then ln(a) > ln(b) ... bigger numbers have bigger logs...so maximizing the product is the same as maximizing the log of the product.... and when we multiply numbers together, we are adding their logarithms to get the log of the product...[ln (bc)=lln (b) + ln(c)] ... now we can use that to think about how the log of the product changes...each time we use a number from the partition of the value, we add x to the sum, and ln(x) to the ln of the product.. so whatever number has the largest ratio of ln(x)/x will be the best choice to use.... If we graph the function (I used winplot, which is a wonderful FREE software.. ) we see the graph below.

It seems to be highest somewhere between 2 and 3, and a calculus student should be able to figure out that happens when f'(x) = 0, and since the numerator of the derivative is 1- ln(x), we know that will happen when ln(x) = 1, or x=e..... but that is NOT an integer... but we do notice the slope is always positive to the left of x=e, and always negative to the right of x=e, so whichever one of the two adjacent integers is most efficient, will be the most efficient.. we find that ln(3)/3 > ln(2)/2 and our question is answered.