## Friday, March 2, 2007

### An Online Pi Day Challenge

Even though pi day is about 2 weeks away, here's an online challenge I gave to all the students in my school back in 2005. Not very 'mathematical' but it engaged them. I posted the problem at 6 PM on March 14th and one student found a web site and submitted her solution in under 7 minutes! Get the feeling that this generation is able to learn a slightly different way!! For those offended by the fact that pi day is celebrated in other countries on July 22nd (22/7), I apologize! I am well aware that many teachers, math departments and schools have created wonderful pi day activities and there are abundant examples of these on the web. I would like others to share their favorites as well. Learning more about pi from MathWorld or Wikipedia is well worth our time, although it does get very technical. The nested radical problem from the other day (with the square roots of 2) relates to pi! Look it up!!

Background: The first 5 places (decimal digits) in pi, after the decimal point, are 14159. Find a web site that will allow you to search millions of places in pi.

The Challenge:

(1) Write down the seven decimal places in pi starting with the digit in the 5,191,306th place.

(2) These digits are a clue to the birthday of a famous mathematician (sorry, it's not Einstein!).

Give the full name of this mathematician and how he is connected to the number pi! Your submission should include the full birth date, the name and the connection.

EXTRA: Now challenge us - find another famous mathematician who is associated with pi and ask your own 'pi' challenge question!

Anonymous said...

After this problem, you should mention the astonishing BBP [Bailey-Borwein-Plouffe] formula for π, which lets one compute an arbitrary hexidecimal digit of π without finding all the previous ones. For example, we can find the quadrillionth base 16 digit of π without knowing any of the others. No decimal version of this can be found.

Dave Marain said...

thank you, eric--
I followed the links for it from wikipedia article on pi. Back in 1997 I'm sure this had a seismic effect on computer scientists and some mathematicians at the time but i must have slept through the earthquake! Considering how significant a result this is, one would think it would have received even more attention. I was able to follow the general idea of Plouffe's algorithm but it's sophisticated and I would need to analyze it much further. I will share the result (not the derivation) with my advanced group.
Now, is this a result you have personally used in your work?

Anonymous said...

No. Then again, I haven't been in academia since 1990, and my field was Riemann Surfaces. The problem here is that the quest to calculate arbitrary digits of π and even to attempt to show whether π is normal are more a nmatter of ego then a matter of science. Oh, it allows computer scientists to check their algorithms too, but it's not clear that the results will lead to anything in particular. On the other hand, this can show your students what mathematicians do. You don't want them thinking that they only learn how to calculate fast.

I wasn't presenting the BBP formula as an itemfor your students to prove; I just consider it an interesting relative to your homework problem.

One further link to mathematical research here is in the theory of Gauss' Arithmetic Geometric Mean. Let a and b be positive numbers, and consider the iteration:

a_0 = a, b_0 = b,
a_{i+1} = (a_i + b_i) / 2, b_{i+1} = (a_i b_i)^{1/2},

The a and b values converge to the AGM of a and b quite rapidly. Combined with an elliptic integral, this allows very fast approximations of π.

David Cox wrote a nice paper in the early 80s (available in π: A Sourcebook, published by Springer), which describes Gauss' research on the AGM of complex numbers (say 1 and z for simplicity). The difficulty is that there are two choices for the square root in each b formula. It turns out that there is one standard choice at each step. If one makes infinitely many nonstandard choices, the values converge to 0. If one always takes the standard choice, one gets a simplest value m. (I know—hahaha.) And if one makes finitely many nonstandard choices, the result turns out to be related to the standard value by a linear fractional transformation: (cm + d)/(em + f), with c, d, e, and f integers. This is an important theme in number theory, algebraic geometry, and complex analysis.