The Vieta formula for \pi suggests the same conclusion, while involving rather less computation! WFL On 11/1/06, Schroeppel, Richard <rschroe@sandia.gov> wrote:

I gave a talk last week on the topic of Numbers, and included a slide about P-adic numbers. I mentioned in passing that a P-adic field includes all rationals, some algebraics, and some of the usual transcendental functions. In particular, those defined by convergent power series, when they converge P-adically. For example, e^4 = 1 + 4 + 16/2! + 64/3! + ... converges 2-adically. Some of the usual identities for exp, log, sin, atan, etc. will hold, because the identities are valid as formal power series. Thus e^log(1+x) = 1+x is 2-adically valid when x is a multiple of 4, since the power series for log(1+x) converges (to a value that's a multiple of 4), and the power series for e^(the log power series) also converges (since the exponent is a multiple of 4). Note that 4/3 qualifies as a multiple of 4 here.

I later began wondering about defining 2-adic pi. Ordinary pi can be computed using atan(1/2) + atan(1/3) = pi/4, and I wondered about extending this to P-adics. For my test case, I needed an arctan formula for pi that converges 2-adically. The atan series is

atan x = x - x^3/3 + x^5/5 - x^7/7 + ...

so any rational x with an even numerator (like 2, or -2/3) will converge 2-adically.

A bit of experimenting turned up the complex number product (1+2i)(1+8i)(3+2i) = -65, which gives the arctan identity

atan 2 + atan 8 + atan(2/3) = pi.

This is valid over the reals; each angle is in the first quadrant. The power series for atan won't converge (on the reals) for 2 and 8, but all three series converge 2-adically.

The arithmetic looks like this:

atan 2, 2-adic 2 = 1o -8/3 = ... 1010 1010 1ooo 32/5 = ... 1001 101o oooo -128/7 = ... 0100 1ooo oooo 512/9 = ... 001o oooo oooo -2048/11 = ... 1ooo oooo oooo ------------------ ... 0010 1100 1010

atan 8, 2-adic 8 = 1ooo -512/3 = ... 101o oooo oooo ------------------ ... 1010 0000 1000

atan(2/3), 2-adic 2/3 = ... 0101 0101 011o -8/81 = ... 1010 0111 1ooo 32/1215 = ... 0111 111o oooo -128/15309 = ... 1101 1ooo oooo 512/177147 = ... 011o oooo oooo -2048/1948617 = ... 1ooo oooo oooo ------------------ ... 0011 0010 1110

[Lower case o is the same as 0; I've used o for low order 0s for readability. To confirm a fraction, just multiply the binary by the denominator, and verify the numerator: For 2/3, ... 0101 0101 011o * 11 = ... 0000 0000 001o = 2.]

However, when I totted up the sum of the three arctangents to compute 2-adic pi, I got an unexpected result:

atan 2 = ... 0010 1100 1010 atan 8 = ... 1010 0000 1000 atan 2/3 = ... 0011 0010 1110 --------- ------------------ 2-adic pi = ... 0000 0000 0000

So the conjectured value for 2-adic pi is 0.

I thought I might have some kind of wraparound problem, but I can convert my arctangent identity to one that's convergent for both reals and 2-adics, and is definitely valid on the reals, although it's a mediocre pi formula. Use the substitutions

atan 2 = atan 2/3 + atan 4/7 atan 8 = atan 2 + atan 6/17

to get

pi = 3 atan 2/3 + 2 atan 4/7 + atan 6/17.

This converges in both reals and 2-adics, and all the angles are in the first quadrant and less than 45 degrees, so it's a plain vanilla pi formula. But the 2-adic version will also give 0 for pi, since the atan substitutions are 2-adically valid. (All the numerators of the atans are even, so the power series all converge, as does the power series proof of the arctan addition formula

atan x + atan y = atan (x+y)/(1-xy). )

This hardly counts as enough numerical work to justify any broad conjecture about P-adic pi, but the result is puzzling.

Rich

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