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It appears to have something to do with the following two equations:
pi/4 = 1 - 1/3 + 1/5 - 1/7+...
pi^2/8 = 1+1/(3^2) + 1/(5^2) + 1/(7^2) + ...
As an experiment -- by no means a formal proof -- let us write out the
terms in pi/4 out to +1/25 and then multiply in the factors that (are
claimed to) contribute to pi/2 -- the idea being that the product should
be pi^2/8 to the same level of accuracy. Each factor of 3/2, 5/6, 7/6,
etc. is to be rendered as a geometric series with 1 as the first term
and whatever common ratio gives the appropriate sum (e.g., +1/3 for 3/2,
-1/5 for 5/6). Terms with denominators greeater than 25 are truncated
off. It's easy to see that all terms in all geometric series will have
a numerator of (+/-) 1 and an odd denominator.
We have:
pi/4 = 1 - 1/3 + 1/5 - .. + 1/25
3/2 = 1 + 1/3 - 1/9
5/6 = 1 - 1/5 + 1/25
7/6 = 1 + 1/7
11/10 = 1 + 1/11
..
23/22 = 1 + 1/23
(all subsequent factors are equal to 1 at the level of truncation.)
Multiplying the pi/4 series by the 3/2 series then gives:
(1- 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 +
1/17 - 1/19 + 1/21 - 1/23 + 1/25) +
(1/3 - 1/9 + 1/15 - 1/21) +
(1/9)
= 1 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 + 1/17 - 1/19 - 1/23 + 1/25
Next we multiply in the 5/6 term, which is this approximation is 1 - 1/5
+ 1/25:
(1 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 + 1/17 - 1/19 - 1/23 + 1/25) -
(1/5 + 1/25) + (1/25)
= 1 - 1/7 + 1/9 - 1/11 + 1/13 + 1/17 - 1/19 - 1/23 + 1/25
The remaining factors up to 23/22 subtract away all the remaining terms
shown with prime denominators. We are then left with the desired
truncation for pi^2/8:
1 + 1/9 + 1/25
Let us look at the key properties of this sieving process that leads
towards the selection of the square denominators in the final sum:
1) All 4n-1 primes are factored in as (4n-1)/(4n-2) = 1 + 1/(4n+1) +
1/(4n+1)^2 + ... . The positive term of 1/(4n-1) in this geometric
expansion cancels out a negative term of 1/(4n-1) in the pi/4 series,
for all terms of 1/(4n-1) are negative in the latter series.
2) All 4n+1 primes are factored in as (4n+1)/(4n+2) = (note signs!) 1 -
1/(4n+1) + 1/(4n+1)^2 - ... . The negative term of 1/(4n+1) introduced
here cancels the corresponding positive term in the pi/4 seroies.
3) Terms with nonsquare, composite denominators fall out as factors
corresponding to the primes are introduced. Thus the factor of 3/2,
which cancels out -1/3 by argument 1, also gets rid of -1/15, for
multiplication of +1/3 (from the geometric series) by +1/5 (from the
pi/4 series) gives +1/15. The signs work out as desired because
factoring a 4n-1 composite like 15 must yield a 4n+1 factor and a 4n-1
factor. A similar mechanism operates against the +1/21 term. Note that
1/15 can't be reintroduced by the 5/6 ( = 1 - 1/5 + 1/25 - ...) factor
because that geomtric series has no terms between 1/5 and 1/25;
similarly for 1/21 relative to the 7/6 factor.
4) Square denominators survive because the casncellation noted in
argument (3) does not work out the same way. The critical differnece is
the introduction of squared denominators with a consistent positive sign
in the geometric series. In our experiment, we see that the 3/2 factor
seems to cancel out +1/9 along with -1/15 and +1/21. But now the next
term in the geometric series for 3/2 reintroduces +1/9, whereas 1/15 and
1/21, involving denominators that aren't multiples of 9, are left
stranded. The astute reader will observe that if 1/27, with a cubed
denominator, were included it would also gets reintroduced; but
subsequent terms in the geometric series compensate for that.
As I said, this is not a complete formal proof. But it demonstrates
clearly how the sieving process works to generate the claimed result.
--OL
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