Some cute divergent series.
- From: w.taylor@xxxxxxxxxxxxxxxxxxxxx
- Date: 30 Aug 2005 23:56:30 -0700
Here is a cute bit of pseudo-math, much of which can be justified
using ideas from formal treatments of divergent series; (cf Hardy).
But we will stick to the wild, informal treatment, being more fun.
The series I want to sum is
S = 1 - 2 + 3 - 4 + 5 - 6 + ... .
A first attempt would be to express it as the sum of the power series
S(x) = 1 - 2x + 3x^2 - 4x^3 + ... "at" x=1.
S(x) = (1 + x)^(-2) hence S = S(1) = 1/4. A nice definite answer!
Further investigation shows that this appears to agree with
the earlier summation obtained last time this topic came up,
(especially if one treats "thinned-out" series appropriately),
namely 1 + 2 + 3 + 4 + ... = -1/12. We leave this as an exercise.
However, we look here at a rather simpler "justification".
There is a current parallel thread on adding up "doubly infinite"
series. We can look here at singly infinite series going infinitely
into the negative indices.
e.g. ...+ 1/8 + 1/4 + 1/2 + 1 + 2 + 4 = 8 , pretty obviously.
Euler first extended this into doubly infinite series to get
the very cute result that SUM[-oo,oo] x^n = 0 for all x =/= 1 !
We will look first at finite series like
4 - 3 + 2 - 1 = 2
and 5 - 4 + 3 - 2 + 1 = 3
and 6 - 5 + 4 - 3 + 2 - 1 = 3 and so on.
The sums are round about half of the leading number,
unsurprisingly for the game theorist.
But there is clearly an "inconsistency" between the result for
odd and even series of this type! But, it is easy to restore
"continuity" here if we extend these series, as they clearly
"want" to be extended, into the infinite negative zone!
4 - 3 + 2 - 1 + 0 - (-1) + (-2) - (-3) + ...
and similarly for other leading terms. And now, the terms
to the right of the zero are just our original series,
in this even case, or its negative, in the odd case.
So, what value should these terms' sum be given,
to achieve our desired continuity here?
YES - you guessed it - it is the above value of 1/4. (!)
If we do this, then all is well, and we get, for any natural n, that
n - (n-1) + (n-2) - ... +/- 0 -/+ (-1) +/- (-2) ... = (2n+1)/4 .
i.e. SUM[n,-oo] (-1)^(n-k) * k = (2n+1)/4 .
Howbout that! :-)
- - - - - - - - - - - - - -
Incidentally, for game players who are used to "progressive" games,
whereby the first player plays 1 move, the second 2, the first 3,
and so on: your sneaking feeling that this still leaves a slight
advantage to the first player, is CORRECT!
He has an advantage of one quarter of a move, still.
To get a truly unbiassed game, one can play either a move series
of 1,2,2,2... moves, (1-2+2-2...=0); or if the progressive mode
is still desired, my favourite, "odd progressive", with move series
1,3,5,7... (1-3+5-7...=0) [exercise for reader]; both perfectly fair.
And incidentaly the latter one also has the cute feature that
the total number of moves played at the end of a turn is always
a perfect square! The "perfect" form for progressive games, indeed!
----------------------------------------------------------------------
Bill Taylor W.Taylor@xxxxxxxxxxxxxxxxxxxxx
----------------------------------------------------------------------
The perfect Go player must be a Zen Buddhist in the opening,
a Taoist in the middle-game, and a Confucian in the endgame!
----------------------------------------------------------------------
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