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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

772.0. "(.9)(.99)(.999)(.9999)(.99999)..." by SQM::HALLYB (I sell too soon) Sat Oct 17 1987 22:30

    This has been rolling around the USENET lately.  No answer yet.
    
    What is the value of the infinite product (0.9)(0.99)(0.999)...

    Formally this can be expressed as:
        
        inf
       -----
        | |           i
        | |    (1 - .1 )
        i=1
    
      John
T.RTitleUserPersonal
Name
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772.10.890010099998999...ENGINE::ROTHMay you live in interesting timesSun Oct 18 1987 14:3028
    It is unclear what one wants by the 'value' of this product...

    In terms of a decimal expansion, I claim the value should look like this:

0     0.89
1	00 1 00 9999 8 999
2	000000 1 0000 99999999 8 99999
3	0000000000 1 000000 999999999999 8 9999999
 ...
k	<4k-2 0's> 1 <2k 0's> <4k 9's> 8 <2k+1 9's>

    It's almost surely transcendental.

    As a function of a complex variable,

	f(z) = PROD(k>0) (1-1/z^k) = PROD(k>0) (z^k-1)/z^k

    is interesting since it has an essential singualarity at zero, and
    zeroes on the unit circle.  This may give a closed form expression in
    terms of some known function.

    Note that you can rearrange the product into the form

	log(f(z)) = - SUM(k>0) 1/(k*(z^k-1))

    at least where this converges.  It doesn't seem to telescope though.

    - Jim
772.2An identity due to EulerENGINE::ROTHMay you live in interesting timesMon Oct 19 1987 05:0137
    I should explain how to obtain the decimal expansion in .1 - it comes
    from plugging into a well known identity involving the generating
    function for the number of partitions of an integer, due to Euler.
    This is in books on number theory, such as Hardy and Wright.  See also
    Knuth, "The Art of Computer Programming".

    The identity is

	f(z) = PROD(k>0) (1-z^k)
	     = 1 + SUM(k>0) (-1)^k * [z^(k*(3*k-1)/2)+z^(k*(3*k+1)/2)]

    With z = 0.1 you can easily fill in a decimal expansion.

    The generating function is actually the reciprocal of f(z),

	F(z) = SUM(k>=0) p(k)*z^k
	     = 1 / f(z) = 1 / PROD(k>0) (1-z^k)

    where p(k) = the number of ways of partitioning an integer.
    Eg. p(5) = 7 since 5 = 4+1 = 3+2 = 3+1+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1,
    7 ways.  There are tables of partions and some other combinatorial
    stuff in AMS55, along with some (hairy) closed form expressions for
    p(n).

    My comment about the singularity of the f(z) in .1 is probably meaningless
    since it doesn't converge when z < 1.

    It's doubtful if there is a simpler way of expressing the sum than this.

    However, products like f(z) are known as theta functions and are actually
    useful since they converge quickly and by taking ratios of them one
    can get the values of elliptic functions.  The latter are nice for
    practical EE work like conformal mapping, synthesizing filters and
    phase splitters, and of course making pretty pictures of minimal
    surfaces.

    - Jim