But does the same happen to the transcendental e?

Discussion in 'Recreational Math' started by Alexm, Dec 21, 2008.

  1. Alexm

    Alexm Guest

    Let e stand for the Euler number (with decimal point removed) and let
    p stand for pi (also withn the decimal point removed). Let e(n)
    represent the nth digit of e counting from the left, and the same for p
    (n). Thus e(4) =8 and p(1) = 3. Let O1 represent an operation which
    compares each e(n) with p(n) and removes e(n) from e if e(n) = p(n)
    for all n = 1,2,3,4, etc. The resulting spaces in e are then closed
    by moving e's remaining digits to the left. For convenience the
    resulting number is still labeled e.. Let O2 represent an operation
    which compares each e(n) with p(n) and p(n+1) and removes e(n) if e
    (n) = p(n) and/or p(n+1). The resulting spaces in e are then closed.
    And so on for O3, O4, O5, etc. Altogether, [ ...etc.....O7.x.(O6 x
    (O5 x (O4 x (O3 x (O2 x (O1 x e))))))) ] = the result. Note that if
    e were replaced by some integer such as 628953364817 then the effect
    of these operations would eliminate this integer. But does the same
    happen to the transcendental e?

    Here is how it goes for an integer, with O1 followed by O2 followed
    by O3, etc., until the integer becomes nothing:

    start integer 628953364817
    after O1 6289334817
    then after O2 6289334817
    then after O3 628334817
    then after O4 628334817
    then after O5 62833481
    then after O6 6834
    then after O7 6834
    then after O8 684
    then after O9 8
    then after O10 8
    then after O11 8
    then after O12 nothing

    alexm
     
    Alexm, Dec 21, 2008
    #1
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  2. Alexm

    Macavity Guest

    pi = 3.1415926535897... i.e. after 14 digits are put down, has all the
    digits from 1 to 9.

    If I understood your question right, if we do O15, on any number y,
    any
    first digit in y will be removed as it will match one of the digits in
    p(1)
    to p(14). In the integer example above, the first digit "6" was
    removed in
    O9 stage itself as p(8) = 6.

    Successive operations, O15, O16, etc. contain O15, and can hence
    remove at
    each such step any "new" first digit which has been "left shifted"
    in. So
    continuing this operation indefinitely should remove all digits
    systematically from any number, including e.

    HTH,
    Mac.
     
    Macavity, Dec 21, 2008
    #2
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  3. Alexm

    Alexm Guest

    Well, there do not appear to me to be any exceptions so you have
    nailed it.

    alexm
     
    Alexm, Dec 21, 2008
    #3
  4. Alexm

    mike Guest

    If I understand what the algorithm is doing, I am not so sure that the
    point has been proven non-trivially by the above.

    True, operations O15 and beyond will always remove the first digit of
    the remaining part of 'e'. But e has an infinite decimal expansion and
    removing the first digit (alone) in each of an infinitely repeated seems
    to result in one of those confusing 'infinite subtraction' type
    arguements. For example, if I start with a simpler digital sequence
    12345678910111213141516171819... formed by concatenating all the natural
    numbers and then subtract digits with an infinite series of operators
    (On) where O1 is 'remove a leading 1', O2 is 'remove a leading 2'...O11
    is 'remove a leading 11' etc...

    Now clearly, after any n operations (i.e. after you have applied On) you
    will still have an infinite string beginning with the decimal expansion
    of n+1, but it is reasonable to argue that after applying the infinite
    set of operations (On) then the string must be empty because all
    possible leading strings have been removed.

    But to me the above is an unsatisfactory arguement. It reminds me of the
    brain-teaser where, at time midnight - 2^(-n), you add balls labeled 10n
    to 10n+10 to a bucket and then remove ball n...with the question being
    "how many balls remain in the bucket at midnight". The answer people
    come up with depends on whether you consider an infinite series of
    actions can be completed on an infinite set.

    The same sort of arguement as often comes up in the discussion of the
    balls in a bucket problem can be used in the 'digits of Euler number
    problem'.

    Getting back to the original operation on the Euler number, assuming pi
    is normal, there will be no finite n such that 'On' clears the final
    digits from e. To show this, consider that if pi is nrmal then at some
    point in its decimal expansion there will be a string of at least m
    concatenated 2s, i.e. ....22222222222[for m digits]2222... (this should
    be found somewhere within the first few times 10^m digits of pi).
    Equally clearly, operation Om will only remove the non-2s from the
    remaining digits of e where it 'overlays' the concatenated 2s. So the
    non-2s will remain after Om. So no Om will ever get rid of the final
    digits of e. Seems therefore that only the infinite application of
    operations O1,O2,O3.... 'might' remove all the digits of e.

    And if the application of an infinite series of operations is
    acceptable, then you could get the same result by applying the trivial
    operation 'remove the first digit' an infinite number of times.

    Mike
     
    mike, Dec 22, 2008
    #4
  5. Ball n is removed at the nth step. End of story. The bucket is empty
    at midnight. The argument that the bucket is "full" is basically that
    there are 9n balls at step n and lim[n-->oo] 9n = oo. But if f(t)
    represents the number of balls at time t, where is it written that
    f(12) = lim(t --> 12) f(t) ? That "continuity" claim needs proof,
    which is tricky, considering it is false by the opposing argument.

    --Lynn

    http://math.asu.edu/~kurtz
     
    [Mr.] Lynn Kurtz, Dec 22, 2008
    #5
  6. Alexm

    Alexm Guest

    Interesting comments. One can also remove 'all' the digits of e once?

    alexm
     
    Alexm, Dec 22, 2008
    #6
  7. Alexm

    Macavity Guest

    If I understand what the algorithm is doing, I am not so sure that the
    point has been proven non-trivially by the above.

    True, operations O15 and beyond will always remove the first digit of
    the remaining part of 'e'. But e has an infinite decimal expansion and
    removing the first digit (alone) in each of an infinitely repeated seems
    to result in one of those confusing 'infinite subtraction' type
    arguements. For example, if I start with a simpler digital sequence
    12345678910111213141516171819... formed by concatenating all the natural
    numbers and then subtract digits with an infinite series of operators
    (On) where O1 is 'remove a leading 1', O2 is 'remove a leading 2'...O11
    is 'remove a leading 11' etc...

    <<rest of post snipped >>

    --
    As long as there are only a countable (or countably infinite) number of
    digits in the sequence, the process you describe should strip it empty. e
    would qualify - after all the original poster started by defining e(n)...

    HTH,
    Mac.



    ** Posted from http://www.teranews.com **
     
    Macavity, Dec 22, 2008
    #7
  8. Alexm

    Jasen Betts Guest

    The bucket is not empty it contains ball 10,20,30 etc,
    because balls with these numbers were added twice but only removed once :)
    Is the problem is chamged to read '10n to 10n+9' then the bucket
    cannot be shown to contain any specific ball.

    Due to the asymptotic behaviour of the ball-count as zero hour is
    approached I don't think it can be proven to be empty either.
     
    Jasen Betts, Dec 23, 2008
    #8
  9. Alexm

    Matt 271829 Guest

    No matter how many operations you perform, there will -- unless the
    decimal expansion of pi has some very unexpected properties -- always
    be an infinite number of digits remaining in "e". So, you'll never
    finish.
     
    Matt 271829, Dec 23, 2008
    #9
  10. Alexm

    Matt 271829 Guest

    Do you still think it would be empty if the balls weren't numbered, so
    that at every step you simply add nine balls? If not, then why should
    numbering the balls make any difference to the answer?
     
    Matt 271829, Dec 23, 2008
    #10
  11. By taking them out in order you establish a specific 1-1
    correspondence that doesn't miss any balls -- a surjection between the
    numbers put in and removed. Of course, you could at each stage remove
    the next even ball and it wouldn't be empty. If N is any positive
    integer, you *could* arrange to have exactly N balls left.

    --Lynn

    http://math.asu.edu/~kurtz
     
    [Mr.] Lynn Kurtz, Dec 23, 2008
    #11
  12. Hey!! No fair he changed the statement of the problem from what I
    expected it to be without warning us to read carefully.

    Merry Christmas everybody.

    --Lynn

    http://math.asu.edu/~kurtz
     
    [Mr.] Lynn Kurtz, Dec 23, 2008
    #12
  13. It's 12:01 and the examiner enters the room. (Nobody denies that 12:01
    occurs, right?) The following conversation ensues:

    Examiner: "The barrel is not empty."

    Me: "Yes it is!"

    Examiner: (reaching into the barrel) "No it isn't! See, I will show
    you a ball that's left." (Examiner draws a ball from the barrel).

    Me: "What number is on that ball?"

    Examiner: "It's number 8."

    Me: Speechless.

    --Lynn

    http://math.asu.edu/~kurtz
     
    [Mr.] Lynn Kurtz, Dec 23, 2008
    #13
  14. Alexm

    Alexm Guest

    A computer would never finish. But the process itself has no inherent
    time element in it.

    alexm
     
    Alexm, Dec 23, 2008
    #14
  15. Alexm

    Matt 271829 Guest

    I see what you mean... yes, I believe you are correct (reserving the
    right to change my mind, of course!)
     
    Matt 271829, Dec 23, 2008
    #15
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