# But does the same happen to the transcendental e?

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

1. ### AlexmGuest

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

2. ### MacavityGuest

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

3. ### AlexmGuest

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

alexm

Alexm, Dec 21, 2008
4. ### mikeGuest

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
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
5. ### [Mr.] Lynn KurtzGuest

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
6. ### AlexmGuest

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

alexm

Alexm, Dec 22, 2008
7. ### MacavityGuest

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
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
8. ### Jasen BettsGuest

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
9. ### Matt 271829Guest

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
10. ### Matt 271829Guest

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
11. ### [Mr.] Lynn KurtzGuest

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
12. ### [Mr.] Lynn KurtzGuest

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
13. ### [Mr.] Lynn KurtzGuest

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
14. ### AlexmGuest

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

alexm

Alexm, Dec 23, 2008
15. ### Matt 271829Guest

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