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James Dow Allen
science forum beginner

Joined: 20 Sep 2005
Posts: 9

Posted: Sun Jul 09, 2006 8:17 am    Post subject: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

This may be an old problem.

Let B(x) be a function such that
If x is not an integer
B(x) = 0
If x is an integer
B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)
Prove (or disprove) that
B(2) = 1

James Dow Allen
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Sun Jul 09, 2006 10:42 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

On Sun, 9 Jul 2006, James Dow Allen wrote:

 Quote: Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 Need you ask? What does B(0) = ?
James Dow Allen
science forum beginner

Joined: 20 Sep 2005
Posts: 9

Posted: Thu Jul 13, 2006 5:59 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

Was my question too well-known to be interesting?
Or just uninteresting?

William Elliot wrote:
 Quote: On Sun, 9 Jul 2006, James Dow Allen wrote: Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 ... What does B(0) = ?

It doesn't matter: When x is a positive integer, the rightside of
the equation will never lead to anything but non-integers or
positive integers. But let's sidestep this bugaboo, by
stipulating
B(x) = 0
whenever x is not a positive integer.

Can you clarify this question? I wasn't *compelled* to post
on Usenet, if that's what you ask.

James
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Thu Jul 13, 2006 11:42 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

James Dow Allen wrote:
 Quote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 James Dow Allen

B(0) = 2^(-0) + B(0) + B(-1/3) + B(2)

which, since B(-1/3) = 0, implies B(2) = -1, as far as I can see...

Is the problem supposed to be more complicated than this?
James Dow Allen
science forum beginner

Joined: 20 Sep 2005
Posts: 9

Posted: Fri Jul 14, 2006 5:25 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

 Quote: B(0) = 2^(-0) + B(0) + B(-1/3) + B(2) which, since B(-1/3) = 0, implies B(2) = -1, as far as I can see...

This assumes that B(0) is defined, and that the definition
is consistent. This is a claim I did not intend to make;
I see now I should have explicitly said B(x) is undefined
for non-positive integers -- apologies.

I defined B(x) for non-integers and positive integers. I believe
these definitions are consistent, though proving that can be
considered part of the problem.

 Quote: Is the problem supposed to be more complicated than this?

I was afraid I'd get responses like "Don't bore us with a
well-known rephrasing of an old problem." Apparently the
rephrasing isn't so well known.

James
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Fri Jul 14, 2006 12:44 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

James Dow Allen wrote:
 Quote: matt271829-news@yahoo.co.uk wrote: B(0) = 2^(-0) + B(0) + B(-1/3) + B(2) which, since B(-1/3) = 0, implies B(2) = -1, as far as I can see... This assumes that B(0) is defined, and that the definition is consistent. This is a claim I did not intend to make; I see now I should have explicitly said B(x) is undefined for non-positive integers -- apologies.

I take this to mean that we ignore, as providing no information, any
candidate equation that has zero or a negative integer as an argument
to B.

 Quote: From what I can glean, I don't think that anything can be deduced about B(2). I think that B(2) can take any value you like, and that B(x) for

other positive integers x can then be defined appropriately so as to
produce a consistent system of equations. This assumes that you really
did mean the equation B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) to
hold only when x is an integer (now known to mean positive integer), as
Eighty
science forum beginner

Joined: 14 Jul 2006
Posts: 4

Posted: Mon Jul 17, 2006 2:12 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

James Dow Allen wrote:
 Quote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 James Dow Allen

No one has said anything in this thread for a while; could you give us
Frederick Umminger
science forum beginner

Joined: 18 Jul 2006
Posts: 1

Posted: Tue Jul 18, 2006 6:45 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

"Eighty" <eightyx@gmail.com> wrote in message
 Quote: James Dow Allen wrote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 James Dow Allen No one has said anything in this thread for a while; could you give us the answer?

If x is an integer,

B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

which implies

B(4x+2) = B(x) - 2^-x - B(4x/3) - B((2x-1)/3)

Let x = 0

B(2) = B(0) - 2^-0 - B(0) - B(-1/3)
B(2) = -1 - B(-1/3)
B(2) = -1

since B(y) = 0 if y is not an integer

-Frederick Umminger
Patrick Coilland
science forum Guru Wannabe

Joined: 29 Jan 2006
Posts: 197

Posted: Tue Jul 18, 2006 6:54 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

Frederick Umminger nous a récemment amicalement signifié :
 Quote: Let x = 0 B(2) = B(0) - 2^-0 - B(0) - B(-1/3) B(2) = -1 - B(-1/3) B(2) = -1 since B(y) = 0 if y is not an integer

Unfortunatly, OP did not accept this solution (first given by william
Elliot) :

<CIT> This assumes that B(0) is defined, and that the definition is
consistent. This is a claim I did not intend to make </CIT>

So, OP wants us to demonstrate that his definition is consistent (or
not).

This is the reason for which we all are interested in his demonstration
of (in)consistancy.

Obviously, you're right : if the definition is consistent, B(2) must
be -1.

--
Patrick
Richard Harter
science forum beginner

Joined: 19 May 2005
Posts: 21

Posted: Thu Jul 20, 2006 8:17 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

On 9 Jul 2006 01:17:59 -0700, "James Dow Allen"
<jdallen2000@yahoo.com> wrote:

 Quote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1

Cute problem (with the correction B(x)=0).

The trick here is to recast the problem to see what is going on.
Consider x modulo 3. We have

x=3p: B(x) = 2^-x + B(4p) + B(12p+2)
x=3p+1: B(x) = 2^-x + B(12p+6)
x=3p+2: B(x) = 2^-x + B(2p+1) + B(12p+10)

What we have is a directed graph G on the integers with
3p -> 4p, 12p+2
3p+1 -> 12p+6
3p+2 -> 2p+1, 12p+10

Moreover every integer except 2 has a unique ancestor, e.g.,

x ancestor
12p 9p
12p+1 18p+2
12p+2 3p
12p+3 18p+5
12p+4 9p+3
12p+5 18p+8
12p+6 3p+1
12p+7 18p+11
12p+8 9p+6
12p+9 18p+14
12p+10 3p+2
12p+11 18p+17

It is evident that one component of G is a rooted tree rooted in 2.
If that is the only component then the sum of all of the contributing
terms is the sum from 1 to inf of 2^-x which is 1. If there is more
than one component then the sum of the contributing terms is less than
one. All of which is to say is that the real problem is whether or
not G is a connected graph.

What other components might G have? As far as I can see there are two
types:

(a) A ring with trees growing off of it, and
(b) a chain with no first element.

I'm not quite sure how to proceed at the moment; perhaps wiser heads
than mine can see a line of attack. I have the horrid suspicion that
it is Collatz Conjecture hard.
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Thu Jul 20, 2006 9:34 pm    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

Richard Harter wrote:
 Quote: On 9 Jul 2006 01:17:59 -0700, "James Dow Allen" jdallen2000@yahoo.com> wrote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1 Cute problem (with the correction B(x)=0). The trick here is to recast the problem to see what is going on. Consider x modulo 3. We have x=3p: B(x) = 2^-x + B(4p) + B(12p+2) x=3p+1: B(x) = 2^-x + B(12p+6) x=3p+2: B(x) = 2^-x + B(2p+1) + B(12p+10) What we have is a directed graph G on the integers with 3p -> 4p, 12p+2 3p+1 -> 12p+6 3p+2 -> 2p+1, 12p+10 Moreover every integer except 2 has a unique ancestor, e.g., x ancestor 12p 9p 12p+1 18p+2 12p+2 3p 12p+3 18p+5 12p+4 9p+3 12p+5 18p+8 12p+6 3p+1 12p+7 18p+11 12p+8 9p+6 12p+9 18p+14 12p+10 3p+2 12p+11 18p+17 It is evident that one component of G is a rooted tree rooted in 2. If that is the only component then the sum of all of the contributing terms is the sum from 1 to inf of 2^-x which is 1. If there is more than one component then the sum of the contributing terms is less than one. All of which is to say is that the real problem is whether or not G is a connected graph. What other components might G have? As far as I can see there are two types: (a) A ring with trees growing off of it, and (b) a chain with no first element. I'm not quite sure how to proceed at the moment; perhaps wiser heads than mine can see a line of attack. I have the horrid suspicion that it is Collatz Conjecture hard.

I still don't understand this problem. These are the equations that I
got when I looked at it before:

B(1) = 2^-1 + B(6)
B(2) = 2^-2 + B(1) + B(10)
B(3) = 2^-3 + B(4) + B(14)
B(4) = 2^-4 + B(1
B(5) = 2^-5 + B(3) + B(22)
B(6) = 2^-6 + B( + B(26)
B(7) = 2^-7 + B(30)
B( = 2^-8 + B(5) + B(34)
B(9) = 2^-9 + B(12) + B(3
B(10) = 2^-10 + B(42)
etc.

I choose B(2) as anything I like. Clearly I can then choose B(1), B(6)
and B(10) somehow so as to satisfy the first two equations. Then taking
equations 3, 4, 5 etc. in turn, in each equation there is always a
B(x), namely the last term, where x is bigger than any number
previously encountered. Whatever values have been previously assigned
to the lower B's I can always choose this B(x) so as to satisfy the
next equation, and I can continue this indefinitely.

Therefore I conclude that nothing can be deduced about the value of
B(2) from the information given.
Paul Abbott

Joined: 19 May 2005
Posts: 99

Posted: Fri Jul 21, 2006 5:00 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

On 9 Jul 2006 01:17:59 -0700, "James Dow Allen" <jdallen2000@yahoo.com>
wrote:

 Quote: This may be an old problem. Let B(x) be a function such that If x is not an integer B(x) = 0 If x is an integer B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Prove (or disprove) that B(2) = 1

Not a proof -- but if one assumes that the sum

Sum[B(i), {i, 1, Infinity}]

exists, then summing over both sides and deleting all B(x) for x not an
integer, one obtains

Sum[B(i), {i, 1, Infinity}] ==

Sum[2^(-i), {i, 1, Infinity}] + Sum[B(4i), {i, 1, Infinity}] +
Sum[B(2i - 1), {i, 1, Infinity}] + Sum[B(4i + 2), {i, 1, Infinity}] ==

1 + Sum[B(2i -1), {i, 1, Infinity}] + Sum[B(2i), {i, 2, Infinity}] ==

1 + Sum[B(i), {i, 1, Infinity}] - B(2)

And hence B(2) = 1.

Cheers,
Paul

_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul
James Dow Allen
science forum beginner

Joined: 20 Sep 2005
Posts: 9

Posted: Fri Jul 21, 2006 7:08 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

After embarassing myself by neglecting to explicitly
exclude non-positive x, I was hoping the thread would
fade away! And I certainly have no proof that the definition
is consistent. In fact, the "intended underlying" B(x) is
a multiset, not a number, with
B(x) = 2^-x + B(4x/3) ...
intended as a sort of shorthand for
B(x) = {n} union B(4x/3) ...
The conjecture
B(2) = 1
corresponds with the conjecture
B(2) = {2,3,4,5,6,7,...} = the integers > 1

Regardless of any flaws in my formulation, I'm fairly
sure that the B(integer)'s into which B(2) expands are
what I intended.

Richard Harter wrote:
 Quote: I have the horrid suspicion that it is Collatz Conjecture hard.

It *is* the Collatz Conjecture! I'm surprised no one
has formulated it like this before -- or maybe someone
has seen it -- but didn't notice this thread.

I didn't set out to "obfuscate" Collatz but rather
the *inverse* Collatz function, exclude even
numbers; my x corresponds with (2x+1); but,
to avoid risk of further embarassment, I'll abstain
from detail.

Pau wrote:
 Quote: Sum[B(i), {i, 1, Infinity}] == ... And hence B(2) = 1.

I also derived something like this, but feared I was improperly
subtracting infinity from both sides of an equation.

If it turns out that Collatz *is* proven this way, please
mention me when you accept the \$Million prize!

James
Paul Abbott

Joined: 19 May 2005
Posts: 99

Posted: Fri Jul 21, 2006 7:50 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

"James Dow Allen" <jdallen2000@yahoo.com> wrote:

 Quote: After embarassing myself by neglecting to explicitly exclude non-positive x, I was hoping the thread would fade away! And I certainly have no proof that the definition is consistent. In fact, the "intended underlying" B(x) is a multiset, not a number, with B(x) = 2^-x + B(4x/3) ... intended as a sort of shorthand for B(x) = {n} union B(4x/3) ... The conjecture B(2) = 1 corresponds with the conjecture B(2) = {2,3,4,5,6,7,...} = the integers > 1 Regardless of any flaws in my formulation, I'm fairly sure that the B(integer)'s into which B(2) expands are what I intended.

Truncating the problem at finite upper limit n I noted that, for an
upper limit of 1000, in your notation

B(2) = {2,3,4,5,6,7,8,9,10,11,12,14,16,17,18,19,21,22,24,25,26,28, ...}

where the smallest missing integer is 13. It would be interesting to
know how far you have to go to obtain 13.

 Quote: I also derived something like this, but feared I was improperly subtracting infinity from both sides of an equation. If it turns out that Collatz *is* proven this way, please mention me when you accept the \$Million prize!

Which \$Million prize? There is a 1000 GBP prize but no Clay institute
prize for Collatz.

Cheers,
Paul

_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Fri Jul 21, 2006 10:45 am    Post subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2)

James Dow Allen wrote:
 Quote: After embarassing myself by neglecting to explicitly exclude non-positive x, I was hoping the thread would fade away! And I certainly have no proof that the definition is consistent. In fact, the "intended underlying" B(x) is a multiset, not a number, with B(x) = 2^-x + B(4x/3) ... intended as a sort of shorthand for B(x) = {n} union B(4x/3) ... The conjecture B(2) = 1 corresponds with the conjecture B(2) = {2,3,4,5,6,7,...} = the integers > 1 Regardless of any flaws in my formulation, I'm fairly sure that the B(integer)'s into which B(2) expands are what I intended. Richard Harter wrote: I have the horrid suspicion that it is Collatz Conjecture hard. It *is* the Collatz Conjecture! I'm surprised no one has formulated it like this before -- or maybe someone has seen it -- but didn't notice this thread. I didn't set out to "obfuscate" Collatz but rather tried to make it easier to work with. I start with the *inverse* Collatz function, exclude even numbers; my x corresponds with (2x+1); but, to avoid risk of further embarassment, I'll abstain from detail. Pau wrote: Sum[B(i), {i, 1, Infinity}] == ... And hence B(2) = 1. I also derived something like this, but feared I was improperly subtracting infinity from both sides of an equation. If it turns out that Collatz *is* proven this way, please mention me when you accept the \$Million prize! James

Can you explain what is wrong with my "answer" that nothing can be
deduced about the value of B(2) (see

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