I started to suspect as much, because I've been trying to devise a
pair of random variables X,Y where E(X+Y)<>E(X) + E(Y), but have
failed to do so.  I guess I feel silly because I've taken courses in
probability and statistics, but it's been over 20 years, so I guess
I forgot that theorem.  I'd to see a proof of it, though.... but as
you say it's a famous theorem, it shouldn't be hard for me to locate
the proof.

--- In mathforfun@yahoogroups.com, jwwarrenva <no_reply@y...> wrote:
> Actually, independence is not required in order to add up the
expected
> values of the various pairs.  E(X+Y) = E(X) + E(Y) even when X and
Y
> are not independent.  I always find this surprising, but the
theorem
> is in all the books, so it must be right.  ;-)

Allo. Alright, here's something that should be easy, but for some
reason, I'm not understanding.

w(t+d)=w(t)-(1/2)(w(t)-144)

Solve for (d), and also explain how it must be true that the change
in (t) is the same for the value of (w) to be halved. (I would
assume, for this, solve for w)
Thanks everyone, soooo tired!

What function is w defining?
Need to find a point t+d and a point t such that both map to the same value
defined by the function w.
     Gary
   ----- Original Message -----
   From: cfgrs
   To: mathforfun@yahoogroups.com
   Sent: Wednesday, October 01, 2003 23:49
   Subject: [MATH for FUN] I'm sleep deprived and not understanding...


   Allo. Alright, here's something that should be easy, but for some
   reason, I'm not understanding.

   w(t+d)=w(t)-(1/2)(w(t)-144)

   Solve for (d), and also explain how it must be true that the change
   in (t) is the same for the value of (w) to be halved. (I would
   assume, for this, solve for w)
   Thanks everyone, soooo tired!


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               ADVERTISEMENT




   To unsubscribe from this group, send an email to:
   mathforfun-unsubscribe@yahoogroups.com



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[Non-text portions of this message have been removed]

I just thought about this on my own and the proof is really simple.
Let X,Y be continous random variables.  (The proof for the discrete
case is analogous.)

Let their joint probability density function be defined by:
f(x,y)

For lack of an integral sign, I'll use Intg(p(x)dx) to denote
integration.

Let the probability distribution of X be denoted as g(x), and that
of Y be denoted as h(y)

Then:
E(X+Y)
= intg(intg(f(x,y)(x+y)dxdy))
= intg(intg(f(x,y)xdxdy)) + intg(intg(f(x,y)ydxdy))
= intg(x*intg(f(x,y)dy)dx) + intg(y*intg(f(x,y)dx)dy)
= intg(x*g(x)dx) + intg(y*h(y)dy)
= E(X) + E(Y)


Intuitively, you could imagine that E(X+Y) = weighted value of X+Y,
where X+Y is measured over the two dimensional domain of (X,Y).  By
the distributive propery, this is the same as taking the sum of the
weighted value of X + weighted value of Y over this same domain,
which intuitively would be the same as E(X) + E(Y)


--- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
wrote:
> I started to suspect as much, because I've been trying to devise a
> pair of random variables X,Y where E(X+Y)<>E(X) + E(Y), but have
> failed to do so.  I guess I feel silly because I've taken courses
in
> probability and statistics, but it's been over 20 years, so I
guess
> I forgot that theorem.  I'd like to see a proof of it, though....
but as
> you say it's a famous theorem, it shouldn't be hard for me to
locate
> the proof.

Hello every body!
Who interest s in Mathematic and Mathematic Olympiad?
Certainly you!  The YahooGroup  [[ MathematicOlympiad ]] eagerly
is desirous [[ you join and enjoy it]]
MathematicOlympiad yahoogroup is for discussing about
MathematicOlympiad and the problems.

Expectations always add.

A random variable is simply a real-valued function on a particular
measure space. Its expectation is simply its integral. Since the
integral of a sum is the sum of the integrals, expectations always
add, regardless of independence.

--- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
wrote:
> How is it you know you're able to add the expected values of each
of
> the 26 events together?  If they were independent events, you
could,
> but in this case, they're not independent.  Still, it's possible
for
> the addition to be allowable even if independence doesn't exist,
but
> only in certain situations.   How is it in this situation, you know
> that the addition is allowed?
>
> --- In mathforfun@yahoogroups.com, jwwarrenva <no_reply@y...> wrote:
> > The expected number of same-colored pairs is 650/51, or
> approximately
> > 12.745.
> >
> > To derive this result, consider one of the 26 pairs drawn from
the
> > deck.  This pair is essentially a sample of size 2 drawn from the
> deck
> > without replacement.  Accordingly, the number of red cards in the
> > sample follows a hypergeometric distribution, and the probability
> that
> > the number of red cards equals one is C(26,1) * C(26,1) / C(52,2)
> =
> > 26/51, where C(n,m) denotes the number of combinations of n items
> > taken m at a time.  Then the probability that the pair of cards
> has
> > the same color is 1 - 26/51 = 25/51.  Since there are 26 pairs,
> the
> > expected number of same-colored pairs is 26 * (25/51) = 650/51.
> >

You can't scare me, tough guy. :)

Actually, I'm just very busy. The moose in Banff says hi.

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> Suppose you have a sequence a_1, a_2, a_3, a_4... and you want to
> multiply a_n to the n'th term of (-x + 1)^n and find the sum for
each
> value of n?
>
> Also, regarding the "^" symbol.  I read that this is a part of
what's
> called arrow notation.  m^n is obviously m to the n'th degree.
m^^n
> = m^m^m^m... n times.  m^^^n= m^^m^^m^^m... n times which equals m
(^m
> (^m(^m(^m... n times ...followed by n ")"  After that it becomes
mind
> boggling complicated for me.  For a short explanation, see
> http://mathworld.wolfram.com/ArrowNotation.html
>
> bq
>
>
> btw, I didn't mean to scare you off jason1990.
> Please come back.

Maybe it has nothing to do with sleep deprivation, and you're just plain
stupid.  Have you considered that?  Let's face it, *everyone* can't be
smart, so some people must be stupid.  Maybe you're stupid.

Hope that helps,

John


>From: cfgrs <no_reply@yahoogroups.com>
>Reply-To: mathforfun@yahoogroups.com
>To: mathforfun@yahoogroups.com
>Subject: [MATH for FUN] I'm sleep deprived and not understanding...
>Date: Thu, 02 Oct 2003 03:49:17 -0000
>
>Allo. Alright, here's something that should be easy, but for some
>reason, I'm not understanding.
>
>w(t+d)=w(t)-(1/2)(w(t)-144)
>
>Solve for (d), and also explain how it must be true that the change
>in (t) is the same for the value of (w) to be halved. (I would
>assume, for this, solve for w)
>Thanks everyone, soooo tired!

_________________________________________________________________
The new MSN 8: smart spam protection and 2 months FREE*
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how can i join to this group?
thanks

tadeh_davtian <tadeh_davtian@...> wrote:
Hello every body!
Who interest s in Mathematic and Mathematic Olympiad?
Certainly you!  The YahooGroup  [[ MathematicOlympiad ]] eagerly
is desirous [[ you join and enjoy it]]
MathematicOlympiad yahoogroup is for discussing about
MathematicOlympiad and the problems.


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[Non-text portions of this message have been removed]

Yeah, that's right John.  In fact, you're such a math guru that you
probably trawl from group to group looking for "stupid" people to
feed you the idea thet you're the best, ever.

--- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...> wrote:
> Maybe it has nothing to do with sleep deprivation, and you're just
plain
> stupid.  Have you considered that?  Let's face it, *everyone* can't
be
> smart, so some people must be stupid.  Maybe you're stupid.
>
> Hope that helps,
>
> John
>
>
> >From: cfgrs <no_reply@yahoogroups.com>
> >Reply-To: mathforfun@yahoogroups.com
> >To: mathforfun@yahoogroups.com
> >Subject: [MATH for FUN] I'm sleep deprived and not understanding...
> >Date: Thu, 02 Oct 2003 03:49:17 -0000
> >
> >Allo. Alright, here's something that should be easy, but for some
> >reason, I'm not understanding.
> >
> >w(t+d)=w(t)-(1/2)(w(t)-144)
> >
> >Solve for (d), and also explain how it must be true that the change
> >in (t) is the same for the value of (w) to be halved. (I would
> >assume, for this, solve for w)
> >Thanks everyone, soooo tired!
>
> _________________________________________________________________
> The new MSN 8: smart spam protection and 2 months FREE*
> http://join.msn.com/?page=features/junkmail

First study John Conway's chained arrow notation and the reference
links http://mathworld.wolfram.com/ChainedArrowNotation.html and
http://mathworld.wolfram.com/ArrowNotation.html and
http://mathworld.wolfram.com/AckermannNumber.html and tell me this
isn't an extrordinarily mind boggling huge number!  This is the
googolplex'th Ackerman number and it needs a name.

googolplex -> googolplex -> googolplex

btw, is there a name for 10^googolplex?  I thought is was called mega
and 10^mega is megaplex, but I guess I was mistaken.  I also hate it
when this form overwrites everything in it.

bq

I reckon that if you take 1 and add a million billion trillion
quillion squillion zillion killion illion [sic] dillion fillion
pillion gillion kazillion zwillion frillion marillion guzillion
brillion chillion crillion yillion spillion jillion zeroes after it,
and then raise 10 to the power of this number, you'll get a really
huge number.

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> First study John Conway's chained arrow notation and the reference
> links http://mathworld.wolfram.com/ChainedArrowNotation.html and
> http://mathworld.wolfram.com/ArrowNotation.html and
> http://mathworld.wolfram.com/AckermannNumber.html and tell me this
> isn't an extrordinarily mind boggling huge number!  This is the
> googolplex'th Ackerman number and it needs a name.
>
> googolplex -> googolplex -> googolplex
>
> btw, is there a name for 10^googolplex?  I thought is was called
mega
> and 10^mega is megaplex, but I guess I was mistaken.  I also hate
it
> when this form overwrites everything in it.
>
> bq

You have two lengths of rope, each of which takes exactly one hour
to burn through from end to end. The ropes burn at an uneven rate.
So you can't, for instance, chop one of the ropes in half in order
to measure 1/2 hour.

You are asked to use the rope(s) and a lighter to measure 45
minutes. Do you have enough information to succeed?

Nick

--- In mathforfun@yahoogroups.com, "nick_hobson" <nick_hobson@y...>
wrote:
> You have two lengths of rope, each of which takes exactly one hour
> to burn through from end to end. The ropes burn at an uneven
rate.
> So you can't, for instance, chop one of the ropes in half in order
> to measure 1/2 hour.
>
> You are asked to use the rope(s) and a lighter to measure 45
> minutes. Do you have enough information to succeed?
>
> Nick

Light 3 ends simultaneously.   The one with 2 ends will extinguish
in 30 minutes.  At that moment, light the 4th end.   That rope will
extinguish 15 minutes after that.

30+15 = 45

--- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
wrote:
> --- In mathforfun@yahoogroups.com, "nick_hobson"
<nick_hobson@y...>
> wrote:
> > You have two lengths of rope, each of which takes exactly one
hour
> > to burn through from end to end. The ropes burn at an uneven
> rate.
> > So you can't, for instance, chop one of the ropes in half in
order
> > to measure 1/2 hour.
> >
> > You are asked to use the rope(s) and a lighter to measure 45
> > minutes. Do you have enough information to succeed?
> >
> > Nick
>
> Light 3 ends simultaneously.   The one with 2 ends will extinguish
> in 30 minutes.  At that moment, light the 4th end.   That rope
will
> extinguish 15 minutes after that.
>
> 30+15 = 45


How do you know that the one with 2 ends will extinguish in 30
minutes?

Nick

--- In mathforfun@yahoogroups.com, "nick_hobson" <nick_hobson@y...>
wrote:
> --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> wrote:
> > --- In mathforfun@yahoogroups.com, "nick_hobson"
> <nick_hobson@y...>
> > wrote:
> > > You have two lengths of rope, each of which takes exactly one
> hour
> > > to burn through from end to end. The ropes burn at an uneven
> > rate.
> > > So you can't, for instance, chop one of the ropes in half in
> order
> > > to measure 1/2 hour.
> > >
> > > You are asked to use the rope(s) and a lighter to measure 45
> > > minutes. Do you have enough information to succeed?
> > >
> > > Nick
> >
> > Light 3 ends simultaneously.   The one with 2 ends will
extinguish
> > in 30 minutes.  At that moment, light the 4th end.   That rope
> will
> > extinguish 15 minutes after that.
> >
> > 30+15 = 45
>
>
> How do you know that the one with 2 ends will extinguish in 30
> minutes?
>
> Nick

If your initial assumtion that each rope takes an hour to burn, then
if you ignite both ends of a rope, it'll take thirty minutes to burn
out... but the rope might be long and burn really fast...


My theorem is that, in order to derive the initial system, you need >
50% of the original composition minus no less than 1 bit of an
entropic energy unit.








































































































































im drunk...  so disregard this part




























































































































oh yeah...



















if you can see this...






































































































































































you've scrolled down way too far.

bq

If you've had the chance to study arrow and chained arrow notatain,
then I'd like to put forward a few definitions.

From what I understand, googolplex is still the largest named number.
So in my lext to last post, I suggested that
googolplex -> googolplex -> googolplex is the largest number that has
no name.  This is the Googolplex'th Ackerman Number which I'd like to
name GAN

Define GAN -> GAN -> GAN as the GAN'th Ackerman Number = "GANAN"

GAN=GAN1
GANAN=GAN2

GANAN -> GANAN -> GANAN = GANANAN = GAN3

So;

GAN(n-1) -> GAN(n-1) -> GAN(n-1) = GAN(n)


which looks and sounds a lot like the ultimate one who has the
Triforce of Power in his right hand...






                                GANON



































































































































































---------
         /
        /
       /
      /
     /
    /
   /
  /
/
--------- ELDA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

This post didn't show up right on the site, but if you look in the
reply form, it'll make more snes.

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> If you've had the chance to study arrow and chained arrow notatain,
> then I'd like to put forward a few definitions.
>
> From what I understand, googolplex is still the largest named
number.
> So in my lext to last post, I suggested that
> googolplex -> googolplex -> googolplex is the largest number that
has
> no name.  This is the Googolplex'th Ackerman Number which I'd like
to
> name GAN
>
> Define GAN -> GAN -> GAN as the GAN'th Ackerman Number = "GANAN"
>
> GAN=GAN1
> GANAN=GAN2
>
> GANAN -> GANAN -> GANAN = GANANAN = GAN3
>
> So;
>
> GAN(n-1) -> GAN(n-1) -> GAN(n-1) = GAN(n)
>
>
> which looks and sounds a lot like the ultimate one who has the
> Triforce of Power in his right hand...
>
>
>
>
>
>
>                                GANON
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> ---------
>         /
>        /
>       /
>      /
>     /
>    /
>   /
>  /
> /
> --------- ELDA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Nick
    Because it will burn towards the middle and meet there.Since the whole rope
burns in an hour,half a rope in half an hour.
      Gary
   ----- Original Message -----
   From: nick_hobson
   To: mathforfun@yahoogroups.com
   Sent: Tuesday, October 07, 2003 17:57
   Subject: [MATH for FUN] Re: Rope burning puzzle


   --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
   wrote:
   > --- In mathforfun@yahoogroups.com, "nick_hobson"
   <nick_hobson@y...>
   > wrote:
   > > You have two lengths of rope, each of which takes exactly one
   hour
   > > to burn through from end to end. The ropes burn at an uneven
   > rate.
   > > So you can't, for instance, chop one of the ropes in half in
   order
   > > to measure 1/2 hour.
   > >
   > > You are asked to use the rope(s) and a lighter to measure 45
   > > minutes. Do you have enough information to succeed?
   > >
   > > Nick
   >
   > Light 3 ends simultaneously.   The one with 2 ends will extinguish
   > in 30 minutes.  At that moment, light the 4th end.   That rope
   will
   > extinguish 15 minutes after that.
   >
   > 30+15 = 45


   How do you know that the one with 2 ends will extinguish in 30
   minutes?

   Nick


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[Non-text portions of this message have been removed]

But that assumes more than the fact that each string takes 1 hour to
burn from end to end.

Consider the following rather weird piece of string.

A->->->->->->C<-<-<-<-<-<-B

Ignite at A: the flame takes 20 minutes to reach C, and a further 40
minutes to reach B.

Ignite at B: the flame takes 20 minutes to reach C, and a further 40
minutes to reach A.

This is consistent with the puzzle statement.

However, if we ignite the string at A and B simultaneously, the
flames meet at C after only 20 minutes!

You may ask why a string should burn faster in one direction than
the other.  Strictly speaking, of course, the reason is irrelevant;
the puzzle is already broken.  But there are plausible reasons for a
preferential burn direction.  Perhaps the string is resting on a
mound, so that it is easier for a flame to progress upward than
downward.

Or maybe the ends of the ">" and "<" symbols, in the diagram above,
represent frayed strands of rope, preferentially sticking out to the
left on the left hand side of the rope, and to the right on the
right hand side of the rope.  They may allow a flame to progress
faster by more easily igniting the frayed strands from one direction.

The bottom line is that there is no purely deductive solution.  We
need to know something about the physical properties of the rope.

Nick


--- In mathforfun@yahoogroups.com, "gflom" <gflom@p...> wrote:
> Nick
>    Because it will burn towards the middle and meet there.Since
the whole rope burns in an hour,half a rope in half an hour.
>      Gary
>   ----- Original Message -----
>   From: nick_hobson
>   To: mathforfun@yahoogroups.com
>   Sent: Tuesday, October 07, 2003 17:57
>   Subject: [MATH for FUN] Re: Rope burning puzzle
>
>
>   --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
>   wrote:
>   > --- In mathforfun@yahoogroups.com, "nick_hobson"
>   <nick_hobson@y...>
>   > wrote:
>   > > You have two lengths of rope, each of which takes exactly
one
>   hour
>   > > to burn through from end to end. The ropes burn at an uneven
>   > rate.
>   > > So you can't, for instance, chop one of the ropes in half in
>   order
>   > > to measure 1/2 hour.
>   > >
>   > > You are asked to use the rope(s) and a lighter to measure 45
>   > > minutes. Do you have enough information to succeed?
>   > >
>   > > Nick
>   >
>   > Light 3 ends simultaneously.   The one with 2 ends will
extinguish
>   > in 30 minutes.  At that moment, light the 4th end.   That rope
>   will
>   > extinguish 15 minutes after that.
>   >
>   > 30+15 = 45
>
>
>   How do you know that the one with 2 ends will extinguish in 30
>   minutes?
>
>   Nick
>
>
>         Yahoo! Groups Sponsor
>               ADVERTISEMENT
>
>
>
>
>   To unsubscribe from this group, send an email to:
>   mathforfun-unsubscribe@yahoogroups.com
>
>
>
>   Your use of Yahoo! Groups is subject to the Yahoo! Terms of
Service.
>
>
>
> [Non-text portions of this message have been removed]

Let AB be your rope, and let X be the point which the rope, if lit at
A, will have burned to in 30 minutes.

It follows from the question that the rope will burn from X to B in
30 minutes as well.  It does not matter where on the rope X is; AX
burns in 30 minutes; BX burns in 30 minutes.  The order of their
burning is irrelevant, and if both ends are lit at the same time,
both will reach X at the same time, meaning the rope is exhausted.

By the same logic, after this half-hour has elapsed, there will be 30
minutes' burning left in Rope 2.  And, by the same logic again,
lighting both ends of the remaining 30 minutes' of rope will cause it
to burn in 15 minutes.

--- In mathforfun@yahoogroups.com, "nick_hobson" <nick_hobson@y...>
wrote:
> --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> wrote:
> > --- In mathforfun@yahoogroups.com, "nick_hobson"
> <nick_hobson@y...>
> > wrote:
> > > You have two lengths of rope, each of which takes exactly one
> hour
> > > to burn through from end to end. The ropes burn at an uneven
> > rate.
> > > So you can't, for instance, chop one of the ropes in half in
> order
> > > to measure 1/2 hour.
> > >
> > > You are asked to use the rope(s) and a lighter to measure 45
> > > minutes. Do you have enough information to succeed?
> > >
> > > Nick
> >
> > Light 3 ends simultaneously.   The one with 2 ends will
extinguish
> > in 30 minutes.  At that moment, light the 4th end.   That rope
> will
> > extinguish 15 minutes after that.
> >
> > 30+15 = 45
>
>
> How do you know that the one with 2 ends will extinguish in 30
> minutes?
>
> Nick

Burn really fast... but so?  The whole rope burns in an hour anyhow.

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> If your initial assumtion that each rope takes an hour to burn,
then
> if you ignite both ends of a rope, it'll take thirty minutes to
burn
> out... but the rope might be long and burn really fast...
>
>
> My theorem is that, in order to derive the initial system, you need
>
> 50% of the original composition minus no less than 1 bit of an
> entropic energy unit.
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
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> im drunk...  so disregard this part
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> oh yeah...
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> if you can see this...
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> you've scrolled down way too far.
>
> bq

Whoops.. my scrollbar has fallen off!  I can't get back to the top of
the page!

So what is the numerical value of GAN?  And what value of n in (GAN
(n) is needed (a negative one, no doubt) to seduce the GAN series
back down to, say, 1?

"Gan", incidentally, is the Irish (Gaelic) word for "without".

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> If you've had the chance to study arrow and chained arrow notatain,
> then I'd like to put forward a few definitions.
>
> From what I understand, googolplex is still the largest named
number.
> So in my lext to last post, I suggested that
> googolplex -> googolplex -> googolplex is the largest number that
has
> no name.  This is the Googolplex'th Ackerman Number which I'd like
to
> name GAN
>
> Define GAN -> GAN -> GAN as the GAN'th Ackerman Number = "GANAN"
>
> GAN=GAN1
> GANAN=GAN2
>
> GANAN -> GANAN -> GANAN = GANANAN = GAN3
>
> So;
>
> GAN(n-1) -> GAN(n-1) -> GAN(n-1) = GAN(n)
>
>
> which looks and sounds a lot like the ultimate one who has the
> Triforce of Power in his right hand...
>
>
>
>
>
>
>                                GANON
>
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> ---------
>         /
>        /
>       /
>      /
>     /
>    /
>   /
>  /
> /
> --------- ELDA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

snes - nice one!

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> This post didn't show up right on the site, but if you look in the
> reply form, it'll make more snes.
>
> --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> > If you've had the chance to study arrow and chained arrow
notatain,
> > then I'd like to put forward a few definitions.
> >
> > From what I understand, googolplex is still the largest named
> number.
> > So in my lext to last post, I suggested that
> > googolplex -> googolplex -> googolplex is the largest number that
> has
> > no name.  This is the Googolplex'th Ackerman Number which I'd
like
> to
> > name GAN
> >
> > Define GAN -> GAN -> GAN as the GAN'th Ackerman Number = "GANAN"
> >
> > GAN=GAN1
> > GANAN=GAN2
> >
> > GANAN -> GANAN -> GANAN = GANANAN = GAN3
> >
> > So;
> >
> > GAN(n-1) -> GAN(n-1) -> GAN(n-1) = GAN(n)
> >
> >
> > which looks and sounds a lot like the ultimate one who has the
> > Triforce of Power in his right hand...
> >
> >
> >
> >
> >
> >
> >                                GANON
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > ---------
> >         /
> >        /
> >       /
> >      /
> >     /
> >    /
> >   /
> >  /
> > /
> > --------- ELDA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

For me, "each of which takes exactly one hour to burn through from
end to end" means "each of which will burn in one hour, regardless of
which end it's burned from".  Otherwise, the question would have said
so.

In that case, if it takes 20 minutes to burn from A to C and 40
minutes to burn from C to B, then it will take 40 minutes to burn
from B to C and 20 minutes to burn from C to A.  Like I said, if thie
weren't the case, the question would say so, I'm sure.



--- In mathforfun@yahoogroups.com, "nick_hobson" <nick_hobson@y...>
wrote:
> But that assumes more than the fact that each string takes 1 hour
to
> burn from end to end.
>
> Consider the following rather weird piece of string.
>
> A->->->->->->C<-<-<-<-<-<-B
>
> Ignite at A: the flame takes 20 minutes to reach C, and a further
40
> minutes to reach B.
>
> Ignite at B: the flame takes 20 minutes to reach C, and a further
40
> minutes to reach A.
>
> This is consistent with the puzzle statement.
>
> However, if we ignite the string at A and B simultaneously, the
> flames meet at C after only 20 minutes!
>
> You may ask why a string should burn faster in one direction than
> the other.  Strictly speaking, of course, the reason is irrelevant;
> the puzzle is already broken.  But there are plausible reasons for
a
> preferential burn direction.  Perhaps the string is resting on a
> mound, so that it is easier for a flame to progress upward than
> downward.
>
> Or maybe the ends of the ">" and "<" symbols, in the diagram above,
> represent frayed strands of rope, preferentially sticking out to
the
> left on the left hand side of the rope, and to the right on the
> right hand side of the rope.  They may allow a flame to progress
> faster by more easily igniting the frayed strands from one
direction.
>
> The bottom line is that there is no purely deductive solution.  We
> need to know something about the physical properties of the rope.
>
> Nick
>
>
> --- In mathforfun@yahoogroups.com, "gflom" <gflom@p...> wrote:
> > Nick
> >    Because it will burn towards the middle and meet there.Since
> the whole rope burns in an hour,half a rope in half an hour.
> >      Gary
> >   ----- Original Message -----
> >   From: nick_hobson
> >   To: mathforfun@yahoogroups.com
> >   Sent: Tuesday, October 07, 2003 17:57
> >   Subject: [MATH for FUN] Re: Rope burning puzzle
> >
> >
> >   --- In mathforfun@yahoogroups.com, slim_the_dude
<no_reply@y...>
> >   wrote:
> >   > --- In mathforfun@yahoogroups.com, "nick_hobson"
> >   <nick_hobson@y...>
> >   > wrote:
> >   > > You have two lengths of rope, each of which takes exactly
> one
> >   hour
> >   > > to burn through from end to end. The ropes burn at an
uneven
> >   > rate.
> >   > > So you can't, for instance, chop one of the ropes in half
in
> >   order
> >   > > to measure 1/2 hour.
> >   > >
> >   > > You are asked to use the rope(s) and a lighter to measure
45
> >   > > minutes. Do you have enough information to succeed?
> >   > >
> >   > > Nick
> >   >
> >   > Light 3 ends simultaneously.   The one with 2 ends will
> extinguish
> >   > in 30 minutes.  At that moment, light the 4th end.   That
rope
> >   will
> >   > extinguish 15 minutes after that.
> >   >
> >   > 30+15 = 45
> >
> >
> >   How do you know that the one with 2 ends will extinguish in 30
> >   minutes?
> >
> >   Nick
> >
> >
> >         Yahoo! Groups Sponsor
> >               ADVERTISEMENT
> >
> >
> >
> >
> >   To unsubscribe from this group, send an email to:
> >   mathforfun-unsubscribe@yahoogroups.com
> >
> >
> >
> >   Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
> >
> >
> >
> > [Non-text portions of this message have been removed]

For me, "each of which takes exactly one hour to burn through from
end to end" means "each of which takes exactly one hour to burn from
end A to end B or from end B to end A."

But the question was, do we have enough information to measure 45
minutes?

Nick


--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> For me, "each of which takes exactly one hour to burn through from
> end to end" means "each of which will burn in one hour, regardless
of
> which end it's burned from".  Otherwise, the question would have
said
> so.
>
> In that case, if it takes 20 minutes to burn from A to C and 40
> minutes to burn from C to B, then it will take 40 minutes to burn
> from B to C and 20 minutes to burn from C to A.  Like I said, if
thie
> weren't the case, the question would say so, I'm sure.
>
>
>
> --- In mathforfun@yahoogroups.com, "nick_hobson"
<nick_hobson@y...>
> wrote:
> > But that assumes more than the fact that each string takes 1
hour
> to
> > burn from end to end.
> >
> > Consider the following rather weird piece of string.
> >
> > A->->->->->->C<-<-<-<-<-<-B
> >
> > Ignite at A: the flame takes 20 minutes to reach C, and a
further
> 40
> > minutes to reach B.
> >
> > Ignite at B: the flame takes 20 minutes to reach C, and a
further
> 40
> > minutes to reach A.
> >
> > This is consistent with the puzzle statement.
> >
> > However, if we ignite the string at A and B simultaneously, the
> > flames meet at C after only 20 minutes!
> >
> > You may ask why a string should burn faster in one direction
than
> > the other.  Strictly speaking, of course, the reason is
irrelevant;
> > the puzzle is already broken.  But there are plausible reasons
for
> a
> > preferential burn direction.  Perhaps the string is resting on a
> > mound, so that it is easier for a flame to progress upward than
> > downward.
> >
> > Or maybe the ends of the ">" and "<" symbols, in the diagram
above,
> > represent frayed strands of rope, preferentially sticking out to
> the
> > left on the left hand side of the rope, and to the right on the
> > right hand side of the rope.  They may allow a flame to progress
> > faster by more easily igniting the frayed strands from one
> direction.
> >
> > The bottom line is that there is no purely deductive solution.
We
> > need to know something about the physical properties of the rope.
> >
> > Nick
> >
> >
> > --- In mathforfun@yahoogroups.com, "gflom" <gflom@p...> wrote:
> > > Nick
> > >    Because it will burn towards the middle and meet
there.Since
> > the whole rope burns in an hour,half a rope in half an hour.
> > >      Gary
> > >   ----- Original Message -----
> > >   From: nick_hobson
> > >   To: mathforfun@yahoogroups.com
> > >   Sent: Tuesday, October 07, 2003 17:57
> > >   Subject: [MATH for FUN] Re: Rope burning puzzle
> > >
> > >
> > >   --- In mathforfun@yahoogroups.com, slim_the_dude
> <no_reply@y...>
> > >   wrote:
> > >   > --- In mathforfun@yahoogroups.com, "nick_hobson"
> > >   <nick_hobson@y...>
> > >   > wrote:
> > >   > > You have two lengths of rope, each of which takes
exactly
> > one
> > >   hour
> > >   > > to burn through from end to end. The ropes burn at an
> uneven
> > >   > rate.
> > >   > > So you can't, for instance, chop one of the ropes in
half
> in
> > >   order
> > >   > > to measure 1/2 hour.
> > >   > >
> > >   > > You are asked to use the rope(s) and a lighter to
measure
> 45
> > >   > > minutes. Do you have enough information to succeed?
> > >   > >
> > >   > > Nick
> > >   >
> > >   > Light 3 ends simultaneously.   The one with 2 ends will
> > extinguish
> > >   > in 30 minutes.  At that moment, light the 4th end.   That
> rope
> > >   will
> > >   > extinguish 15 minutes after that.
> > >   >
> > >   > 30+15 = 45
> > >
> > >
> > >   How do you know that the one with 2 ends will extinguish in
30
> > >   minutes?
> > >
> > >   Nick
> > >
> > >
> > >         Yahoo! Groups Sponsor
> > >               ADVERTISEMENT
> > >
> > >
> > >
> > >
> > >   To unsubscribe from this group, send an email to:
> > >   mathforfun-unsubscribe@yahoogroups.com
> > >
> > >
> > >
> > >   Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> > Service.
> > >
> > >
> > >
> > > [Non-text portions of this message have been removed]

Just so long as burn direction is irrelevant, the answer is yes.  If you set
each end to burn on
one rope, the rope will stop burning when the end meet, which is necessarily at
the 30-minute
mark.  If it isn't (either longer or shorter than 30-minutes), then the rope
wasn't an hour-long
rope.  With a second rope having burnt from one end for 30 minutes, the rope is
now a 30-minute
rope, with the same properties as the hour-long rope.  With burn direction being
irrelevant, this
rope will necessarily take 15 minutes to burn when the second end is lit (using
the same
argument).

If I remember the original question correctly, you asked that initial
assumptions be included with
the solution.  I think the only initial assumption you need is that burn
direction of the ropes is
irrelevant.


--- nick_hobson <nick_hobson@...> wrote:
> For me, "each of which takes exactly one hour to burn through from
> end to end" means "each of which takes exactly one hour to burn from
> end A to end B or from end B to end A."
>
> But the question was, do we have enough information to measure 45
> minutes?
>
> Nick
>
>
> --- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> > For me, "each of which takes exactly one hour to burn through from
> > end to end" means "each of which will burn in one hour, regardless
> of
> > which end it's burned from".  Otherwise, the question would have
> said
> > so.
> >
> > In that case, if it takes 20 minutes to burn from A to C and 40
> > minutes to burn from C to B, then it will take 40 minutes to burn
> > from B to C and 20 minutes to burn from C to A.  Like I said, if
> thie
> > weren't the case, the question would say so, I'm sure.
> >
> >
> >
> > --- In mathforfun@yahoogroups.com, "nick_hobson"
> <nick_hobson@y...>
> > wrote:
> > > But that assumes more than the fact that each string takes 1
> hour
> > to
> > > burn from end to end.
> > >
> > > Consider the following rather weird piece of string.
> > >
> > > A->->->->->->C<-<-<-<-<-<-B
> > >
> > > Ignite at A: the flame takes 20 minutes to reach C, and a
> further
> > 40
> > > minutes to reach B.
> > >
> > > Ignite at B: the flame takes 20 minutes to reach C, and a
> further
> > 40
> > > minutes to reach A.
> > >
> > > This is consistent with the puzzle statement.
> > >
> > > However, if we ignite the string at A and B simultaneously, the
> > > flames meet at C after only 20 minutes!
> > >
> > > You may ask why a string should burn faster in one direction
> than
> > > the other.  Strictly speaking, of course, the reason is
> irrelevant;
> > > the puzzle is already broken.  But there are plausible reasons
> for
> > a
> > > preferential burn direction.  Perhaps the string is resting on a
> > > mound, so that it is easier for a flame to progress upward than
> > > downward.
> > >
> > > Or maybe the ends of the ">" and "<" symbols, in the diagram
> above,
> > > represent frayed strands of rope, preferentially sticking out to
> > the
> > > left on the left hand side of the rope, and to the right on the
> > > right hand side of the rope.  They may allow a flame to progress
> > > faster by more easily igniting the frayed strands from one
> > direction.
> > >
> > > The bottom line is that there is no purely deductive solution.
> We
> > > need to know something about the physical properties of the rope.
> > >
> > > Nick
> > >
> > >
> > > --- In mathforfun@yahoogroups.com, "gflom" <gflom@p...> wrote:
> > > > Nick
> > > >    Because it will burn towards the middle and meet
> there.Since
> > > the whole rope burns in an hour,half a rope in half an hour.
> > > >      Gary
> > > >   ----- Original Message -----
> > > >   From: nick_hobson
> > > >   To: mathforfun@yahoogroups.com
> > > >   Sent: Tuesday, October 07, 2003 17:57
> > > >   Subject: [MATH for FUN] Re: Rope burning puzzle
> > > >
> > > >
> > > >   --- In mathforfun@yahoogroups.com, slim_the_dude
> > <no_reply@y...>
> > > >   wrote:
> > > >   > --- In mathforfun@yahoogroups.com, "nick_hobson"
> > > >   <nick_hobson@y...>
> > > >   > wrote:
> > > >   > > You have two lengths of rope, each of which takes
> exactly
> > > one
> > > >   hour
> > > >   > > to burn through from end to end. The ropes burn at an
> > uneven
> > > >   > rate.
> > > >   > > So you can't, for instance, chop one of the ropes in
> half
> > in
> > > >   order
> > > >   > > to measure 1/2 hour.
> > > >   > >
> > > >   > > You are asked to use the rope(s) and a lighter to
> measure
> > 45
> > > >   > > minutes. Do you have enough information to succeed?
> > > >   > >
> > > >   > > Nick
> > > >   >
> > > >   > Light 3 ends simultaneously.   The one with 2 ends will
> > > extinguish
> > > >   > in 30 minutes.  At that moment, light the 4th end.   That
> > rope
> > > >   will
> > > >   > extinguish 15 minutes after that.
> > > >   >
> > > >   > 30+15 = 45
> > > >
> > > >
> > > >   How do you know that the one with 2 ends will extinguish in
> 30
> > > >   minutes?
> > > >
> > > >   Nick
> > > >
> > > >
> > > >         Yahoo! Groups Sponsor
> > > >               ADVERTISEMENT
> > > >
> > > >
> > > >
> > > >
> > > >   To unsubscribe from this group, send an email to:
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--- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@y...>
wrote:
> Just so long as burn direction is irrelevant, the answer is yes.
If you set each end to burn on
> one rope, the rope will stop burning when the end meet, which is
necessarily at the 30-minute
> mark.

Nick's example shows this is not necessary.

> If it isn't (either longer or shorter than 30-minutes), then the
rope wasn't an hour-long
> rope.  With a second rope having burnt from one end for 30 minutes,
the rope is now a 30-minute
> rope, with the same properties as the hour-long rope.

This assumption is not part of the problem. Only the original rope
has the property that the burn time is the same no matter which end
is lit. The problem does not say that portions of the rope also have
this property.

> With burn direction being irrelevant, this
> rope will necessarily take 15 minutes to burn when the second end
is lit (using the same
> argument).
>
> If I remember the original question correctly, you asked that
initial assumptions be included with
> the solution.  I think the only initial assumption you need is that
burn direction of the ropes is
> irrelevant.

The way I understand Nick's example is this: take an ordinary rope
(which burns at a constant rate) and lay it atop a traditional
triangular (nonflammable) rooftop so that half the rope hangs down
each side of the peak. Since heat rises, the rope will burn quickly
at first as the flame climbs to the peak of the roof. Then as the
flame tries to crawl down the other side of the roof, the rate at
which the rope burns will decrease. By symmetry, the total burn time
is the same no matter which end of the rope is lit first (i.e. burn
direction is irrelevant). On the other hand, if we only consider a
portion of the rope (e.g. the part that goes from one end to the
peak), the total burn time will depend on which end we light first.

By choosing the length of the rope appropriately, we can set it up so
that this rope burns in one hour, regardless of which end we light.
If we then light both ends, the rope will burn up in less than 30
minutes. If we only light one end and let it burn for 30 minutes, we
will be left with a portion of rope that burns in 30 minutes from one
direction but burns in less than 30 minutes from the other, so that
the remaining portion does not have the "irrelevant direction"
property.

>
>
> --- nick_hobson <nick_hobson@y...> wrote:
> > For me, "each of which takes exactly one hour to burn through
from
> > end to end" means "each of which takes exactly one hour to burn
from
> > end A to end B or from end B to end A."
> >
> > But the question was, do we have enough information to measure 45
> > minutes?
> >
> > Nick
> >
> >
> > --- In mathforfun@yahoogroups.com, clooneman <no_reply@y...>
wrote:
> > > For me, "each of which takes exactly one hour to burn through
from
> > > end to end" means "each of which will burn in one hour,
regardless
> > of
> > > which end it's burned from".  Otherwise, the question would
have
> > said
> > > so.
> > >
> > > In that case, if it takes 20 minutes to burn from A to C and 40
> > > minutes to burn from C to B, then it will take 40 minutes to
burn
> > > from B to C and 20 minutes to burn from C to A.  Like I said,
if
> > thie
> > > weren't the case, the question would say so, I'm sure.
> > >
> > >
> > >
> > > --- In mathforfun@yahoogroups.com, "nick_hobson"
> > <nick_hobson@y...>
> > > wrote:
> > > > But that assumes more than the fact that each string takes 1
> > hour
> > > to
> > > > burn from end to end.
> > > >
> > > > Consider the following rather weird piece of string.
> > > >
> > > > A->->->->->->C<-<-<-<-<-<-B
> > > >
> > > > Ignite at A: the flame takes 20 minutes to reach C, and a
> > further
> > > 40
> > > > minutes to reach B.
> > > >
> > > > Ignite at B: the flame takes 20 minutes to reach C, and a
> > further
> > > 40
> > > > minutes to reach A.
> > > >
> > > > This is consistent with the puzzle statement.
> > > >
> > > > However, if we ignite the string at A and B simultaneously,
the
> > > > flames meet at C after only 20 minutes!
> > > >
> > > > You may ask why a string should burn faster in one direction
> > than
> > > > the other.  Strictly speaking, of course, the reason is
> > irrelevant;
> > > > the puzzle is already broken.  But there are plausible
reasons
> > for
> > > a
> > > > preferential burn direction.  Perhaps the string is resting
on a
> > > > mound, so that it is easier for a flame to progress upward
than
> > > > downward.
> > > >
> > > > Or maybe the ends of the ">" and "<" symbols, in the diagram
> > above,
> > > > represent frayed strands of rope, preferentially sticking out
to
> > > the
> > > > left on the left hand side of the rope, and to the right on
the
> > > > right hand side of the rope.  They may allow a flame to
progress
> > > > faster by more easily igniting the frayed strands from one
> > > direction.
> > > >
> > > > The bottom line is that there is no purely deductive
solution.
> > We
> > > > need to know something about the physical properties of the
rope.
> > > >
> > > > Nick

Yip

--- In mathforfun@yahoogroups.com, "nick_hobson" <nick_hobson@y...>
wrote:
> For me, "each of which takes exactly one hour to burn through from
> end to end" means "each of which takes exactly one hour to burn
from
> end A to end B or from end B to end A."
>
> But the question was, do we have enough information to measure 45
> minutes?
>
> Nick
>

How do you gain access to this group?
  Thanks

     Gary
   ----- Original Message -----
   From: tadeh_davtian
   To: mathforfun@yahoogroups.com
   Sent: Thursday, October 02, 2003 20:07
   Subject: [MATH for FUN] MathematicOlympiad [[ YahooGroup ]] .


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