Conference rusure::math

The one that I have trouble with is the claim that (and I hope I get it
right):

Tarski proved that a single sphere could be sectioned into pieces in such a
way that two spheres identical to the original could be formed.

I'll post the source when I find it.  I don't have the proof, but would be
intrigued to see one.

    ref .-1
    you must mean a solid sphere, and when you say idetical, you mean 
    two new spheres, but the two new spheres not all solid (will be
    partially hollow inside) , but they have same surface area as original, 
    that would be possible offcourse, but no way one could make 2 spheres 
    both solid from one solid, and both still have same surface area as 
    original !
    my intuituition would not accept that, never mind the law of
    conversation of mass ! 
    
    /naser
    
    
    
    
    

RE: .2 (/naser)

    No, this is not impossible -- just not practical in the real world.
    Yes, the initial "sphere" is a solid one (its actually about the union
    of a sphere and its interior, rather than a sphere), but no the final
    spheres "really" are  identical -- they do not have any hollow places
    inside.  The key is that the pieces that you remove and reassemble are
    not "ordinary" -- they are disconnected fractals.  The proof of this
    requires the Axiom of Choice which, though intuitively obvious (in
    non-technical language, it just says that if you "have" a set you can
    always "choose" or reference one of its elements), and consistent with
    other set theoretic axioms, leads to many counterintutive results so
    some mathematians feel that it "aught" not to be used.  For this
    reason, theorems which depend on it for their proof are usually labeled
    that way.

    It simply turns out that, in three dimensional space, you can cut up
    any shape into some number of pieces (generally very large) -- many
    consisting of fractal dust as I understand it -- and then move those
    pieces around and rotate them (with each disconnected part of a piece
    holding its relation to the other parts of the piece throughout the
    movement/rotation) and slide them together again so that the gaps
    in one fill up the gaps in the others.  The result is that (there may
    be some restrictions) any solid can be turned into any other solid
    or set of solids, regardless of volume.

    What about mass conservation?  That concerns discrete matter, not
    infinitely divisible points.  The number of *points* in two spheres
    of equal volume is the same as the number of points in one of those
    two spheres so there is no extra points appearing out of nowhere.

    There was a good "popular" article about this recently.  I'll try
    to find it.

					Topher

        The theorem as I remember it is that you can take a solid
        sphere (treated as the set of points (x,y,z) in R^3
        satisfying x^2 + y^2 + z^2 <= 1), partition it into a
        finite number of sets, and move those (translations,
        rotations) sets so that they recombine into two solid
        spheres of the same radius as the original.
        
        Some of the pieces can best be described as "sets of
        points."
        
        Dan

The following is (of course) not a proof, but a claim that there is a proof:

{ Extracted w.o. permission from 'Tomorrow's Math' by C. Stanley Ogilvie,
c. 1962, Oxford University Press }

p. 7
' .. Other difficulties arise in connection with the volume of solids.
Intuitively, it would seem that if you start with a given finite solid object,
cut it up into a finite number of pieces, and then reassemble the same pieces in
any way, provided there are no spaces anywhere between the reassembled pieces
the new solid should have the same volume as the original.

In 1924 Banach and Tarski showed that this is not necessarily so by proving
the following extraordinary theorem: it is possible to cut a solid sphere into
a finite number of pieces and reassemble them by rigid motions (no distortion)
to form two solid spheres (no holes), each of the same size as the original one.
It might be supposed that the number of pieces in the dissection would have to
be very large.  But Raphael Robinson has shown that there need be no more than
5 pieces !  These unbelievable results show that we must revise our fundamental
notion of volume.  There cannot be any general definition which will preserve
volume under rigid motion -- something always previously accepted as "obvious"
..'

Here are the pointers:

'7. The Banach-Tarksi theorem was published in Fundamenta Mathematicae, Vol. 6
(1924), p. 244.

7. Raphael Robinson, Fundamenta Mathematicae, Vol. 34(1947), p. 246.
See also Amer. Math. Monthly, Vol. 55(1948), p. 459.'

I could never ever believe this even if i see a proof for it ,
(assuming i understood the proof :-)
this sounds like proving the grand unified field theory, except
we have to accept we live in 10 dimensional space-time continuum.

i've been struggling with a homework in Riemann surfaces, where you 
connect two or more sheets along the z-values (cut branch) where the 
branches of the original multi-valued complex function becomes discontinuous, 
so as to construct one continuous function by jumping from sheet (domain) 
to another when you hit the cut branch, however these Riemann surfaces 
cannot be constructed physically since the surfaces cross over along the cut,
and the author says , only mathematically this is possible, physically you 
cant do it, this sphere deal sounds something like Riemann surfaces, 
out of this world :-)

/naser

    

    See also topic 325.
    
    
    				-- edp

RE: .3 (me)

    The article is:

	Stewart, Ian; "The Ultimate Jigsaw Puzzle" _New Scientist_, Vol 130,
	#1764 (13 April, 1991) pp 30-33.

    This is the same Ian Stewart who has been writing Mathematical Games
    for Sci. Am. of late.

    The article isn't "really" about the Banach-Tarski paradox/theorem but
    about a related 2D problem referred to in the article as "Tarski's
    circle-squaring" problem.  This asks whether a circle (including its
    interior) can be divided into a finite number of pieces which can be
    rigidly moved around and be assembled into a square of equal area (or,
    of course, vice versa).  Most of the article, however, is devoted to
    providing background -- which is mostly the Banach_Tarski theorem and
    its resulting paradox.  (The answer, by the way, is that it has
    recently been found that the answer is that you *can* disect the circle
    into a square -- and you can simply rearrange the pieces without
    rotations to accomplish it.  It does require about 10^50 pieces to
    do the trick, however.  The result is general, any two shapes whose
    boundaries consist of smoothly curved pieces can be disected into each
    other.)

				    Topher

    Re: .8 (Topher)
    I guess I must see the article to see how they go around that
    fact that squaring the circle contradicts the fact the Pi is
    irrational. 

>> that squaring the circle contradicts the fact that Pi is irrational.
        
        That Pi isn't reached from the rationals by successive
        quadratic extensions doesn't contradict the result.  It
        just means that the pieces can't be cut out of the circle
        with standard "Euclidean" compass, straightedge, and
        scissors.
        
        Dan

    this little one might belong in here too:
    the series {1/n} as n->oo , diverges. 
    i think most students who see this first time, and without any analysis 
    done, would say it converges, not diverges. 
    

    Here's one that I found surprising when I first encountered it.
    
    If buses arrive at a bus stop such that the average time between
    arrivals is, say, 10 minutes then the expected time you'll have to wait
    for a bus is not necessarily 5 minutes. In fact it can be made arbitrarily
    large. I guess I have a certain empathy with this result because I catch
    the bus to work each day.

    If average time between arrivals is X, should not the expected time
    to wait for a bus be X too?
    (pepole dont use buses much in the states, so i could be wrong :-)

    RE .13
    
      The intuitive 5 minutes is because one assumes average means the
      busses come at uniform 10 minute intervals.  If this were the case
      the five minutes makes sense because you would have a long wait of
      ten minutes and a short wait of zero averageing out at five.
    
      What I assumed the author was getting at was that the ten minute
      average time does not disclose information about the distribution
      of bus arivals.  If in every hour you get six busses, at the hour
      and the next five one minute intervals you average a bus every
      ten minutes but might have to wait as long a 55 minutes.
    
    -Joe
    

    re .13,.14
    
    Once you realise that the expected waiting time might depend on more
    than just the mean of the inter-arrival time distribution, it still isn't
    intuitively obvious what the effect of differing distributions will be.
    
    The way it was explained to me was that you are more likely to arrive
    during a long gap between buses and thus have to wait a long time (while
    several buses close together are pointless because if you miss one you'll
    miss several). Thus any departure from a rigid cycle of a bus every 10
    minutes increases the expected waiting time.
    
    To pick up the example of .14, if k buses arrive every 10k minutes in a
    fixed cycle of buses one every minute from the start of the 'cycle'
    until the k buses have arrived and then nothing more for the rest of
    the cycle, then assuming I got my calculation right the expected
    waiting time is (81k+19)/20. Setting k=1 gives the intuitive answer of
    5 minutes for that case. For the actual example of .14, k=6 gives an
    expected waiting time of 25.25 minutes.
    
    Fixed cycles of buses are all very well but any smart passenger will
    find out the cycle and then can at least wait at home.
    
    Suppose the inter-arrival time is a continuous random variable with
    probability density function (pdf) p(x) = k.exp(-kx) then it can easily
    be shown that the average time between buses is 1/k. This distribution
    has the interesting property of being memoriless i.e. roughly speaking,
    no matter how long since the last bus the average time still remaining
    before the next bus is the same (I know the feeling).
    
    Consequently the expected waiting time is 1/k i.e. twice what intuition
    would suggest (and it's as if you always just missed one).
    
    (I hope I got the details right. It's been a long time. I stand ready
     to be corrected.)

    That's as good an explanation as I ever got. :-)
    
    Of course, the NMWTP (no memory waiting time process) is just one of
    the many alternatives you might assume.  Only in that case does the
    logic hold true.  On the other hand, did you ever hear of a bus with a
    memory?  :-)
    
    Dick

    re .-1
    
>    Of course, the NMWTP (no memory waiting time process) is just one of
>    the many alternatives you might assume.
    
    Certainly.
    
    As an alternative, suppose that the inter-arrival time is uniformly
    distributed on [0,20] (so the mean is still 10), then it can be shown
    that the expected time you'll have to wait for the bus is 10 * 2/3
    i.e. somewhere between the 'intuitive' answer of 5 and the 'memoriless'
    answer of 10.
    
>    Only in that case does the logic hold true.
    
    Yes. The given pdf is a special case... a case of at least theoretical
    interest (but rather inappropriate for this problem).
    
    Note that even though I just asserted that the given pdf was
    'memoriless' and hand-waved that this meant that mean waiting time =
    mean inter-arrival time, the mean waiting time can be found directly
    from the pdf (if you like integration by parts for this particular pdf).
    
>    On the other hand, did you ever hear of a bus with a memory?  :-)
    
    The buses may not have memory but they are controlled and scheduled by
    an organisation that does.

    re 1456.17 by ELIS::GARSON 

>The buses may not have memory but they are controlled and scheduled by
>an organisation that does.

    <acid rain on>
    
    Sorry, my parochial experience is limited to bus authorities run by
    political hacks who wouldn't know a schedule if it bit them.  :-(
    
    <acid rain off>
    
    
    Dick

================================================================================
Note 1472.0                  Also true for elevators                   2 replies
HGOVC::JOELBERMAN                                    15 lines  21-JUL-1991 02:44
--------------------------------------------------------------------------------
    I think elevators (lifts) exhibit the same symptoms.  Most modern
    elevator controls have scheduling algorithms that try to keep them on
    the ground floor when people are coming to work in the morning and
    after lunch, and then keep them spaced vertically during the lunch hour and
    quitting time.  
    
    I go to many taller buildings and have noticed an enormous difference
    in waiting times.  SOme of this may be explained by different speed
    elevators, but I suspect that some companies did not apply any
    mathematice to the scheduling algorithms.
    
    Is there a general solution to the minimum latency elevator ( or bus)
    problem?
    
    /joel	

================================================================================
Note 1472.1                  Also true for elevators                      1 of 2
CIVAGE::LYNN "Lynn Yarbrough @WNP DTN 427-5663"       5 lines  25-JUL-1991 14:27
                                    -< Eh? >-
--------------------------------------------------------------------------------
I think the elevator problem is discussed in Knuth TAOCP Vol I.

I also think I detect a *reply* disguised as a *write*. If the moderator 
could straighten this out (and delete this note) we might get back on 
track. Or am I missing something?

this might be trivial to some of you, but i read something that
if the ratio of the nth term of two sequences is +1 as n increases with no 
limit, and one of the sequences if convergent, the other could still be 
divergent !

seemed counter intuitive to me , but they give example : (hit CR)



 (-1)^n
 --------     other sequence    (-1)^n
  SQRT(n)                      ------- ( 1+ (-1)^n   )
                               SQRT(n)      ------
                                             SQRT(n)

 the first in convergent, the other is divergent, yet ratio of their 
 n terms is +1 as n->oo .

 this is interesting since it is counter intutive at first, right? 
 i think in convergence problems one cant talk anything for granted !

/nasser
p.s reference is "George Conter, his mathematical and philosphy of the
infinite" by joseph Dauben, princeton univ. press, 1979
 
    
    

    
     
    my cranky intution told me that the fourier series aproximation to
    COS(t) should come out to be COS(t) (i.e. one term), but i keep getting
    the a(k) coeff. zero, and the b(k) coeff. zero, so the Fourier 
    series is 0. (no terms)
    
    f(t)= a0/2  +  sum(1..oo)(  a(k)cos(kt) + b(k)sin(kt) )
    
    a0= 1/Pi Intgral(-Pi..Pi) of cos(t)  = 0
    
    a(k)= 1/Pi Integral(-Pi..Pi) of  ( cos(t) cos(kt) ) = 0
    
    b(k)= 1/Pi Integral(-Pi..Pi) of ( cos(t) sin(kt) ) = 0
    
    so f(t) = 0
    
    i either done something wrong in calcualtions even though i double
    checked few times, or completely confused ,or both (most likely).
    
    any one see where i went wrong?
    thanks,
    /nasser
    p.s. i must reinstall MAPLE on my workstation again sometime !

re .21
    
>if the ratio of the nth term of two sequences is +1 as n increases with no 
>limit, and one of the sequences if convergent, the other could still be 
>divergent !

You mean *series* not sequences.

If lim   a  exists, call it a, and lim   b / a  exists and equals 1
   n->oo  n                        n->oo  n   n

Then lim   b  exists and equals a
     n->oo  n

as is reasonably intuitive.



If lim   b / a  exists and equals 1 and the series a converges, say to s, then
   n->oo  n   n

you are correct in observing that the series b need not converge (and if it
does converge the limit need not be s).

>seemed counter intuitive to me

I wouldn't trust my intuition. Try some algebra...

  n          n                    n
sigma b  = sigma a � (b / a ) = sigma a � (1 + b / a  - 1)
 i=1   i    i=1   i    i   i     i=1   i        i   i

             n          n
         = sigma a  + sigma a � (b / a  - 1)
            i=1   i    i=1   i    i   i

Now by assumption the series a converges, say to s, and it is a well known
result that therefore the sequence a converges to 0. Also by assumption the
sequence b / a  converges to 1.
          n   n

Thus a � (b / a  - 1) converges to 0 since both factors do. This does not
      n    n   n

however mean that the series converges, the most well-known example be 1/n.
[a sequence converging to 0 is a necessary but not sufficient condition for
the series converging]
If this series converges to a non-zero value then the series b will actually
converge but to a value other than s.

>, but they give example : (hit CR)
>
> (-1)^n
> --------     other sequence    (-1)^n
>  SQRT(n)                      ------- ( 1+ (-1)^n   )
>                               SQRT(n)      ------
>                                             SQRT(n)

And surprise surprise evaluating a � (b / a  - 1) for these sequences gives
                                  n    n   n

1/n.

> i think in convergence problems one cant take anything for granted !

Agreed.

If you can't prove it rigorously then it might not be correct (which of course
applies to all mathematics).

    ref .23 
    thanks, your right and my faith in fourier series has been restored.
    the dumb mistake i did was during evaluating a(k) one gets this
    expression
    
    a(K)= 1/(1-k) ( sin(1-k)Pi + sin(1-k)Pi ) + ..same but with (1+k)..
    
    so my brain lied to me and said :for all k=0..oo the sin(1-k) are zero!
    so a(k) is zero for all k, offcourse i did not notice that k=1 is
    speical case because of 0/0 term, and k=1 must be evaluated separtly.
    and i fell in this before, gee wiz, i'll tie something on my finger from 
    now on to remind me about this one.
    well, this goes to show that even great mathematicians make mistakes :-)
    
    /nasser
    

    
    ref .24, thanks, i'll go over what you wrote.
    
    i'll blupper again something from my cranky intution without
    tinking too much about it: if the *sequence* is convergent then the
    the partial sum *series* must also be convergent. this got to be
    correct, but i'll take your advice and try to prove rigo'rsly first.
    
    i just like to solve problems intutivly first, then see if it is 
    correct, this way may be i dont have to do the thing i fear most in my 
    life: proofs ! 
    
    /nasser

                      <<< Note 1456.26 by STAR::ABBASI >>>
                          -< intution on convergence >-

...
   
>    i just like to solve problems intutivly first, then see if it is 
>    correct, this way may be i dont have to do the thing i fear most in my 
>    life: proofs ! 

...

My approach is different.  When my intuition at one level is weak, I use 
calculation or proof to change to a different level of abstraction where my
intuition is stronger.  I _never_ trust my intuition where infinities or 
existence proofs are concerned.

Dick

    this is interesting are, to see how math people think about problems,
    i got this book that talks about this, some famouse maths people are
    investigated and asked questions to analyse their process of thinking.
    
    i'll put the reference here later, it is a good reading material.
    
    but i think, one must have an intution on what the solution looks like,
    befor doing the calculations, right?.
    maybe be this is too general topic to discuess with any concrete ideas.
    (my intution tells me so :-)
    
    /nasser

    this is a nice little book that touches on intuition in math among
    other things:
    "the psychology of invention in the mathematical field" By Jacques
    Hadamard. Dover edition.1954. (QA 9 H25)
    contents: General views and inquiries
              discussions on unconsiousness "they talk about how
    the unconsiouse mind could be working on the problem without you
    reliazing it"
              the unconscious and discovery
              the preparation stage. logic and chance
              the later concious work.  
              discovery as a synthesis. the help signs
              Different Kinds of mathematical minds
              paradoxical cases of intuition
              general direction of research
    
    the general theme of the book is "how creativity is tapped in science.
    the unconscious mind and discovery. intuition vs. Verbal resoning.
    POINCARE's forgetting hypothesis. creative techniques of EINSTEIN,
    PASCAL, WIENER and others"
    
    i figured, when i finish reading this book, may i'll finally figure the 
    trick to how be good at math... nothing else seems to help...
    
    /nasser
           
    

re .29

I enjoyed that book myself many years ago.  To improve your math
performance, I would recommend Polya's "How to Solve It".  Sorry, I don't
have any publisher information.  I've found that the first hurdle to solving
math problems is just as Polya describes it, "having the problem".

What he means is that you must feel that you "own" the problem. You are
responsible, not to someone else, but to yourself for the solution.  In a
sense, you have to become obsessed with the problem.  And yet, you must
approach it without thinking of the consequences of success or failure.  

So Polya's prescription (and mine, for what its worth) is

   1) Own the problem like your life depended on it.
   
   2) Remember its a game, not life or death.
   
I know it's paradoxical, but it works.

Dick

       
        / 
     I= ) 1/x dx 
       /
    
    integration by parts, let u=1/x , dv=dx
    then du= -1/x^2 dx, v=x
    
                 /             /
    then I = uv- ) v du  = 1 - ) x(-1/x^2) dx= 1+ I
                /             /
    
    so I= 1+I => 1=0
    
    /Nasser
    

        2
       /          
    I= ) 1/x dx 
      /           
      0
    
    let t= x/2, then x=2t, dx=2dt =>  
    
        2          1            1        1
       /           /           /        /
    I= ) 1/x dx  = ) 2 dt/2t = ) dt/t = ) dx/x (since dt=dx/2 & t=x/2)
       /          /            /        /
      0           0            0        0
    
        2        1
       /        /
    so ) dx/x - ) dx/x = 0
       /        /
      0         0
    
                  2                      2
                 /                      |
    this means   ) dx/x = 0  i.e   ln x |  = 0 , i.e.  ln 2 =0,
                 /                      |
                 1                       1
    
    i.e. e^0=2, i.e. 1=2   
    
    /nasser

    if a function is bounded => it is integrable (Riemman)
    
    true or false?
    
    if false, can you give an example of a bounded function that is
    not integrable?
    
    /Nasser

    
    No,
    
    let f(x)=0 if x ix irrational, 1 if x is rational. Bounded, but
    not integrable.
    
    /�ke

    Eh, I mean not Riemann integrable. Lebesgue integrable, yes.
    

    yes, that is the one example i knew off too, any one knows of other
    examples?
    thanks,
    /nasser

        A bounded function f:[a,b] -> R is Riemann integrable if
        and only if it is continuous almost everywhere, i.e.,
        except for a set of Lebesgue measure zero.
        
        Dan