HDTenEightyP/NetNews huggingface dataset:
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Newsgroups: rec.puzzles
Subject: Re: Random Points on a Sphere
Date: 22 Nov 1992 23:12 CST
Organization: Texas A&M University, Academic Computing Services
Distribution: world
NNTP-Posting-Host: zeus.tamu.edu
News-Software: VAX/VMS VNEWS 1.41    

matt@severian.chi.il.us (Matt Crawford) writes...
>That's two questions in a row dealing with randomly selected points
>in or on a sphere and neither one stated the distribution.  Assuming
>uniform distribution, I have a half-answer to this one:

I get 21/20 for the other one, but the solution is not enlightening.

>>Four points are randomly selected from the surface of a sphere.
>>What is the probability that all four lie in the same hemisphere?
> 
>If A is the solid angle subtended by 3 points on the sphere, the
>probability that a 4th point chosen uniformly on the sphere is, uh,
>cohemispheric with the 3 is 1-A/4pi if A!=2pi; 1 if A=2pi.

Neat. So we need the expectation value of A. I think it is pi/2.
Let B be chosen along the equator and along the prime
meridian. Let C be chosen along the equator, and let D be chosen in
the northern hemisphere. The great circles BD and CD divide the hemisphere
into 4 regions which can be shown to have equal expected areas. But the
expected areas must add to 2pi.

So the final probability is: 7/8

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: ncramer@bbn.com (Nichael Cramer)
Newsgroups: rec.puzzles
Subject: Re: Special sentence
Date: 19 Nov 92 22:51:02 GMT
Reply-To: ncramer@labs-n.bbn.com (Nichael Cramer)
Organization: Bolt Beranek and Newman Inc., Cambridge MA
NNTP-Posting-Host: bbn.com

In article <1992Nov16.233117.20615@hellgate.utah.edu> tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>What is special about these sentences?
>
>  Simon now is!
>  Axis six v.
>  3 moss is some.
>
>What are others?

One example I posted when I discussed these a while back was:

    NO SIX IS ON

(for which someone else invented a context that turned this into a
meaningful sentence; as I recall it had something to do with the game
cricket.)

Also, as was pointed out at the time, one can expand the possibilities
enourmously by choosing the right font.  As case in point, see the many
examples in the work of Scott Kim.

N
From: Dennis.Matney@f3.n8120.z86.toadnet.org (Dennis Matney)
Newsgroups: rec.puzzles
Subject: Re: songs containing stat
Date: Mon, 16 Nov 92 13:01:00 cdt
Organization: FidoNet node 86:8120/3 - Sometimes BBS, Portage IN

 To: RVESTERM@vma.cc.nd.edu (<RVESTERM@vma.cc.nd.edu>)

Cruising thru the ionosphere, <rvesterm shocked All by stating:

 <r> there's a song which is currently being played on pop stations called
 <r> "tenessee."  unfortunately, i don't know who does this song.  but
 <r> here's (a vague semblance of) the chorus, if that's any help in
 <r> identification:  "take me to another place/take me to another land/
 <r> let me forget all that hurts me/let me understand your plans".

That song is by Arrested Development-falling out of the Top Ten about now...



... Small world, but I wouldn't want to paint it.
=== Blue Wave/QWK v2.10


--- TosScan 1.00
 * Origin: The Sometimes BBS (219-763-2031) (86:8120/3)
--  
Dennis Matney - via Toadnet/UUCP Gateway 86:86/200
UUCP: ....!wupost!8120!3!Dennis.Matney
INTERNET: Dennis.Matney@f3.n8120.z86.toadnet.org
From: ath@linkoping.trab.se (Anders Thulin)
Newsgroups: rec.puzzles
Subject: Re: Songs containing states
Date: 20 Nov 92 13:56:35 GMT
Organization: Telia Research AB, Teknikringen 2B, S-583 30 Linkoping, Sweden

In article <92318.203137RVESTERM@vma.cc.nd.edu> RVESTERM@vma.cc.nd.edu writes:
>
>                          "sweet georgia brown," "georgia on my mind,"
>and "the devil went down to georgia" (not sure who the first two are
>by - i think the second is ray charles -

Hoagey Charmichael.

-- 
Anders Thulin       ath@linkoping.trab.se 	013-23 55 32
Telia Research AB, Teknikringen 2B, S-583 30 Linkoping, Sweden
From: M.Willis@ee.surrey.ac.uk (Mike Willis)
Newsgroups: rec.puzzles
Subject: Re: Elements ending in H
Date: 16 Nov 92 14:56:05 GMT
Organization: University of Surrey, Guildford, Surrey, UK. GU2 5XH

In article <1992Nov14.214052.1264@Princeton.EDU>, jacobw@phoenix.Princeton.EDU (Jacob Solomon Weinstein) writes:
|> 
|> I was just looking over the FAQ list, and I couldn't help noticing an
|> incomplete answer. In english/element.s, the question is raised:
|> What element ends in h? The answer given is-- well, I won't spoil
|> it, but only one answer is given. There is ANOTHER element, in addition
|> to the one in the FAQ, that ends in "h." Can anybody tell me what it
|> is?

Well lets see....Earth, Sky, Fire and Water....One is Earth.
From: @ee.surrey.ac.uk (null)
Newsgroups: rec.puzzles
Subject: Too clever by half
Date: 16 Nov 92 16:50:51 GMT
Organization: University of Surrey, Guildford, England

Not exactly a puzzle, but an amusing story about that Feynmann bloke (about 
whom many amusing stories circulate). He was showing off to his easily impressed
colleagues that he could calculate, mentally, any number to within 1% (or some
such) that could be specified in one sentence. He had no problem with things like
ln(10) (he probably knew it) and other easy problems that people were asking him.
His undoing came about when Hans Bethe walked past. "What's tan(10^100) then?" he
asked. Ahhh...

P.S. -.4116229629
From: kenrolph@peg.pegasus.oz.au
Newsgroups: rec.puzzles
Date: 23 Nov 92 01:46 EST
Subject: Palindromes
Nf-ID: #N:peg:290100001:000:798
Nf-From: peg.pegasus.oz.au!kenrolph    Nov 23 01:46:00 1992


RE: Palindromes

I am looking to create a palindrome to set at the heart of a story I am 
writing. It has occurred to me that there must be some information on the 
subject around somewhere. There must be principles in English for creating 
such sets of characters, such as frequently useable sets of letters or 
words.

Does anyone know of any books on this subject?

Information on palindromic dates or numbers of significance would also be 
welcome.

Thanks

_____________________________________________________________________

Ken Rolph                                  Blacktown, NSW Australia
                                           peg:kenrolph
                                           kenrolph@peg.pegasus.oz.au
_____________________________________________________________________


Newsgroups: rec.puzzles
From: fgg@gemini.gsfc.nasa.gov (Frank G. Gomez)
Subject: expected distance between points in a sphere
Organization: NASA GSFC CDP VLBI
Date: Fri, 20 Nov 1992 07:25:01 GMT

Two small problems:

Given two random points within a sphere of unit radius, what is
the expected straight-line distance between them ?

Given a square sheet of paper, make three folds (nothing fancy, just a normal
fold).  At the end you will get a polygon.  What is the maximum number of sides
this polygon can have ?  Can you generalize to n folds (assuming the paper has
no thickness) ?

Frank
--
Frank G. Gomez           ___O__
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Billion-year survivability
Date: 17 Nov 92 03:39:39 GMT
Reply-To: warwick@cs.uq.oz.au

jhsegal@wiscon.weizmann.ac.il (Livian) writes:

>  The crazy thing I thought about,is to send a very very very powerful signal
>in space in such a(calculated) way that the echo reflected by some astroligical
>body will arrive back to the earth in 1 bilion years.

Which astronomical body?  Any one you pick might have ceased to existed 499 milion
years ago (and you'll find out in 1 million years that your 1 billion year message
is heading for spongy super-nova remnants... or a black hole... or an ET mine.


--
Warwick
--
  _-_|\      warwick@cs.uq.oz.au            /Disclaimer:
 /     * <-- Computer Science Department,  /  
 \_.-._/     University of Queensland,    /   C references are NULL && void*
      v      Brisbane, Australia.        /
From: cd866@cleveland.Freenet.Edu (Marc Sira)
Newsgroups: rec.puzzles
Subject: Re: Mastermind - some EASY questions.
Date: 20 Nov 1992 07:28:21 GMT
Organization: Case Western Reserve University, Cleveland, OH (USA)
Reply-To: cd866@cleveland.Freenet.Edu (Marc Sira)
NNTP-Posting-Host: hela.ins.cwru.edu


In a previous article, wft@math.canterbury.ac.nz (Bill Taylor) says:

>1. What is the best thing to do when following a very slow car ?
>
>2. What is the term for a single run through a compiler ?
>
>3. What is a common slang term for an implicit sexual suggestion ?
>
>4. What is the most usual method of obtaining the ball in basketball ?
>
>5. What is the most commonly heard call in contract bridge ?

I'm stumped...guess I'll pass on this one.

B-)

Marc
-- 
                                      30
From: varhus@meringue.seas.upenn.edu (Kristofor A Varhus)
Newsgroups: rec.puzzles
Subject: Our favorite series
Date: 17 Nov 92 18:24:43 GMT
Organization: University of Pennsylvania

Here's everybody's favorite series again:

	1
	11
	21
	1211
	111221
	...

1. Which is the first term that has a "4" in it?

2. Prove your answer to number 1.

--
-------------------------------------------------------------------------------
Kristofor A. Varhus			|       THIS SPACE AVAILABLE
varhus@eniac.seas.upenn.edu		|        E-MAIL IMMEDIATELY!
-------------------------------------------------------------------------------
From: warwick@cs.uq.oz.au (Warwick Allison)
Newsgroups: rec.puzzles
Subject: Re: Billion-year survivability
Date: 19 Nov 92 00:18:32 GMT
Reply-To: warwick@cs.uq.oz.au

dcarroll@oucsace.cs.ohiou.edu (Dana Carroll) writes:

>So the answer is simple: simulatenously send out the radio signal in all
>directions, powerful enough so that it will still be around in a billion
>years, and hope that it will eventually be reflected back to earth.

Considering how far the energy must travel (recalling that you're energy
will disperse into a 22,500,000,000,000,000,000,000,000,000,000,000 square
metre area, the reflectivity of any likely surface, and the fact that it
must make its way back, what are you suggesting as a transmitter?

IMO, the best solution so far is to simply throw things into space
with the message written on it in mathematical form.  Pretend you're
communicating with aliens, since that's what 1Byear hence "humans"
will be.  Make the objects self replicating.  Throw them in thousands
of orbits.

Of course, when the Galactic Admin finds out, the Human Race is in for
one hell of a littering fine.  Anything that's going to be around in
1 billion years if definately NOT biodegradible!


--
Warwick
--
  _-_|\      warwick@cs.uq.oz.au            /Disclaimer:
 /     * <-- Computer Science Department,  /  
 \_.-._/     University of Queensland,    /   C references are NULL && void*
      v      Brisbane, Australia.        /
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: A Simple Puzzle
Date: 21 Nov 1992 05:52:27 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <1992Nov20.230941.20679@EE.Stanford.EDU>, you write:
=One day a merchant presented him with a bag full of diamonds. The king
=announced that he will equally distribute them amongst his wives later. 
=
=However, the queens were of somewhat impatient nature. That night,
=one queen stole one-third the number of diamonds. Later that night,
=the second queen came and stole one-third the number left and as you 
=can guess now, the third queen also came and stole one-third the
=number left. (They were all pretty honest, not stealing all). 
=
=Next morning, the king had the bag brought to him, divided into three 
=portions with one diamond left over. He gave one portion to each of the queens
=and kept the lone diamond for himself. 
=
=What was the number of diamonds in the bag to start with?
=
=(To be technically right, what is the smallest such number?) 

<SPOILER>

Well, the bag ends up, in the morning, with 3n+1 diamonds, so it started out
with (3n+1)*(3/2)*(3/2)*(3/2) = (3n+1)*27/8 diamonds.  So we simply need to find
the smallest n for which 3n+1 is an integral multiple of 8.  That is n = 5.  So
in the morning, there were 16 diamonds, meaning it started with 54 diamonds.
So:
	Originally:		54 diamonds
	After 1st wife:		36 diamonds
	After 2nd wife:		24 diamonds
	After 3rd wife:		16 diamonds
In the morning, each wife gets 5 diamonds and the king keeps one.  So:
The 1st wife ends up with 23 diamonds.
The 2nd wife ends up with 17 diamonds.
The 3rd wife ends up with 13 diamonds.
The king ends up with      1 diamond.
For a total of            54 diamonds.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: Demoel@Fwva.Saic.Com (Ed de Moel)
Newsgroups: rec.puzzles
Subject: Re: Mastermind - some EASY questions.
Date: Fri, 20 Nov 1992 20:58:12 GMT
Organization: Science Applications Int'l, Corp. - Computer systems operation

cvadsavi@vmsa.is.csupomona.edu (Carl R. Knecht) writes:
>In article <BxzoJH.7Gn@cantua.canterbury.ac.nz>, wft@math.canterbury.ac.nz (Bill Taylor) writes...
>>These questions aren't just easy - you can hardly get them wrong!
>>  
>>1. What is the best thing to do when following a very slow car ?
>
>..er, pass it!
 
well, that's what the other questions suggest...
But if you're following someone, getting ahead is the last thing you
would want to do. However slow your guide is, if you want to follow,
you have to stay behind.
 
I'd say: Keep following it.
Newsgroups: rec.puzzles
From: mjfreem@hubcap.clemson.edu (Matthew J. Freeman)
Subject: Re: Random Points on a Sphere
Organization: Clemson University
Date: Sun, 22 Nov 1992 02:01:05 GMT

isaak@aurora.com (Mark Isaak) writes:

>This problem has me stumped:

>Four points are randomly selected from the surface of a sphere.
>What is the probability that all four lie in the same hemisphere?
>How about for 5 points?  6?  more?
>-- 
>Mark Isaak           "Every generation thinks it has the answers, and every
>isaak@aurora.com      generation is humbled by nature." - Philip Lubin

Think of placing 4 points on an otherwise featureless sphere.
The first three points will define a hemisphere no matter
where they are placed, and the probability that the fourth is
in the same hemisphere is 1/2. For 5 points, the probability
goes down to (1/2)*(1/2). Unless all of the points lie on a
great circle ...

-Matt Freeman
 hubcap.clemson.edu

Newsgroups: rec.puzzles
From: jacobw@phoenix.Princeton.EDU (Jacob Solomon Weinstein)
Subject: Re: Kit Williams' old new book
Originator: news@nimaster
Organization: Princeton University
Date: Wed, 18 Nov 1992 14:51:06 GMT

strgh@csv.warwick.ac.uk (J E H Shaw) writes:
>DKSmith@msor.exeter.ac.uk (David Smith) writes:
>...
>>
>>The book was called "The Bee on the Comb"
>>but it never got as much publicity as the
>>book "Masquerade" ... after all, the media
>>were more interested in the valuable prize
>>in Masquerade.
>>
>...
>As I remember, the book was called `Bees Only Sting'.

Nope- you fell pray to the same red herring that my friends and I did.
If you take a look at the front cover of the book, you'll notice that
you can see the edge of the piece of honeycomb pattern that covers the
title. From this, you can figure out how many characters are in
the title, and "Bees Only Sting" doesn't fit.

I know this, by the way, because Kit Williams himself told me at a book signing
shortly after the book was released. So it comes from a pretty
unimpeachable source.

Still, the question remains: What were some of the clues? And what was
the winning entry like? With hindsight, I now realize that one of the 
seasons had a comb stuck in her hair, with an image of a bee on it.

I also know that the solution involved Morse code, and finding the
location of the hidden animal in each picture. Again, this comes from
Kit Williams. Given this, and the answer, can anybody work backwards and
figure the thing out?
From: ralf@ark.abg.sub.org (Ralf Stephan)
Newsgroups: rec.puzzles
Subject: Re: Random Points on a Sphere
Date: 22 Nov 92 18:08
Organization: Not Very Ltd.

mjfreem@hubcap.clemson.edu (Matthew J. Freeman) writes:
> isaak@aurora.com (Mark Isaak) writes:
> >Four points are randomly selected from the surface of a sphere.
> >What is the probability that all four lie in the same hemisphere?
> >How about for 5 points?  6?  more?
> 
> Think of placing 4 points on an otherwise featureless sphere.
> The first three points will define a hemisphere no matter
> where they are placed, ...        ^^^
                                     |
Really?  For example, Rome, Berlin, and Prague are all on the northern
hemisphere --- but surely they are as well not much east of Greenwich,
so there's another hemisphere --- the one with the border through Greenwich
and the North Pole.  On the other hand, not all hemispheres are equally
likely to contain these three points and the likeliness changes drastically
if the points are far away!

It seems that one has to look on the distances between points...

ralf
--
***********************************************************************
PLEASE NOTE: Some Quantum Physics Theories Suggest That When the Reader
Is Not Directly Observing This Message, It May Cease to Exist or Will
Exist Only in a Vague and Undetermined State.
From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger)
Newsgroups: rec.puzzles
Subject: Re: Trig. inequality
Date: 22 Nov 92 03:08:01 GMT
Organization: Dartmouth College, Hanover, NH
X-Posted-From: InterNews1.0a5@newshost.dartmouth.edu

In article <1ekddfINNi96@uwm.edu>
radcliff@csd4.csd.uwm.edu (David G Radcliffe) writes:

> Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4.
> 
> --
> David Radcliffe                                
> radcliff@csd4.csd.uwm.edu  

Simple math.  Remember that (sin x)^2+(cos x)^2=1.  An identity from my
trig class back in high school.  Setting sin x equal to cos x, we get
(sin x)^2=1/2.  The answer to this is x=pi/4.  And since sin x
increases from 0 to 1 as x increases from 0 to pi/2, it's obvious that
cos x will be greater than sin x.  As neither is negative in this
range, (cos x)^2 will be greater than (sin x)^2 from 0<x<pi/4.  I
haven't taken any math since freshman year of college, but this was
easy!          


--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."
From: goddard@NeXTwork.Rose-Hulman.Edu (Bart E. Goddard)
Newsgroups: rec.puzzles
Subject: Re: Billion-year survivability
Date: 22 Nov 1992 22:19:20 GMT
Organization: Computer Science Department at Rose-Hulman
Reply-To: goddard@NeXTwork.Rose-Hulman.Edu (Bart E. Goddard)
NNTP-Posting-Host: g214-1.nextwork.rose-hulman.edu

In article <1992Nov22.214532.26970@newshub.ccs.yorku.ca>  
tony@nexus.yorku.ca (Tony Wallis) writes:

> What's it called when 
> [4 easy steps deleted]>
 
Er, a waste of bandwidth?  
:-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-)
:-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-):-)

Bart

P.S. What do I win?
From: pirard@vm1.ulg.ac.be (Andre' Pirard)
Newsgroups: rec.puzzles
Subject: Sequence
Date: 16 Nov 92 17:06:15 GMT
Organization: Universite' de Lie`ge (Belgium)
X-UserAgent: Nuntius v1.1.1d11
X-XXMessage-ID: <A72D85FF3101180A@ampa.segi.ulg.ac.be>
X-XXDate: Mon, 16 Nov 92 16:08:15 GMT

1
11
21
1211
111221
?

-----
Andre Pirard          SEGI, Univ. de Li`ge         139.165 IP coordinator
B26 - Sart Tilman     B-4000 Li`ge 1 (Belgium)            +32 (41) 564932
pirard@vm1.ulg.ac.be  alias PIRARD@BLIULG11.BITNET
Organization: University of Maine System
Date: Tuesday, 17 Nov 1992 13:12:53 EST
From: Betty Johnson <BETTYJ@MAINE.MAINE.EDU>
Newsgroups: rec.puzzles
Subject: Re: Songs containing states
 <92318.203137RVESTERM@vma.cc.nd.edu> <92318.205554RVESTERM@vma.cc.nd.edu>
 <1992Nov16.231918.8942@alw.nih.gov>

Gary, Indiana  from Music Man
Maine Stein Song
Missouri Waltz
Tennessee Waltz
Louisiana Saturday Night
New York, New York  sung by Frank Sinatra
Jersey Bounce    (a bit far afield, I'll grant)
Pennsylvania 65000  by Glen Miller
Carolina Moon keep shinin'
Stars Fell on Alabama

Many of these are pretty old dating back to WWII and before.  Most of
you 'young squirts' wouldn't have heard of them.  Just us 'old farts'
remember them.
Newsgroups: rec.puzzles
From: is@groucho.dev.uga.edu (Bob Stearns)
Subject: Re: Random Points on a Sphere
Organization: University of Georgia
Date: Mon, 23 Nov 1992 22:34:53 GMT

In article <1erk03INNntn@roundup.crhc.uiuc.edu> sarwate@uicsl.csl.uiuc.edu (Dilip V. Sarwate) writes:
>isaak@aurora.com (Mark Isaak) writes:
>
>>Four points are randomly selected from the surface of a sphere.
>>What is the probability that all four lie in the same hemisphere?
>>How about for 5 points?  6?  more?
>
>One simple-minded solution (which doesn't worry about boundaries) is as follows.
>The first two points picked, together with the center of the sphere, define a
>plane that divides the sphere into two hemispheres.  The two points are on the
>boundary dividing the hemispheres.  With probability 1, the third point lies
>strictly in the interior of one of the hemispherical surfaces, and this becomes
>our chosen hemisphere (the first two points, being on the boundary, are included
>by courtesy in the chosen hemisphere.)  Thus, assuming that the points are
>picked with a uniform distribution over the surface of the sphere, the
>probability that the fourth point lies in the chosen hemisphere is 1/2, the
>probability that the fifth point also lies in the chosen hemisphere is 1/4, etc.

This makes the first two points in some way distinguished. As I
interpret the question, is there no way to chose one pair of the four to
use in the manner you outline such that the other two fall into one of
the two hemispheres created by the first chosen two. This makes the
problem much more complex. I have no simple solution to propose, but 
merely wish to point out that your solution is too simple by far. 


-- 
Bob Stearns
University of Georgia
is@groucho.dev.uga.edu
(404)542-5110
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Trig. inequality
Date: 22 Nov 1992 16:26:40 GMT
Organization: HST Wide Field/Planetary Camera
Distribution: world
Reply-To: carl@SOL1.GPS.CALTECH.EDU
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <1992Nov22.030801.4773@dartvax.dartmouth.edu>, J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
=> Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4.
=
=Simple math.  Remember that (sin x)^2+(cos x)^2=1.  An identity from my
=trig class back in high school.  Setting sin x equal to cos x, we get
=(sin x)^2=1/2.  The answer to this is x=pi/4.  And since sin x
=increases from 0 to 1 as x increases from 0 to pi/2, it's obvious that
=cos x will be greater than sin x.  As neither is negative in this
=range, (cos x)^2 will be greater than (sin x)^2 from 0<x<pi/4.  I
=haven't taken any math since freshman year of college, but this was
=easy!          

And wrong.  Fine; you've proven that sin(x)^2 < cos(x)^2 for 0<x<pi/4. The
problem, however, was to show that
	sin(x)^sin(x) < cos(x)^cos(x)
Not the same thing at all.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.