HDTenEightyP/NetNews huggingface dataset:
From: kyee@bnlux1.bnl.gov (kenton yee)
Newsgroups: sci.math
Subject: a more general constraint integral (was:help me, please
Date: 12 Sep 92 19:43:56 GMT
Organization: Brookhaven National Laboratory, Upton, NY 11973

>kyee@bnlux1.bnl.gov (kenton yee) writes:
>#  int dx_1 dx_2 ... dx_N  1    =   ??
>#where  0 < x_i < 1    subject to the condition  
>#  x_1 + x_2 + ... + x_N  is less than or equal to 1.  

9.4738@cs.brown.edu> dzk@cs.brown.edu (Danny Keren) writes:
>S1 = {(x1,x2....xn)/xi>=0, x1+x2+...xn<=1}
>S2 = {(x1,x2....xn)/0<=x1<=x2...<=xn<=1}
>What you are asking is the volume of S1. Now, look at the following
>transformation from S1 to S2:
>(x1,x2,....xn) ----> (x1,x1+x2,x1+x2+x3,...x1+x2+...+xn)
>it is trivial that it is 1-1 and onto; also, its determinant is 1,
>so it preserves volume. Its also trivial that the volume of
>S2 is 1/n! (symmetry considerations) so your integral is equal
>to 1/n! .

Thanks.  Can you extend this to do integrals of the form:

 int dx1 dx2 ... dxn 1   =  ???

where   x1+...+x4 <=1   AND
        x4+...+x8 <=1   AND
        x8+...+x12 <=1  etc... until xn ?  (The double occurence
          of x4 and x8 inside two constraints is not a typo.)
Newsgroups: sci.math
From: ilan@leland.Stanford.EDU (ilan vardi)
Subject: Re: Folding numbers
Organization: DSG, Stanford University, CA 94305, USA
Date: Sun, 6 Sep 92 00:22:07 GMT

In article <4958@balrog.ctron.com> wilson@web.ctron.com (Dave Wilson) writes:
>
>    Suppose we have a leaflet of N connected pages, e.g:
>
>        +-------+-------+-------+-     -+-------+-------+
>        |       |       |       |       |       |       |
>        | TITLE |   2   |   3   |  ...  |  N-1  |   N   |
>        |       |       |       |       |       |       |
>        +-------+-------+-------+-     -+-------+-------+
>
>    Let f(N) be the number of ways to fold the leaflet along its
>    perforations to the size of a single page so that the title page
>    appears on top of the folded leaflet.  I have computed the

I think that this is known as the stamp folding problem and was 
studied by Touchard.

-ilan
From: lucas@WATSON.IBM.COM ("Bruce Lucas")
Newsgroups: sci.math
Subject: computable reals
Date: 12 Sep 92 04:24:12 GMT

In a related vein to the recent discussion about the "computability of
the universe", I have always found it in some way a little disturbing
that we feel compenent to reason about uncountable sets such as the reals
using a language that, being based on discrete symbols, can actually only
represent a countable subset of them.

So, suppose we decided to limit the discussion to the set of "computable"
reals.  This set could be defined in a number of ways which I suspect
are all equivalent: for example a real is computable if there is a
computable function f(n) that produces its nth digit; or, a real is
computable if there are computable functions f(n) and g(n) that produce
an approximation f(n)/g(n) to within 1/n.

This gives us a countable set that forms a field and contains all the
"useful" reals (pi, e, sqrt(2), ..., and in fact probably all the reals
anyone would want to talk about except bizarre examples described by
diagonalization to prove that the set of reals isn't countable).  It
has the satisfying property that we could actually write down a
representation of any member of the set.

My question is is this in any way an interesting thing to do?  For example,
what properties of the reals does this set have and which does it lack?

Cheers,
Bruce Lucas
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 15 Sep 92 15:26:55 GMT
 <1992Sep13.003856.13264@infodev.cam.ac.uk> <1992Sep15.032439.19637@ll.mit.edu>

In <1992Sep15.032439.19637@ll.mit.edu> shoham@ll.mit.edu (Daniel Shoham) writes:

>>In article <1992Sep12.174545.23214@ima.isc.com>, karl@ima.isc.com (Karl Heuer) writes:
>>
>>> Here's a more advanced chessboard question.
>>> 
>>> A white king, a black king, and a white pawn are randomly placed on an NxN
>>> chessboard.  (Independent, uniform distributions.)  What is the limit, as
>>> N -> infinity, of the probability that the position is a win for white?
>>> 
>>> (Note that it doesn't matter what you do about collisions or illegal
>>> positions, since they occur with probability zero in the limit.)
>>

>We will use an x-y system to describe the position of each square. The last
>(queening) row is defined as y=0, whereas the first row is y=1. x, is similarly
>ranged between 0 and 1.
>(px,py) = pawn's starting position
>(bx,by) = black (king) starting position
>(wx,wy) = white (king) starting position

>px, py, bx, by, wx, wy are iidrv with uniform (0,1) distribution.

>Strategy 1 works if:
>py < by  or  py < |px-bx|

>Strategy 2 works if:
>d(p,w) < d(p,b)      where d(i,j)=max(|ix-jx| , |iy-jy|)  (i.e. chess-distance)

>We could now use brute force to integrate over all cases, or (if your religion
>permits it) do a Monte Carlo simulation.

>As my religion does not forbid me from doing a Monte Carlo after I have found
>an analytic algorithm (that I am too lazy to integrate), I did it:

>P(Draw) =~ 12.5%  (40,000 runs, i.e. +/- .5%)

Funny, I get 43/192 for the analytic answer, which is 22.4%.


>Dan Shoham
>shoham@ll.mit.edu
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.
Newsgroups: sci.math,news.answers
From: neuhaus@vier.informatik.uni-kl.de (Stephan Neuhaus (HiWi Mattern))
Subject: Re: Jokes in papers (was Re: Einstein's Erdos number)
Organization: University of Kaiserslautern, Germany
Date: Wed, 9 Sep 1992 11:50:45 GMT

I vaguely recall from a math course that there was a paper published
by three soviet mathematicians, the third of which having signed the
paper only to make the three mathematicians' names sound like ``alpha,
beta, gamma''.  The third mathematician's name was Gamow.  Does
anybody know the names of the other two?


-- 
Stephan <neuhaus@informatik.uni-kl.de>
sig closed for inventory.  Please leave your pickaxe outside.
Newsgroups: sci.math
From: matmcinn@nuscc.nus.sg (Mcinnes B T (Dr))
Subject: Re: Warm and Fuzzy?  Koalas vs. Math
Organization: National University of Singapore
Date: Mon, 14 Sep 1992 05:30:34 GMT

The string theorists used to talk about "SU[3] holonomy", this being the
holonomy group of a 3D Kahler manifold which is Ricci-flat. This is
true, however, only if the Ricci tensor is EXACTLY zero. If you take a
3D Kahler manifold with scalar curvature s, then the holonomy is not
SU[3] but rather U[3], NO MATTER HOW SMALL s is. I wonder if this has
any bearing on the cosmological constant problem, he speculated idly.
Newsgroups: sci.math
From: steele@isis.cgd.ucar.edu (Alfred Steele)
Subject: Re: Question on divergence of series given condition
Organization: Climate and Global Dynamics Division/NCAR, Boulder, CO
Date: Sat, 5 Sep 1992 05:58:55 GMT

In article <1992Sep4.174051.3223@eagle.lerc.nasa.gov> sshws@convx1.lerc.nasa.gov (Herb Schilling) writes:
>>
>>We think there is a theorem ( By Weierstrauss ? ) on this subject
>>but we cannot find
>>it. Can anyone out there tell us if the following is true? 
>>
>>=======================================================================
>>
>>  If the terms, a  , of a series have the following property:
>>                 n
>>
>>         a                               A
>>          n+1                1            n
>>        ------ =    1   -  -----   +   -----    as n -> oo
>>         a                   n           n
>>          n
>>
>>
>>
>>    where A    ->   0 as n -> oo
>>           n
>>
>>
>>
>>
>>  then does sum of the a    diverge  ?
>>                        n 
>>
>>
>> Thanks in advance for the help.
>>-- 
>>Herb Schilling , NASA Lewis Research Center , 21000 Brookpark Road, MS 142-5
>>Cleveland, Ohio 44135 . (216) 433-8955       sshws@convx1.lerc.nasa.gov



The formula is not the same but you can find more that you need in:
Knopp "Theory and Application of Infinite Series" Blackie && Son  pg. 288 #172
J. Pierpont "Functions of a Complex Variable" Dover pg. 52 article27 of 
chapter II

Alfred T. Steele
Newsgroups: sci.math
From: kwhyte@dent.uchicago.edu (Kevin Whyte)
Subject: Foliations ...
Organization: Dept. of Mathematics, U. of Chicago
Date: Wed, 9 Sep 1992 16:31:23 GMT

  A question:  does anyone know an example of a manifold which
occurs as a leaf in a foliation of a compact manifold, but
which doesn't cover anything compact?

 Kevin


A great many people think they are thinking       | Kevin Whyte 
when they are merely rearranging their prejudices.| kwhyte@math.uchicago.edu 
--William James                                   | (312) 702-6045
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Proof of God's Existence
Date: 8 Sep 92 16:56:44 GMT
Organization: U of Cambridge, England

In article <1992Sep8.143447.6958@st-andrews.ac.uk>, ajtd@honey.st-and.ac.uk (Tony Davie) writes:
> I think this one is due to Smullyan:
> 
> Consider the sentence "If this sentence is true then God exists"
> 
> Proof: 1. The sentence is true                            Hypothesis
>        2. If this sentence is true then God exists        From 1
>        3. God exists                                      Modus Ponens 1,2
> 
>        4. If this sentence is true then God exists        Deduction Theorem 1,3
> 
> The sentence is proved (and therefore true) QED
> 
> Corollary: By the theorem and the fact that it is true and Modus Ponens,
> God exists.
> 
> OR, of course, by a similar argument, God doesn't exist or My aunty's a banana
> or anything you like.

It becomes clearer why this works if you rewrite "If this sentence is true
then I am the Pope" as "Either this sentence is false or I am the Pope".
Here's another version, which is basically the same:

Consider the sentence "This sentence is false".
Assume for contradiction that I am not the Pope.
<insert Epimenides paradox here>
So we have a contradiction; hence I am the Pope.

-- 
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@cus.cam.ac.uk  Cambridge University, England.    [Research student]
Newsgroups: sci.math
From: pa@verano.sba.ca.us (Pierre Asselin)
Newsgroups: sci.math,sci.math.num-analysis
Subject: Cycles in a triangulation
Keywords: graph tree chain cycle boundary basis
Date: 13 Sep 92 21:44:56 GMT
Followup-To: sci.math
Organization: None.  Santa Barbara, CA

Are there combinatorial algorithms to find a basis for the k-cycles in a
triangulation of an n-dimensional manifold?

The intended application is the solution of partial differential
equations by finite elements in R^2 and R^3, particularly
electromagnetic field computations.  There is a well-known method for
k=1 and a recent publication probably takes care of k= n-1.

Has anybody solved the general case on a computer?  I don't think the
solution is known in my field.  References, anyone?
-- 

--Pierre Asselin, Santa Barbara, California
  pa@verano.sba.ca.us
From: jlam@ariel.ucs.unimelb.edu.au (Lord Automato)
Newsgroups: sci.math
Subject: Re: XAY+XB+CY+D=0, HELP
Date: 11 Sep 92 01:51:54 GMT
Organization: University of Melbourne


In article <1992Sep10.140346.5187@cronkite.ocis.temple.edu>, neilfp@astro.ocis.temple.edu (Neil F. Palumbo) writes:
|> I am trying to solve the matrix equation XAY+XB+CY+D=0 analytically for the
|> unknown matrices {X,Y}. Anyone have any ideas, comments, suggestions...?


The solutions can be parametrized as follows: 

With  Y  such that   AY+B  is nonsingular, then  X = -(AY+B)^{-1} (CY+D) .

This is a fractional transformation on Y .

Lin Can
=========================================================================
.-- .... --- . ...- . .-. -.. . -.-. --- -.. . ... - .... .. ...
--. . - ... -. --- - .... .. -. --.        
.-.. .. ..-. . .. ... ... .... --- .-. - .-- --- .-. -.- .... .- .-. -..
                                                   .-.. .. -. -.-. .- -.
=========================================================================
From: kewe@bskewe.atr.bso.nl (Cornelis Wessels)
Newsgroups: sci.math
Subject: Fast Fourier Transform 
Date: Mon, 14 Sep 92 19:03:09 GMT
Organization: Me? Organized?


In article <1992Sep14.111815.12937@phillip.edu.au> t9014186@phillip.edu.au writes:

  > 
  > Does anybody know where I can find, or does anybody have, a source listing of
  > how to perform a Fast Fourier Transform ??
  > 
  > A listing in C would be preferable but not essential.
  > 
  > 
  > 
  > -- 
  > =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
  > =-= Steven Corral                                                       =-=
  > =-=                                                                     =-=
  > =-= T9014186@phillip.edu.au                                             =-=
  > =-= scorral@arcadia.mc.phillip.edu.au                                   =-=
  > =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
  > 
Cornelis Wessels
kewe@bskewe.atr.bso.nl
From: D.J.Nettleton@newcastle.ac.uk (D J Nettleton)
Newsgroups: sci.math
Subject: Simple Matrix Question !!
Date: 15 Sep 92 10:16:32 GMT
Organization: University of Newcastle upon Tyne, UK NE1 7RU

Hello I wonder if anyone can spare a second or two to
help with this anoyingly easy problem.

  If you have a 2*2 matrix representing a contraction mapping,
with coefficients a,b,c,d say, how do you check (quickly) that
when applied to a shape it:

1) reflects in both x and y axis

2) reflects in just one

3) neither.

  All irrespective of any contraction.

  Is it simply to do with the values of the eigenvalues?
i.e. both -'ve then 1) above, one -'ve then 2) above ...
(from decomposition of A^{n}=PD^{n}P^{-1} where D e-value
matrix and P matrix of corresponding e-vectors).

 Many thanks,

  Dave

e-mail:  D.J.Nettleton@durham.ac.uk
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math
Subject: Re: Partitioning of uncountable sets
Date: 8 Sep 92 14:58:25 GMT
Article-I.D.: mentor.Bu9LLE.Av8
Organization: Purdue University Statistics Department

In article <1992Sep8.182706.90039@vaxc.cc.monash.edu.au> kevin@vaxc.cc.monash.edu.au writes:
>A proof that every uncountable set can be partioned into two uncountable
>sets.
>Let X be an uncountable set. Consider the set of ordered pairs, (x,0),(x,1)
>where x is in X. Call this set Y. Then Y is also uncountable, moreover,
>{(x,0) with x in X}=Y(0) and {(x,1) with x in X}=Y(1) are both uncountable.
>But the cardinaltiy of X is the cardinality of Y. Thus, there is a bijection
>f from Y to X. f(Y(0)) and f(Y(1)) form the desired partition of X.
>The proof is of course easier if one assumes A.C, which I have avoided.

If finite sets are not included in countable, the theorem is trivially true.
But otherwise, you have not avoided all of AC, although what you have used
is less.  It is not necessarily the case that the union of two disjoint
uncountable sets of the same cardinality has that cardinality.  In fact,
the theorem itself is false, counterexample as follows.

There are Fraenkel-Mostowski models in which there is an infinite set A with
all subsets finite or co-finite.  This set is already a counterexample.  But
if one insists on transfinite, we adjoin such a set to the integers, and we
obtain a transfinite set such that every partition has a finite or countable
element.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
Newsgroups: sci.math
From: schissel@mace.Princeton.EDU (Eric Schissel)
Subject: Gnat
Organization: Princeton University
Date: Sun, 13 Sep 1992 04:01:49 GMT

The U.S. seems to be organized in such a manner that status is inversely
related with actual usefulness.  So the President has a very high salary,
and mathematicians of the 'pure' persuasion don't do too badly, and
neither do sports heroes; what sort of salary is the person who takes out
your garbage getting?  Not bad (if unionized) but nowhere near good, i'll
bet.

schissel@math.cornell.edu  (do not r, i'll no longer have this
Princeton account in a day or so)
 
Newsgroups: sci.math
From: ceblair@ux1.cso.uiuc.edu (Charles Blair)
Subject: My programming experiences
Organization: University of Illinois at Urbana
Date: Mon, 14 Sep 1992 18:53:22 GMT

   I have a math Ph.D., learned Fortran as undergrad, switched to Pascal
about six years ago, began writing in C three years ago.  It still takes
me huge amounts of time to get something to work--- a recent program to
print text next to a chess diagram took about 12 hours.  It is certainly
possible to program badly in any language!  I have never even tried to
do anything that involved numerical analysis.

   Even though I haven't really learned it, I can see that there is a skill
in programming that goes beyond a clear understanding of the algorithm one
is trying to implement.  Perhaps those who are curious about this might
look at Kernighan's ELEMENTS OF PROGRAMMING STYLE (examples mostly in Fortran).
If you can guess what the criticisms of the examples will be before reading
the commentary, you already know quite a lot!  (the examples are taken from
programming texts, indicating that the authors make mistakes too)
From: duentsch@dosuni1.rz.uni-osnabrueck.de
Newsgroups: sci.math
Subject: Aliquot
Date: 14 Sep 92 13:48:02 GMT
Organization: University of Osnabrueck, FRG

A slighly revised version of 'aliquot' is available via anonymous ftp from
 
              tethys.rz.uni-osnabrueck.de (131.173.17.10)
 
in pub/msdos/math as aliquot.zip.
 
'aliquot' is a utility for computing aliquot chains and keeping a record
of the results. The archive also contains a file of the targets (or
whatever they are called) of the numbers up to 111,335 with about 13%
unresolved.
 
Ivo Duentsch
Univ. Osnabrueck
Bitnet:    duentsch@dosuni1
Internet:  duentsch@dosuni1.rz.uni-osnabrueck.de
Newsgroups: sci.math
From: las@cunyvms1.gc.cuny.edu
Subject: Re: Proof of God's Existence
Reply-To: las@cunyvms1.gc.cuny.edu
Organization: Computer Center, CUNY Graduate School, New York NY 10036
Date: Sat, 12 Sep 1992 04:14:25 GMT

In article <17ui6kINNsft@matt.ksu.ksu.edu>, bubai@matt.ksu.ksu.edu (P.Chatterjee) writes:
>I was told by somebody that there is a mathematical proof of God's existence; was wondering if somebody could shed some light on the same.
>
>Thanks. Looking forward to the response.


S: I can furnish a proof provided you or anyone else can prove that they
   exist.  In other words, forget about God for now. First prove that 
   you exist and then we are in bussiness. No! You can't assume your 
   existence. You have to prove it, and don't use the  "I think therefore 
   I'm "  trick. I suggest you begin by defining who your "I" is. Procceed
   from there.
  



Newsgroups: sci.math
From: achim@unibi.uni-bielefeld.de (Achim Flammenkamp)
Subject: Re: nth prime
Date: Fri, 11 Sep 92 13:31:29 GMT
Organization: Universitaet Bielefeld

In article <1992Sep10.190905.10741@waikato.ac.nz> bill@waikato.ac.nz writes:
>Sorry to ask such a FAQ, but what's the quickest algorithm to
>find the nth prime? Is it O(nlogn) or something else?
>
I think it is not a FAQ :)

You have to calculate the "inverse function" of PI(n), the number of primes 
less than n. There are clever algorithmns like the telescop method of Meissel
to calculate PI(n) in time O(n^2/3). Look for a recent calculation in Math.
of Computation about 1985 from Odlytzko (probably spelled wrong, sorry). They
calculated PI(x) upto 4*10^16. Maybe today there are algorithmns availible
which have a running time of n^k with 1/3<k<2/3. To get the nth-prime you have
to too a binary search on PI(n) roughly spoken , order log(n). So you will
get the nth-prime in about n^k which k <= 2/3.
My personally opinion is that the theoritical bound is less than n^eps for
each postive eps , but to find such an algorithmn you have to understand the
prime distribution totally :)
achim
From: rickert@mp.cs.niu.edu (Neil Rickert)
Newsgroups: sci.math
Subject: Re: Lebesgue integral (was: Couple of questions
Date: 15 Sep 92 02:38:01 GMT
Organization: Northern Illinois University

In article <1992Sep15.020445.26398@unixg.ubc.ca> ramsay@unixg.ubc.ca (Keith Ramsay) writes:
>
>I've heard that much (nearly all, perhaps) of what is done with L^2,
>regarded as Lebesgue square-integrable functions, modulo functions
>supported on sets of measure zero, is just as nicely done (if not
>better) by regarding L^2 as the formal completion (relative to the L^2
>norm) of a convenient dense subset (where the Lebesgue measure theory
>is not needed). I'm not enough of an analyst to confirm this.

This leaves the physicist with a function that he can't be sure he can
get a Fourier series for, and a solution to a differential equation
which has a Fourier series but is not a function since it is in some
mysterious formal completion space.
Newsgroups: sci.math
From: rusin@mp.cs.niu.edu (David Rusin)
Subject: Re: A tarnished Gaussian sum?
Organization: Northern Illinois University
Date: Thu, 10 Sep 1992 19:36:34 GMT

In article <1992Sep10.151807.25693@tdb.uu.se> matsa@tdb.uu.se (Mats Andersson) writes:
>
>Let p be a prime and w a pth root of unity. Does there always exist a k
>such that
>
>    (1-w)(1-w^2)...(1-w^((p-1)/2)) = w^k * Gaussian sum ?

Let Z be the left side. Then if Z* is its complex conjugate, 

Z*=(1-w^(-1))...(1-w^((p+1)/2))=(1-w^((p+1)/2)...(1-w^(p-1)),

so  Z.Z* is the product of all conjugates of (1-w); that is, you let
x=1 in the polynomial \Prod (x-w^j) = 1+x+...+x^(p-1): Z.Z*=p.
Thus certainly  Z  is \sqrt(p) times a complex number of norm 1.

On the other hand, (1-w^(2j)) = w^j . (w^(-j)-w^j)  and similarly
(1-w^(2j+1))=(w^p-w^(2j+1))=w^p.(1-w^(2j+1-p)) can be factored in the
form w^something.(w^a-w^(-a)). So if you collect all the factors in
Z you get a big power of w times a product of terms (w^a-w^(-a)),
all of which are purely imaginary. So  Z  is of the form
(power of  w ) . (i ^ (p-1)/2)) . real.

So up to powers of  w  this product  Z  is just  i^((p-1)/2) . sqrt(p).

dave  rusin@math.niu.edu

PS - yes, I think there is a way to decide on the signs in the square root
but I forget what it is.
 

Newsgroups: sci.math
From: sperry@convex.csd.uwm.edu (Robert Joseph Sperry)
Subject: HELP! Polynomials modulo irreducible polynomial
Followup-To: sperry@convex.csd.uwm.edu 
Organization: Computing Services Division, University of Wisconsin - Milwaukee
Date: Mon, 7 Sep 1992 21:19:19 GMT


Judging from what I've read from this newsgroup, this question/problem
will be beneath most of you, but I desparately need the help of you
wonderful "mathematics" people :-)

I trying to figure out a nice,systematic algorithm for modulating (?) 
one polynomial by another.

This is going to be (hopefully) part of a computer program. The program
will hold polynomials modulo an irreducible polynominal.

Example:
The irreduciable polynomial is x^2 + 1.  The user enters x^5 + 4x^3 +
3x^2 + 9.  What I need is a nice algorithm to find the result of 
this user-entered number modulo the irreducible polynomial...

Sort of like Galois field p^m....

*Any* help would be greatly appreciated! Thanks!


Newsgroups: sci.math
From: tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman)
Subject: Retarded log question
Date: 10 Sep 92 22:51:34 MDT
Organization: University of Utah, CompSci Dept


If you are given a number, such as 2^X, how can you determine what
the same number would be with a different base, such as 10?

For example, 2^100,000,000 = 10^ ??


Thanks a lot