Subject: Re: comp.speech Frequently Asked Questions - part 2/3
From: jot@hplb.hpl.hp.com (Jonathan Thompson)
Date: Fri, 20 Dec 1996 15:59:49 GMT
James Salsman (jsalsman@bovik.org) wrote:
: Jonathan Thompson wrote:
: >
: > ...You mean to say that a scaling factor has been added to
: > the definition of the cepstrum for the sake of military secrets.
: No, I mean to say that the cepstrum was redefined for the
: sake of military secrets. What makes you say that the
: difference between DFT(log|DFT(.)|) and IDFT(log|DFT(.)|)
: is a scaling factor?
Yes, If you look in any DSP book you will see the only difference in the
DFT and IDFT algorithms is that the IDFT is scaled by 1/N, where N is the
length of the DFT. So what effect does this have on the cepstrum?
1) The 'Oppenheim-Schafer' coefficients are 1/N smaller than
`Bogert' cepstrum.
2) If you tried to decode `Bogert' cepstra with the `Oppenheim-Schafer'
algorithm, you would get a speech signal with a dynamic range of
10000dB!! So you'd soon realise something was wrong there.
: > I think you'll probably find that the reason for all subsequent
: > references to to the cepstrum using an IDFT instead of a DFT is
: > because the original paper was wrong.
: How can the original use of a novel term be wrong? I've been
: studying this topic for over a year, and I have no doubt that the
: Schafer-Oppenheim cepstrum is very nearly the identity; it's only
: a slight convolution. If you would read Appendix 2 of the
: Bogert, Healy, and Tukey paper, it will be quite clear that their
: definition is rock-solid.
I think the quote in Herve Broulard and Nelson Morgan's book is quite apt -
`nothing is invented and perfected at the same time' - need I say more.
: > ... far more plausible than some conspiracy theory.
: What's implausible about a government at war (c. 1970) and with
: much earlier levels of encryption technology wanting to protect
: cockpit and other high-noise-environment voice radio encryption
: schemes from automated attacks which require a way to determine
: when the plaintext speech has been recovered?
The cepstrum in either guise is not a military secret, they both have been
published!! Having worked on military equipment design (a number of years ago
before joining H.P.) I can tell you that everything that is to be published
has to be approved, so the cepstrum according to your argument would never have
seen the light of day. Also virtually all algorithms, used in military equipment
are fairly standard and can be found in papers and books, it's the way they
use them, and the exact parameters of the system that is secret. If you want
to know how the British military CODEC of that era worked, then see John Holmes
paper on the JSRU filter-bank (IEE Proceedings 1980). There you
will find that it is not the coder that is secret, but the way they
encrypt the filter bank coefficients. The same will apply to the US
cepstrum system (if it exists).
: Could it be that H-P, one of the founders of which was
: Secretary of Defense in that era, might have been involved?
: See 37 USCFR 5.2, http://www.kuesterlaw.com/lawrule/rules9.htm#52
: Sincerely,
: :James Salsman
:
As my signature says, I speak for myself not for H.P. I am not a spokes-
person for the Industrial-Military machine that some people think rules
this world (Oops the X-Files is creeping in again).
To find out why Oppenheim and Schafer define the cepstrum in the way they did,
have you ever considered asking them? E.g. write a letter to Messieurs
Oppenheim and Schafer c/o IEEE Signal Processing Society. I think you'll get
an answer that has some mathematical reasoning behind it, rather than some
blarney about them having to change it at the behest of the DoD.
Jonathan Thompson
--
The views represented here are mine own and not necessarily those of my
employer, Hewlett Packard.
Subject: Re: Help with Math Problem
From: "E.M."
Date: Fri, 20 Dec 1996 11:46:45 -0500
Sam Bozman wrote:
>
> Could anyone help me with some Math. I have tried to find the answer in
> our local Library, but I am unable to locate the information I need. I
> only have basic grade school Math and perhaps this question is beyond my
> capabilities, but if anyone could give me the formula’s or techniques to
> figure this and other similar problems, I would be grateful.
>
> There are two moons orbiting a planet, both on the same plane. The
> circumference of both orbits is 21 miles. At the beginning of this
> scenario both moons are lined up with the planet at a position I will call
> mile "0". Moon "A" orbits the planet once every day. Moon "B" is slowing
> down at a rate of 2 miles a day. Therefore on the first day moon "B"
> orbits 19 miles, 2 miles short of a full orbit. On the second day moon "B"
> only orbits 17 miles and on the third only orbits 15 miles.
> The question is, how many days will pass before both moons are again lined
> up at mile "0".
>
> This problem looks fairly simple, and in fact I was able to figure it out
> by trial and error. However, I would like to solve similar problems of the
> same type with more complex orbits.
>
> Thank You
> Sam Bozman
> --
To solve this (and similar) problem we need some elementary algebra
and a concept of rotation and angular velocity. Let the first moon's
velocity is N1 (=1 orbit per day). That means that it will make
A = N1*T full cycles for the time T (=number of days). The second
moon's rotational speed is not constant, and at the time T it equals
N2 = N1-T*(2/21) (decelerating and then rotating in the opposite
direction since we are not told that this moon STOPS). From the time
moment 0 until time T this moon will make
(use the average speed, (N1+N2)/2):
B = T*(N1+N2)/2 = T*(N1+N1-T*2/21)/2 = T*(N1-T/21) cycles.
Pay attention to the fact that the latter value first increases and then
decreases and becomes negative (as the 2nd moon starts to rotate
backwards)
The moons are again lined when A = n*B, n being an integer, because
the first 21 day the second moon's absolute velocity is less than that
of the first.
That gives N1*T = n*T*(N1-T/21), or
N1 = n*(N1-T/21) or 21/n = 21 - T. We have N1 = 1, T is unknown, and n
may be 1, 3, 7 or 21 as these numbers divide 21.
(1) n=1 fitted with T=0 (trivial)
(2) n=3 gives T=14
(3) n=7 gives T=18
(4) n=21 gives T=20.
To proceed, you should give negative integer values to n, as follows:
N1 = -n*(N1-T/21), n=21, 7, 3, 1,
and as B becomes more than A, then
-n*N1 = N1-T*21, or T=(N1+n)/21, where n=1,2,3,4...
Merry Christmas!
Eugene.
(2) n=2
N1*T = T*(N1-T/21), or N
Subject: Re: Average
From: Jim Smith UT
Date: Fri, 20 Dec 1996 10:01:51 -0700
> Steven Johnson wrote in article <01bbeded$439b3da0$894277a1@pc66-137.state.ut.us>...
> | Is an average always a mean in mathamatical terms. Or can a average be a
> | median or a mode.
I have seen only one statistics book that contained an obtuse statement
that the median could also be a "kind of average". Otherwise, I have
never seen a definition of "average" in a mathematical or scientific
text or context that was not identical to the definition of "mean". I
have never seen or heard the word "average" used in a scientific,
technical, educational or professional journal, book, presentation, or
acticle where it's usage indicated anything other than a "mean".
However, especially over the past few decades, the term "mean" does seem
to have become the preferred term.
When I was being taught basic math in elementary school, I was taught to
take an "average" of a set of numbers by summing the values of the
elements of the set and dividing by the number of elemants of the set.
Sounds identical to a "mean" to me!
The problem is more of popular perception than of definition. Ask the
person-on-the-street what an "average" value, i.e. the average age of
death is X, means (pun not intended). The response will probably be
something along the line of "half die before age X and half die after".
or "X is the age at which people most commonly die or can expect to
die".
The education most people receive in statistics is similar to my
elemantary education: they know how to average a set of numbers. They
then see one "bell" curve after another and no one bothers to tell them
(or maybe they just don't pay attention) that not all distributions are
normal and that the average (or mean), median, and mode are different
characteristics that can be used to describe a set of numbers, samples,
etc. and that the average, median, and mode are not always the same or
coincident.
Finally, I am fully confident that if the use of "mean" were to become
more common and the-person-on-the-street were to begin to use "mean" in
place of "average", (batting-mean, earned-run-mean?) the same popular
confusion between "mean" and "median" and "mode" will appear as now
exists between "average" and "median" and "mode".
James D. SMITH
Utah Div of Oil, Gas, and Mining
jsmith@sarek.osmre.gov
Subject: Re: Why can't 1/0 be defined???
From: Jon Haugsand
Date: 20 Dec 1996 19:44:24 +0100
pausch@electra.saaf.se (Paul Schlyter) writes:
>
> > 1=1, so 1^(.5)=1^(.5), so -1=1.
> >
> > Similar argument. But that doesn't mean square roots don't exist.
>
> Square roots are different. We all know that taking the square root
> of a number produces two possible results. But subtracting one
> number from some other number should always produce only ONE result.
> Infinity - infinity does not meet this requirement -- therefore
> infinity is not a number.
This may be a matter of definition, but the most common definition of
taking square root of a positive real number x is to take the
*positive* number a such that a*a = x.
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Pho/fax: +47-22852441/+47-22852401
Subject: Re: bincancel:22 large binaries:AR784:@@NCM
From: red@redpoll.mrfs.oh.us (Richard E. Depew)
Date: Fri, 20 Dec 1996 17:41:14 GMT
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--
Richard E. Depew, Munroe Falls, OH red@redpoll.mrfs.oh.us (home)
It's over, and can't be helped, and that's one consolation, as they
always say in Turkey, when they cut the wrong man's head off''
-- Charles Dickens, _The Pickwick Papers_
Subject: Re: Why can't 1/0 be defined???
From: pausch@electra.saaf.se (Paul Schlyter)
Date: 20 Dec 1996 18:34:19 +0100
In article ,
Travis Kidd wrote:
> I've already replied to this through e-mail,
> but briefly:
>
> pausch@electra.saaf.se (Paul Schlyter) writes:
>>infinity + 5 = infinity, OK?
>>infinity + 8 = infinity, OK?
>>Thus:
>> infinity + 5 = infinity + 8
>>If we subtract infinity from both sides of this equation, we get:
>> 5 = 8
> But you can't simply subtract infinity from infinity and expect
> to get zero. This is where your argument is flawed. Infinity
> minus infinity is indeterminate, and could be anything.
Correct. And the reason you cannot consider infinity - infinity
the same as zero is that infinity is not a number.
> 1=1, so 1^(.5)=1^(.5), so -1=1.
>
> Similar argument. But that doesn't mean square roots don't exist.
Square roots are different. We all know that taking the square root
of a number produces two possible results. But subtracting one
number from some other number should always produce only ONE result.
Infinity - infinity does not meet this requirement -- therefore
infinity is not a number.
>> oo = oo + x = oo + y = oo = oo + oo
>>Subtract oo:
>> 0 = x = y = 0 = oo
>
> Same flawed reasoning.
True -- and the flaw is considering infinity to be a number.
>>> So 0*oo=1! Nothing wrong with that.
>>0*oo = any number, not only 1....
>
> But it does equal 1.
It also equals 2, 3 or any other number. Are 1 = 2 = 3 = .... ????
>>These adjustments would have to be not only for special cases but also
>>for everyday cases such as addition and subtraction --- see above. If
>>infinity is a number, then any number equals any other number.
>
> Now: a-a=0 for all a.
> Adjusted: a-a=0 for all finite a.
>
> Not too hard.
It gets damned hard if you need to introduce special cases like that
even for subtraction (and addition).
Why do you want to define "infinity" as a number? Probably because
you don't want to have to treat division by zero as a special case.
But by doing this you've opened a can of worms -- OK, you can now
divide by any number without exception -- but you can no longer add
or subtract any numbers without accounting for special cases.
And even if you define 1/0 = +infinity, you'll have to treat division
with caution. An example:
1/(1/A) = A for all A OK?
Well not for infinities! Consider this:
1/(1/(-infinity)) = 1/0 = +infinity ......
and -infinity is different from +infinity, isn't it?
So you still have to treat division by zero as a special case. So what
did you win? Nothing. What did you lose? A lot -- you now also need
special cases in addition, subtraction, and probably multiplication
as well.
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pausch@saaf.se psr@net.ausys.se paul@inorbit.com
WWW: http://www.raditex.se/~pausch/ http://spitfire.ausys.se:8003/psr/
Subject: Re: generalized associative operations
From: beigel@cs.umd.edu (Richard Beigel)
Date: 20 Dec 1996 14:50:22 -0500
Thanks to everyone who responded. I wrote a program to try all mappings
from {0,1}^3 to {0,1}. It turned up one additional example:
* [a,b,c] = not (a xor b xor c)
I wrote:
: I am looking for a mapping [,,] from {0,1}^3 to {0,1} satisfying
: [[a,b,c],d,e] = [a,[b,c,d],e] = [a,b,[c,d,e]]
: I know that the following mappings have that property:
: * [a,b,c] = a or b or c
: * [a,b,c] = a xor b xor c
: * [a,b,c] = a and b and c
: * [a,b,c] = a
: * [a,b,c] = c
: * [a,b,c] = 0
: * [a,b,c] = 1
--
Richard Beigel telephone: (301)405-2764
Dept. of Computer Science email: beigel@cs.umd.edu
University of Maryland fax: (301)405-6707
College Park, MD 20742-3251 office: A. V. Williams Bdg. #3171
Subject: Re: Why can't 1/0 be defined???
From: yuggoth
Date: Fri, 20 Dec 1996 13:59:03 -0600
Andreas Neubacher wrote:
>
> In article <58ndvv$oh4@newshound.csrv.uidaho.edu>,
> foster@cs.uidaho.edu (James Foster) writes:
> JF> Isn't the easy answer that if 1/0 = a, then a*0 = 1, and this can't be
> JF> true for any a>0. So 1/0 = 0. But then b/0 = 0, too, by the same
>
> Yes, 1/0 = a cannot hold for any real number `a'.
>
It won't hold for ANY real number 'a', but it will hold for ALL real
numbers 'a'. Think about it this way...
If you have a pie, and cut it into no parts.. how much of a pie do you
have left? Either way you look at it ypou have no pie left. If you
kept doing this you could end up with either NO parts of the pie left,
but isn't division how many times you can remove something from
something else? That would mean you could remove an infinite amount of
zeroes from that pie.
> The point is that one could define a *new* constant `a' that is different
> from any real or complex number and has the property that 1/0 = a. Then
> one can start defining additional properties, axioms, etc. and see what
> kind of structure this yields.
>
> Andreas.
>
If you start defining new types of constants you would have a new type
of math... useless to anyone that wasn't trying to divide something by
zero. That would mean you have to define ALL other elements in that new
math. No rules would apply to it that you could not prove using only
YOUR constants.
***************************************************
Just my two cents..
Tom Adam
yuggoth@geocities.com
Subject: Re: Vietmath War: need help on physics history
From: c.street2@ukonline.co.uk (Chris Street)
Date: Fri, 20 Dec 1996 20:17:09 GMT
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) wrote:
>In article <32B9BC0A.63A3@mindspring.com>
>Richard Mentock writes:
>> What about the old trisection of angle using compass and straightedge?
>> That was long considered impossible, and was proven impossible in the
>> nineteenth century. However, if you change the rules just a little bit
>> (allow the solver to mark on the straightedge, or even to have had a
>> mark on the straightedge), then the problem is solvable.
>The purpose of analogies is to devise them to help you think. Not to
>turn the switch off. For every 1 helpful analogy, a person can dream up
>an infinity of useless analogies. But I am talking to deaf and dumb
>ears here, can you read lips?
>>
>> That's what happens with FLT and the p-adics. Now, the bigger question,
>> whether the p-adics are or should be the set of numbers that we use to
>> view the world, seems to be a question in the domain of physics or
>> metaphysics.
> You use in wasting anymore of my time on you. You sound like a blue
>collar worker who is faking to be a learned intellectual. Dumbo, your
>above implies you believe that mathematics is utterly distinct from
>physics. And you probably even think there are three distinct subjects
>of math, physics and metaphysics. But you are not alone, for the
>majority of people think that these subjects are distinct. I myself
>realize that all subjects are somewhere inside of physics, everything,
>from poetry to biology to physics itself. And there ain't no thing as
>metaphysics. And your precious mathematics is just a subdepartment of
>physics.
> I don't write this reply for your edification Mentock, I write it for
>everyone except you. I want you to stay in the weeds.
I guess we can see here the true scientist volunteering information to
all free of charge, and the alchemist hiding his arcane knowledge from
those who are not worthy. I know who I would trust.
The Original Mad alCHEMIST
The sender(s) of unsolicited Email to this address agree to pay £250 for proofreading services.
"79.84% of all statistics are made up on the spot."
Subject: Re: Geometry Question
From: "E.M."
Date: Fri, 20 Dec 1996 15:08:01 -0500
Dave Robinson wrote:
> I've got a hypersphere defined as:
> B^2= x1^2+x2^2+... the radius B is known.
> I have a general surface defined in terms of the x vector and some
> constant z: for example:
> g(x,z)= z + a0+a1*x1^2 + a2*x2 + ..... the constants: a_i are known.
>
> the function will typically be a 2nd order polynomial in X
>
> Problem (finally): how to find the X_i such that the surface is
> tangent to the sphere?
In general we can formulate the problem as follows: find X=X0(vector)
such that surfaces f1(X)=0 and f2(X)=0 ar tangent at X0.
In other words we have to satisfy the system of equations
f1(X0)=0, f2(X0)=0 and Df1(X0) = q*Df2(X0), where D is a symbol for
gradient, Df(X) = (df/dX1, df/dX2, ..., df/dXn) and q is a
proportionality coefficient. If z is to be considered as unknown
then we obtain n+2 equations for n+2 unknowns which may (or may not)
have solution(s).
Eugene.
Subject: Number Theory Puzzle
From: blackj@toadflax.cs.ucdavis.edu (John R. Black)
Date: 20 Dec 1996 20:22:41 GMT
I have a puzzle to which I don't know the solution, and unfortunately am not
even sure I correctly remember the problem statement. I'm hoping someone
here recognizes the puzzle and knows the solution. Here it is:
---------------------------
There are 2 mathematicians, call them A and B.
X and Y are positive integers > 1.
A knows the sum of X and Y, i.e., A knows the value of S=X+Y.
B knows the product of X and Y, i.e., B knows the value of P=X*Y.
A and B have the following conversation:
A : You CAN'T know the sum.
B : I still don't know the sum.
A : I don't know the product yet.
B : Now I know the sum.
Find X and Y.
---------------------------
It's easy to see that S cannot be even (if you believe the Goldback
conjecture) or a prime+2, but I've gotten no further.
john//
Subject: A flow problem
From: "Phil"
Date: 20 Dec 1996 20:45:28 GMT
Can anyone help? I have a container of high strength effluent, into which I
plan to pour (at constant rate, and assuming perfect mixing) a lower
strength effluent. The resulting mix is to overflow to sewer. I know the
volumes of the container, original effluent, both strengths, and the flow
rate. I need to be able to calculate the strength at a given time. Is
this a simple 1st order differential equation (y=ae^kx), using the 1/2 life
method, or would that be too easy? I would appreciate any thoughts.
Thanks.
Subject: Re: Why can't 1/0 be defined???
From: marnix@worldonline.nl (Marnix Klooster)
Date: Fri, 20 Dec 1996 21:40:57 GMT
tkidd@hubcap.clemson.edu (Travis Kidd) wrote:
> Gene Eberts writes:
> > To divide a number A by a number B is to find a number Q such that
> > A = B * Q
> I totally agree.
>
> > A is called the dividend, B is called the divisor and Q is called
> >the quotient.
> Yes, since I speak English.
>
> > Two natural requirements are imposed on the quotient: it must exist
> >and it must be unique.
> Why? What does this get us? Can't we merely use the above definition
> and then *prove* that Q exists and is unique if B is not zero? And if
> B=0, then those requirements are not met. What's the harm?
Nothing, as long as you don't try to define `the' quotient of
something and zero. If A and B are zero, then any Q satisfies
the above equation. So what is 0/0 then: an expression with
multiple values? This gets you into trouble real quick. If you
define division so that
a/b = q == a = b*q (1)
(where == is logical equivalence) holds for all `numbers' a, b,
and q, you're in for trouble. (I assume there are distinct
`numbers' 0, 1, and 2, with 0*1=0*2=0.) Taking a=0 and b=0 you
find that
0/0 = q
for every `number' q with 0*q=0. Thus 0/0 is equal to both 1 and
2, and by transitivity of equality 1 and 2 are equal to each
other: a contradiction. In mathematical parlance, definition (1)
above is not sound, since it violates the uniqueness requirement.
As another example, elsewhere in this thread you argue that
0*oo (zero times infinity) is equal to any number. Naively,
transitivity then gives
0*oo = 1 and 0*oo = 2, therefore 1 = 2
and you have again introduced a contradiction.
Conclusion: current mathematics does not really like
expressions that can have more than one value. This is
understandable, really, since it requires major rethinking of the
formal underpinnings of mathematics: either every expression has
at most one value, or you need to modify your notion of equality.
I believe this can be done, but not in the context of current
mainstream mathematics. If you find an elegant and consistent
way of using expressions with multiple values within conventional
mathematics, let me know. (BTW -- does anyone have pointers to
research about formal systems that support this kind of
`nondeterminism'?)
* * *
As an aside, there are formal theories that have a quantified
expression (tau x| P) ("an x such that P") with the following
proof rule:
(Ex:: P) == P[x:=(tau x| P)]
(where (Ex:: P) is existential quantification of P over x, and
P[x:=...] means to substitute for x in P.) For example, using
(x=3 or x=7) for P we can prove
(tau x| x=3 or x=7) = 3 or (tau x| x=3 or x=7) = 7
Thus we have an expression that is equal to either 3 or 7, but
*we have no way of knowing which*.
Back to division: for a,b in any set S which has an element 0
and a multiplication operator * with zero 0, we can now define
division unconditionally as
a/b = (tau q| q in S and a=b*q)
Using the above rule then gives
(Eq:: q in S and a=b*q) == (a/b) in S and a=b*(a/b)
For b=0 one can prove this to be equivalent to
a=0 => (a/0) in S
Thus according to this definition, basically all we know of 0/0
is that it is an element of S, and we know nothing of a/0 for
a=/=0.
More on this tau-operator (sometimes called epsilon, and it
has a friend called iota) can be found in a book by A.C.
Leisenring (I don't have a more precise reference on hand --
sorry.)
> -Travis
Groetjes,
<><
Marnix
--
Marnix Klooster | If you reply to this post,
marnix@worldonline.nl | please send me an e-mail copy.
Subject: Re: comp.speech Frequently Asked Questions - part 2/3
From: James Salsman
Date: 20 Dec 1996 21:01:48 GMT
Jonathan Thompson wrote:
> James Salsman (jsalsman@bovik.org) wrote:
> : Jonathan Thompson wrote:
> : >
> : > ...You mean to say that a scaling factor has been added to
> : > the definition of the cepstrum for the sake of military secrets.
> :
> : No, I mean to say that the cepstrum was redefined for the
> : sake of military secrets. What makes you say that the
> : difference between DFT(log|DFT(.)|) and IDFT(log|DFT(.)|)
> : is a scaling factor?
>
> Yes, If you look in any DSP book you will see the only difference in the
> DFT and IDFT algorithms is that the IDFT is scaled by 1/N, where N is the
> length of the DFT.
If that were the case, why would FFT-IFFT code libraries have
similar-length definitions for each function, instead of a
simple loop to rescale the outputs after a call to the other
more lengthy function?
I've been through FFT code and chapter 33 of _Intro._to_Algorithms_
by Cormen, Leiserson, and Rivest. I believe there is a grain
of truth here, but the scalling in question occurs deep inside
the inner loop of some implementation of the {I,}DFT, and does
not apply simply to the outputs.
Would you please cite (or better yet excerpt) a source showing
this difference in scaling between the DFT and IDFT?
Sincerely,
:James Salsman
Subject: Re: Why can't 1/0 be defined???
From: electronic monk
Date: Fri, 20 Dec 1996 15:53:54 -0800
Travis Kidd wrote:
> >lim 3^x = oo
> >x->oo+
> This makes no sense.
>
> >lim 3^x = 0
> >x->oo-
> No, this limit is oo.
i wasn't trying to confuse. I meant
lim 3^x = oo
x->+oo
and
lim 3^x = 0
x->-oo
except that i was trying to keep with the idea that +oo and -oo are the
same thing. that was why i used oo+ instead of +oo, so that instead of
"approaching positive infinity" is said "approach infinity from the
right." i just didn't want to contradict myself by saying first that
+oo is the same as -oo and then separating them again.
> >oo isn't a set number and doesn't equal itself, as has been proven in
> >previous messages, so 3^oo does not nessasarily have to equal 3^oo.
> oo indeed is a set number that indeed equals itself. Every number
> equals itself.
oo != oo because oo-oo != 0 and oo/oo != 1. numbers that are the same
(excluding 0) would make all these true. zero is an exception because
it is related to oo very closely. on a polar grid, r = 0 is one pole
and r = oo can be thought of as the "other" pole, for just like on a
globe, if your on the north pole, no matter which direction you walk in,
your heading tward the south pole, of course, on a globe, you can reach
the other pole.
> You are correct that 3^oo does not necessarily have
> to equal 3^oo. After all, 1^(1/2) does not necessarily have to
> equal 1^(1/2).
if you think of 1^(1/2) as one number your right, but it is two numbers
and you can pick from the two which ever you like. ex:
1^(1/2) = x
1 = x^2
0 = x^2 - 1
0 = (x-1)(x+1)
so one picks 1, or -1, for which ever use he finds.
1^(1/2) is not just one number, it is two, and those numbers are always
the same. so 1^(1/2) = 1^(1/2) if you take into account both roots at
the same time.
> But if you take an infinite power as infinite mul-
> tiplication and a negative infinite power as infinite division,
> then certainly there is a clear difference between oo and -oo.
yes, that is because the path you take tward oo leads you tward 0 if you
go around the left side, and if you go right, the graph tends upward,
leading you toward oo. i'll have to think of another example of this
probably.
>
> >this is very interesting. in some cases infinity could be sensitive to
> >which direction that you approach it from. in this case, if you
> >approach it from the negative side, you'll begin to slant downward,
> >because you'll take the reciprical with the negative exponent. it's
> >tough to try and prove all this though.
> I don't think it is a matter of proof. Rather, it is axiomatic.
> I suppose a proof that oo exists can be given by answering the
> question "How many numbers are there?".
that's very true.
> >I don't really know if this is true, but it makes a little sense to me.
> >i'll see if i can find any literature on this because i think it is a
> >pretty cool idea.
> Yeah. I do too.
>
> -Travis
electronic monk
Subject: Re: holiday group question
From: Fred Galvin
Date: Fri, 20 Dec 1996 15:37:40 -0600
On 19 Dec 1996, Dr D F Holt wrote:
> If by F_n you mean the free group of rank n, then F_n and F_m are not
> isomorphic for n != m, as is easily seen from the fact that the number
> of homomorphisms from F_n to the group of order 2 is exactly 2^n.
> If n and m are greater than 1, then F_n and F_m contain each other
> as subgroups, but it is misleading to say that they are in some sense
> isomorphic.
This argument also generalizes to any variety (of universal algebras)
having a nontrivial finite model.
> A related question which occurred to recently.
> Suppose n and m are distinct infinite cardinal numbers.
> Is it always true that F_n and F_m are not isomorphic?
Yes, because F_n has cardinality n if n is infinite.
> For those of you who find the original problem too easy, try:
>
> Let G be a countable group. Can G be embedded in a 2-generator group?
I seem to recall that there was an expository note about this in the
Monthly, a couple years ago.
Subject: Re: Why can't 1/0 be defined???
From: scottb@wolfram.com (Scott Brown)
Date: 20 Dec 1996 20:51:10 GMT
tkidd@hubcap.clemson.edu (Travis Kidd) writes:
>I've already replied to this through e-mail,
>but briefly:
>pausch@electra.saaf.se (Paul Schlyter) writes:
>>infinity + 5 = infinity, OK?
>>infinity + 8 = infinity, OK?
>>Thus:
>> infinity + 5 = infinity + 8
>>If we subtract infinity from both sides of this equation, we get:
>> 5 = 8
>But you can't simply subtract infinity from infinity and expect
>to get zero. This is where your argument is flawed. Infinity
>minus infinity is indeterminate, and could be anything.
But I thought you said infinity is a number? So, what kind
of number did you have in mind? The well-definedness of
subtraction is for most numbers a most useful characteristic.
One of us is confused.
Scott
------------------------------------------------------------------------
| Scott Brown | "I may be speaking from Wolfram Research, Inc., |
| scottb@wolfram.com | but that doesn't mean I'm speaking for them." |
------------------------------------------------------------------------
Subject: Re: A question about an operator and its spectrum
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 20 Dec 1996 14:52:16 -0500
In article ,
Alexander E. Gutman wrote:
:> From: dutour@jonque.ens.fr (Mathieu Dutour)
:> Date: Thu, 19 Dec 96 18:31:45 GMT
:> Subject: Re: A question about an operator and its spectrum
:>
:> the trivial counterexample is:
:> [1 1]
:> [ ]
:> [0 1]
:> which has norm equal to 1 and spectrum equal to {1}
:
:I cannot find a norm in C^2 under which this operator has norm 1.
:Can you help me?
It cannot exist: If an operator T has an induced norm equal to 1 (or
less) then the sequence {T^n : n>0 integer} is norm bounded. However, the
operator I+J above (J being the 2x2 Jordan cell) makes the sequence
n |-> (I+J)^n = I + n*J unbounded in every norm. (We're in finite
dimension, and all norms are Lipschitz equivalent. Also, every norm in a
Banach space X which is dominated by the original norm and makes X
complete is Lipschitz equivalent to the original norm (the "two norms
lemma", provable from the Closed Graph Theorem).)
The poster of the alleged counterexample might have had in mind the
maximum absolute value norm of a matrix, but this is not even
submultiplicative, much less an operator norm.
I'm still working on the original question. As you can see from an
extension of my explanation, the answer is YES in finite dimension: if
spectrum(T)={1} and norm(T)=1 then T=I.
Good luck, ZVK (Slavek).
