mm-434
Subject: Re: trig problemsemailed me seperatly, I don't care
for the tone.kika_chuck>2. Find all values of y such that the
distance is 12>(3,y) (-2,9)> What is the formula for the
distance between two points, in> rectangular coordinates?
Apply that to the given pair of points, set> it equal to 12,
and solve for y.> -- > Stan Brown, Oak Road Systems, Cortland
County, New York, USA> http://OakRoadSystems.com> Fortunately,
I live in the United States of America, where we are> gradually
coming to understand that nothing we do is ever our> fault,
especially if it is really stupid. --Dave Barry
===
Subject: Re:
trig problems>emailed me seperatly, I don't care for the
tone.If you have a problem with that person perhaps you should
(a) take it up with him or her directly and (b) not quote my
public follow-up since it has no relevance to your comment
_and_ I didn't e-mail you.-- Stan Brown, Oak Road Systems,
Cortland County, New York, USA
http://OakRoadSystems.comFortunately, I live in the United
States of America, where we aregradually coming to understand
that nothing we do is ever ourfault, especially if it is
really stupid. --Dave Barry
===
Subject: Re: trig problems Well,
I wasn't talking about you, in fact I found your message
helpful, sojust relax.kika_chuckwho>emailed me seperatly, I
don't care for the tone.> If you have a problem with that
person perhaps you should (a) take> it up with him or her
directly and (b) not quote my public follow-up> since it has
no relevance to your comment _and_ I didn't e-mail you.> -- >
Stan Brown, Oak Road Systems, Cortland County, New York, USA>
http://OakRoadSystems.com> Fortunately, I live in the United
States of America, where we are> gradually coming to
understand that nothing we do is ever our> fault, especially
if it is really stupid. --Dave Barry
===
Subject: Re: Need help
with proof of: sum(1...n^2) = n^3/3 + n^2/2 + n/6
===
Subject:
find the value of a and bI'm stuck with this problem:find de
value of a en b for which the following equations are
consistent -for every value of x the three equations are
valida^2.x =(a+b)x + a + 2ba^2.x + 1 =b^2.x+a+b(a^2 + 2ab).x
+a +b + 1=3b^2.xI can't find the right angle to tackle this
one.Who can help meMany thanks in advance,Harry
===
Subject: Re:
find the value of a and bContent-transfer-encoding: 8bit> I'm
stuck with this problem:> find de value of a en b for which
the following equations are consistent -> for every value of x
the three equations are valida^2.x =(a+b)x + a + 2b> a^2.x + 1
=b^2.x+a+b> (a^2 + 2ab).x +a +b + 1=3b^2.xI can't find the
right angle to tackle this one.> Who can help meMany thanks in
advance,> HarrySo, you have:[1] a^2*x = (a + b)*x + a + 2*b;[2]
a^2*x + 1 = b^2*x + a + b;[3] (a^2 + 2*a*b)*x + a + b + 1 =
3*b^2*x.You want values of a and b such that..A. [1] and [2]
and [3] are all true for all x for the same a and b orB. An a
and b such that [1] is true for all x; an a and b such that
[2] is true for all x; an a and b such that [3] is true for
all x.C. [1] and [2] and [3] all represent the same equation
orD. [1] and [2] and [3] all have the same solution orE. There
is at least one x such that [1] and [2] and [3] are all true
orF. Something else.You have seen solutions to some of these
for your previous questionwhich you should be able to adapt to
this one (with the new [2]).I am not sure anyone knew exactly
what the question was - I know I didnot.Perhaps if you picked
one of A - F (and in the case of F told us whatthe something
else was), someone would try again to help you out.-- Paul
SperryColumbia, SC (USA)
===
Subject: Re: find the value of a
and bContent-transfer-encoding: 8bitI'm stuck with this
problem:> find de value of a en b for which the following
equations are consistent -> for every value of x the three
equations are valida^2.x =(a+b)x + a + 2b> a^2.x + 1
=b^2.x+a+b> (a^2 + 2ab).x +a +b + 1=3b^2.xI can't find the
right angle to tackle this one.> Who can help meMany thanks in
advance,> HarrySo, you have:> [1] a^2*x = (a + b)*x + a + 2*b;>
[2] a^2*x + 1 = b^2*x + a + b;> [3] (a^2 + 2*a*b)*x + a + b + 1
= 3*b^2*x.You want values of a and b such that..A. [1] and [2]
and [3] are all true for all x for the same a and b or> B. An
a and b such that [1] is true for all x; an a and b such that
> [2] is true for all x; an a and b such that [3] is true for
all x.> C. [1] and [2] and [3] all represent the same equation
or> D. [1] and [2] and [3] all have the same solution or> E.
There is at least one x such that [1] and [2] and [3] are all
true or> F. Something else.You have seen solutions to some of
these for your previous question> which you should be able to
adapt to this one (with the new [2]).I am not sure anyone knew
exactly what the question was - I know I did> not.Perhaps if
you picked one of A - F (and in the case of F told us what>
the something else was), someone would try again to help you
out.OP e-mailed me a response. My reply to him bounced, so
here it is:OP: I mean A. The answer is a = -3/2 and b = 1/2 or
a = -1/2 and b = -1/2 The problem is how to find these values
for a and bMe:To solve A you would need at least one a and b
so that _all_ ofa^2 = a + b; 0 = a +2*b;a^2 = b^2, 1 = a +
b;a^2 + 2*a*b = 3*b^2, a + b + 1 =0are true.Six equations with
two unknowns obtained by equating coefficients in[1], [2] and
[3]. That system has no solution since, for example, 1 = a + b
and a + b + 1 = 0 is not possible.However, substitute a = -3/2
and b = 1/2 into [3]. You get (9/4 -3/2)*x - 3/2 + 1/2 + 1 =
(3/4)*x which reduces to 0 = 0.Substitute a = b = -1/2 in [3]
and you will get (1/4 +1/2)*x -1/2 - 1/2 + 1 = (3/4)*x which
also reduces to 0 = 0.Working backward, equating coefficients
of [3] gives a^2 + 2*a*b = 3*b^2 and a + b + 1 = 0. The first
equation gives 0 = a^2 + 2*a*b - 3*b^2 = (a + 3*b)*(a - b). So
a = -3*b or a = b.Substituting a = -3*b into a + b + 1 = 0
gives b = 1/2 and a = -3/2.Substituting a = b gives a = b =
-1/2.So, I think the problem was really BSimilarly, for [2],
a^2 = b^2 and 1 = a + b gives a = b = 1/2.For [1], a^2 = a + b
and 0 = a + 2*b gives a = b = 0 or a = 1/2 and b = -1/4.a =
1/2, b = -1/4 or a = b = 0 will make [1] be an identity - true
forall x.a = b = 1/2 will make [2] be an identity.a = -3/2 b =
1/2 or a = b = -1/2 will make [3] be an identity.-- Paul
SperryColumbia, SC (USA)
===
Subject: Re: find the value of a
and bI'm stuck with this problem:> find de value of a en b for
which the following equations are consistent -> for every value
of x the three equations are valida^2.x =(a+b)x + a + 2b> a^2.x
+ 1 =b^2.x+a+b> (a^2 + 2ab).x +a +b + 1=3b^2.xI can't find the
right angle to tackle this one.> Who can help meMany thanks in
advance,> HarryThe techniques applicable are those described in
alt.algebra.help whenyou asked a similar question.--
G.C.
===
Subject: Re: The trouble with calculators (Re: tricky
factorization)>>Further, I think this indicates that *in
practice* it's not an essential >>idea. Not that I'm defending
using that as a gold standard (*in >>practice* all that is
essential is fulfilling biological urges).>> >Perhaps I
misunderstand. Are you saying that substitution is not an
>essential idea? If so I disagree strongly. don't), I disagree
strongly too. I was tired, and these exercises are part of a
series of tricky substitutions (well, this one [u=x+1/x] looks
like one). That is, it's a substitution that makes a difficult
calculation easier -- it doesn't really promote understanding.
Is this the best use of our time? But I don't have the full
details of the syllabus, the students, etc., so maybe I was
just being cranky.>>The reason that calculators/computers were
introduced into the >>curriculum was to counteract an emphasis
on calculation over >>understanding. How does knowing tricks
of extremely limited >>applicability aid understanding?>There was no emphasis on calculation in any of the
mathematics courses I >ever took.There was some emphasis on
calculation in your high school courses and, unless you were
pretty fortunate, in your calculus course. It's the nature of
the beast (in calculus) that the calculation required taxes
the abilities of a subset of students. This was noticed, and
the claim was made that1) ideas are more important than
calculation and2) machines calculate better than people
anyway.So the idea behind calculators is that we can emphasize
ideas rather than getting hung up on the calculations.Part of
the idea is that any sort of reasonable application gets to
calculations that are beyond the interest of everyone
(including the instructor) long before the model gets
realistic enough to even approach practicality. (Why do we
ignore drag in calculus ballistics problems? Because the level
of detail is too involved for a calculus course. But how do the
students get some sort of idea that they can use calculus in
real life, if we never show them anything even remotely
resembling real life?)Anyway, that's the idea behind it. I
have to admit that the idea is right -- why bother with
calculus at all if we're not going to either do it right
(rigorous) or do it useful (applications)? Is it just a
weed-out course for the engineering college?> There is now, in
the wake of calculators and the onslaught of >the push-button
world. I watched this disaster as it occurred. Its main
>function, in my experience, has been to give teachers with
weak insight >into mathematics something to do.I am baffled
why (even good) teachers don't discuss the limitations of what
we're teaching. Differentials for approximations? Show some
examples where it doesn't work well, and show why. Show them
functions where the just push the buttons answer is wrong.
Show them how to find out when the calculator gives wrong
answers. (4x3=1. It's what the calculator says, write it
down.)> If their students are typing and looking at LED's and
blinking lights all day, at least they're doing something.> To
be honest, I still think teaching calculus in high school is a
bad idea. Not because the students can't handle it, but
because so many of the teachers can't. But they can teach bad
habits.Jon Miller
===
Subject: Re: Math Tricksays...> Here is a
Math Trick I invented.http://w4u.sytes.net/MagicMathLet me
know how you find it. gonzalep@.84scientist.com no .84> it's
hosedmike
===
Subject: Regular GrammarsHi everyone,Suppose A and
B are regular languages over some set I (i.e. they havethe same
alphabet). I am to show that the set 'A intersection B' isalso
a regular language. Call this set C.There are two ways I can
go about this problem. I can either show thatthere is a
regular grammar G such that C = L(G) or I can show that
afinite-state automaton M exists such that C = Ac(M) where
Ac(M) is theset of accepting input strings for M.Unfortunately
I can't think of any solid argument to show that C is aregular
language. If C is finite, I could very well create a
regulargrammar to generate every string in C. However, I can't
really assumethat C is finite. Any hints or
suggestions?Bernd
===
Subject: Re: Regular GrammarsLet the
following be given:G1: terminal alphabet T; nonterminal
alphabet N1; start symbol S1G2: terminal alphabet T;
nonterminal alphabet N2; start symbol S2G1 generates AG2
generates BN1 and N2 assumed disjoint without loss of
generalityNow let G be the grammar with terminal alphabet T,
nonterminal alphabetpowerSet(N1 union N2), and start symbol
S={S1,S2}, whose productions are...(i) all productions P->xQ
for which Q = union_i {Qi in Ni: Pi->xQi is aproduction of Gi
for some Pi in P}(ii) all productions P->(empty) such that,
for each i, Pi->(empty) is aproduction of Gi for some Pi in
P(iii) nothing elseAn obvious induction shows that x1...xnP is
deriveable from S using theproductions of G iff P=union_i {Pi
in Ni: x1...xnPi is deriveable from Si usingthe productions of
Gi}. Hence L(G)=(A intersection B).| Hi everyone,|| Suppose A
and B are regular languages over some set I (i.e. they have|
the same alphabet). I am to show that the set 'A intersection
B' is| also a regular language. Call this set C.|| There are
two ways I can go about this problem. I can either show that|
there is a regular grammar G such that C = L(G) or I can show
that a| finite-state automaton M exists such that C = Ac(M)
where Ac(M) is the| set of accepting input strings for M.||
Unfortunately I can't think of any solid argument to show that
C is a| regular language. If C is finite, I could very well
create a regular| grammar to generate every string in C.
However, I can't really assume| that C is finite. Any hints or
suggestions?|| Bernd
===
Subject: Re: Regular Grammars>Hi
everyone,>Suppose A and B are regular languages over some set
I (i.e. they have>the same alphabet). I am to show that the
set 'A intersection B' is>also a regular language. Call this
set C.>There are two ways I can go about this problem. I can
either show that>there is a regular grammar G such that C =
L(G) or I can show that a>finite-state automaton M exists such
that C = Ac(M) where Ac(M) is the>set of accepting input
strings for M.>Unfortunately I can't think of any solid
argument to show that C is a>regular language. If C is finite,
I could very well create a regular>grammar to generate every
string in C. However, I can't really assume>that C is finite.
Any hints or suggestions?Try the second approach. Start with
M(A), a finite-stateautomaton that accepts A, and M(B), a
finite-state automaton thataccepts B, and try to construct a
finite-state automaton M thataccepts C. It's a standard
construction, but it may prove a bittricky the first time you
see it. The basic idea is to run M(A)and M(B) simultaneously,
accepting the input if and only if bothmachines accept it; the
trick is to figure out how to make asingle machine do this.
Hint: Let the state set of M by theCartesian product of the
state sets of M(A) and M(B). That is,the states of M are going
to be the pairs (s, t) such that s is astate of M(A), and t is
a state of M(B).Brian
===
Subject: Re: Short FLT, more
information> I've been told that I haven't included enough
information for the > mathematicians. So, here's as much as I
can get out with this format > with an attempt at more
rigor.1> x^n + y^n = z^n with n odd prime. no variables zero.
x,y,z E N2> Assuming a solution, another solution (if it
existed) could be > reached by the following: (x+ delta x)^n +
(y+ delta y)^n = (z+ delta z)^nHow do you know this? Couldn't
the solution (x,y,z) be unique for n?
===
Subject: Re: Short
FLT, more information> I've been told that I haven't included
enough information for the > mathematicians. So, here's as
much as I can get out with this format > with an attempt at
more rigor.1> x^n + y^n = z^n with n odd prime. no variables
zero. x,y,z E N2> Assuming a solution, another solution (if it
existed) could be > reached by the following: (x+ delta x)^n +
(y+ delta y)^n = (z+ delta z)^nHow do you know this? Couldn't
the solution (x,y,z) be unique for n?No... if you have one
solution, you can multiply both sides by, say, 2^nand
construct another solution from it, which then could be
representedin the above form. If there's one solution, there
are an infinity ofother ones related to it as well.
===
Subject:
Re: questions about the Dedekind construction of R
<hale-1508032320000001@dialup-5-144.tulane.edu>
<9d88b76e.0308161722.14696a35@posting.google.com>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate:
SPA(GIS)X-Tinguish: Mark Griffith
<markgriffith@rocketmail.com>X-Treme: C&C,DWS at 06:22 PM,
u4knuj@hotmail.com (Stuck) said:>By equivalence, I mean that
these sets>are equal, not just isomorphic. By that definition
they're not equivalent. They are however, allisomorphic. More
than that; the isomorphisms are unique. EveryMathematician
I've heard of would consider that to be equivalence.-- Shmuel
(Seymour J.) Metz, SysProg and JOATAny unsolicited bulk E-mail
will be subject to legal action. I reserve theright to publicly
post or ridicule any abusive E-mail.Reply to domain Patriot dot
net user shmuel+news to contact me. Do not replyto
spamtrap@library.lspace.org
===
Subject: Re: questions about the
Dedekind construction of RX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate:
SPA(GIS)X-Tinguish: Mark Griffith
<markgriffith@rocketmail.com>X-Treme: C&C,DWS at 05:22 PM,
u4knuj@hotmail.com (Stuck) said:>1) R was defined as the set
of all cuts. Like most people, I>normally think of a real
number as something with an infinite decimal>expansion. I
can't speak for most people, but that is not a productive way
tothink of real numbers and any school system that teaches
that way isbadly broken.>It appears to me that a decimal
expansion is not the same as a cut.Correct.>How is this
problem resolved?Don't let anybody teach in the public schools
unless they know thesubject that they are supposed to be
teaching. Close all of theschools of education and replace the
current certification laws withlaws requiring competence.>2)
Rudin showed that there is an isomorphism between Q, and
the>ordered field Q*, whose elements are the rational cuts,
which he>defined. He then argued that therefore Q is a
subfield of R. Are you sure that he didn't argue that Q* is a
subfield and that thereis an natural isomorphism between Q and
Q*? Or perhaps he was justbeing sloppy, although I consider
that less likely. Could you quotethe exact wording?-- Shmuel
(Seymour J.) Metz, SysProg and JOATAny unsolicited bulk E-mail
will be subject to legal action. I reserve theright to publicly
post or ridicule any abusive E-mail.Reply to domain Patriot dot
net user shmuel+news to contact me. Do not replyto
spamtrap@library.lspace.org