mm-451
Subject: Re: My quick math quide> One of the reasons I made
the post Quick Math Guide to core error> issues was to have a
handy reference. Having found myself floundering> a bit, I
found myself looking back at it, wch is when I realized a> wle
back that I'd wondered far astray, and I posted that I'd made
a> mistake with my addition of y to an example from Rick
Decker.> I then sat back, thought for a wle, and realized that
possibly my> problem had to do with difficult concepts that so
push the human brain> that most are not up to handling the
mathematics as I'd been giving> it.> So I'm starting again,
and looking for better ways to to explain.> P(m) = f^2((m^3
f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)> The form
of the polynomial allows me to factor P(m) into> non-polynomial
factors, and the factorization with those factors is> P(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)> where the a's are roots
of the following cubic:> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 -
3m^2 f^2 + 3m).> I've adjusted those expressions several ways
to try and make it easier> to underd.> So far those adjustment
apparently haven't been very successful, and> now I'm beginning
to realize that I need to underd human beings> better as
abstractions. I need to figure out how their brains work.>
That puts me back into study mode. I have to underd your
mental> wiring to an even greater degree than I do now.> I
have to figure you out.> Harris!Are you an old student of
sosiology? Never stopped studying? Eight years?Throw a
matamatical problem to sci.math, with lots of errors and sit
backand study what happens?OK, but now is the time to stop,
because you have been caught by your ownproblem. You have been
obcessed by your problem stated and can't get out.I will state
a really simple problem for you (your answer is crucial for
meto continue looking in/on your treads)If A is a positive
real number, what is sqrt(A^2)?
===
Subject: Re: Confusing
natural logarithms> I'm just wondering, is it possible to
solve ts equation without the> use of some number cruncng
software?4.551 = X + 2e^(-0.5X)We only have one variable here,
but I just can't seem to isolate it. > Avvfter computing away,
i got a rough value of X being equal to 4.32Any help with ts
or any resources you could send me to would be> greatly
appreciated. .According to one person I know there *is* a
closed form solution ifyou can put the equation in the formx =
ye^y -- in ts case, AFAICT you'll want y=-0.5Xtheny = W(x)
where W(x) is the Lambert W function also known
ashttp://mathworld.wolfram.com/LambertsW-Function.htmlWhat is
closed form, anyway?You can see an easy example
at:http://www.anasoft.co.uk/EE/widlarlambert/
widlarlambert.htmlI'm about to post on an equation that's in
the same form as yours,it just looks worse, so check out the
thread. I have the solutionsfrom Maple and Mathematica, I just
can't get to the x = ye^y form.--
===
Subject: Re: Confusing
natural logarithmsCorrection<snip>y = W(x) where W(x) is the
Lambert W function also known as ^^^^^^^ sorry,
ProductLog
===
Subject: Topology helpI would be most
appreciative if someone could please help with the
followingLet h: S^1 -> S^1 (where S^1 is the unit circle in
R^2) be continuous andantipode-preserving (i.e., h(-x) = -h(x)
for all x in S^1) with h(b_0) =b_0. Show that h_* (h_*: [the
fundamental group of S^1 at base pointb_0] -> [the fundamental
group of S^1 at base point b_0] such that h_*([f])=[h of f],
where [f] is the path homotopy class of f) carries agenerator
of the fundamental group of S^1 at base point b_0 to an odd
powerof itself.If it helps any, the Borsuk-Ulam theorem for
S^2 is covered up to ts point
===
Subject: Lambert's W function
(AKA ProductLog) problem:Please view in a fixed-width font
like Courier.I've beat myself to death trying to get ts
solved. I've got theMaple solution from a post where all
algebraic manipulation stepswere skipped and I get the same
result from Mathematica.A simple example can be found
athttp://www.anasoft.co.uk/EE/widlarlambert/
widlarlambert.htmlNote, ts solution is not a common EE
solution. Only a couple ofdesigners have pointed out the use
of Lambert and only one of thosefigured it out mself. The
other used Maple.Here's the bomb, though:Ie --> I sub eIs -->
I sub sVbe --> V sub beVt --> V sub t4600 Ie + 2 Vbe = 6 / Vbe
Ie = Is |exp(----) - 1| Vt /so Ie + Is Vbe = ln(--------) Vt
Is2300 Ie + ln[(Ie + Is)/Is] Vt = 3 eq1the solution is: /
(3/Vt + (2300 Is)/Vt) |2300 e Is| -2300 Is + ProductLog
|-------------------------------| Vt Vt / Ie =
-----------------------------------------------------------
2300or / (3/Vt + (2300 Is)/Vt) |2300 e Is| ProductLog
|-------------------------------| Vt - Is Vt / Ie =
-----------------------------------------------------------
2300where ProductLog is Lambert's W function.So, not satisfied
with taking yes for an answer, and knowing that myalgebra
skills need work, I set about beating myself to get eq1 inthe
formx = ye^y so I could apply ts technique in the future
withoutneeding special software.2300 Ie + ln[(Ie + Is)/Is] Vt
= 3 eq1ln[(Ie + Is)/Is] Vt = 3 - 2300Ie 3 - 2300Ieln[(Ie +
Is)/Is] = ------------ eq2 Vt***notice at ts early stage that
the solution contains3/Vt + 2300Is/Vt = (3 + 2300 Is)/Vt ?I
still haven't figured out how I'm going to get rid of that
-sign in eq2. Moving on...Can't remember what to call ts other
than inverse function --taking the inverse natural log of both
sides, i.e., raising e to thepower of the lhs and rhs (??) / 3
- 2300 Is Ie + Is | ----------- |------- = Vt / Is eSo, using
the link I supplied above as an example I figures that ifI set
3 - 2300 Is y = ----------- Vtand rearranged tngs, I'd get a
solution. But when I rearrangetngs and substitute tngs, I end
up getting y in the denominatorof the rhs. Inverting it just
changes thex = y^-1.e^y form to x = ye^-y wch just swaps the
terms in theexponent / 3 - 2300 Is | ----------- | Vt / to /
2300 Is - 3 | ----------- | Vt /and ts leads me to believe I'm
heading down the garden pathbecause the final solution contains
no negative sign in theexponent. If that's not an issue, I'm
still hung up because I'vegone further and tngs just aren't
looking right. I'd post my next2 steps, but I'm not sure if
they're useful and that might just foultngs up.Would someone
please help me get home on ts?-- 
===
Subject: Re: Lambert's W
function (AKA ProductLog) problem> 2300 Ie + ln[(Ie + Is)/Is]
Vt = 3 eq1> the solution is:> / (3/Vt + (2300 Is)/Vt) > |2300
e Is|> -2300 Is + ProductLog |-------------------------------|
Vt> Vt /> Ie =
----------------------------------------------------------->
2300Correct above. But below there is a small mistake: The
last term in yournumerator shouldn't be in the numerator, so
that the answer would correctlybe / (3/Vt + (2300 Is)/Vt)
|2300 e Is| ProductLog |-------------------------------| Vt Vt
/ Ie = ----------------------------------------------- - Is
2300> or> / (3/Vt + (2300 Is)/Vt) > |2300 e Is|> ProductLog
|-------------------------------| Vt - Is> Vt /> Ie =
----------------------------------------------------------->
2300> where ProductLog is Lambert's W function.> So, not
satisfied with taking yes for an answer, and knowing that my>
algebra skills need work, I set about beating myself to get
eq1 in> the form> x = ye^y so I could apply ts technique in
the future without> needing special software.> 2300 Ie +
ln[(Ie + Is)/Is] Vt = 3 eq1> Would someone please help me get
home on ts?OK, I'll show you what I did, in unpolished
form.Begin by letting u denote the argument of the logarithm,
Ie/Is + 1.Then in terms of u, the equation becomes2300 Is (u -
1) + Vt ln(u) = 32300 Is u + Vt ln(u) = 3 + 2300 Is2300 Is/Vt u
+ ln(u) = (3 + 2300 Is)/Vtwch is in the form a u + ln(u) =
bwhere a = 2300 Is/Vt and b = (3 + 2300 Is)/Vt.To solve a u +
ln(u) = b, apply the natural exponential function, exp, toboth
sides:exp(a u + ln(u)) = exp(b)u exp(a u) = exp(b)Now, letting
y = a u, we gety/a exp(y) = exp(b)y exp(y) = a exp(b), wch can
now be solved using the Lambert Wfunction (wch I'll denote
simply as W), givingy = W(a exp(b)).We're basically finished
then. It's just a simple matter to work backthrough to obtain
Ie. Anyway, here are the steps: a u = W(a exp(b)) u = 1/a W(a
exp(b))Ie/Is + 1 = 1/a W(a exp(b)) Ie + Is = Is/a W(a exp(b))
Ie = Is/a W(a exp(b)) - IsFinally, just replace a by 2300
Is/Vt and b by (3 + 2300 Is)/Vt, andsimplify Is/a to Vt/2300,
and you'll have the corrected form of theanswer I'd mentioned
earlier.
===
Subject: Re: JSH: Does the math win yet?> It looks
like there was more to adding in that y as a variable than>
was realized as I saw the usual suspects replying--yet
again--trying> to attack the mathematics, attacking algebra,
as if they could still> convince all of you.However, the other
shoe fell as I set y=0 givinga^2 - xa + 7x^2 = 0,soa =
(1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I
then have(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2
+ c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)from my modification
of Rick Decker's example.And as I mentioned in another post in
ts thread, that modification doesn't work.Now I guess that
Decker is a good enough mathematician to recognize> that now
it's completely over, but I wonder about the others.Will they
*dare* step out now despite the mathematics clearly showing>
that I've been right all along?Oh well, that certainly is
embarrassing given that my modification was wrong.You can add
another variable, but it's not that simple to do.
===
Subject:
Re: JSH: Does the math win yet?> It looks like there was more
to adding in that y as a variable than> was realized as I saw
the usual suspects replying--yet again--trying> to attack the
mathematics, attacking algebra, as if they could still>
convince all of you. However, the other shoe fell as I set y=0
giving a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and
introducing c_1(x,y) and c_2(x,y) I then have
(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 +
c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)> > from my
modification of Rick Decker's example.> And as I mentioned in
another post in ts thread, that modification doesn't work. Now
I guess that Decker is a good enough mathematician to
recognize> that now it's completely over, but I wonder about
the others. Will they *dare* step out now despite the
mathematics clearly showing> that I've been right all along?
Oh well, that certainly is embarrassing given that my
modification was wrong.> You can add another variable, but
it's not that simple to do.> Your so-called 'embarrassment'
sounds suspicious. Too bad you discovered your error *after*
condemning the'usual suspects' for attacking
mathematics.--There are two tngs you must never attempt to
prove: the unprovable -- and the obvious.--
--http://www.crbond.com
===
Subject: Re: JSH: My quick math
quide Adjunct Assit Professor at the University of
Montana.>>One of the reasons I made the post Quick Math Guide
to core error>>issues was to have a handy reference. >Huh - I
didn't see the Quick Math Guide.back in October.-- 
===
Subject:
Re: JSH: My quick math quide> One of the reasons I made the
post Quick Math Guide to core error> issues was to have a
handy reference. Having found myself floundering> a bit, I
found myself looking back at it, wch is when I realized a> wle
back that I'd wondered far astray, and I posted that I'd made
a> mistake with my addition of y to an example from Rick
Decker.I then sat back, thought for a wle, and realized that
possibly my> problem had to do with difficult concepts that so
push the human brain> that most are not up to handling the
mathematics as I'd been giving> it.> Here's a different view
on ts. Decker presented an examplewch showed in terms simple
enough for you to underd thatyour method did not work. You
agonized over ts for a wle,then decided to add in a new
variable to your polynomial to avoid Decker's math. Ts failed.
You were wrong and you admittedit in one of your infamous Oops!
statements. So after an extended argument you admitted you were
wrong. So then you sat back, thought for a wle and decided
thatthe problem was, we mathematicians are not up to handling
the mathematics. Doesn't ts strike you as just a wee bit
strange? You are proved wrong about the math, you admit it,
and then you conclude that *we* - not you - are not up to
handling it ??? I mean, if maybe it had not happened
previously many times over in the past 8years - if ts was an
exception - it might be a reasonableconclusion. But it HAS
happened repeatedly. You are provenwrong, you admit it, and
then you conclude that WE are not capable of handling the
math. Isn't your keyboard still measurably warm from typing in
theword Oops! ???> So I'm starting again, and looking for
better ways to to explain.> True, you are not good at
explaining. But that is not the problem. As with Decker's
example, we all understood what youwere saying. The problem
was - and is - what you were saying was wrong. You don't need
a better way to explain. You need abetter way to avoid
mathematical errors.P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f)The form of the polynomial allows me
to factor P(m) into> non-polynomial factors, and the
factorization with those factors isP(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)where the a's are roots of the following
cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).I've
adjusted those expressions several ways to try and make it
easier> to underd.So far those adjustment apparently haven't
been very successful, and> now I'm beginning to realize that I
need to underd human beings> better as abstractions. I need to
figure out how their brains work.> The lesson of the Decker
example and many, many others is, mathematicians'brains work
logically and tend to come to correct conclusions. Yoursdoes
not. You are deluding yourself into tnking you are some kindof
superhuman mutant. You're not. As proven repeatedly here,
yourabilities, and certainly your knowledge, are well below
that of theAVERAGE mathematician.> That puts me back into
study mode. I have to underd your mental> wiring to an even
greater degree than I do now.> Again you are pretending that
you have a godlike, advanced intellect. If you did you would
not make the mistakes that youdo, ranging from careless little
algebra errors to deep misunderdingsof mathematical structures.
A superhuman intellect would not havebeen proven wrong dozens
of times over an 8-year period. Youare going more and more
deeply into delusional tnking. > I have to figure you out.>
OK, let's take your present polynomial example. I will
assumehere that f is a prime, m and u are integers coprime to
f, and xis a polynomial variable. You are factoring P(m) with
respectto x with a factorization of the form P(m) = (a1*x +
uf)*(a2*x + uf)*(a3*x + uf),where a1, a2, and a3 are algebraic
integers wch are dependenton m. You now know, and we agree
(since we said it in the first place)that in general a1/f is
not an algebraic integer. After that we diverge. You conclude
from ts that there isno way to factor f^2 out of P(m) so that,
when the piecesare distributed among the factors above, the
resulting expression has algebraic integer coefficients. You
tnk theonly way to factor f^2 that will work is in the form
f^2 = f*f*1.Ts causes you to believe that you must create a
new ring of numbers to accommodate tngs like a1/f. We don't
agree. a1/f is simply an algebraic number with nospecial
properties. No new ring is necessary. The real core of the
disagreement, however, is not abouta1/f. We agree that
factoring f^2 out of the expression sothat one of the terms
(ai + uf) is divided by f causes a problem. The f*f*1
factorization just does not work. The real core of the
disagreement is that we say there is ANOTHER way to factor f^2
out of (a1*x + uf)*(a2*x + uf)*(a3*x + uf)in such a way that
the resulting coefficients are all algebraic integers. Ts
means that it is quite unnecessaryto create some new ring of
numbers. You however do not accept that a factorization of the
kind we describe can be done.I tnk it is again a matter of your
not knowing enough mathto underd what we are saying. You refuse
to learn thenecessary math. Then you blame us for being
inadequatelywired or whatever. Wake up. The problem again, as
it has always been, is notwith our wiring. It's with yours.
Wake up and learn somemath! 
===
Subject: Re: JSH: My quick math
quideSo I'm starting again, and looking for better ways to to
explain.> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2
)x u^2 + u^3 f)> Ts example is overly complicated. Consider
tssimpler presentation of the core error proof.Lemma 1: Let
P(m) be a polynomial with coefficients inA, the algebraic
integers. Let each coefficient bedivisible (in A) by the
algebraic integer f.Let P(m) = g_1(m) * g_2(m), with g_1(m)
and g_2(m)functions from A to A. Then there exist s and tin A
such that s*t=f; and for all m in A,s divides g_1(m) and t
divides g_2(m). Core Error Proof Consider the function
P(m,f,x) = mfx^2 +2fx + f^2 (m,f,x in A) [Among other
possibilities We can consider ts as a polynomial in m, with
parameters f and x, or a polynomial in x with parameters m and
f. Ts only matters if we want to motivate the following]
Consider the quadratic t^2 + 2t + mf (*) Let a_1(m,f) = 1 -
sqrt(1-mf) and a_2(m,f) = 1 + sqrt(1-mf) be the negatives of
the roots of (*). Some simple algebra gives P(m,f,x) = (
a_1(m,f) x + f ) ( a_2(m,f) x + f ) or (with a minor abuse of
notation)letting P(m) = P(m,f,x), g_1(m) = a_1(m,f) x + fand
g_2(m) = a_2(m,f) x + f P(m) = g_1(m) * g_2(m) Noting that
P(m) can be considered a polynomial in mwith coefficients in
A, all coefficients divisible by fwe apply lemma 1 and obtain:
There exists s and t, such that s*t=f; andfor all m, s divides
g_1(m) = a_1(m,f) x + ft divides g_2(m) = a_2(m,f) x + f Let m
= 0. g_1(0) = f and g_2(0) = 2x + f.The only choice for s and t
(up to units)is s=f and t=1. thus f divides (a_1(m,f) x +
f)thus f divides a_1(m,f) Now specialize by letting m=1, f =3
3 divides a_1(1,3) = 1-sqrt(1-(1)(3)) = 1 - i sqrt(2) but (1-
i sqrt(2))/3 is a root of 9x^2 - 6x + 4 At ts point the
mathematical universe collapses Or does it, The problem is
with Lemma 1.Wle ts holds for g_1 and g_2 polynomialsit is not
true in general. Counterexamples areeasy to construct if g_1
and g_2 are not continuous.We get a more interesting example
by lettingx=1 and f = 3 above. Let P(m) = 3m + 15, ts is
divisible by 3. By simple algebra P(m) = 3m+15 =
(4-sqrt(1-3m)) (4+sqrt(1-3m)) so let g_1(m) = (4-sqrt(1-3m))
and g_2(m) = (4+sqrt(1-3m)) P( -1) = 12, g_1( -1) = 2, g_2(
-1) = 6P( -5) = 0, g_1( -5) = 0. g_2( -5) = 8P( -8) = -9, g_1(
-8) = -1. g_2( -8) = 9P(-16) = -33, g_1(-16) = -3. g_2(-16) =
11 We note that wle P(m) is always divisible by3, the factor
of 3 switches back and forthbetween g_1 and g_2. 
===
Subject:
Re: JSH: My quick math quide> So I'm starting again, and
looking for better ways to to explain.> P(m) = f^2((m^3 f^4 -
3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)> Ts example is
overly complicated. Consider ts> simpler presentation of the
core error proof.Well at least you're trying so I guess I can
look over your attempt tounderd. Lemma 1:> > Let P(m) be a
polynomial with coefficients in> A, the algebraic integers.
Let each coefficient be> divisible (in A) by the algebraic
integer f.> Let P(m) = g_1(m) * g_2(m), with g_1(m) and
g_2(m)> functions from A to A. Then there exist s and t> in A
such that s*t=f; and for all m in A,> s divides g_1(m) and t
divides g_2(m).That's not what I've been saying, and
represents some interpretationson your part, wch, for the
moment with you in particular, I'm readyto try and underd. In
the past I'd get frustrated and angry, nowI just feel
frustrated as I don't know how often I have to explain thesame
tng, over, and over, and over again.You keep trying to *stay*
in algebraic integers, when my point is thatyou're pushed out,
as your claims for your Lemma 1 don't make sensein context with
the actual argument that *I* have presented.So you lose it from
the start.I need you to explain to me if you can accept the
possibility of beingpushed out of the ring of algebraic
integers, that is, can youcontemplate the possibility that
given factors g_1(m) and g_2(m) ofP(m) like in your example
that g_1(m) can NOT have any factors otherthan 1 and -1 in
common with f *in the ring of algebraic integers*;although,
P(m) has f as a factor?That is, can that idea make any sense
to you at any level so that youcan start from that
possibility?
===
Subject: Re: JSH: My quick math quide> That's
not what I've been saying, and represents some
interpretations> on your part, wch, for the moment with you in
particular, I'm ready> to try and underd. In the past I'd get
frustrated and angry, now> I just feel frustrated as I don't
know how often I have to explain the> same tng, over, and
over, and over again.Until you get it right!
===
Subject: Re:
JSH: My quick math quideThat's not what I've been saying, and
represents some interpretations> on your part, wch, for the
moment with you in particular, I'm ready> to try and underd.
In the past I'd get frustrated and angry, now> I just feel
frustrated as I don't know how often I have to explain the>
same tng, over, and over, and over again.It's easy to figure
out how often you'll have to explain. You'll needto explain
over and over until you realize your explanation is *wrong*--
wch probably never will happen. So you'd best get used to
beingfrustrated, since it's likely to be a permanent condition
for you.-- 
===
 Subject: Re: JSH: My quick math quide> One of
the reasons I made the post Quick Math Guide to core error>
issues was to have a handy reference. Having found myself
floundering> a bit, I found myself looking back at it, wch is
when I realized a> wle back that I'd wondered far astray, and
I posted that I'd made a> mistake with my addition of y to an
example from Rick Decker.I then sat back, thought for a wle,
and realized that possibly my> problem had to do with
difficult concepts that so push the human brain> that most are
not up to handling the mathematics as I'd been giving> it.So
I'm starting again, and looking for better ways to to
explain.> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2
)x u^2 + u^3 f)The form of the polynomial allows me to factor
P(m) into> non-polynomial factors, and the factorization with
those factors isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)>
Bad notation. a_1, a_2 and a_3 are complicated
functions.Better is a_1(x), a_2(x) and a_3(x). (even betteris
a_1(x,m,f), a_2(x,m,f) a_3(x,m,f) )> where the a's are roots
of the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 -
3m^2 f^2 + 3m).I've adjusted those expressions several ways to
try and make it easier> to underd.> Since no one is disputing
the results to ts point, youdon't have to make it easier to
underd.Your problem is that you note that f^2 divides P(m) and
correctlyconclude that f^2 must divide (a_1(x) + uf)(a_2(x) +
uf)(a_3(x) + uf)Noting that a_1(0) = a_2(0) = 0 and a_3(0) !=
0 you correctly note that**at x=0** the first and second terms
are divisible by f and the trd termis coprime to f. You then
incorrectly conclude that **for all x** the first and second
termsare divisible by f and the trd term is coprime to
f.Unsurprisingly, ts incorrect conclusion leads to a
contradiction. -
===
Subject: Re: JSH: My quick math quide
Adjunct Assit Professor at the University of Montana.P(m) =
f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3
f)The form of the polynomial allows me to factor P(m) into>>
non-polynomial factors, and the factorization with those
factors isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)>> >Bad
notation. a_1, a_2 and a_3 are complicated functions.>Better
is a_1(x), a_2(x) and a_3(x). (even better>is a_1(x,m,f),
a_2(x,m,f) a_3(x,m,f) )Actually, they depend on m and f (but,
I tnk, not on u), but they donot depend on x; in the above
formulation, for each value of m, f, u,we take the expression
as a cubic polynomial in x and factor it into aproduct of
linear terms with algebraic integer coefficients. There issome
bit of work to justify the choice of cont terms, but that canbe
done with the current polynomial... [.snip.]-- 
===
Subject: Re:
JSH: My quick math quide> One of the reasons I made the post
Quick Math Guide to core error> issues was to have a handy
reference. Having found myself floundering> a bit, I found
myself looking back at it, wch is when I realized a> wle back
that I'd wondered far astray, and I posted that I'd made a>
mistake with my addition of y to an example from Rick Decker.I
then sat back, thought for a wle, and realized that possibly
my> problem had to do with difficult concepts that so push the
human brain> that most are not up to handling the mathematics
as I'd been giving> it.So I'm starting again, and looking for
better ways to to explain.> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)The form of the polynomial
allows me to factor P(m) into> non-polynomial factors, and the
factorization with those factors isP(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)where the a's are roots of the following
cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).I've
adjusted those expressions several ways to try and make it
easier> to underd.So far those adjustment apparently haven't
been very successful, and> now I'm beginning to realize that I
need to underd human beings> better as abstractions. I need to
figure out how their brains work.That puts me back into study
mode. I have to underd your mental> wiring to an even greater
degree than I do now.I have to figure you out.> That is your
first mistake! The only brain whose operation you need worry
about is your own.
===
Subject: Re: JSH: My quick math quide> I
then sat back, thought for a wle, and realized that possibly
my> problem had to do with difficult concepts that so push the
human brain> that most are not up to handling the mathematics
as I'd been giving> it.It's *your* brain that's the problem,
not the *human brain*.> So I'm starting again, and looking for
better ways to to explain.> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)> The form of the
polynomial allows me to factor P(m) into> non-polynomial
factors, and the factorization with those factors is> P(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf)> where the a's are roots
of the following cubic:> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 -
3m^2 f^2 + 3m).> I've adjusted those expressions several ways
to try and make it easier> to underd.> So far those adjustment
apparently haven't been very successful, and> now I'm beginning
to realize that I need to underd human beings> better as
abstractions. I need to figure out how their brains
work.CORRECTION: You need to figure out how to get *your*
brains to work.> That puts me back into study mode. I have to
underd your mental> wiring to an even greater degree than I do
now.CORRECTION: You need to get *your* wiring fixed. It has
numerous faultyconnections. Do a Google search on and Oops!.>
I have to figure you out.No you don't. You simply need to look
at yourself. As Robert BurnsO wad some Pow'r the giftie gie
us.To see oursels as others see us!--There are two tngs you
must never attempt to prove: the unprovable --and the
obvious.----http://www.crbond.com