mm-400
Subject: Re: middle term expansionMathematics is about
recognising patterns; here are three clues:Pascal's Triangle -
wch I'll bet you've heard of - gives coefficients 1,6, 15, 20,
15, 6, 1There are seven terms, the largest being x^6 and the
smallest being 1/(x^6),so the middle term is just a number.The
minus sign in the original expression means that the terms
alternate insign, the first being positive.Putting these
together the result is - 20.You have enough information to
write out the full expansion but I tnk yourteacher is trying
to show you that all that work is not necessary.I hope that
makes it easier - Ian Hutcheson
===
Subject: Re: what expression
represents the nth term of the sequence?> If the first term of
a geometric sequence is 3/2 and the second and> trd terms are
-3/4 and 3/8 respectively, what is the expression for> the nth
term of the sequence? Geometric sequence: a, a*r, a*r^2, a*r^3,
...with a = 3/2, a*r = -3/4 and a*r^2 = 3/8, and so
on
===
Subject: Re: What is the value of k? ===Subject: Re: What
is the value of k?If ending in minus 15 gave remainder 7 then
ending in minus 22 wouldfactorise exactly.The key information
that helps solve the problem then, is that for somevalue p,
say(x - 2) (x^2+px+11) = x^3+kx^2 - 3x - 22. Equating
coefficients of x squaredand x:- 2 + p = k11 - 2p= - 3p=7 so k
= 5I hope that makes it easy to follow - Ian
Hutcheson
===
Subject: Re: What is the value of k?If ending in
minus 15 gave remainder 7 then ending in minus 22
wouldfactorise exactly.The key information that helps solve
the problem then, is that for somevalue p, say(x - 2)
(x^2+px+11) = x^3+kx^2 - 3x - 22. Equating coefficients of x
squaredand x:- 2 + p = k11 - 2p= - 3p=7 so k = 5I hope that
makes it easy to follow - Ian Hutcheson
===
Subject: Re: What is
the value of k?If ending in minus 15 gave remainder 7 then
ending in minus 22 wouldfactorise exactly.The key information
that helps solve the problem then, is that for somevalue p,
say(x - 2) (x^2+px+11) = x^3+kx^2 - 3x - 22. Equating
coefficients of x squaredand x:- 2 + p = k11 - 2p= - 3p=7 so k
= 5I hope that makes it easy to follow - Ian
Hutcheson
===
Subject: Re: What is the value of k?> If the
remainder is 7 when x^3+kx^2-3x-15 is divided by x-2, then k=
?> Would be gratefull if you drop a few words about the way
the solution goes.> ,Given any polynomial, P(x) in x, the
remainder theorem says that the remainder when P(x) is divided
by x-a equals the value of the polynomial when x = a, i.e.,
P(a) So let P(x) = x^3+kx^2-3x-15 and solve P(2) = 7 for
k.
===
Subject: Re: Derivatives Help> I am taking a class called
Circuit Analysis for Engineering.> It requires alot of Calculus
concepts that I honestly forgot and the bookis> no help at all
and doesnt even have any examples..> The problem that I am
currently having problem with is sometng likets.> i(t) = C *
dv(t)/dt> The given info is C = 0.2 F (farads) and v(t) = 4(1
- e^-10t) for t >0> They want you to determine the expression
for i(t)> The answer is 8 x 10^-6 e^-10t> Now I have been
trying to figure out how to do that and how they got that>
answer for a long time and I am just going crazy. I was tnking
that10^-6> is a conversion between Farads and MicroFarads..> I
atleast want to find out what the derivative of v(t) is.> Any
help is appreciated > Sports Talk Forum -
http://www.houstonsportstalk.comdv/dt=40e^-10ti(t)=.2(40e^-10t
)=8e^-10tI am guessing that the 10^-6 is for microfarads, as
you suggest.
===
Subject: Re: Derivatives Help Its starting to
come back to me now....slowly :)So when you have a function
such as: f(x) = n(1 - e^y) ....the derivative would always be
f'(x) = yn* e^y ?-- Free PC Tech Support -
http://www.webzila.comSports Talk Forum -
http://www.houstonsportstalk.com>> I am taking a class called
Circuit Analysis for Engineering.>> It requires alot of
Calculus concepts that I honestly forgot and the book> is>> no
help at all and doesnt even have any examples..>> The problem
that I am currently having problem with is sometng like> ts.>>
i(t) = C * dv(t)/dt>> The given info is C = 0.2 F (farads) and
v(t) = 4(1 - e^-10t) for t >0>> They want you to determine the
expression for i(t)>> The answer is 8 x 10^-6 e^-10t>> Now I
have been trying to figure out how to do that and how they got
that>> answer for a long time and I am just going crazy. I was
tnking that> 10^-6>> is a conversion between Farads and
MicroFarads..>> I atleast want to find out what the derivative
of v(t) is.>> Any help is appreciated>> -- >> Steve, I-Net+>>
Free PC Tech Support - http://www.webzila.com>> Sports Talk
Forum - http://www.houstonsportstalk.com> dv/dt=40e^-10t>
i(t)=.2(40e^-10t)=8e^-10t> I am guessing that the 10^-6 is for
microfarads, as you suggest.
===
 Subject: Re: Derivatives Help>
Its starting to come back to me now....slowly :)> So when you
have a function such as: f(x) = n(1 - e^y) ....the derivative>
would always be f'(x) = yn* e^y ?The derivative of e^x is e^x.
For a general function e^(ax), the derivativewill be
ae^(ax).Wle you're at it, it might be a good idea to learn
sometng calledlogarithmic differentiation if you're good with
manipulating logs. Ts willmake differentiating some functions
easier and for some functions,logarithmic differentiation is
the only way to go.> -- > Free PC Tech Support -
http://www.webzila.com> Sports Talk Forum -
http://www.houstonsportstalk.com> I am taking a class called
Circuit Analysis for Engineering.>> It requires alot of
Calculus concepts that I honestly forgot and thebook> is>> no
help at all and doesnt even have any examples..>> The problem
that I am currently having problem with is sometng like> ts.>>
i(t) = C * dv(t)/dt>> The given info is C = 0.2 F (farads) and
v(t) = 4(1 - e^-10t) for t >0>> They want you to determine the
expression for i(t)>> The answer is 8 x 10^-6 e^-10t>> Now I
have been trying to figure out how to do that and how they
gotthat>> answer for a long time and I am just going crazy. I
was tnking that> 10^-6>> is a conversion between Farads and
MicroFarads..>> I atleast want to find out what the derivative
of v(t) is.>> Any help is appreciated>> -- >> Steve, I-Net+>>
Free PC Tech Support - http://www.webzila.com>> Sports Talk
Forum - http://www.houstonsportstalk.com>> dv/dt=40e^-10t
i(t)=.2(40e^-10t)=8e^-10t I am guessing that the 10^-6 is for
microfarads, as you suggest. 
===
Subject: Re: Derivatives
Helpd/dt of 4(1-e^(-10t)) = +40e^(-10t) (take the trouble to
find out why)You have told me the rest yourself, i.e. C = 0.2
x 10^6Multiply out and you are home and dryI hope that helps -
Ian Hutcheson
===
Subject: Re: JSH: Research question answered
One of the reasons I've been so fascinated by the Decker
example is that it revealed to me that I wasn't exactly
certain about methodologies that I'd discovered, as it seems I
didn't underd exactly how everytng worked. An understatement of
monumental proportions ... If the Decker example is so
fascinating to you, why is it > that, in the 15-20 separate
threads you have started on ts> topic, you have never ONCE
quoted Decker's conclusion ? Are> you so terrified that other
people might read it and realize> that Decker was totally,
obviously right, and your claims to> the contrary were
totally, obviously so much hot air?I will repeat that you
should off.If you wish to continue replying to me, you will
find you are giving> me more opportunities to tell you to
OFF!!!I know now that you're not a woman. What woman would so
*publicly*> stalk a guy like ts and repeatedly be told to off
and keep at> it? A man comes into my neighborhood and begins
screaming at the top of s lungs that my neighbors and I are
morons, incompetents,welfare queens, that we are conspiring to
cheat m out of s rightful recognition, that he is going to call
in the FBI andthe U.S. Army, s Congressman, the President, that
he has *proof* that we have been lying, etc.. I point out that
s facts are wrong and give m ample opportunity to refute ts.
Every day he comes back with more shrieking, wning, accusing -
vicious, nasty stuff, and endless repetitions of s
fundamentally unsound proof, alongwith claims that he is a
misunderstood genius, etc., but no answers to what I say. I
reply with additional evidence that he is wrong. He ignores ts
and continues to shriek and insult the entire neighborhood. Now
he gets drunk and starts howling obscenities and says I
am*stalking* m !!! I'm not backing off. You are wrong and I
tnk you actually know it. > Who are you really Nora Baron?>
Here is another example related to that of Rick Decker,with a
couple of additional twists. Let P(x) = 7*(25*x^2 + 60*x -
133). Ts can be re-written as [*] P(x) = 7*{25*(x^2 + x) +
7*(x - 4)*5 + 7}. Now assume P(x) is factored as a polynomial
in the variable 5 in the form[**] P(x) = (5 a_1(x) + 7)*(5
a_2(x) + 7),where a_1(x) and a_2(x) are roots of[***] a^2 -
7*(x - 4)*a + 7*(x^2 + x). Note that when x = 0, the roots are
a_1(0) = 0 and a_2(0) = 28. Both are divisible by 7. As it
turns out, when x = 2, these two roots are a_1(2) = -7 +
sqrt(7) and a_2(2) = -7 - sqrt(7). Both a_1(2) and a_2(2) are
divisible in the algebraicintegers by sqrt(7), but neither is
divisible by 7. In the factorization [**], sqrt(7) can be
divided out of eachfactor to produce P(2)/7 = (5*(-sqrt(7)+1)
+ sqrt(7))*(5*(-sqrt(7)-1) + sqrt(7)),a factorization in wch
*all the individual terms* are algebraicintegers. Note that
the left side, from [*], has the value P(2)/7 = (25*4 + 60*2 -
133) = 87. You can easily check that the right side has the
same value.It's gh-school arithmetic. Note that the
coefficients of the last two terms in [***]have the same
residue mod 7 (i.e., 0), regardless of the value of x. The
value x = 2 is not an exception. I chose ts examplebecause the
arithmetic is easy. An analogous factorizationexists for *most*
integer values of x. The existence of suchis guaranteed by a
theorem of Dedekind. You have seen ts and you know it is
correct. At least you have never arguedwith it. You say ts
factorization is impossible. You keep dancing and twisting
around, trying to find a counterargument, based on your
invalid ideas about cont terms. You have not confronted the
arithmetic in Decker's original example for the obvious reason
that you cannot possibly refute it. The same is true here.

===
Subject: Re: JSH: Research question answered> Actually,
JSH's proofs work, once you underd s terminology. For>
example, when he says> X, so Y> that doesn't mean Y follows
from X like it does in dard terminology.> In JSH terminology,
it means X is true, and now, for sometng completely>
different, here is Y, wch has notng to do with X, and probably
has a> counterexample in small integers, and I'm a genius.>
Once you realize ts, and a couple other places where he has s
own,> special, terminology, s proofs make sense, in the sense
that everytng in> them is a correct statement. :-)> --> More
Harristotelian logic:If A = B and B = C, then I am the
greatest mathematician of all time and you areall evil
subhuman conspirators. .----http://www.crbond.com
===
Subject:
Re: JSH: Research question answered>> One of the reasons I've
been so fascinated by the Decker example is>> that it revealed
to me that I wasn't exactly certain about>> methodologies that
I'd discovered, as it seems I didn't underd>> exactly how
everytng worked.>> An understatement of monumental proportions
... If the Decker example is so fascinating to you, why is it
>> that, in the 15-20 separate threads you have started on
ts>> topic, you have never ONCE quoted Decker's conclusion ?
Are>> you so terrified that other people might read it and
realize>> that Decker was totally, obviously right, and your
claims to>> the contrary were totally, obviously so much hot
air?>> > off.>I will repeat that you should off.Oops. Another
nt: when you reply like ts it looks,even to people who aren't
following the math, that yousimply can't answer the question.
Bad strategy.>If you wish to continue replying to me, you will
find you are giving>me more opportunities to tell you to
OFF!!!>I know now that you're not a woman. What woman would so
*publicly*>stalk a guy like ts and repeatedly be told to off
and keep at>it?>Who are you really Nora
Baron?>************************
===
Subject: Re: JSH: Research
question answered> Who are you really Nora Baron?I heard a
really juicy rumour that she works at Hamilton College. Does>
that interest you?>And me tnking it was OK State.Interesting
theory, but I don't tnk so. There are certainly expertsin
algebraic number theory here, but none of them are regular
sci.math participants. (Hmm, as far as I know - maybe I need
toask ___ whether he's adopted a
pseudonym...)************************
===
Subject: Re: JSH:
Research question answered>> One of the reasons I've been so
fascinated by the Decker example is>> that it revealed to me
that I wasn't exactly certain about>> methodologies that I'd
discovered, as it seems I didn't underd>> exactly how everytng
worked.>> An understatement of monumental proportions ... If
the Decker example is so fascinating to you, why is it >>
that, in the 15-20 separate threads you have started on ts>>
topic, you have never ONCE quoted Decker's conclusion ? Are>>
you so terrified that other people might read it and realize>>
that Decker was totally, obviously right, and your claims to>>
the contrary were totally, obviously so much hot air? ****
off.I will repeat that you should **** off.If you wish to
continue replying to me, you will find you are giving> me more
opportunities to tell you to **** OFF!!!I know now that you're
not a woman. What woman would so *publicly*> stalk a guy like
ts and repeatedly be told to **** off and keep at> it?What
decent man would speak that way to a woman (or another
man)?But then, you're neither decent nor a man.> Who are you
really Nora Baron?Clearly, she's someone who's got your
number. Otherwise she wouldn'tmake you so angry by asking
questions you can't answer.-- 
===
Subject: Re: JSH: Research
question answered One of the reasons I've been so fascinated
by the Decker example is that it revealed to me that I wasn't
exactly certain about methodologies that I'd discovered, as it
seems I didn't underd exactly how everytng worked. An
understatement of monumental proportions ... If the Decker
example is so fascinating to you, why is it > that, in the
15-20 separate threads you have started on ts> topic, you have
never ONCE quoted Decker's conclusion ? Are> you so terrified
that other people might read it and realize> that Decker was
totally, obviously right, and your claims to> the contrary
were totally, obviously so much hot air?> off.I will repeat
that you should off.If you wish to continue replying to me,
you will find you are giving> me more opportunities to tell
you to OFF!!!I know now that you're not a woman. What woman
would so *publicly*> stalk a guy like ts and repeatedly be
told to off and keep at> it?Who are you really Nora Baron?She
could be someone who is more interested in truth than in your
petty temper tantrums. They only show everyone how little you
are worth.
===
Subject: JSH: Pattern argumentSo I've put out the
Decker quadratic (source information at bottom),but replies
from various people indicate there's still room forconfusion,
so I'll give a pattern argument.The original quadratic is
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where s a's
are roots of a^2 - (x - 1)a + 7(x^2 + x).You can modify it
easily enough to getb^2 - (x - 1)b + 17(x^2 + x), where you
then have(5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 25x) +
17(5)(x-1) + 17^2 wch is(5b_1(x) + 17)(5b_2(x) + 17) =
17(25x^2 + 30x + 12) or you can go down toc^2 - (x - 1)c +
2(x^2 + x), wch gives you(5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2
+ 25x) - 2(5)(x-1) + 2^2 wch is(5c_1(x) + 2)(5c_2(x) + 2) =
2(25x^2 + 30x - 3).So I have(5a_1(x) + 7)(5a_2(x) + 7) =
7(25x^2 + 30x + 2) (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 +
30x + 12) and(5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x -
3).The more astute of you should see a pattern. Clearly the
contfactor on the right, wch shows up as a coefficient on the
left mustequal 2 mod 5 for ts particular setup.But why?It
turns out that I'm exploiting a symmetry with
*integer*coefficients in the factorization.Imagine that in
each case there's a base equation(5y_1(x) + 1)(5y_2(x) +
2)where you can linearly tranform the second factor (5y_2(x) +
2) rathereasily leaving you free to multiply the first factor
by variousintegers as long as they equal 2 mod 5.Now if you
beliee that *any* of the explanations given by variousposters
out there can relate that to reducibility over Q or
GaloisTheory, then you're not good with a basic pattern.The
problem is that y_1(x) and y_2(x) can't both be algebraic
integerfunctions.Now if you *still* are having a problem
underding, ask yourself,how can anyone force a cont factor
like 7, or 2, or 17 to onlywork with *one* factorization?Like
consider againb^2 - (x - 1)b + 17(x^2 + x), where you
have(5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12).Why
does that 17 lead to *one* factorization on the left? How is
itpossible for it to be limited?The answer is that there's
only one way to distribute it and have*integer* coefficients
in the factorization on the left.Mathematicians here need to
be at least a little curious, or you'llnever figure it
out!!!Tnk logically, and force people disagreeing with me to
answer basicquestions, like how do they explain the
pattern:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
(5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 12)
and(5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x - 3)?Decker
Quadratic Source Information---------------------Recently Rick
Decker, a professor at Hamilton College, apparentlytrying to
refute my research came up with a quadratic example, wch Ilike
because it's a quadratic, and easier to manipulate than
thecubics I've used before.If you wish to see s original post
here are some headers wch alsoshow that he posts from Hamilton
College:
===
Subject: Re: Mathematical consistency, courageDecker
put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x + 2) where s a's are roots of a^2 - (x - 1)a + 7(x^2 + x).