mm-1052
> Um, if the numbers are constant, and so are independent of
x, then,
> duh, why should it matter what value x has?
If they are independent of ÔxÕ, why did you have 
to set ÔxÕ
to zero to
uncover them? Evaluating a univariate polynomial at any
numeric value of
the variable produces a numeric result. Are all these results
ÔconstantsÕ?
> The logic is inescapable.
> LifeÕs too short.
I think yours is too long.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
--
http://www.crbond.com
===
Subject: Re: JSH: Simply fascinating
> The math here is so readily understandable that itÕs
actually almost
> as interesting watching how people react to it, as anything
else.
> For instance, IÕve given a polynomial P(x) repeatedly 
where
I factor
> it into three factors.
> I point out that the factors must include factors of the
constant term
> of the polynomial P(x).
> I note that the factors of the constant term are
independent of x, as
> they are, in fact, constant.
> Mathematically itÕs easy to show:
> g_1(x) g_2(x) g_3(x) = P(x)
> and
> g_1(0) g_2(0) g_3(0) = P(0)
> as P(0) gives the constant term of the polynomial, and
since the
> polynomial results from multiplying together the three
factors which
> IÕve called g_1(x), g_2(x), and g_3(x), then you can get
the factors
> of the constant term, by setting x=0.
> ItÕs that simple.
Yes. Right so far. Trivial, but right.
> Notice (1) you have the factors of P(x), (2) the constant
term of P(x)
> is determinable by setting x=0, (3) you also get the
factors of the
> constant term.
> Mathematically, itÕs simple to the point of trivial.
Correct.
> Now then, if you have *constants* which are factors of
another
> constant then why would anyone try to argue that they are
actually
> variables?
No one does. If as you say, you assume that your factors are
constant, of course they are not variables. That is nothing
but
a dimwit tautology.
And of course that is not what we say. We say that if you
factor
49 out of P(x) in a *variable* way, as functions w_1(x),
w_2(x),
and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and
g_3(x)
by those variable functions, then you can obtain algebraic
integer
factors on both sides of the resulting equation. That is,
g_1(x)/w_1(x), etc., are all algebraic integers. That is all
you
require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x),
and
g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to
g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0).
No problem with that either. And the product of these three
constant terms equals P(0)/49.
It all works out. Factoring out 49 as a product of variable
algebraic integer functions does not lead to ANY
contradiction. It
is only when you try to use a CONSTANT factorization that you
arrive at what you think is a problem with algebraic
integers. If
you do the factorization in the right way there is no such
problem.
Suppose z is an integer variable, and you consider
Q(z) = (x + 5)(x + 6).
You notice that Q(z) is alway an even integer. Therefore
you can always divide it by 2.
You notice that when z = 0, Q(z) = Q(0) = 5 * 6.
You notice that Q(0)/2 = 5 * (6/2).
That is, when you divide by 2 to get an integer, you divide
the constant term of the second factor, 6, by 2. If you tried
to divide 5 by 2 you would not have an integer quotient.
So by your logic,
Q(z)/2 = (z + 5)*(z/2 + 6/2) = (z + 5)*(z/2 + 3)
would be the only right, CONSTANT way to factor out 2. This is
analogous to your division by 49 = 7*7*1:
(5 a_1(x)/7 + 7/7)(5 a_2(x)/7 + 7/7)(5 b_3(x)/1 + 22/1)
But, back to Q(z)/2 = (z + 5)(z/2 + 3). The problem is, z/2
is not always an integer. If you want integer quotients in
both
factors, you have to divide out 2 in a NONCONSTANT way. When z
is even, you divide 2 out of (z + 6). When z is odd, you
divide
2 out of (z + 5).
Get it? Factoring 2 out in a constant fashion does not work.
Factoring 2 out in a *variable* fashion does work. No problem
arises with constant terms.
> You have x, as the variable. Besides x there are just these
numbers.
> If you clear out x, then whatÕs left are constants. 
Letting
x=0,
> clears it out, leaving the constants visible.
> Some may say, yeah, sure, at x=0, but what about when x
doesnÕt equal
> 0?
Some may say.
> Um, if the numbers are constant, and so are independent of
x, then,
> duh, why should it matter what value x has?
You MUST divide by a variable factorization of 49. If you
assume a constant factorization you get into trouble, because
a_1(x)/7 in general is not an algebraic integer. You know
this,
yet you persist in thinking that the factorization has to be
constant. You conclude *not* that your own thinking is wrong
(as
you should) but that there is something fundamentally wrong
with
algebraic number theory. You try to justify using the constant
7*7*1 factorization by repeating your mantra that the constant
terms must be constant. This is based on your incorrect
conclusion that, if you divide (a_1(x) + 7) by w_1(x), then
the constant term is
7/w_1(x).
And right there is your central mistake. BY YOUR OWN
DEFINTION,
the constant term of (a_1(x) + 7)/w_1(x) is not 7/w_1(x). It
is instead 7/w_1(0). You want to conclude that 7/w_1(x) must
be equal to the constant, 7, because 1 * 1 * 22 = 22, the
constant
term of P(x)/49. If you do the *correct factorization*, you
still
get 22, but it is in the form
(7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22
because w_1(x)*w_2(x)*w_3(x) = 49.
Now, you may howl, YES BUT 22/w_3(x) IS NOT AN ALGEBRAIC
INTEGER!!!
To which the reply is : yes, youÕre right. BUT THERE IS NO
REASON
IT SHOULD BE.
All that is required is that (5 a_3(x) + 7)/w_3(x) is an
algebraic
integer, and this follows from the correct choice of w_1(x),
w_2(x),
and w_3(x). As it turns out, both 5 a_3(x)/w_3(x) AND
7/w_3(x) are
algebraic integers.
And as above: the constant term of
(5 a_3(x) + 7)/w_3(x)
is NOT 22/w_3(x), as you believe. It is merely 22/w_3(0) =
22. These
two are not the same, unless you assume what you want to
prove,
i.e., that w_3(x) is a constant function. And you know it is
not.
Everything works out. There is no contradiction or problem
involving
the constant terms. It is YOU who has been trying to claim
that the
constant terms are not constant, e.g., when you say that
22/w_3(x) is
the constant term of (5 a_3(x) + 7)/w_3(x).
You are making such a rookie mistake. Why canÕt you see it?
> The logic is inescapable.
True enough.
> In terms of difficulty, my proof is about as easy as it gets
in
> algebraic number theory, in terms of the actual mathematics.
Easy, yes. Correct, no.
> But the concepts are where there is a problem, and the
social
[tiresome pompous rant about social factors etc. deleted]
Nora B.
> LifeÕs too short.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: JSH: Simply fascinating
>The math here is so readily understandable that itÕs
actually almost
>as interesting watching how people react to it, as anything
else.
For instance, IÕve given a polynomial P(x) repeatedly where 
I
factor
>it into three factors.
I point out that the factors must include factors of the
constant term
>of the polynomial P(x).
I note that the factors of the constant term are independent
of x, as
>they are, in fact, constant.
Mathematically itÕs easy to show:
g_1(x) g_2(x) g_3(x) = P(x)
and
g_1(0) g_2(0) g_3(0) = P(0)
as P(0) gives the constant term of the polynomial, and since
the
>polynomial results from multiplying together the three
factors which
>IÕve called g_1(x), g_2(x), and g_3(x), then you can get 
the
factors
>of the constant term, by setting x=0.
ItÕs that simple.
Yes. Right so far. Trivial, but right.
>Notice (1) you have the factors of P(x), (2) the constant
term of P(x)
>is determinable by setting x=0, (3) you also get the factors
of the
>constant term.
Mathematically, itÕs simple to the point of trivial.
Correct.
>Now then, if you have *constants* which are factors of
another
>constant then why would anyone try to argue that they are
actually
>variables?
No one does. If as you say, you assume that your factors are
> constant, of course they are not variables. That is nothing
but
> a dimwit tautology.
> And of course that is not what we say. We say that if you
factor
> 49 out of P(x) in a *variable* way, as functions w_1(x),
w_2(x),
> and w_3(x) of x, and corresponding divide g_1(x), g_2(x)
and g_3(x)
> by those variable functions, then you can obtain algebraic
integer
> factors on both sides of the resulting equation. That is,
> g_1(x)/w_1(x), etc., are all algebraic integers. That is
all you
> require. The constant terms of g_1(x)/w_1(x),
g_2(x)/w_2(x), and
> g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to
> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0).
> No problem with that either. And the product of these three
> constant terms equals P(0)/49.
Ok, IÕm going to answer the Nora Baron poster yet again.
And I want readers to understand that IÕve replied to this
poster who
you know lies anyway as itÕs a guy posting as a woman using 
a
name
thatÕs a palindrome MANY TIMES explaining in detail.
What happens is that when I shoot down these objections, the
poster
either just repeats later, or replies with nonsense. One
telling time
when I carefully refuted point-by-point the poster replied
deleting
out everything IÕd said.
Just deleted out everything, and you know what? I STILL so
people
replying about how supposedly I donÕt answer the objections
from Nora
Baron.
ItÕs partly a game for some people on sci.math, 
IÕm sure, and
partly a
case where many readers are hoodwinked as *they* donÕt know
itÕs a
game. They seem incapable of realizing that there are people
in this
world willing to behave in such a way, on such a level.
So why reply to this poster? Maybe IÕm hopelessly naive, but
I just
have to believe that eventually people will just get tired of
being
made fools of by this poster and others in the group with him
(or
her). It hasnÕt happened yet from what IÕve 
seen, but I keep
hoping.
Ok, so whatÕs wrong with the posterÕs 
assertion?
Well, the wÕs the posters gives are factors of 49. But 49
comes into
the picture because 49 is a multiple of the polynomial. But
the wÕs
STILL REMAIN after 49 has been divided off--after the
multiple is
divided off--as the poster actually claims.
(Notice Nora Baron gives you clues, but somehow sci.math
readers
donÕt seem to get it.)
Notice the posterÕs actually gives that when you divide off
49, you
get
g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x)
which are TRACES, left of 49--after it has been divided off.
So here we have mathematics. Here in an area where the truth
can
actually be determined, a poster who you canÕt be sure is a
guy or a
girl has confused many of you into questioning whether or not
a
multiple of a polynomial divides off as a variable or not.
Did it ever occur to any of you that for some people that
might be a
great lark?
Are you so trusting as to not consider that a poster who
otherwise
would probably be unknown can get off on confusing you not
only on his
or her gender, but on some of the most basic concepts in
mathematics?
Look now, even after the post where Nora Baron ended with a
male
name, people are STILL supporting the poster!
It has occurred to me that this poster did so deliberately,
ended that
post with a male name, because the person knew you better
than you
still seem to know yourselves.
In a way itÕs sad as youÕre giving up so much 
for nothing in
return.
James Harris
===
Subject: Re: JSH: Simply fascinating
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4JZQl20264;
>>The math here is so readily understandable that itÕs
actually
almost
>>as interesting watching how people react to it, as anything
else.
>>For instance, IÕve given a polynomial P(x) repeatedly 
where
I
factor
>>it into three factors.
>>I point out that the factors must include factors of the
constant
term
>>of the polynomial P(x).
>>I note that the factors of the constant term are
independent of
x, as
>>they are, in fact, constant.
>>Mathematically itÕs easy to show:
>>g_1(x) g_2(x) g_3(x) = P(x)
>>and
>>g_1(0) g_2(0) g_3(0) = P(0)
>>as P(0) gives the constant term of the polynomial, and
since the
>>polynomial results from multiplying together the three
factors
which
>>IÕve called g_1(x), g_2(x), and g_3(x), then you can get 
the
factors
>>of the constant term, by setting x=0.
>>ItÕs that simple.
>> Yes. Right so far. Trivial, but right.
>>Notice (1) you have the factors of P(x), (2) the constant
term of
P(x)
>>is determinable by setting x=0, (3) you also get the
factors of
the
>>constant term.
>>Mathematically, itÕs simple to the point of trivial.
>> Correct.
>>Now then, if you have *constants* which are factors of
another
>>constant then why would anyone try to argue that they are
actually
>>variables?
>> No one does. If as you say, you assume that your factors
are
>> constant, of course they are not variables. That is
nothing but
>> a dimwit tautology.
>> And of course that is not what we say. We say that if you
factor
>> 49 out of P(x) in a *variable* way, as functions w_1(x),
w_2(x),
>> and w_3(x) of x, and corresponding divide g_1(x), g_2(x)
and g_3(x)
>> by those variable functions, then you can obtain algebraic
integer
>> factors on both sides of the resulting equation. That is,
>> g_1(x)/w_1(x), etc., are all algebraic integers. That is
all you
>> require. The constant terms of g_1(x)/w_1(x),
g_2(x)/w_2(x), and
>> g_3(x)/w_3(x) are BY YOUR OWN DEFINITION equal to
>> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_3(0).
[misprints corrected here]
>> No problem with that either. And the product of these three
>> constant terms equals P(0)/49.
>Ok, IÕm going to answer the Nora Baron poster yet again.
>And I want readers to understand that IÕve replied to this
poster who
>you know lies anyway as itÕs a guy posting as a woman using
a name
>thatÕs a palindrome MANY TIMES explaining in detail.
>What happens is that when I shoot down these objections, the
poster
>either just repeats later, or replies with nonsense. One
telling
time
>when I carefully refuted point-by-point the poster replied
deleting
>out everything IÕd said.
Because it was the SOS, repeated for the nth time. I replied
with a detailed rigorous proof. You were enraged but never
really provided any refutation.
>Just deleted out everything, and you know what? I STILL so
people
>replying about how supposedly I donÕt answer the objections
from
Nora
>Baron.
You STILL so people replying ??? Really!
>ItÕs partly a game for some people on sci.math, 
IÕm sure,
and partly
a
>case where many readers are hoodwinked as *they* donÕt know
itÕs a
>game. They seem incapable of realizing that there are people
in this
>world willing to behave in such a way, on such a level.
>So why reply to this poster? Maybe IÕm hopelessly naive, 
but
I just
>have to believe that eventually people will just get tired
of being
>made fools of by this poster and others in the group with
him (or
>her). It hasnÕt happened yet from what IÕve 
seen, but I keep
hoping.
Whine, whine, whine. Why not go straight to the math rather
and
skip the propaganda ?
>Ok, so whatÕs wrong with the posterÕs 
assertion?
>Well, the wÕs the posters gives are factors of 49. But 49
comes into
>the picture because 49 is a multiple of the polynomial. But
the wÕs
>STILL REMAIN after 49 has been divided off--after the
multiple is
>divided off--as the poster actually claims.
I claim no such thing. The product of the wÕs is 49. You
divide P(x) by 49. You divide the product of the gÕs by
the product of the wÕs. The 49 is gone from both sides.
Note that here
g_1(x) = (5 a_1(x) + 7),
g_2(x) = (5 a_2(x) + 7),
and g_3(x) = (5 a_3(x) + 7) = (5 b_3(x) + 22),
where a_1, a_2, a_3 are all algebraic integer functions of x,
and b_3(x) = a_3(x) - 3. Note that a_1(0) = a_2(0) = 0, and
a_3(0) = 3. Also, note that w_1(0) = w_2(0) = 7, and w_3(0) =
1.
When I divide the product of the gÕs by the product of the
wÕs,
the result is
(5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)),
where c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x).
Here is what Mr. Harris is squawking about. The product of
the dÕs is 7. (That happens to be true for any x.) Harris
thinks
that is a problem, because it then looks like the product of
the
constant terms of
(5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x))
is 7, whereas the constant term of P(x)/49 is 22.
That is, d_1(x)*d_2(x)*d_3(x) = 7, whereas P(0)/49 = 22.
Puzzling! Is Harris right?
I will now give my deceptive, lying, sleight-of-hand
explanation.
Watch very carefully, because as Mr. Harris says, I am going
to
try very hard to mislead you!
So here is the explanation. The dÕs are NOT the constant 
terms
of the factors that contain them. That is, d_1(x) is NOT the
constant term of (5 c_1(x) + d_1(x)). Mr. Harris thinks it is.
Certainly it is in the right position for being a constant
term.
Since Mr. Harris understands and mostly relies on only
high-school
level mathematics, he detects constant terms by inspection.
The irony here is, he has provided a perfectly good, rigorous
definition of constant term of a function. It is the value
that the function takes on when the argument is 0. Thus the
constant term of (5 c_1(x) + d_1(x)) is NOT d_1(x) = 7/w_1(x).
It is instead d_1(0) = 7/w_1(0) = 1. This is directly from
HarrisÕs own definition of constant term. Why is 
he so
reluctant to apply that definition? Why does he instead rely
on inspection, which gives the incorrect answer? What HE
thinks
is the constant term, namely d_1(x) = 7/w_1(x) is not even
constant (unless he assumes what he wants to prove) !
Similary, the constant term of (5 c_2(x) + d_2(x)) is 1 also.
However, the constant term of
(5 c_3(x) + d_3(x))
is (5 c_3(0) + d_3(0)) = (5 a_3(0)/w_3(0) + 7/w_3(0))
= 5 * 3/1 + 7/1 = 15 + 7 = 22.
Wow, look at that! The product of the constant terms is
the constant term of the product! That is,
d_1(0)*d_2(0)*(5 a_3(0) + d_3(0)) = 1*1*22 = 22 = P(0)/49.
Amazing! Did I successfully trick you?
WhereÕs the trick?
Everything works out just as it should: IF you correctly
identify the constant terms.
WhereÕs the trick, Mr. Harris ?
>(Notice Nora Baron gives you clues, but somehow sci.math
readers
>donÕt seem to get it.)
I think not. I think that, with perhaps one exception,
sci.math
readers DO get it.
In fact, I even think there is no exception. I think Mr.
Harris
actually gets it also, but he cannot stand to admit it. There
is too much at stake. He clings to a delusion on the one-in-a
trillion chance that he will be proven right.
>Notice the posterÕs actually gives that when you divide off
49, you
>get
>g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x)
>which are TRACES, left of 49--after it has been divided off.
See above. There is no mystery. After division, the product
of the constant terms is the constant term of the product,
just
as it should be. That is, IF you correctly identify what the
constant terms actually are. If you donÕt, then you get the
wrong answer. This is Mr. HarrisÕs mistake.
This is not rocket science. This is not Galois theory, or
algebraic number theory, or field theory, or group theory. 
This
is not abstract algebra at all. This is not even high-school
level
algebra. The mistake Mr. Harris is making is really a
low-level one.
A ROOKY mistake, as Mr. Harris used to say (about me and
Arturo
Magidin and others). He is simply not using his own 
definition.
He is substituting in x when he should be substituting in 0.
And incredibly, he seems unable to understand it. Math genius
Harris, boy wonder Harris - from this you would think he
needed
help changing his own soiled underwear.
[more propaganda deleted]
Nora B.
>James Harris
===
Subject: Re: JSH: Simply fascinating
>>The math here is so readily understandable that itÕs
actually almost
>>as interesting watching how people react to it, as anything
else.
>>For instance, IÕve given a polynomial P(x) repeatedly 
where
I factor
>>it into three factors.
>>I point out that the factors must include factors of the
constant term
>>of the polynomial P(x).
>>I note that the factors of the constant term are
independent of x, as
>>they are, in fact, constant.
>>Mathematically itÕs easy to show:
>>g_1(x) g_2(x) g_3(x) = P(x)
>>and
>>g_1(0) g_2(0) g_3(0) = P(0)
>>as P(0) gives the constant term of the polynomial, and
since the
>>polynomial results from multiplying together the three
factors which
>>IÕve called g_1(x), g_2(x), and g_3(x), then you can get
the factors
>>of the constant term, by setting x=0.
>>ItÕs that simple.
>> Yes. Right so far. Trivial, but right.
>>Notice (1) you have the factors of P(x), (2) the constant
term of P(x)
>>is determinable by setting x=0, (3) you also get the
factors of the
>>constant term.
>>Mathematically, itÕs simple to the point of trivial.
>> Correct.
>>Now then, if you have *constants* which are factors of
another
>>constant then why would anyone try to argue that they are
actually
>>variables?
>> No one does. If as you say, you assume that your factors
are
>> constant, of course they are not variables. That is
nothing but
>> a dimwit tautology.
>> And of course that is not what we say. We say that if you
factor
>> 49 out of P(x) in a *variable* way, as functions w_1(x),
w_2(x),
>> and w_3(x) of x, and corresponding divide g_1(x), g_2(x)
and g_3(x)
>> by those variable functions, then you can obtain algebraic
integer
>> factors on both sides of the resulting equation. That is,
>> g_1(x)/w_1(x), etc., are all algebraic integers. That is
all you
>> require. The constant terms of g_1(x)/w_1(x),
g_2(x)/w_2(x), and
>> g_3(x)/w_3(x) are BY YOUR OWN DEFINITION equal to
>> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_3(0).
[a couple of misprints corrected here]
>> No problem with that either. And the product of these three
>> constant terms equals P(0)/49.
>Ok, IÕm going to answer the Nora Baron poster yet again.
>And I want readers to understand that IÕve replied to this
poster who
>you know lies anyway as itÕs a guy posting as a woman using
a name
>thatÕs a palindrome MANY TIMES explaining in detail.
>What happens is that when I shoot down these objections, the
poster
>either just repeats later, or replies with nonsense. One
telling time
>when I carefully refuted point-by-point the poster replied
deleting
>out everything IÕd said.
Because it was the SOS, repeated for the nth time. I replied
with a complete detailed and rigorous proof. You were enraged
but
never really provided any refutation.
>Just deleted out everything, and you know what? I STILL so
people
>replying about how supposedly I donÕt answer the objections
from Nora
>Baron.
You STILL so people replying ??? Really!
>ItÕs partly a game for some people on sci.math, 
IÕm sure,
and partly a
>case where many readers are hoodwinked as *they* donÕt know
itÕs a
>game. They seem incapable of realizing that there are people
in this
>world willing to behave in such a way, on such a level.
>So why reply to this poster? Maybe IÕm hopelessly naive, 
but
I just
>have to believe that eventually people will just get tired
of being
>made fools of by this poster and others in the group with
him (or
>her). It hasnÕt happened yet from what IÕve 
seen, but I keep
hoping.
Whine, whine, whine. Why not go straight to the math rather
and
skip the propaganda ?
>Ok, so whatÕs wrong with the posterÕs 
assertion?
>Well, the wÕs the posters gives are factors of 49. But 49
comes into
>the picture because 49 is a multiple of the polynomial. But
the wÕs
>STILL REMAIN after 49 has been divided off--after the
multiple is
>divided off--as the poster actually claims.
I claim no such thing. The product of the wÕs is 49. You
divide P(x) by 49. You divide the product of the gÕs by
the product of the wÕs. The 49 is gone from both sides.
Note that here
g_1(x) = (5 a_1(x) + 7),
g_2(x) = (5 a_2(x) + 7),
and g_3(x) = (5 a_3(x) + 7) = (5 b_3(x) + 22),
where a_1, a_2, a_3 are all algebraic integer functions of x,
and b_3(x) = a_3(x) - 3. Note that a_1(0) = a_2(0) = 0, and
a_3(0) = 3. Also, note that w_1(0) = w_2(0) = 7, and w_3(0) =
1.
When I divide the product of the gÕs by the product of the
wÕs,
the result is
(5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)),
where c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x).
Here is what Mr. Harris is squawking about. The product of
the dÕs is 7. (That happens to be true for any x.) Harris
thinks
that is a problem, because it then looks like the product of
the
constant terms of
(5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x))
is 7, whereas the constant term of P(x)/49 is 22.
That is, d_1(x)*d_2(x)*d_3(x) = 7, whereas P(0)/49 = 22.
Puzzling! Is Harris right?
I will now give my deceptive, lying, sleight-of-hand
explanation.
Watch very carefully, because as Mr. Harris says, I am going
to
try very hard to mislead you!
So here is the explanation. The dÕs are NOT the constant 
terms
of the factors that contain them. That is, d_1(x) is NOT the
constant term of (5 c_1(x) + d_1(x)). Mr. Harris thinks it is.
Certainly it is in the right position for being a constant
term.
Since Mr. Harris understands and mostly relies on only
high-school
level mathematics, he detects constant terms by *inspection*.
The irony here is, he has provided a perfectly good, rigorous
definition of constant term of a function, but he 
doesnÕt use
it.
It is the value that the function takes on when the argument
is 0. Thus the constant term of (5 c_1(x) + d_1(x)) is NOT
d_1(x) = 7/w_1(x). It is instead d_1(0) = 7/w_1(0) = 1. This
is
directly from Mr. HarrisÕs own definition of 
constant term. Why
is he so reluctant to apply that definition? Why does he
instead rely
on inspection, which gives the incorrect answer? What HE
thinks
is the constant term, namely d_1(x) = 7/w_1(x) is not even
constant (unless he assumes what he wants to prove) !
Similary, the constant term of (5 c_2(x) + d_2(x)) is 1 also.
However, the constant term of
(5 c_3(x) + d_3(x))
is (5 c_3(0) + d_3(0)) = (5 a_3(0)/w_3(0) + 7/w_3(0))
= 5 * 3/1 + 7/1 = 15 + 7 = 22.
Wow, look at that! The product of the constant terms is
the constant term of the product! That is,
d_1(0)*d_2(0)*(5 a_3(0) + d_3(0)) = 1*1*22 = 22 = P(0)/49.
Amazing! Did I successfully trick you?
WhereÕs the trick?
Everything works out just as it should: IF you correctly
identify the constant terms.
WhereÕs the trick, Mr. Harris ?
>(Notice Nora Baron gives you clues, but somehow sci.math
readers
>donÕt seem to get it.)
I think not. I think that, with perhaps one exception,
sci.math
readers DO get it.
In fact, I even think there is no exception. I think Mr.
Harris
actually gets it also, but he cannot stand to admit it. There
is too much at stake. He clings to a delusion on the one-in-a
trillion chance that he will be proven right.
>Notice the posterÕs actually gives that when you divide off
49, you
>get
>g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x)
>which are TRACES, left of 49--after it has been divided off.
See above. There is no mystery. After division, the product
of the constant terms is the constant term of the product,
just
as it should be. That is, IF you correctly identify what the
constant terms actually are. If you donÕt, then you get the
wrong answer. This is Mr. HarrisÕs mistake.
This is not rocket science. This is not Galois theory, or
algebraic number theory, or field theory, or group theory. 
This
is not abstract algebra at all. This is not even high-school
level
algebra. The mistake Mr. Harris is making is really a
low-level one.
A ROOKY mistake, as Mr. Harris used to say (about me and
Arturo
Magidin and others). He is simply not using his own 
definition.
He is substituting in x when he should be substituting in 0.
And incredibly, he seems unable to understand it. Math genius
Harris, boy wonder Harris - from this you would think he
needed
help changing his own soiled underwear.
[more propaganda deleted]
Nora B.
>James Harris
===
Subject: Re: JSH: Simply fascinating
>> The math here is so readily understandable that itÕs
actually almost
>> as interesting watching how people react to it, as
anything else.
>> For instance, IÕve given a polynomial P(x) repeatedly
where I factor
>> it into three factors.
>> I point out that the factors must include factors of the
constant
term
>> of the polynomial P(x).
>> I note that the factors of the constant term are
independent of x, as
>> they are, in fact, constant.
>> Mathematically itÕs easy to show:
>> g_1(x) g_2(x) g_3(x) = P(x)
>> and
>> g_1(0) g_2(0) g_3(0) = P(0)
>> as P(0) gives the constant term of the polynomial, and
since the
>> polynomial results from multiplying together the three
factors which
>> IÕve called g_1(x), g_2(x), and g_3(x), then you can get
the factors
>> of the constant term, by setting x=0.
>> ItÕs that simple.
> Yes. Right so far. Trivial, but right.
>> Notice (1) you have the factors of P(x), (2) the constant
term of
P(x)
>> is determinable by setting x=0, (3) you also get the
factors of the
>> constant term.
>> Mathematically, itÕs simple to the point of trivial.
> Correct.
>> Now then, if you have *constants* which are factors of
another
>> constant then why would anyone try to argue that they are
actually
>> variables?
> No one does. If as you say, you assume that your factors are
>constant, of course they are not variables. That is nothing
but
>a dimwit tautology.
> And of course that is not what we say. We say that if you
factor
>49 out of P(x) in a *variable* way, as functions w_1(x),
w_2(x),
>and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and
g_3(x)
>by those variable functions, then you can obtain algebraic
integer
>factors on both sides of the resulting equation. That is,
>g_1(x)/w_1(x), etc., are all algebraic integers. That is all
you
>require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x),
and
>g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to
> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0).
>No problem with that either. And the product of these three
>constant terms equals P(0)/49.
> Ok, IÕm going to answer the Nora Baron poster yet again.
> And I want readers to understand that IÕve replied to this
poster who
> you know lies anyway as itÕs a guy posting as a woman 
using
a name
> thatÕs a palindrome MANY TIMES explaining in detail.
Please provide a link to a message where nora baron says that
she is a
woman.
<... Look now, even after the post where Nora Baron ended
with a male
> name, people are STILL supporting the poster!
Please provide a link to a message where nora baron says that
he is a
man.
===
Subject: Re: JSH: Simply fascinating
> ... 49 is a multiple of the polynomial.
You keep saying this kind of stuoid stuff. But 49 is *not* a
multiple of the
polynomial. Read the definition of the word 
ÔmultipleÕ.
> Look now, even after the post where Nora Baron ended with a
male
> name, people are STILL supporting the poster!
That is because nobody really cares what NoraÕs geneder
really is. They care
about what he says... i.e. his math. Is that not what you
claim is the only
thing that counts? If Nora was a ßaming tranvestite
hemaphroditic
cross-dresser, it would not make his math incorrect or your
math correct.
And it is not a lie to use a pseudonym, it is a personal
choice. Most of us
do not use our real names here.
===
Subject: Re: JSH: Simply fascinating
posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe
What does NoraÕs gender have to do with math. You only point
this out
because you are immature, trying to create a distraction, and
you know
your work is wrong. Get a life.
Dave
===
Subject: Re: JSH: Simply fascinating
> What does NoraÕs gender have to do with math. You only
point this out
> because you are immature, trying to create a distraction,
and you know
> your work is wrong. Get a life.
> Dave
The poster lies repeatedly. IÕve argued with this poster for
some
time and noted the lying, and the gender lying is just
another example
of a trend.
Some people donÕt like being lied to, and take it seriously.
My position is that I can explain in detail, and with very
simple
concepts how my argument works, but that mathematicians might
believe
they have lots of reasons for avoiding the truth--all social
ones.
Unfortunately, they are aided by the poster who manages to
maintain
confusion about the issues, and maintain the lie that there
is any
vagueness or area of real mathematical doubt about my work,
as many
people trust.
They trust that if I were right mathematicians wouldnÕt
disagree with
me, and there wouldnÕt be so much opposition to my work.
So the poster Nora Baron can just lie for the sake of showing
opposition, which helps to create the illusion of uncertainty
about my
work, helping to hide the truth from people who donÕt know
better.
The people who are well-trained mathematicians though, they
are not
fooled, and I know from my own experiences talking with
well-trained
mathematicians about my work.
So whatÕs happening now is sad in many ways, but part of it
has to do
with a basic contempt for the public, and a belief that they
can be
lead astray indefinitely by rather basic tactics.
I say, the strategy is about to die, and the passive-aggessive
strategy of hoping that indefinitely the public will be fooled
will be
shown to be one of professional suicide.
James Harris
===
Subject: Re: JSH: Simply fascinating
>What does NoraÕs gender have to do with math. You only 
point
this out
>because you are immature, trying to create a distraction,
and you know
>your work is wrong. Get a life.
>Dave
> The poster lies repeatedly. IÕve argued with this poster
for some
> time and noted the lying, and the gender lying is just
another example
> of a trend.
Bullcrap. Nora Baron has repeatedly shown errors in the
poster James
Harrismath and THAT is why he hates her. (Oh he does! He
said so not
long ago). He appears to be particularly infuriated because
she keeps
pinning him back to the more involved polynomials from which
his simpler
examples are derived, and showing where his error lies (no
pun intended).
This poster James Harris historically responds to math
rebuttals with
diatribes like this when he is unable to respond successfully
to the math
with math.
Perhaps what REALLY pisses him off is that a _female_ could
so effectively
demolish his erroneous math.
> Some people donÕt like being lied to, and take it 
seriously.
> My position is that I can explain in detail, and with very
simple
> concepts how my argument works, but that mathematicians
might believe
> they have lots of reasons for avoiding the truth--all
social ones.
Here it is:
> Unfortunately, they are aided by the poster who manages to
maintain
> confusion about the issues, and maintain the lie that there
is any
> vagueness or area of real mathematical doubt about my work
What this poster James Harris calls Ôliesare 
any math
rebuttals that show
that there is any vagueness or area of real mathematical
doubt about his
work. Since his work is correct they _must_ be lies.
KeithK
>, as many
> people trust.
<Rest of diatribe snipped James Harris
===
Subject: Re: JSH: Simply fascinating
> The people who are well-trained mathematicians though, they
are not
> fooled, and I know from my own experiences talking with
well-trained
> mathematicians about my work.
So whatÕs your problem? If well-trained mathematicians have
discussed
your work with you, surely you donÕt need to keep
banging your head against the lame brains at sci.math or any
other
newsgroup. Just have some of these well-trained
mathematicians promote your work and publications, accolades,
Field
medals, parades, etc. will follow.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
===
Subject: Re: JSH: Simply fascinating
> The math here is so readily understandable that itÕs
actually almost
> as interesting watching how people react to it, as anything
else.
> For instance, IÕve given a polynomial P(x) repeatedly 
where
I factor
> it into three factors.
> I point out that the factors must include factors of the
constant term
> of the polynomial P(x).
> I note that the factors of the constant term are
independent of x, as
> they are, in fact, constant.
> Mathematically itÕs easy to show:
> g_1(x) g_2(x) g_3(x) = P(x)
> and
> g_1(0) g_2(0) g_3(0) = P(0)
> as P(0) gives the constant term of the polynomial, and
since the
> polynomial results from multiplying together the three
factors which
> IÕve called g_1(x), g_2(x), and g_3(x), then you can get
the factors
> of the constant term, by setting x=0.
> ItÕs that simple.
> Notice (1) you have the factors of P(x), (2) the constant
term of P(x)
> is determinable by setting x=0, (3) you also get the
factors of the
> constant term.
> Mathematically, itÕs simple to the point of trivial.
Yes if
P(x) = g_1(x) g_2(x) g_3(x)
then
P(0) = g_1(0) g_2(0) g_3(0)
why did you surround this trivial fact with all the verbiage.
No
one is disputing this.
> Now then, if you have *constants* which are factors of
another
> constant then why would anyone try to argue that they are
actually
> variables?
> You have x, as the variable. Besides x there are just these
numbers.
> If you clear out x, then whatÕs left are constants. 
Letting
x=0,
> clears it out, leaving the constants visible.
> Some may say, yeah, sure, at x=0, but what about when x
doesnÕt equal
> 0?
> Um, if the numbers are constant, and so are independent of
x, then,
> duh, why should it matter what value x has?
> The logic is inescapable.
And by this logic the constant term of (a(0)=0, w(0)=1)
h(x) = (a(x)/w(x) + 7/w(x)), a(0)=0, w(0)=1
is 7. The constant term is not 7/w(x). Indeed you can write
h(x)
so you can see the constant. Let
t(x) = (a(x) -7w(x) +7)/w(x)
Note t(0) = 0 and
h(x) = (t(x) + 7)
So the constant term of h(x) is 7. And yes
7 divides the constant term of P(x). No one is arguing that
7/w(x) divides the constant term of P(x).
<standard diatribe snipped>
-William Hughes
===
Subject: Re: Simply fascinating
Having fun following this thread, but I just want to make
sure I understand
it.
So, James has the expression:
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
and he has shown that each a_*(x) evaluates to an algebraic
integer for
integer x.
And he has shown that a_*(0) is an integer.
Then he makes this leap by saying that
a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer
x.
Is that correct how I stated it, or am I missing something?
Darren
===
Subject: Re: Simply fascinating
> Then he makes this leap by saying that
> a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all
integer x.
> Is that correct how I stated it, or am I missing something?
You got it all correctly. You see, 7 is a constant. It does
not depend on x.
So 7 cannot change just because x changes. So a_1(x) is still
7, since 7 has
not changed..... get it? And by the way, the above equations
have no memory
of the 49 that got properly divided off as a multiple, or
something like
that. Get it?
===
Subject: Re: Simply fascinating
Meant:
> a_1(0) = 7 implies a_1(x) = 7.b (b algebraic Integer) for
all integer x.
===
Subject: Re: Simply fascinating
> But the concepts are where there is a problem, and the
social hang-up
> is in accepting that thereÕs this simple technique that
shows a BIG
> problem, which can invalidate claims of proof for, well,
over a
> hundred plus years.
That is not the hang-up. The hang-up is that you are 100% ed
up.
I would have no problem accepting something that would show a
big problem,
invalidating proofs for over a hundred years. I would just
love it. I would
think, wow, that is so cool! It would be a thrill. But you
have not
convinced me or anyone else that you have found anything
interesting or
correct. Just ßuffy bogus claims. Lots of claims. Little
valid math. Little
valid logic. Lots of paranoia. Lots of grandiosity. Lots of
non-standard
terminology. Lots of smoke screen crap, like the gender of
Nora.
Look James... lots of people here would love to see the next
crisis/revolution in math... another Russell shakeup, another
G.9adel
shakeup.
So tell me... why has nobody been interested in any of your
ideas? Could it
be that your ideas are ? Or do you think that it is more
likely that
everyone else is nuts. Apply OccamÕs Razor here! Just a
thought.
===
Subject: Graphing functions using rules for shifting
stretching and
reßecting
f(x)= 1 / x+2
is the first problem...
Sketch the graph of each of the following functions by using
the rules for
shifting stretching and reßecting.
another problem is:
g(x)=f(x+3)
My teacher really didnÕt explain this and my book is majorly
abridged.
Any help on how to get the points and how to sketch these
would be greatly
appreciated.
Kenny Chunn
kennyz79@gmail.com
jkc014@latech.edu
===
Subject: Re: Graphing functions using rules for shifting
stretching and
reßecting
| f(x)= 1 / x+2
| is the first problem...
|
| Sketch the graph of each of the following functions by
using the rules
for
| shifting stretching and reßecting.
|
| another problem is:
| g(x)=f(x+3)
|
| My teacher really didnÕt explain this and my book is
majorly abridged.
| Any help on how to get the points and how to sketch these
would be
greatly
| appreciated.
|
| Kenny Chunn
| kennyz79@gmail.com
| jkc014@latech.edu
|
Try experiment yourself :)
Here are a few examples.
Always start from with the most basic form of the formula.
Plot this: f(x) = x
Then plot f(x) = x + 1, what happens?
Then f(x) = x + 2, what happens?
Then the following, f(x) = x - 1 ; x - 2 etc.
What is, then, f(x) = x + n ??
Make your conclusion! What have you learnt?
Then try your problem: f(x) = 1/x with the following,
f(x) = 1/x + 1 ; 1/x +2 ... 1/(x + 1); 1/(x + 2) ... 1/(x +
1) + 1 ...etc
What is f(x) = 1/(x + m) + n?
Pay attention to where the y or x intercepts are on this
graphs. Also note
how
the gradient changes.
It helps if you have graph plotting software. If not use this
basic one
http://www.shodor.org/interactivate/activities/sketcher/
index.html
===
Subject: JSH: So is this him?
http://www.iomas.com/gina/ultrahiq/mega-society/noesisarchive
/harris.jpg
===
Subject: Re: JSH: So is this him?
>
http://www.iomas.com/gina/ultrahiq/mega-society/noesisarchive
/harris.jpg
Yes, at least according to him. HeÕs had the same picture on
other
sites he created.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: JSH: Without a trace
Some of you may have noticed that IÕm talking about
abstractions now
at higher and higher levels, which I think should help.
For instance, given a multiple of a polynomial it has been
well
accepted that you can divide that multiple off without
leaving a
trace.
Consider
P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10
and it is true that you can divide 5 from that factorizaton
giving
(x+1)(x+2) = x^2 + 3x + 2
without leaving a trace. That is, there is no indication left
that
the polynomial was ever multiplied by 5, as how could there
be?
Rationally there are an *infinity* of potential multiples that
you
could use, so why mathematically should a factorization of
x^2 + 3x +
2, have traces of one particular multiple, like traces of 5.
Now thatÕs obvious with polynomial factors but some
sci.mathÕers
clearly have many of you convinced that things change if you
have
P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10
and NOW divide the 5 off, as many of you seem convinced that
NOW if
the aÕs and bÕs are somehow complex, or weird, 
or otherwise
different
from what you get with polynomial then maybe, hmmm, possibly,
you
know? Maybe there IS a trace, right?
But how?
If I divide the 5 off, then I have a factorization of
x^2 + 3x + 2
just as before, and there is no rational reason to suppose
that a
trace of 5 is left, as why 5? Why not 7? Or 293874983?
Logically, it doesnÕt matter how complicated that 
aÕs and bÕs
are in
that example, when the 5 is divided off, it goes--without a
trace.
The situation now where posters argue with me, with arguments
that
have to boil down to some trace being left by a multiple, so
that they
can argue that the multiple divides through dependent on some
variable, is not unlike an argument between a scientist and
people who
donÕt believe in evolution, but worse.
In my case I have precedent from thousands of years of
mathematics
that a multiple can be divided off, absolute logic, and just
the plain
oddity of the notion that a multiple has to leave a trace
when itÕs
divided off, but STILL people have argued with me quite
successfully,
and I figure many of you, despite what I say here, remain
unconvinced
that IÕm right.
And you are no different than Creationists arguing with
scientists
against evolution. Or people who donÕt believe man landed on
the
moon.
You are no different, and in fact worse, as here itÕs
mathematics,
with absolute proof.
You people who canÕt accept mathematics are no different 
from
those
other people who canÕt accept science. You may think you 
like
or even
love mathematics, but you cannot when you refuse to accept
even the
most basic concepts in mathematics, with a result that
clearly you
donÕt like.
I understand the *desire* to have certain things be true, but
in
mathematics that doesnÕt matter. At the end of the day, you
put away
your emotions, and go with whatÕs true--if you truly value
mathematics.
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that IÕm talking about
abstractions now
> at higher and higher levels, which I think should help.
> For instance, given a multiple of a polynomial it has been
well
> accepted that you can divide that multiple off without
leaving a
> trace.
> Consider
> P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10
> and it is true that you can divide 5 from that factorizaton
giving
> (x+1)(x+2) = x^2 + 3x + 2
> without leaving a trace. That is, there is no indication
left that
> the polynomial was ever multiplied by 5, as how could there
be?
> Rationally there are an *infinity* of potential multiples
that you
> could use, so why mathematically should a factorization of
x^2 + 3x +
> 2, have traces of one particular multiple, like traces of 5.
> Now thatÕs obvious with polynomial factors but some
sci.mathÕers
> clearly have many of you convinced that things change if
you have
> P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10
> and NOW divide the 5 off, as many of you seem convinced
that NOW if
> the aÕs and bÕs are somehow complex, or 
weird, or otherwise
different
> from what you get with polynomial then maybe, hmmm,
possibly, you
> know? Maybe there IS a trace, right?
> But how?
> If I divide the 5 off, then I have a factorization of
> x^2 + 3x + 2
> just as before, and there is no rational reason to suppose
that a
> trace of 5 is left, as why 5? Why not 7? Or 293874983?
> Logically, it doesnÕt matter how complicated that 
aÕs and
bÕs are in
> that example, when the 5 is divided off, it goes--without a
trace.
> The situation now where posters argue with me, with
arguments that
> have to boil down to some trace being left by a multiple,
so that they
> can argue that the multiple divides through dependent on
some
> variable, is not unlike an argument between a scientist and
people who
> donÕt believe in evolution, but worse.
Here yet again, lacking a proof, you try to convince
people with an oversimplified toy example. Examples are
not proofs. As usual, your example is a reducible polynomial.
As you well know, at the heart of this controversy is the
difference between reducible and irreducible polynomials.
So consider an *irreducible* quadratic: a little simpler
than your cubic example, but more complicated than the
quadratic example you just gave.
Q(x) = 3 * (4*x^2 + 2*x - 1).
= 3 * (4*x^2 + 2*(x - 2) + 3).
Consider factoring this in the form
Q(x) = (2*a1(x) + 3)*(2*a2(x) + 3).
I will call this a Harristotelian factorization:
that is, you are factoring as if 2 is a polynomial
variable.
The aÕs satisfy the equation
a^2 - (x - 2)*a + 3 x^2 = 0.
Note that the roots of this equation are
always algebraic integers.
For example, when x = 0, the aÕs are
a = 0 and a = -2.
This gives the factorization
Q(0) = 3 * (-4 + 3) = -3.
The obvious way to factor 3 out of both sides of
this is thus:
Q(0)/3 = (2*0 + 3)*(2*(-2) + 3)/3
= (0/3 + 3/3)*(-4 + 3) = -1.
That is, a1(0) = 0 and is divisible by 3,
and a2(0) = -1 and is NOT divisible by 3.
Now consider Q(1). In this case, the aÕs
satisfy the equation
a^2 + a + 3 = 0.
This is irreducible. The two roots are
a_1(1) = (-1 + sqrt(-11))/2 and
a_2(1) = (-1 - sqrt(-11))/2.
You can easily check that a_1(1) * a_2(1) = 3,
as it should.
Therefore both a_1(1) and a_2(1) are not coprime
to 3. Moreover, neither a_1(1) nor a_2(1) are divisible
by 3. You can easily check that too.
So how should Q(1) be factored?
Recall from above that
Q(1) = (2 a_1(1) + 3)*(2 a_2(1) + 3).
Now I will divide both sides by 3. But I
am NOT going to divide the first term by 3
and the second term by 1. That will not work,
because 2 a_1(1)/3 is not an algebraic integer.
Instead I am going to divide the first term
by w_1(1) = a_1(1) and the second term by
w_2(1) = a_2(1) !
Does this work? LetÕs see ---
The first term becomes
(2 a_1(1) + 3)/a_1(1) = (2 + a_2(1)).
The second term becomes
(2 a_2(1) + 3)/a_2(1) = (2 + a_1(1)).
Both of these are algebraic integers! ThatÕs
good!
But I know from the definition of Q(x) that
Q(1) = 15, that is,
Q(1)/3 = 5.
So, is it true that
(2 + a_2(1)) * (2 + a_1(1)) = 5?
Check it out! It works! Everything comes out
right!
So whatÕs the moral of all this?
The moral is, contrary to what Harris says,
the RIGHT way to divide both sides of such equations,
so that you get algebraic integer factors after the
division, has to be dependent on x . When x = 0,
you divide the first and second terms respectively
by 3 and 1. When x = 1, you divide the first and
second terms respectively by w_1(1) = a_1(1) and
w_2(1) = a_2(1), that is, by two DIFFERENT factors
of 3. All the terms after the division are algebraic
integers.
Finally: no, examples are not proofs. Examples are
disproofs.
[non-math blather deleted]
Nora B.
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that IÕm talking about
abstractions now
> at higher and higher levels, which I think should help.
> For instance, given a multiple of a polynomial it has been
well
> accepted that you can divide that multiple off without
leaving a
> trace.
> Consider
> P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10
> and it is true that you can divide 5 from that factorizaton
giving
> (x+1)(x+2) = x^2 + 3x + 2
> without leaving a trace. That is, there is no indication
left that
> the polynomial was ever multiplied by 5, as how could there
be?
> Rationally there are an *infinity* of potential multiples
that you
> could use, so why mathematically should a factorization of
x^2 + 3x +
> 2, have traces of one particular multiple, like traces of 5.
Without a trace has no precise meaning, (and not much meaning
in any sense). You are retreating into not even wrong
territory.
However, something that has disappeared without a trace is any
mention of the constant term. This makes perfect sense. Each
time you brought up the constant term, everyone would ask:
isnÕt the constant term of (a(x)/w(x) + 7/w(x)) equal to 7?
Unfortunately for you, you could not answer this question
without
aknowleging that your entire argument was fallacious.
And too many people were asking the question for you to
comfortably
ignore it. So you could either admit you were wrong or retreat
into not even wrong territory. No one is surprised by your
choice.
- William Hughes
> Now thatÕs obvious with polynomial factors but some
sci.mathÕers
> clearly have many of you convinced that things change if
you have
> P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10
> and NOW divide the 5 off, as many of you seem convinced
that NOW if
> the aÕs and bÕs are somehow complex, or 
weird, or otherwise
different
> from what you get with polynomial then maybe, hmmm,
possibly, you
> know? Maybe there IS a trace, right?
> But how?
> If I divide the 5 off, then I have a factorization of
> x^2 + 3x + 2
> just as before, and there is no rational reason to suppose
that a
> trace of 5 is left, as why 5? Why not 7? Or 293874983?
> Logically, it doesnÕt matter how complicated that 
aÕs and
bÕs are in
> that example, when the 5 is divided off, it goes--without a
trace.
> The situation now where posters argue with me, with
arguments that
> have to boil down to some trace being left by a multiple,
so that they
> can argue that the multiple divides through dependent on
some
> variable, is not unlike an argument between a scientist and
people who
> donÕt believe in evolution, but worse.
> In my case I have precedent from thousands of years of
mathematics
> that a multiple can be divided off, absolute logic, and
just the plain
> oddity of the notion that a multiple has to leave a trace
when itÕs
> divided off, but STILL people have argued with me quite
successfully,
> and I figure many of you, despite what I say here, remain
unconvinced
> that IÕm right.
> And you are no different than Creationists arguing with
scientists
> against evolution. Or people who donÕt believe man landed
on the
> moon.
> You are no different, and in fact worse, as here itÕs
mathematics,
> with absolute proof.
> You people who canÕt accept mathematics are no different
from those
> other people who canÕt accept science. You may think you
like or even
> love mathematics, but you cannot when you refuse to accept
even the
> most basic concepts in mathematics, with a result that
clearly you
> donÕt like.
> I understand the *desire* to have certain things be true,
but in
> mathematics that doesnÕt matter. At the end of the day, 
you
put away
> your emotions, and go with whatÕs true--if you truly value
> mathematics.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that IÕm talking about
abstractions now
> at higher and higher levels, which I think should help.
IÕm begging for enlightnment.
> For instance, given a multiple of a polynomial it has been
well
> accepted that you can divide that multiple off without
leaving a
> trace.
You are obviously referring to those
wicked but rational criminal rings
who go about stripping integer polynomials of
their factors leaving no finger prints,
DNA or other forensic evidence.
ItÕs always a shock when your polynomials are
tampered with in this way.
However, more brutal irrationals tend
to leave polys hideously mutilated
and because of the resistance of the victim
the perpetratorÕs blood and other bodily
ßuids can be left at the scene. LetÕs hope the
forces of law and order can at least capture these
irrational miscreants.
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that IÕm talking about
abstractions now
> at higher and higher levels, which I think should help.
> For instance, given a multiple of a polynomial it has been
well
> accepted that you can divide that multiple off without
leaving a
> trace.
CanÕt we just agree on this one bit of terminology?
If you consider the number 12, say ...
3 is a FACTOR of 12, and
24 is a MULTIPLE of 12.
Surely itÕs not against your principles to use that bit of
correct
terminology?
===
Subject: Re: JSH: Without a trace
posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe
If you really understood mathematics, youÕd see where your
errors are.
YouÕre just too pompous to see that.
Dave
===
Subject: Re: JSH: Without a trace
> If you really understood mathematics, youÕd see where your
errors are.
> YouÕre just too pompous to see that.
> Dave
Yet IÕve made quite a few mistakes over the years trying out
different
ideas, and dropping those that failed.
Here posters need only do one thing: prove me wrong.
And itÕs a sad lie for some of them to just repeat that they
have or
that I donÕt answer objections as IÕve 
answered objections
many, many,
many times over a period of years.
These people simpy will not listen to reason.
What do they appear to accomplish? My guess is that leading
mathematicians who are aware of my work, pay attention to
them, as
evidence that they can rely on the passive-aggessive strategy
of
waiting.
So then sci.math may actually be having an impact after all.
You may be convincing certain people who do know that my work
is
correct they have the luxury of waiting.
James Harris
===
Subject: Re: JSH: Without a trace
> Here posters need only do one thing: prove me wrong.
> And itÕs a sad lie for some of them to just repeat that
they have or
> that I donÕt answer objections as IÕve 
answered objections
many, many,
> many times over a period of years.
Yes, but your answers are consistently off base. First, you
misrepresent the
objections; second, you ignore the
counter-examples or disproofs entirely; third, you repeat
your own
previously ßawed argument. In your mind, that disposes
of the matter. Everyone else, however, sees that you have
*not* answered the
objection, but only substituted a loud noise.
Of all the p-baked cranks, where Ôpequals 1/2, 
you take the
q, where ÔqÕ
equals cake.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
===
Subject: Re: Mechanics problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4GMNn03882;
>A stone is thrown vertically upwards with a speed of 20 m/s.
One
>second later a second stone is thrown vertically upwards
with a speed
>of 25 m/s. At what height above the ground do they collide?
>I canÕt get my head round this one. Can you help?
If you are doing problems like this then you surely know the
height formula: h= -4.9 t^2+ v0 t where t is the time since
the
object was thrown up and v0 is the initial speed.
Taking t= 0 when the first stone is thrown,
h= -4.9 t^2+ 20t for the first stone.
time since it was thrown is t-1 rather than t.
h= -4.9(t-1)^2+ 25t.
When the stones collide, both those hÕs will be the same for
the
same t. Set them equal and solve for t.
===
Subject: Re: Mechanics problem
> time since it was thrown is t-1 rather than t.
>
> h= -4.9(t-1)^2+ 25t.
ThatÕs actually
h = -4.9(t - 1)^2 + 25(t - 1)
> When the stones collide, both those hÕs will be the same
for the
> same t. Set them equal and solve for t.
And when youÕll do that, youÕll get two values 
of t that make
it so.
Figure out what the two answers physically mean and which
one(s) is
correct.
To the original poster -- I recommend you leave the
acceleration and
initial velocities as symbols, and then substitute them back
in at the
end of the problem. That way you do as little arithmetic as
possible,
making error less likely.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: Mechanics problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4MdVQ02657;
s=20t - 4.9 t^2
s=25(t-1)- 4.9 (t-1)^2
25t - 25 - 4.9 (t^2 - 2t + 1)
25t - 25 - 4.9t^2 + 9.8t -4.9
20t - 4.9 t^2 = 25t - 25 - 4.9t^2 + 9.8t -4.9
20t - 25t - 9.8t = -25 - 4.9
14.8t = 29.9
t= 2.020
Which is not the bookÕs answer (20.4 m) nor does it agree
with RichÕs
coment about getting two answers (probably by using the
quadratic
formula.
So IÕm still stuck!
Jo
===
Subject: Re: Mechanics problem
> s=20t - 4.9 t^2
> s=25(t-1)- 4.9 (t-1)^2
> 25t - 25 - 4.9 (t^2 - 2t + 1)
> 25t - 25 - 4.9t^2 + 9.8t -4.9
> 20t - 4.9 t^2 = 25t - 25 - 4.9t^2 + 9.8t -4.9
> 20t - 25t - 9.8t = -25 - 4.9
> 14.8t = 29.9
> t= 2.020
ThatÕs the correct answer --- for the TIME of the collision.
> Which is not the bookÕs answer (20.4 m)
h = 20(2.02) - 4.9(2.02)^2 = ?
h = 25(2.02 - 1) - 4.9(2.02 - 1)^2 = ?
> nor does it agree with RichÕs coment about getting two
answers
I was wrong about that. I donÕt know what I was thinking.
Consider the height vs. time graphs -- the parabolas have
the same curvature, so they can only intersect once.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: question, fiddling around inequalities
hello everyone I have an inequality question which I think is
relatively
easy if you know the right trick, however I have not managed
to find the
right trick as of yet.
Ok this is the question.
I have 2 critcal points X and Y where 0 < X < Y, in
particular;
X = (1/2)(A-sqrt(A^2-4(B^2))), Y = (1/2)(A+sqrt(A^2-4(B^2))).
Note also given that 0 < B < (1/2)A.
I know that (2A(B^2)Y)/((B^2+Y^2)^2) < 1, but I was required
to show this,
and did so using the following method:
Consider ((B^2+Y^2)^2)-(2A(B^2)Y). Expanding out using the
definition of Y
above gives;
(1/2)(A^4)-2(A^2)(B^2)+(1/2)(A^3)sqrt(A^2-4(B^2))-A(B^2)sqrt(
A^2-4(B^2)).
(*)
since B < (1/2)A this is > (1/2)(A^4)-2(A^2)(B^2). then again
> 0
so we see ((B^2+Y^2)^2)-(2A(B^2)Y) > 0 and indeed
(2A(B^2)Y)/((B^2+Y^2)^2) <
1, as required.
Now heres the bit I cannot do. Show (2A(B^2)X)/((B^2+X^2)^2)
> 1. For this I
considered showing:
(2A(B^2)X)-((B^2+X^2)^2)>0. Expanding out did not lead to the
result as
before, however it may be possible to combine the expanded
out result with
Either way I have
tried for a long time now and would appreciate someone
steering me in the
correct direction.
X.
===
Subject: Re: question, fiddling around inequalities
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> hello everyone I have an inequality question which I think
is
relatively
> easy if you know the right trick, however I have not
managed to find
the
> right trick as of yet.
> Ok this is the question.
> I have 2 critcal points X and Y where 0 < X < Y, in
particular;
> X = (1/2)(A-sqrt(A^2-4(B^2))), Y =
(1/2)(A+sqrt(A^2-4(B^2))).
> Note also given that 0 < B < (1/2)A.
> I know that (2A(B^2)Y)/((B^2+Y^2)^2) < 1, but I was
required to show
this...
Since (B^2+Y^2)^2 > 1, you can rewrite the inequality as
2A(B^2)Y < (B^2+Y^2)^2
Then substitute for Y and expand out both sides to get
A^2*B^2 + A*B^2*sqr(A^2-4*B^2) < A^4/2 - A^2*B^2 +
A^3/2*sqr(A^2-4*B^2)
Now, because sqr(A^2-4*B^2) > 0, this inequality is
guaranteed true if
both
A^2*B^2 < A^4/2 - A^2*B^2
and
A*B^2 < A^3/2
both of which follow easily from 0 < B < (1/2)A.
===
Subject: Re: question, fiddling around inequalities
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Since (B^2+Y^2)^2 > 1, you can rewrite the inequality as ...
Of course I meant to say Since (B^2+Y^2)^2 > 0
===
Subject: Re: question, fiddling around inequalities
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Since (B^2+Y^2)^2 > 1, you can rewrite the inequality as ...
Of course, I meant to say Since (B^2+Y^2)^2 > 0
===
Subject: Revolutionary Mathematics: Non-polynomial
Factorization
IÕm an amateur mathematician who has found that using some
simple
ideas I can show a remarkable error in thinking which
unfortunately
underpins much of whatÕs thought to be known in the
discipline of
algebraic number theory.
The mathematics which proves the problem is extrarodinary in
that it
is very simple, relying on some of the most basic concepts in
algebra.
It is revolutionary in that it so upsets the status quo,
changes the
historical positions of so many mathematicians, and is just
plain
surprising in many ways.
Essentially what I do is relate one polynomial to a family of
polynomials, using what I call a non-polynomial
factorization, which I
call that as the factoring of the primary polynomial is into
factors
that are themselves not polynomials.
For simplicity in explaining I donÕt initially give a
tremendous
amount of detail about where the polynomial comes from as
years of
experience talking about this on Usenet has shown me how
easily
posters can confuse people with detail of that sort, but I
have no
problem going into detail if real curiosity emerges.
So I say I use a polynomial. So letÕs see it.
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
where x is an integer.
It is, as you can see, a polynomial, and itÕs distinctive in
an
important way, as it has 49 as a multiple as
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
and IÕll relate it to a family of polynomials, where that
multiple 49
is very important.
Now hereÕs where a basic idea steps in, as years ago I
discovered the
idea of separating out a polynomial in a special way so that
you can
factor it as if itÕs a polynomial in a *different* variable
from the
polynomial variable itself.
Here that gives
P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
where if you multiply it all out and simplify, youÕll still
get
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
and the pattern I use next might not quite be visible, so
IÕll make a
substitution using Y=5 and Z=7, to get
P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2 +
Z^3
and you can hopefully see how that can factor as
P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z)
and going ahead and putting back in their values, as I used Y
and Z
just as a bit of help (a dangerous bit of help as sci.math
posters
have routinely jumped in at this point in the past to claim
that
actually x is not the polynomial variable at all!!!), I have
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
and the aÕs are easily determined by using
(5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7) =
49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3
as you get three simultaneous equations, for instance,
a_1(x) a_2(x) a_3(x) = 49(2401 x^3 - 147 x^2 + 3x)
is one of the three.
Solving for the aÕs, you get a cubic, which is
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
where the roots of that cubic are a_1(x), a_2(x), and a_3(x),
and that
cubic is the family of polynomials that are related to P(x),
as for
any given value of x, you get a polynomial.
Notice that the multiple of P(x), 49 is locked into the
family of
cubics. But itÕs not a multiple of the cubic, as you have 
the
term
3(-1 + 49x)
which is of course, coprime to 49.
ThatÕs important, as somehow with this technique of 
factoring
P(x) in
a special way, IÕve related that factorization of a
polynomial that
has 49 as a multiple, to a family of polynomials that do not.
ItÕs one of the most important relations in math history.
Why? Well, for P(x), 49 is just a multiple, so I can divide
it off.
That gives me
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and reasonably, dividing 49 from the factorization of P(x),
gets rid
of it, without a trace, but that results in the factorization
P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
and so much arguing, over a period of years with this
technique,
settles on what happens next with such an example.
Before I go further though, letÕs stop to consider what 
IÕve
done:
1. I have a polynomial P(x) that has 49 as a multiple.
2. I factor that polynomial in a special way to get
non-polynomial
factors.
3. I solve for those factors giving me a cubic family of
polynomials.
4. I now decide to divide 49 from P(x), and am at the point of
considering how that affects its factorization.
Here P(x) is key. It has a multiple 49, and its factorization
provides the relation to the cubic family that defines the
aÕs. It is
a basic concept in algebra that a multiple can be divided off
without
problem, and necessarily that must be true, as consider, how
can a
factorization of
300125x^3 - 18375 x^2 - 360 x + 22
have some impact from the polynomial resulting from dividing
off a
multiple?
The equation has no memory. After all, if it did, why 49? Why
not 3
or 11, or 83947397, or an infinity of other numbers?
I belabor that point as if you accept it, then what follows is
obvious.
So I have
P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
and I want to know how the 49 divides through the factors of
P(x),
while I already know that
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
so 49 divides off of P(x) itself, without a trace, which is
not a
surprise.
The simplest way to find out is to check directly. However,
the aÕs
are rather complex being defined by the cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and it might seem extraordinarily difficult to 
find out
anything about
them, so is the cause lost?
No, because I can simplify by focusing on the constant term
of P(x).
Why the constant term? Because unlike the other terms it is
independent of x itself, and much of the complexity washes
out.
The constant term of P(x) is given by setting x=0, as that
sets the
terms that have x as a variable to 0, leaving just the
constant term:
P(0) = 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = 49(22).
Now what about the factors of P(x)? Well, they become a LOT
easier to
manage as well, as I then have
a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0) )
which is
a^3 - 3a^2 = 0 and a^3 - 3a^2 = 0, is a^2(a - 3) = 0
so despite all the early complexity, I have now the simple
result that
two of the aÕs equal 0, at x=0, while one equals 3.
So I need to pick the aÕs to proceed, and my usual 
convention
is to
let
a_1(0) = 0, a_2(0) = 0, and a_3(0) = 3
so with
P(0)/49 = (5a_1(0) + 7)(5a_2(0)+ 7)(5a_3(0) + 7)/49
I have
P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = (7)(7)(22)/49 =
22
which is correct as we already found out earlier.
But wait, arenÕt I just checking at a *single* value, for
something
thatÕs terribly complicated, where maybe things are 
different
at a
different value?
Well, sure, things are different for terms that have x as a
factor, as
thatÕs how algebra works. As x varies, you get different
things
happening.
Well, yes, for terms that vary with x, that is correct. But if
something is constant, then it doesnÕt vary.
Checking at x=0 clears out those terms that vary, leaving
those that
do not, revealing that for two of the terms, whatÕs left
over, is 7,
while for one, whatÕs left over is 22.
ThatÕs just a fact. ItÕs such a simple fact 
that one of the
more
remarkable things over the years IÕve found is the ability 
of
some
people to argue around it.
If you accept that constants are not variable, and that 7 is
just a
number that does not change with x, and you accept that
setting x=0
reveals constants by eliminating the terms that vary, then
you should
accept that the constants are constant without regard to the
value of
x.
So if I could look at the constants at x=39473987, then
thatÕd be
fine, but with that value, the terms with x get in the way,
but at
x=0, they do not.
But with P(x)/49, I already have that 49 is gone, without a
trace.
So, if the constants for factors of P(x) are 7, 7 and 49, who
MUST the
49 divide through?
Like this
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
with indices arbitrary, as remember, I picked the first two
aÕs to be
those that go to 0, when x=0.
So far, so good, and you may wonder how something so simple
can be
revolutionary.
Well, remember I related the factorization of P(x) to a
*family* of
cubics given by the roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and now IÕve shown a result that indicates that two of the
aÕs have 7
as a factor, without regard to the value of x.
ThatÕs HUGE as it turns out that a long time ago (like over 
a
hundred
years ago) msthematicians decided that you couldnÕt make 
such
a
determination if the roots of a polynomial were all
irrational and
didnÕt all have the same factor.
(Like x^2 - 3 has sqrt(3) and -sqrt(3) as roots.)
That belief is the basis for the modern usage of group theory
including Galois Theory.
YouÕve just seen a basic result showing it to be a false
belief.
Like stick in x=1, and you have
S(a) = a^3 + 3(48)a^2 - 49(2257)
and without solving for the roots you know already that two
of its
roots should have 7 as a factor, but now thereÕs another
problem.
Over a hundred years ago, back in the late 1800Õs
mathematicians
studied polynomials special in that they had a leading
coefficient of
1 or -1, and integer coefficients. Polynomials with a leading
coefficient of 1 or -1 are called monic, so more technically,
they
studied monic polynomials with integer coefficients.
They called the roots of these polynomials algebraic integers.
Those roots form a group of numbers called the ring of
algebraic
integers.
And it turns out that for
S(a) = a^3 + 3(48)a^2 - 49(2257)
if you take its roots, you canÕt find any that 
when divided by
7, give
an algebraic integer.
It gets more complicated, as you can prove that with the
factorization
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
its possible to find algebraic integers w_1, w_2, and w_3,
such that
w_1 w_2 w_3 = 49
where the wÕs are the respective factors of
(5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
when I just said that you canÕt get algebraic integers with
x=1, and
the roots of
a^3 + 3(48)a^2 - 49(2257)
wher 7 is a factor of *any* of those roots, let alone two.
Hmmm...problem, right? Is it all lost? Does that mean
everything
before was wrong? Yuck, did I just waste your time with
claims of
revolutionary mathematics and all of that, when thereÕs this
weird
result that seems to show it all must be wrong?
But wait, for my result I used some very basic concepts. Like
I rely
on 49 as a multiple just dividing off, without a trace. And I
focused
on constant terms because they are, well, constant, and
simpler to
work with than terms that include x, where all the complexity
comes
into the picture.
Well, there was that weird technique of factoring a
polynomial some
odd way.
But, sure, itÕs different, but all of the mathematical
operations are
valid ones.
So what gives?
LetÕs focus on the result again, as I said that with
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
you can have algebraic integers w_1, w_2, and w_3 that are
factors of
the factors of P(x).
But I also showed that
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
but that canÕt work with these numbers called algebraic
integers!
Well, consider u_1 u_2 u_3 = 1, where multiplying through
gives
P(x)/49 =
(5a_1(x)u_1/7 + u_1)(5a_2(x) u_2/7 + u_2)(5a_3(x) u_3 + 7u_3)
where now it all *does* work with algebraic integers.
That is, for some reason, while two of the aÕs do not have 7
as a
factor in the ring of algebraic integers, they *do* have 7u_1
and
7u_2, respectively, as factors, where u_1 and u_2 are units,
in that
they are factors of 1.
But, hereÕs where itÕs rather strange, as u_1 
and u_2 are
algebraic
integers i.e. roots of a monic polynomial with integer
coefficients,
while u_3 is not.
So, for that reason, u_1 and u_2 are NOT units in the ring of
algebraic integers.
HereÕs an example that I hope helps you see how it works.
In integers you can have
S(x) = (3x + 1)(x + 1) = 3x^2 + 4x + 1
but now consider
SÕ(x) = (3x + u_1)(x + u_2) = 3x^2 + kx + 1
where u_1 u_2 = 1, and k is an integer.
Here, u_1 CAN be an algebraic integer, but u_2 CANNOT be an
algebraic
integer.
So, in the ring of algebraic integers, u_1 cannot be a unit.
Well, maybe itÕs some kind of fraction, right? Well, yes,
possibly,
it is, and like, for k=0, you can see that it IS some kind of
fraction. But, have you covered all the possibilities?
The answer is, no, you canÕt have and have it all be
mathematically
consistent.
So why all this talk about algebraic integers? They seem kind
of
messy at this point.
Like you get this nifty result with some basic concepts and a
special
factoring technique, where it was all rather straightforward,
and then
suddenly focusing on roots of monic polynomials with integer
coefficients gives all kinds of head-scratching trouble!
Well, the reason for the focus is that mathematicians for
over a
hundred years focused on the ring of algebraic integers not
understanding that it was, IÕll say, quirky.
Actually itÕs worse than quirky, as not understanding the
weirdness
that can arise from focusing on the roots of monic
polynomials with
integer coefficients you can prove lots of things with the
ring of
algebraic integers that are in fact mathematically false.
ItÕs a nasty little bug.
To give you some perspective, the tools used by Wiles in his
work that
purportedly proves the Taniyama-Shimura Conjecture, to most
people,
work that supposedly proves FermatÕs Last Theorem, 
donÕt work.
They donÕt work because theyÕre based on an 
improper
understanding of
the ring of algebraic integers. Works like WilesÕs cannot be
rescued
from this bug.
So now maybe you understand the controversy!
Is my work actually complicated for a trained mathematician?
No. My guess is that a trained mathematician can go over the
entire
thing, and understand the implications in about an hour.
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
Well, I typed up this long post yesterday, looked it over a
couple of
times, and felt good enough to post it, but noticed a minor
error near
the end which IÕm fixing here.
I may as well ask and answer a question that might be
pertinent at
this time:
Why make such a freaking long post?
Answer is thatÕs what it takes to cover all the ground and
describe
the issue as well as the basis for controversy, like why are
all these
people always replying with such motivated interest in my
work? Well,
read the post and you can understand what theyÕre 
fighting to
defend,
which is why they have all that obsessive energy. And youÕll
understand why IÕm working so hard to get this work past the
people
trying to block it.
IÕm an amateur mathematician who has found that using some
simple
ideas I can show a remarkable error in thinking which
unfortunately
underpins much of whatÕs thought to be known in the
discipline of
algebraic number theory.
The mathematics which proves the problem is extrarodinary in
that it
is very simple, relying on some of the most basic concepts in
algebra.
It is revolutionary in that it so upsets the status quo,
changes the
historical positions of so many mathematicians, and is just
plain
surprising in many ways.
Essentially what I do is relate one polynomial to a family of
polynomials, using what I call a non-polynomial
factorization, which I
call that as the factoring of the primary polynomial is into
factors
that are themselves not polynomials.
For simplicity in explaining I donÕt initially give a
tremendous
amount of detail about where the polynomial comes from as
years of
experience talking about this on Usenet has shown me how
easily
posters can confuse people with detail of that sort, but I
have no
problem going into detail if real curiosity emerges.
So I say I use a polynomial. So letÕs see it.
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
where x is an integer.
It is, as you can see, a polynomial, and itÕs distinctive in
an
important way, as it has 49 as a multiple as
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
and IÕll relate it to a family of polynomials, where that
multiple 49
is very important.
Now hereÕs where a basic idea steps in, as years ago I
discovered the
idea of separating out a polynomial in a special way so that
you can
factor it as if itÕs a polynomial in a *different* variable
from the
polynomial variable itself.
Here that gives
P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
where if you multiply it all out and simplify, youÕll still
get
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
and the pattern I use next might not quite be visible, so
IÕll make a
substitution using Y=5 and Z=7, to get
P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2 +
Z^3
and you can hopefully see how that can factor as
P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z)
and going ahead and putting back in their values, as I used Y
and Z
just as a bit of help (a dangerous bit of help as sci.math
posters
have routinely jumped in at this point in the past to claim
that
actually x is not the polynomial variable at all!!!), I have
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
and the aÕs are easily determined by using
(5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7) =
49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3
as you get three simultaneous equations, for instance,
a_1(x) a_2(x) a_3(x) = 49(2401 x^3 - 147 x^2 + 3x)
is one of the three.
Solving for the aÕs, you get a cubic, which is
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
where the roots of that cubic are a_1(x), a_2(x), and a_3(x),
and that
cubic is the family of polynomials that are related to P(x),
as for
any given value of x, you get a polynomial.
Notice that the multiple of P(x), 49 is locked into the
family of
cubics. But itÕs not a multiple of the cubic, as you have 
the
term
3(-1 + 49x)
which is of course, coprime to 49.
ThatÕs important, as somehow with this technique of 
factoring
P(x) in
a special way, IÕve related that factorization of a
polynomial that
has 49 as a multiple, to a family of polynomials that do not.
ItÕs one of the most important relations in math history.
Why? Well, for P(x), 49 is just a multiple, so I can divide
it off.
That gives me
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and reasonably, dividing 49 from the factorization of P(x),
gets rid
of it, without a trace, but that results in the factorization
P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
and so much arguing, over a period of years with this
technique,
settles on what happens next with such an example.
Before I go further though, letÕs stop to consider what 
IÕve
done:
1. I have a polynomial P(x) that has 49 as a multiple.
2. I factor that polynomial in a special way to get
non-polynomial
factors.
3. I solve for those factors giving me a cubic family of
polynomials.
4. I now decide to divide 49 from P(x), and am at the point of
considering how that affects its factorization.
Here P(x) is key. It has a multiple 49, and its factorization
provides the relation to the cubic family that defines the
aÕs. It is
a basic concept in algebra that a multiple can be divided off
without
problem, and necessarily that must be true, as consider, how
can a
factorization of
300125x^3 - 18375 x^2 - 360 x + 22
have some impact from the polynomial resulting from dividing
off a
multiple?
The equation has no memory. After all, if it did, why 49? Why
not 3
or 11, or 83947397, or an infinity of other numbers?
I belabor that point as if you accept it, then what follows is
obvious.
So I have
P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
and I want to know how the 49 divides through the factors of
P(x),
while I already know that
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
so 49 divides off of P(x) itself, without a trace, which is
not a
surprise.
The simplest way to find out is to check directly. However,
the aÕs
are rather complex being defined by the cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and it might seem extraordinarily difficult to 
find out
anything about
them, so is the cause lost?
No, because I can simplify by focusing on the constant term
of P(x).
Why the constant term? Because unlike the other terms it is
independent of x itself, and much of the complexity washes
out.
The constant term of P(x) is given by setting x=0, as that
sets the
terms that have x as a variable to 0, leaving just the
constant term:
P(0) = 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = 49(22).
Now what about the factors of P(x)? Well, they become a LOT
easier to
manage as well, as I then have
a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0) )
which is
a^3 - 3a^2 = 0 and a^3 - 3a^2 = 0, is a^2(a - 3) = 0
so despite all the early complexity, I have now the simple
result that
two of the aÕs equal 0, at x=0, while one equals 3.
So I need to pick the aÕs to proceed, and my usual 
convention
is to
let
a_1(0) = 0, a_2(0) = 0, and a_3(0) = 3
so with
P(0)/49 = (5a_1(0) + 7)(5a_2(0)+ 7)(5a_3(0) + 7)/49
I have
P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = (7)(7)(22)/49 =
22
which is correct as we already found out earlier.
But wait, arenÕt I just checking at a *single* value, for
something
thatÕs terribly complicated, where maybe things are 
different
at a
different value?
Well, sure, things are different for terms that have x as a
factor, as
thatÕs how algebra works. As x varies, you get different
things
happening.
Well, yes, for terms that vary with x, that is correct. But if
something is constant, then it doesnÕt vary.
Checking at x=0 clears out those terms that vary, leaving
those that
do not, revealing that for two of the terms, whatÕs left
over, is 7,
while for one, whatÕs left over is 22.
ThatÕs just a fact. ItÕs such a simple fact 
that one of the
more
remarkable things over the years IÕve found is the ability 
of
some
people to argue around it.
If you accept that constants are not variable, and that 7 is
just a
number that does not change with x, and you accept that
setting x=0
reveals constants by eliminating the terms that vary, then
you should
accept that the constants are constant without regard to the
value of
x.
So if I could look at the constants at x=39473987, then
thatÕd be
fine, but with that value, the terms with x get in the way,
but at
x=0, they do not.
But with P(x)/49, I already have that 49 is gone, without a
trace.
So, if the constants for factors of P(x) are 7, 7 and 49, who
MUST the
49 divide through?
Like this
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
with indices arbitrary, as remember, I picked the first two
aÕs to be
those that go to 0, when x=0.
So far, so good, and you may wonder how something so simple
can be
revolutionary.
Well, remember I related the factorization of P(x) to a
*family* of
cubics given by the roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and now IÕve shown a result that indicates that two of the
aÕs have 7
as a factor, without regard to the value of x.
ThatÕs HUGE as it turns out that a long time ago (like over 
a
hundred
years ago) msthematicians decided that you couldnÕt make 
such
a
determination if the roots of a polynomial were all
irrational and
didnÕt all have the same factor.
(Like x^2 - 3 has sqrt(3) and -sqrt(3) as roots.)
That belief is the basis for the modern usage of group theory
including Galois Theory.
YouÕve just seen a basic result showing it to be a false
belief.
Like stick in x=1, and you have
S(a) = a^3 + 3(48)a^2 - 49(2257)
and without solving for the roots you know already that two
of its
roots should have 7 as a factor, but now thereÕs another
problem.
Over a hundred years ago, back in the late 1800Õs
mathematicians
studied polynomials special in that they had a leading
coefficient of
1 or -1, and integer coefficients. Polynomials with a leading
coefficient of 1 or -1 are called monic, so more technically,
they
studied monic polynomials with integer coefficients.
They called the roots of these polynomials algebraic integers.
Those roots form a group of numbers called the ring of
algebraic
integers.
And it turns out that for
S(a) = a^3 + 3(48)a^2 - 49(2257)
if you take its roots, you canÕt find any that 
when divided by
7, give
an algebraic integer.
It gets more complicated, as you can prove that with the
factorization
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
its possible to find algebraic integers w_1, w_2, and w_3,
such that
w_1 w_2 w_3 = 49
where the wÕs are the respective factors of
(5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
when I just said that you canÕt get algebraic integers with
x=1, and
the roots of
a^3 + 3(48)a^2 - 49(2257)
wher 7 is a factor of *any* of those roots, let alone two.
Hmmm...problem, right? Is it all lost? Does that mean
everything
before was wrong? Yuck, did I just waste your time with
claims of
revolutionary mathematics and all of that, when thereÕs this
weird
result that seems to show it all must be wrong?
But wait, for my result I used some very basic concepts. Like
I rely
on 49 as a multiple just dividing off, without a trace. And I
focused
on constant terms because they are, well, constant, and
simpler to
work with than terms that include x, where all the complexity
comes
into the picture.
Well, there was that weird technique of factoring a
polynomial some
odd way.
But, sure, itÕs different, but all of the mathematical
operations are
valid ones.
So what gives?
LetÕs focus on the result again, as I said that with
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
you can have algebraic integers w_1, w_2, and w_3 that are
factors of
the factors of P(x).
But I also showed that
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
but that canÕt work with these numbers called algebraic
integers!
Well, consider u_1 u_2 u_3 = 1, where multiplying through
gives
P(x)/49 =
(5a_1(x)u_1/7 + u_1)(5a_2(x) u_2/7 + u_2)(5a_3(x) u_3 + 7u_3)
where now it all *does* work with algebraic integers.
That is, for some reason, while two of the aÕs do not have 7
as a
factor in the ring of algebraic integers, they *do* have 7u_1
and
7u_2, respectively, as factors, where u_1 and u_2 are units,
in that
they are factors of 1.
But, hereÕs where itÕs rather strange, as u_1 
and u_2 are not
algebraic integers i.e. roots of a monic polynomial with
integer
coefficients, while u_3 is.
So, for that reason, neither u_1, u_2, nor u_3 is a unit in
the ring
of algebraic integers.
HereÕs an example that I hope helps you see how it works.
In integers you can have
S(x) = (3x + 1)(x + 1) = 3x^2 + 4x + 1
but now consider
SÕ(x) = (3x + u_1)(x + u_2) = 3x^2 + kx + 1
where u_1 u_2 = 1, and k is an integer.
Here, u_1 CAN be an algebraic integer, but u_2 CANNOT be an
algebraic
integer.
So, in the ring of algebraic integers, u_1 cannot be a unit.
Well, maybe itÕs some kind of fraction, right? Well, yes,
possibly,
it is, and like, for k=0, you can see that it IS some kind of
fraction. But, have you covered all the possibilities?
The answer is, no, you canÕt have and have it all be
mathematically
consistent.
So why all this talk about algebraic integers? They seem kind
of
messy at this point.
Like you get this nifty result with some basic concepts and a
special
factoring technique, where it was all rather straightforward,
and then
suddenly focusing on roots of monic polynomials with integer
coefficients gives all kinds of head-scratching trouble!
Well, the reason for the focus is that mathematicians for
over a
hundred years focused on the ring of algebraic integers not
understanding that it was, IÕll say, quirky.
Actually itÕs worse than quirky, as not understanding the
weirdness
that can arise from focusing on the roots of monic
polynomials with
integer coefficients you can prove lots of things with the
ring of
algebraic integers that are in fact mathematically false.
ItÕs a nasty little bug.
To give you some perspective, the tools used by Wiles in his
work that
purportedly proves the Taniyama-Shimura Conjecture, to most
people,
work that supposedly proves FermatÕs Last Theorem, 
donÕt work.
They donÕt work because theyÕre based on an 
improper
understanding of
the ring of algebraic integers. Works like WilesÕs cannot be
rescued
from this bug.
So now maybe you understand the controversy!
Is my work actually complicated for a trained mathematician?
No. My guess is that a trained mathematician can go over the
entire
thing, and understand the implications in about an hour.
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> Well, I typed up this long post yesterday, looked it over a
couple of
> times, and felt good enough to post it, but noticed a minor
error near
> the end which IÕm fixing here.
OOPS! The original was correct, so my correction was wrong.
ThatÕs the second time IÕve done something 
like this which
is, yes,
worrisome.
I guess IÕm so used to making mistakes that I now 
find
mistakes where
there are none.
James Harris
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> Well, I typed up this long post yesterday, looked it over a
couple of
> times, and felt good enough to post it, but noticed a minor
error near
> the end which IÕm fixing here.
> I may as well ask and answer a question that might be
pertinent at
> this time:
> Why make such a freaking long post?
> Answer is thatÕs what it takes to cover all the ground and
describe
> the issue as well as the basis for controversy, like why
are all these
> people always replying with such motivated interest in my
work? Well,
> read the post and you can understand what theyÕre 
fighting
to defend,
> which is why they have all that obsessive energy. And 
youÕll
> understand why IÕm working so hard to get this work past
the people
> trying to block it.
Better question: why wasnÕt it longer? References to the
precise rings
would have been nice. References to properties that you
wished to
preserve would have been nice. If you had made it about 10%
longer with
those few details, it would have been far clearer. As it
stands now,
you are doing manipulations that show no goal, no respect for
any rings,
etc. You might as well be doing all your work in the reals and
functions from the reals to the reals for all youÕve said. 
In
those
cases, all your arguments become silly, at best.
In other words, at every step make sure we know what each
thing is *in
detail*.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
Discussion, linux)
> Well, I typed up this long post yesterday, looked it over a
couple of
> times, and felt good enough to post it, but noticed a minor
error near
> the end which IÕm fixing here.
> I may as well ask and answer a question that might be
pertinent at
> this time:
> Why make such a freaking long post?
I think more pertinent is: Why make *two* such freaking long
posts?
There are better ways to correct errors. Ways that would even
let the
reader see *what* the error was with yesterdayÕs post, why 
it
was
wrong and what is correct.
HereÕs the only differences between the two posts:
,----[ Yesterday ]
| But, hereÕs where itÕs rather strange, as 
u_1 and u_2 are
algebraic
| integers i.e. roots of a monic polynomial with integer
coefficients,
| while u_3 is not.
|
| So, for that reason, u_1 and u_2 are NOT units in the ring
of
| algebraic integers.
`----
,----[ Today ]
| But, hereÕs where itÕs rather strange, as 
u_1 and u_2 are
not
| ^^^
| algebraic integers i.e. roots of a monic polynomial with
integer
| coefficients, while u_3 is.
| ^^^
|
| So, for that reason, neither u_1, u_2, nor u_3 is a unit in
the ring
| ^^^^^^^ ^^^^^^^^^^^^^^^^^
| of algebraic integers.
`----
Think about your archivists, dammit.
--
These mathematicians are worse than communists, as how do you
explain
their behavior? I *am* the American Dream, fighting for what
should be
mine, having to get past weak-minded academics who are
fighting to
block my success. But I shall prevail!!! -- James S. Harris
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> ... ad infinitum ...
What I want to know is if itÕs mathematically possible for
Jim to write
something with less than 10,000 words.
I should introduce you to my sister. She talks all the time
too and
likewise never says anything.
--
Oppie the Bear
aka TOJ (The Other John)
ÔRemoveMYWORRIES to email me!
Calulus and alcohol donÕt mix.
So remember ...
DonÕt drink and derive!
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
Discussion, linux)
>> ... ad infinitum ...
[...]
> I should introduce you to my sister.
YouÕre weird.
--
Jesse F. Hughes
Leaving things always seems to fix me,
Running seems to ease my worried mind.
-- Bad Livers, Honey, IÕve Found a Brand New Way
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> IÕm an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whatÕs thought to be known in the
discipline of
> algebraic number theory.
> The mathematics which proves the problem is extrarodinary
in that it
> is very simple, relying on some of the most basic concepts
in algebra.
> It is revolutionary in that it so upsets the status quo,
changes the
> historical positions of so many mathematicians, and is just
plain
> surprising in many ways.
> Essentially what I do is relate one polynomial to a family
of
> polynomials, using what I call a non-polynomial
factorization, which I
> call that as the factoring of the primary polynomial is
into factors
> that are themselves not polynomials.
> For simplicity in explaining I donÕt initially give a
tremendous
> amount of detail about where the polynomial comes from as
years of
> experience talking about this on Usenet has shown me how
easily
> posters can confuse people with detail of that sort, but I
have no
> problem going into detail if real curiosity emerges.
> So I say I use a polynomial. So letÕs see it.
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> where x is an integer.
> It is, as you can see, a polynomial, and itÕs distinctive
in an
> important way, as it has 49 as a multiple
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
--
Mike Headon
e-mail: mike dot headon at enn tee ell world dot com
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> IÕm an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whatÕs thought to be known in the
discipline of
> algebraic number theory.
Which theorem is wrong? Which axioms leads to the
inconsistency?
> So I say I use a polynomial. So letÕs see it.
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> where x is an integer.
So P(x) is a member of Z[x] where x ranges over the integers.
Got it.
> It is, as you can see, a polynomial, and itÕs distinctive
in an
> important way, as it has 49 as a multiple as
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> and IÕll relate it to a family of polynomials, where that
multiple 49
> is very important.
Ok, so it can be factored in Z[x].
> Now hereÕs where a basic idea steps in, as years ago I
discovered the
> idea of separating out a polynomial in a special way so
that you can
> factor it as if itÕs a polynomial in a *different* 
variable
from the
> polynomial variable itself.
> Here that gives
> P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
> where if you multiply it all out and simplify, youÕll 
still
get
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> and the pattern I use next might not quite be visible, so
IÕll make a
> substitution using Y=5 and Z=7, to get
> P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2
+ Z^3
> and you can hopefully see how that can factor as
> P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z)
Ok, so you want to view P as a polynomial in Z[x,Y,Z] with
Y=5, Z=7.
> and going ahead and putting back in their values, as I used
Y and Z
> just as a bit of help (a dangerous bit of help as sci.math
posters
> have routinely jumped in at this point in the past to claim
that
> actually x is not the polynomial variable at all!!!), I have
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> and the aÕs are easily determined by using
> (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7) =
> 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3
> as you get three simultaneous equations, for instance,
> a_1(x) a_2(x) a_3(x) = 49(2401 x^3 - 147 x^2 + 3x)
> is one of the three.
> Solving for the aÕs, you get a cubic, which is
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> where the roots of that cubic are a_1(x), a_2(x), and
a_3(x), and that
> cubic is the family of polynomials that are related to
P(x), as for
> any given value of x, you get a polynomial.
So the aÕs are not polynomials, but algebraic functions of 
x.
Got it.
> Notice that the multiple of P(x), 49 is locked into the
family of
> cubics. But itÕs not a multiple of the cubic, as you have
the term
> 3(-1 + 49x)
> which is of course, coprime to 49.
Huh? In which of the several rings? In all of them?
> ThatÕs important, as somehow with this technique of
factoring P(x) in
> a special way, IÕve related that factorization of a
polynomial that
> has 49 as a multiple, to a family of polynomials that do
not.
But what you factored P(x) into are not polynomials. If you
had done
your factorization in Z[x], you would have found that 7 would
be a
factor of two of the terms, or 49 would be a factor of one of
the two
terms. Since you are not doing your factorization in Z[x],
this is no
longer guaranteed.
> ItÕs one of the most important relations in math history.
> Why? Well, for P(x), 49 is just a multiple, so I can divide
it off.
Or, more formally, multiply it by 1/49.
> That gives me
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> and reasonably, dividing 49 from the factorization of P(x),
gets rid
> of it, without a trace, but that results in the
factorization
> P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
> and so much arguing, over a period of years with this
technique,
> settles on what happens next with such an example.
It depends on what properties you wish to preserve. Right
now, you
havenÕt stated what properties the a_i(x) have, so 
itÕs
difficult to say
for sure which of those properties you wish to preserve.
Similarly for
the factors (5a_i(x) + 7).
> Before I go further though, letÕs stop to consider what
IÕve done:
> 1. I have a polynomial P(x) that has 49 as a multiple.
> 2. I factor that polynomial in a special way to get
non-polynomial
> factors.
> 3. I solve for those factors giving me a cubic family of
polynomials.
> 4. I now decide to divide 49 from P(x), and am at the point
of
> considering how that affects its factorization.
I can answer 4 quite easily: dividing by 49 has no effect on
the
how can I write a factorization of P(x)/49 such that itÕs
factors
satisfy [insert desired properties here]?
> Here P(x) is key. It has a multiple 49, and its
factorization
> provides the relation to the cubic family that defines the
aÕs. It is
> a basic concept in algebra that a multiple can be divided
off without
> problem, and necessarily that must be true, as consider,
how can a
> factorization of
> 300125x^3 - 18375 x^2 - 360 x + 22
> have some impact from the polynomial resulting from
dividing off a
> multiple?
What possible problems are you referring to? The most obvious
ones I
can think of are dealt with by the definition of a
factorization within
a ring.
> The equation has no memory. After all, if it did, why 49?
Why not 3
> or 11, or 83947397, or an infinity of other numbers?
Why would you even suggest anthropomorphizing an equation
this way?
> I belabor that point as if you accept it, then what follows
is
> obvious.
> So I have
> P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
> and I want to know how the 49 divides through the factors
of P(x),
> while I already know that
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> so 49 divides off of P(x) itself, without a trace, which is
not a
> surprise.
Since P(x) has 49 as a factor, it is natural to refer to the
other
factor as P(x)/49, yes. However, it seems that what you
really want is
to find a factorization of P(x)/49 that is somehow related to
the
peculiar way you chose to factor P(x).
> The simplest way to find out is to check directly. However,
the aÕs
> are rather complex being defined by the cubic
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and it might seem extraordinarily difficult to 
find out
anything about
> them, so is the cause lost?
Back up. The fact that they are roots of a cubic is, in
itself,
something weÕve found out about them quite simply. The
difficulty of
discovery lies in knowing what you wish to discover. For
example, it
should be easy to see that they do not represent mappings
from Z -> Z.
So, what properties are we discussing?
> No, because I can simplify by focusing on the constant term
of P(x).
> Why the constant term? Because unlike the other terms it is
> independent of x itself, and much of the complexity washes
out.
However, since the a_i(x)Õs are not polynomials, they do not
necessarily
have constant terms. Also, since we have no stated goal yet,
it is
difficult to see what, if anything, your hypothetical constant
terms
will have to do with our unstated goal. At this point you
really need
to back up and ask yourself some questions. What are you
trying to do?
What properties do you care about? Of these properties, which
do you
wish to preserve in the factor P(x)/49? Right now you are
performing
analysis that may be irrelevent.
> The constant term of P(x) is given by setting x=0, as that
sets the
> terms that have x as a variable to 0, leaving just the
constant term:
> P(0) = 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) =
49(22).
> Now what about the factors of P(x)? Well, they become a LOT
easier to
> manage as well, as I then have
> a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0) )
> which is
> a^3 - 3a^2 = 0 and a^3 - 3a^2 = 0, is a^2(a - 3) = 0
> so despite all the early complexity, I have now the simple
result that
> two of the aÕs equal 0, at x=0, while one equals 3.
> So I need to pick the aÕs to proceed, and my usual
convention is to
> let
> a_1(0) = 0, a_2(0) = 0, and a_3(0) = 3
> so with
> P(0)/49 = (5a_1(0) + 7)(5a_2(0)+ 7)(5a_3(0) + 7)/49
> I have
> P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = (7)(7)(22)/49
= 22
> which is correct as we already found out earlier.
Ok, but how is this relevent to anything? Better yet, what, if
anything, is it relevent to?
> But wait, arenÕt I just checking at a *single* value, for
something
> thatÕs terribly complicated, where maybe things are
different at a
> different value?
It depends on what things you care about. By the way, what
things *do*
you care about?
> Well, sure, things are different for terms that have x as a
factor, as
> thatÕs how algebra works. As x varies, you get different
things
> happening.
Sometimes. It depends on what those things are. Usually
mathematicians
are interested in discovering the things that are independant
of the
value of x. This is why they talk about factoring polynomials
(in the
ring of polynomials), among other things.
> Well, yes, for terms that vary with x, that is correct. But
if
> something is constant, then it doesnÕt vary.
Pretty much by definition. The issue then becomes: how does
the thing
that is constant interact with that which is not, if at all?
> Checking at x=0 clears out those terms that vary, leaving
those that
> do not, revealing that for two of the terms, whatÕs left
over, is 7,
> while for one, whatÕs left over is 22.
> ThatÕs just a fact. ItÕs such a simple fact 
that one of the
more
> remarkable things over the years IÕve found is the ability
of some
> people to argue around it.
ItÕs a fact, but what, if anything, is it relevent to?
> If you accept that constants are not variable, and that 7
is just a
> number that does not change with x, and you accept that
setting x=0
> reveals constants by eliminating the terms that vary, then
you should
> accept that the constants are constant without regard to
the value of
> x.
HereÕs the problem, setting x=1 reveals a different set of
contants.
Setting x=2 reveals a different set of constants. Each set of
constants
simply is. So, what makes the constants revealed by setting
x=0
special, and what do they allow us to do?
> So if I could look at the constants at x=39473987, then
thatÕd be
> fine, but with that value, the terms with x get in the way,
but at
> x=0, they do not.
HereÕs a problem: in your non-polynomial factorization it 
may
not be
appropriate to talk about terms at all.
> But with P(x)/49, I already have that 49 is gone, without a
trace.
> So, if the constants for factors of P(x) are 7, 7 and 49,
who MUST the
> 49 divide through?
> Like this
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> with indices arbitrary, as remember, I picked the first two
aÕs to be
> those that go to 0, when x=0.
And certainly, multiplying one side by 1/49 is equivalent to
multiplying
the other side by 1/7 and 1/7, which you can distribute in.
However,
aside from doing something convenient when x=0, what else
does this do?
> So far, so good, and you may wonder how something so simple
can be
> revolutionary.
Easy to wonder, since you have claimed no properties or
conclusions in
all this.
> Well, remember I related the factorization of P(x) to a
*family* of
> cubics given by the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and now IÕve shown a result that indicates that two of the
aÕs have 7
> as a factor, without regard to the value of x.
How do you figure that? You have shown that two of the 
aÕs are
0 when
x=0, but that is far from showing the aÕs have 7 as a factor
in their
ring. Granted, the most obvious ring to me for the aÕs to
exist in is
the ring of algebraic functions, in which *every* function
has 7 as a
factor. Did you want the aÕs to be in a more restricted 
ring?
If so,
which?
> ThatÕs HUGE as it turns out that a long time ago (like 
over
a hundred
> years ago) msthematicians decided that you couldnÕt make
such a
> determination if the roots of a polynomial were all
irrational and
> didnÕt all have the same factor.
Huh? 1) Where did you arrive at this belief? 2) How have you
tied this
to any previous work?
> (Like x^2 - 3 has sqrt(3) and -sqrt(3) as roots.)
> That belief is the basis for the modern usage of group
theory
> including Galois Theory.
If you want to tie your result to group theory, wouldnÕt it
be useful to
refer to it a little earlier in your discussion?
> YouÕve just seen a basic result showing it to be a false
belief.
How can that be? You have discussed functions in unknown
rings and done
multiplication in unknown rings. How can you conclude from
that
something about group theory? First of all, youÕre doing 
ring
theory.
Second, you have not connected your work to it.
> Like stick in x=1, and you have
> S(a) = a^3 + 3(48)a^2 - 49(2257)
> and without solving for the roots you know already that two
of its
> roots should have 7 as a factor, but now thereÕs another
problem.
And perhaps they do. It depends on what ring you believe
these roots
live in. If the aÕs are merely algebraic functions, then it
would be
natural to view the roots as algebraic numbers. If you view
them as
something else, then whether they have 7 as a factor needs
some
additional work earlier on.
> Over a hundred years ago, back in the late 1800Õs
mathematicians
> studied polynomials special in that they had a leading
coefficient of
> 1 or -1, and integer coefficients. Polynomials with a 
leading
> coefficient of 1 or -1 are called monic, so more
technically, they
> studied monic polynomials with integer coefficients.
> They called the roots of these polynomials algebraic
integers.
> Those roots form a group of numbers called the ring of
algebraic
> integers.
They for a set of numbers or a ring. To say they form a group
suggests
you wish to work only with addition or multiplication.
Calling them a
ring right after that is not clear.
> And it turns out that for
> S(a) = a^3 + 3(48)a^2 - 49(2257)
> if you take its roots, you canÕt find any that 
when divided
by 7, give
> an algebraic integer.
This suggests that your earlier work failed to take this
possible goal
into account. ThatÕs ok, it had no stated goal and no 
special
efforts
were taken to insure any properties were preserved. Given
that, itÕs
hardly a surprise that none of these roots are divisible by 7.
> It gets more complicated, as you can prove that with the
factorization
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> its possible to find algebraic integers w_1, w_2, and w_3,
such that
> w_1 w_2 w_3 = 49
> where the wÕs are the respective factors of
> (5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
> when I just said that you canÕt get algebraic integers 
with
x=1, and
> the roots of
> a^3 + 3(48)a^2 - 49(2257)
> wher 7 is a factor of *any* of those roots, let alone two.
Again, why should this be a surprise? However, it might have
been nice
if you had noted earlier that you were thinking of the aÕs 
as
functions
from Z -> Algebraic Integers. Then this issue could have been
addressed
much sooner by noting that 1/7 is not an algebraic integer
and that
multiplying a factor by 1/7 is therefor not necessarily going
to yield
an algebraic integer result.
> Hmmm...problem, right? Is it all lost? Does that mean
everything
> before was wrong? Yuck, did I just waste your time with
claims of
> revolutionary mathematics and all of that, when thereÕs
this weird
> result that seems to show it all must be wrong?
Well, it certainly seems to indicate you could stand to add a
few notes
about your goals and environment as you work. When you do a
bunch of
algebraic manipulation with no attention paid to any
properties, itÕs
not surprising that a property of interest should be violated
in the
course of working. This is how people have been proving that
0=1 for ages.
> But wait, for my result I used some very basic concepts.
Like I rely
> on 49 as a multiple just dividing off, without a trace. And
I focused
> on constant terms because they are, well, constant, and
simpler to
> work with than terms that include x, where all the
complexity comes
> into the picture.
You never asserted which properties you wanted to preserve,
however, nor
did you indicate any connection between these constant terms
and
anything else. You were doing manipulations without respect
for a goal
or properties.
> Well, there was that weird technique of factoring a
polynomial some
> odd way.
> But, sure, itÕs different, but all of the mathematical
operations are
> valid ones.
> So what gives?
Perhaps the constant terms didnÕt give you as much useful
information
as you thought they did. Perhaps you got to hasty in
introduction
algebraic numbers into your calculations on algebraic
integers. Who knows?
> LetÕs focus on the result again, as I said that with
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> you can have algebraic integers w_1, w_2, and w_3 that are
factors of
> the factors of P(x).
And this does a nice job of preserving our discussion in
functions from
Z -> Algebraic Integers.
> But I also showed that
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> but that canÕt work with these numbers called algebraic
integers!
Well, since you introduced non-algebraic integers with those
1/7Õs, why
should this be surprising?
> Well, consider u_1 u_2 u_3 = 1, where multiplying through
gives
But wait: what do these numbers represent? Or are they
functions?
Whatever else they may be, they are clearly units in whatever
ring they
are in.
> P(x)/49 =
> (5a_1(x)u_1/7 + u_1)(5a_2(x) u_2/7 + u_2)(5a_3(x) u_3 +
7u_3)
> where now it all *does* work with algebraic integers.
Back up: whatÕs it all that works now? I donÕt 
know what the
uÕs are,
but for it all to work, there must be at least one of them
that is not
working with algebraic integers. I would guess that u_3 is
actually a
function from Z -> algebraic numbers such that the
7*(codomain of u_3)
is a subset of the algebraic integers. Moreover, it looks
like u_1 and
u_2 must be functions from Z -> 7*(Algebraic Integers), which
would then
suggest that they were derived from the wÕs above.
> That is, for some reason, while two of the aÕs do not have
7 as a
> factor in the ring of algebraic integers, they *do* have
7u_1 and
> 7u_2, respectively, as factors, where u_1 and u_2 are
units, in that
> they are factors of 1.
Actually, they would have 7/u_1 and 7/u_2 as factors.
> But, hereÕs where itÕs rather strange, as 
u_1 and u_2 are
algebraic
> integers i.e. roots of a monic polynomial with integer
coefficients,
> while u_3 is not.
Strange indeed, since I thought they were functions. Making
them be
numbers would require a revisiting of the above to see what
all happens.
> So, for that reason, u_1 and u_2 are NOT units in the ring
of
> algebraic integers.
Correct. They donÕt even appear to be numbers. They appear 
to
be
functions. Did you mean for them to be numbers?
> HereÕs an example that I hope helps you see how it works.
Back up a moment: what is supposed to have happened that is
special?
> In integers you can have
> S(x) = (3x + 1)(x + 1) = 3x^2 + 4x + 1
> but now consider
> SÕ(x) = (3x + u_1)(x + u_2) = 3x^2 + kx + 1
> where u_1 u_2 = 1, and k is an integer.
> Here, u_1 CAN be an algebraic integer, but u_2 CANNOT be an
algebraic
> integer.
Ok, and?
> So, in the ring of algebraic integers, u_1 cannot be a unit.
And? Perhaps you thought that declaring u_1 u_2=1
automatically made
both of them units? It does, *in any ring where both exist*.
Since u_2
is not an algebraic integer, the fact that u_1 u_2=1 does not
imply that
u_1 is a unit in the ring of algebraic integers, but rather
that it is
not. Where is the strangeness you are waiting for?
> Well, maybe itÕs some kind of fraction, right? Well, yes,
possibly,
> it is, and like, for k=0, you can see that it IS some kind
of
> fraction. But, have you covered all the possibilities?
You mean like the algebraic numbers? How about the (algebraic
integers)[u_2]? There are many possiblities. Which exhibits a
property
you care about?
> The answer is, no, you canÕt have and have it all be
mathematically
> consistent.
However, the second example will be a subring of any ring
where
u_1 u_2=1 is defined.
> So why all this talk about algebraic integers? They seem
kind of
> messy at this point.
No they donÕt, but carry on.
> Like you get this nifty result with some basic concepts and
a special
> factoring technique, where it was all rather
straightforward, and then
> suddenly focusing on roots of monic polynomials with integer
> coefficients gives all kinds of head-scratching trouble!
This suggests that you should have started by focussing on
algebraic
integers or never looked at them at all.
> Well, the reason for the focus is that mathematicians for
over a
> hundred years focused on the ring of algebraic integers not
> understanding that it was, IÕll say, quirky.
> Actually itÕs worse than quirky, as not understanding the
weirdness
> that can arise from focusing on the roots of monic
polynomials with
> integer coefficients you can prove lots of things with the
ring of
> algebraic integers that are in fact mathematically false.
Care to give an example? It may not do what you expect, but
thatÕs not
a big deal.
> ItÕs a nasty little bug.
No, itÕs math.
> To give you some perspective, the tools used by Wiles in
his work that
> purportedly proves the Taniyama-Shimura Conjecture, to most
people,
> work that supposedly proves FermatÕs Last Theorem, 
donÕt
work.
Huh? Care to fill in a detail or three?
> They donÕt work because theyÕre based on an 
improper
understanding of
> the ring of algebraic integers. Works like WilesÕs cannot
be rescued
> from this bug.
Start by showing the bug, then by showing how it relates to
his work.
> So now maybe you understand the controversy!
No. All youÕve shown is that if you want to work within a
particular
ring, you cannot reliably achieve it by working with elements
that are
not in the ring. This is hardly surprising.
> Is my work actually complicated for a trained mathematician?
> No. My guess is that a trained mathematician can go over
the entire
> thing, and understand the implications in about an hour.
It didnÕt take long to read over. The implications are less
stunning
than you think. HereÕs a synopsis.
Take some functions that map from Z to the algebraic integers.
Observe that 0 maps to 0 for two of them.
Multiply those to functions by 1/7.
Observe that the resulting functions are not maps from Z to
the
algebraic integers.
Claim that this suggests a problem with the algebraic
integers.
Now, what made you think that the algebraic integers had a
property that
they were all multiples of 7?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
James, I (like you) am an amateur mathematician, but I just
cannot see how
you use:
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
to show that (a/7) is an algebraic integer.
Trivially,
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0
implies:
7^3(a/7)^3 + 37^2(-1 + 49x)(a/7)^2 - 49(2401 x^3 - 147 x^2 +
3x) = 0
(a/7)^3 + 3{(-1 + 49x)/7}(a/7)^2 - (1/7)(2401 x^3 - 147 x^2 +
3x) = 0
And {(-1 + 49x)/7} is not an integer for any x>0.
So, if you are going to show that (a/7) is an algebraic
integer for some
integer x>0, you are going to have to show us a (different)
monic
polynomial
with integer coeffiecients that it solves. If it really is as
trivial as
you
say, then why donÕt you simply tell us what value of x to
use, and which
polynomial (a/7) is a root of.
IÕm not looking at this with any bias or slant. I (like you)
would find it
very interesting if you really found something important, but
I just donÕt
get your argument, and no amount of hand-waving about
constants has yet
convinced me otherwise.
-Darren
> IÕm an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whatÕs thought to be known in the
discipline of
> algebraic number theory.
> The mathematics which proves the problem is extrarodinary
in that it
> is very simple, relying on some of the most basic concepts
in algebra.
> It is revolutionary in that it so upsets the status quo,
changes the
> historical positions of so many mathematicians, and is just
plain
> surprising in many ways.
> Essentially what I do is relate one polynomial to a family
of
> polynomials, using what I call a non-polynomial
factorization, which I
> call that as the factoring of the primary polynomial is
into factors
> that are themselves not polynomials.
> For simplicity in explaining I donÕt initially give a
tremendous
> amount of detail about where the polynomial comes from as
years of
> experience talking about this on Usenet has shown me how
easily
> posters can confuse people with detail of that sort, but I
have no
> problem going into detail if real curiosity emerges.
> So I say I use a polynomial. So letÕs see it.
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> where x is an integer.
> It is, as you can see, a polynomial, and itÕs distinctive
in an
> important way, as it has 49 as a multiple as
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> and IÕll relate it to a family of polynomials, where that
multiple 49
> is very important.
> Now hereÕs where a basic idea steps in, as years ago I
discovered the
> idea of separating out a polynomial in a special way so
that you can
> factor it as if itÕs a polynomial in a *different* 
variable
from the
> polynomial variable itself.
> Here that gives
> P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
> where if you multiply it all out and simplify, youÕll 
still
get
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> and the pattern I use next might not quite be visible, so
IÕll make a
> substitution using Y=5 and Z=7, to get
> P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2
+ Z^3
> and you can hopefully see how that can factor as
> P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z)
> and going ahead and putting back in their values, as I used
Y and Z
> just as a bit of help (a dangerous bit of help as sci.math
posters
> have routinely jumped in at this point in the past to claim
that
> actually x is not the polynomial variable at all!!!), I have
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> and the aÕs are easily determined by using
> (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7) =
> 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3
> as you get three simultaneous equations, for instance,
> a_1(x) a_2(x) a_3(x) = 49(2401 x^3 - 147 x^2 + 3x)
> is one of the three.
> Solving for the aÕs, you get a cubic, which is
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> where the roots of that cubic are a_1(x), a_2(x), and
a_3(x), and that
> cubic is the family of polynomials that are related to
P(x), as for
> any given value of x, you get a polynomial.
> Notice that the multiple of P(x), 49 is locked into the
family of
> cubics. But itÕs not a multiple of the cubic, as you have
the term
> 3(-1 + 49x)
> which is of course, coprime to 49.
> ThatÕs important, as somehow with this technique of
factoring P(x) in
> a special way, IÕve related that factorization of a
polynomial that
> has 49 as a multiple, to a family of polynomials that do
not.
> ItÕs one of the most important relations in math history.
> Why? Well, for P(x), 49 is just a multiple, so I can divide
it off.
> That gives me
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> and reasonably, dividing 49 from the factorization of P(x),
gets rid
> of it, without a trace, but that results in the
factorization
> P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
> and so much arguing, over a period of years with this
technique,
> settles on what happens next with such an example.
> Before I go further though, letÕs stop to consider what
IÕve done:
> 1. I have a polynomial P(x) that has 49 as a multiple.
> 2. I factor that polynomial in a special way to get
non-polynomial
> factors.
> 3. I solve for those factors giving me a cubic family of
polynomials.
> 4. I now decide to divide 49 from P(x), and am at the point
of
> considering how that affects its factorization.
> Here P(x) is key. It has a multiple 49, and its
factorization
> provides the relation to the cubic family that defines the
aÕs. It is
> a basic concept in algebra that a multiple can be divided
off without
> problem, and necessarily that must be true, as consider,
how can a
> factorization of
> 300125x^3 - 18375 x^2 - 360 x + 22
> have some impact from the polynomial resulting from
dividing off a
> multiple?
> The equation has no memory. After all, if it did, why 49?
Why not 3
> or 11, or 83947397, or an infinity of other numbers?
> I belabor that point as if you accept it, then what follows
is
> obvious.
> So I have
> P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
> and I want to know how the 49 divides through the factors
of P(x),
> while I already know that
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> so 49 divides off of P(x) itself, without a trace, which is
not a
> surprise.
> The simplest way to find out is to check directly. However,
the aÕs
> are rather complex being defined by the cubic
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and it might seem extraordinarily difficult to 
find out
anything about
> them, so is the cause lost?
> No, because I can simplify by focusing on the constant term
of P(x).
> Why the constant term? Because unlike the other terms it is
> independent of x itself, and much of the complexity washes
out.
> The constant term of P(x) is given by setting x=0, as that
sets the
> terms that have x as a variable to 0, leaving just the
constant term:
> P(0) = 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) =
49(22).
> Now what about the factors of P(x)? Well, they become a LOT
easier to
> manage as well, as I then have
> a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0) )
> which is
> a^3 - 3a^2 = 0 and a^3 - 3a^2 = 0, is a^2(a - 3) = 0
> so despite all the early complexity, I have now the simple
result that
> two of the aÕs equal 0, at x=0, while one equals 3.
> So I need to pick the aÕs to proceed, and my usual
convention is to
> let
> a_1(0) = 0, a_2(0) = 0, and a_3(0) = 3
> so with
> P(0)/49 = (5a_1(0) + 7)(5a_2(0)+ 7)(5a_3(0) + 7)/49
> I have
> P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = (7)(7)(22)/49
= 22
> which is correct as we already found out earlier.
> But wait, arenÕt I just checking at a *single* value, for
something
> thatÕs terribly complicated, where maybe things are
different at a
> different value?
> Well, sure, things are different for terms that have x as a
factor, as
> thatÕs how algebra works. As x varies, you get different
things
> happening.
> Well, yes, for terms that vary with x, that is correct. But
if
> something is constant, then it doesnÕt vary.
> Checking at x=0 clears out those terms that vary, leaving
those that
> do not, revealing that for two of the terms, whatÕs left
over, is 7,
> while for one, whatÕs left over is 22.
> ThatÕs just a fact. ItÕs such a simple fact 
that one of the
more
> remarkable things over the years IÕve found is the ability
of some
> people to argue around it.
> If you accept that constants are not variable, and that 7
is just a
> number that does not change with x, and you accept that
setting x=0
> reveals constants by eliminating the terms that vary, then
you should
> accept that the constants are constant without regard to
the value of
> x.
> So if I could look at the constants at x=39473987, then
thatÕd be
> fine, but with that value, the terms with x get in the way,
but at
> x=0, they do not.
> But with P(x)/49, I already have that 49 is gone, without a
trace.
> So, if the constants for factors of P(x) are 7, 7 and 49,
who MUST the
> 49 divide through?
> Like this
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> with indices arbitrary, as remember, I picked the first two
aÕs to be
> those that go to 0, when x=0.
> So far, so good, and you may wonder how something so simple
can be
> revolutionary.
> Well, remember I related the factorization of P(x) to a
*family* of
> cubics given by the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and now IÕve shown a result that indicates that two of the
aÕs have 7
> as a factor, without regard to the value of x.
> ThatÕs HUGE as it turns out that a long time ago (like 
over
a hundred
> years ago) msthematicians decided that you couldnÕt make
such a
> determination if the roots of a polynomial were all
irrational and
> didnÕt all have the same factor.
> (Like x^2 - 3 has sqrt(3) and -sqrt(3) as roots.)
> That belief is the basis for the modern usage of group
theory
> including Galois Theory.
> YouÕve just seen a basic result showing it to be a false
belief.
> Like stick in x=1, and you have
> S(a) = a^3 + 3(48)a^2 - 49(2257)
> and without solving for the roots you know already that two
of its
> roots should have 7 as a factor, but now thereÕs another
problem.
> Over a hundred years ago, back in the late 1800Õs
mathematicians
> studied polynomials special in that they had a leading
coefficient of
> 1 or -1, and integer coefficients. Polynomials with a 
leading
> coefficient of 1 or -1 are called monic, so more
technically, they
> studied monic polynomials with integer coefficients.
> They called the roots of these polynomials algebraic
integers.
> Those roots form a group of numbers called the ring of
algebraic
> integers.
> And it turns out that for
> S(a) = a^3 + 3(48)a^2 - 49(2257)
> if you take its roots, you canÕt find any that 
when divided
by 7, give
> an algebraic integer.
> It gets more complicated, as you can prove that with the
factorization
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> its possible to find algebraic integers w_1, w_2, and w_3,
such that
> w_1 w_2 w_3 = 49
> where the wÕs are the respective factors of
> (5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
> when I just said that you canÕt get algebraic integers 
with
x=1, and
> the roots of
> a^3 + 3(48)a^2 - 49(2257)
> wher 7 is a factor of *any* of those roots, let alone two.
> Hmmm...problem, right? Is it all lost? Does that mean
everything
> before was wrong? Yuck, did I just waste your time with
claims of
> revolutionary mathematics and all of that, when thereÕs
this weird
> result that seems to show it all must be wrong?
> But wait, for my result I used some very basic concepts.
Like I rely
> on 49 as a multiple just dividing off, without a trace. And
I focused
> on constant terms because they are, well, constant, and
simpler to
> work with than terms that include x, where all the
complexity comes
> into the picture.
> Well, there was that weird technique of factoring a
polynomial some
> odd way.
> But, sure, itÕs different, but all of the mathematical
operations are
> valid ones.
> So what gives?
> LetÕs focus on the result again, as I said that with
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> you can have algebraic integers w_1, w_2, and w_3 that are
factors of
> the factors of P(x).
> But I also showed that
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> but that canÕt work with these numbers called algebraic
integers!
> Well, consider u_1 u_2 u_3 = 1, where multiplying through
gives
> P(x)/49 =
> (5a_1(x)u_1/7 + u_1)(5a_2(x) u_2/7 + u_2)(5a_3(x) u_3 +
7u_3)
> where now it all *does* work with algebraic integers.
> That is, for some reason, while two of the aÕs do not have
7 as a
> factor in the ring of algebraic integers, they *do* have
7u_1 and
> 7u_2, respectively, as factors, where u_1 and u_2 are
units, in that
> they are factors of 1.
> But, hereÕs where itÕs rather strange, as 
u_1 and u_2 are
algebraic
> integers i.e. roots of a monic polynomial with integer
coefficients,
> while u_3 is not.
> So, for that reason, u_1 and u_2 are NOT units in the ring
of
> algebraic integers.
> HereÕs an example that I hope helps you see how it works.
> In integers you can have
> S(x) = (3x + 1)(x + 1) = 3x^2 + 4x + 1
> but now consider
> SÕ(x) = (3x + u_1)(x + u_2) = 3x^2 + kx + 1
> where u_1 u_2 = 1, and k is an integer.
> Here, u_1 CAN be an algebraic integer, but u_2 CANNOT be an
algebraic
> integer.
> So, in the ring of algebraic integers, u_1 cannot be a unit.
> Well, maybe itÕs some kind of fraction, right? Well, yes,
possibly,
> it is, and like, for k=0, you can see that it IS some kind
of
> fraction. But, have you covered all the possibilities?
> The answer is, no, you canÕt have and have it all be
mathematically
> consistent.
> So why all this talk about algebraic integers? They seem
kind of
> messy at this point.
> Like you get this nifty result with some basic concepts and
a special
> factoring technique, where it was all rather
straightforward, and then
> suddenly focusing on roots of monic polynomials with integer
> coefficients gives all kinds of head-scratching trouble!
> Well, the reason for the focus is that mathematicians for
over a
> hundred years focused on the ring of algebraic integers not
> understanding that it was, IÕll say, quirky.
> Actually itÕs worse than quirky, as not understanding the
weirdness
> that can arise from focusing on the roots of monic
polynomials with
> integer coefficients you can prove lots of things with the
ring of
> algebraic integers that are in fact mathematically false.
> ItÕs a nasty little bug.
> To give you some perspective, the tools used by Wiles in
his work that
> purportedly proves the Taniyama-Shimura Conjecture, to most
people,
> work that supposedly proves FermatÕs Last Theorem, 
donÕt
work.
> They donÕt work because theyÕre based on an 
improper
understanding of
> the ring of algebraic integers. Works like WilesÕs cannot
be rescued
> from this bug.
> So now maybe you understand the controversy!
> Is my work actually complicated for a trained mathematician?
> No. My guess is that a trained mathematician can go over
the entire
> thing, and understand the implications in about an hour.
> James Harris
> http://mathforprofit.blogspot.com/
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> James, I (like you) am an amateur mathematician, but I just
cannot see
how
> you use:
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> to show that (a/7) is an algebraic integer.
I believe James currently admits that (a/7) is not
necessarily an algebraic
integer for all integers x. His problem is that he has proved
by his own
ßawed methods that two of the aÕs are divisible by 7,
although you
will
never find out from him exactly what he means by this.
ThatÕs to say, I believe he acknowledges that a/7 may not be
an algebraic
integer, but none the less, the algebra proves the two of the
aÕs are
divisible by 7. He might say they are properly divisible by 7
or
something, although I donÕt know if he has used this exact
phrase. (He
uses
phrases like properly a unit for factors of 1 which he
acknowledges are
not algebraic integers, without specifying any larger ring to
which they
belong.)
To James this all proves there is a problem with core
mathematics (the
ring of algebraic integers) because he knows two of the aÕs
are
divisible
by 7, but he knows a/7 may not be an algebraic integer and it
really should
be in his opinion...
If you ask James exactly what he means by divisible properly
a unit
etc., he will never properly define these terms. When people
ask specific
questions he answers by simply reposting his original proof
and later
claims IÕve shot down all objections to my proofs! ;-)
> Trivially,
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0
> implies:
> 7^3(a/7)^3 + 37^2(-1 + 49x)(a/7)^2 - 49(2401 x^3 - 147 x^2
+ 3x) = 0
> (a/7)^3 + 3{(-1 + 49x)/7}(a/7)^2 - (1/7)(2401 x^3 - 147 x^2
+ 3x) = 0
> And {(-1 + 49x)/7} is not an integer for any x>0.
> So, if you are going to show that (a/7) is an algebraic
integer for some
> integer x>0, you are going to have to show us a (different)
monic
polynomial
> with integer coeffiecients that it solves. If it really is
as trivial as
you
> say, then why donÕt you simply tell us what value of x to
use, and which
> polynomial (a/7) is a root of.
> IÕm not looking at this with any bias or slant. I (like
you) would find
it
> very interesting if you really found something important,
but I just
donÕt
> get your argument, and no amount of hand-waving about
constants has yet
> convinced me otherwise.
Yes, James is mightily confused about constants too! I think
he believes
that all functions (not just polynomials) have constant terms
in the
same
way that polynomials do. Everybody apart from James can see
this isnÕt
entirely logical, but for Jamessake they go along with the
following
precise definition:
DEFINITION: the constant term of f(x) is f(0).
Nora and others even believe this is Jamesown 
definition,
and for sure he
may even have said this at some point, but I donÕt believe
this is whatÕs
going on in his head. He uses phrases like
setting x to 0 CLEARS OUT the variables, REVEALING the
constant in
the
function
and
the function f_1(x) CONTAINS the constant f_1(0)
I donÕt think he would use this wording if he thought the
constant was
simply (and NOTHING MORE THAN) the value of the function at
x=0. For
instance, would he say an arbitrary function f(x) contains the
constants
f(1), f(2), f(3) etc.? Clearly he believes the constant of an
arbitrary
function f is actually in there somehow, whatever this means!
Otherwise
he would instantly recognise that statements about constant
values of
his
various expressions is telling him absolutely nothing more
than the
behaviour of his expressions for the case x=0...
arbitrary function f (e.g. one of your non-polynomial
factors) tells
you
NOTHING MORE about f than the value of f at zero (i.e. f(0)).
This is a
bit
like your mantra the fact that c is an algebraic integer
tells you
NOTHING
MORE than that c is the root of an irreducible monic
polynomial with
integer
coefficients, so IÕm hoping this will help you 
understand what
IÕm
saying.
It will be a real test of you as a mathematician, but IÕm
hoping youÕll
come
through and be able to follow - itÕs not 
difficult - in the
end itÕs just
algebra and algebra is TRUTH and NEVER LIES blah blah
condescending blah
blah blah.
Mike.
> -Darren
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
>James, I (like you) am an amateur mathematician, but I just
cannot see
how
>you use:
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>to show that (a/7) is an algebraic integer.
I donÕt.
> I believe James currently admits that (a/7) is not
necessarily an
algebraic
> integer for all integers x. His problem is that he has
proved by his own
> ßawed methods that two of the aÕs are divisible by 7,
although you
will
> never find out from him exactly what he means by this.
ThatÕs all covered in my original post and yes, for those 
who
wondered, itÕs clear that some people donÕt 
bother to read
through my
posts, as theyÕre just here to argue.
For instance, I donÕt say divisible by 7 in the post. And 
the
accusation that my methods are ßawed is, youÕll notice, 
given
without any supporting mathematics.
That arguing for the sake of arguing is the worst feature of
Usenet.
If you read through my post youÕll find that I 
answered all
these
issues, in detail.
James Harris
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
>James, I (like you) am an amateur mathematician, but I just
cannot see
how
>you use:
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>to show that (a/7) is an algebraic integer.
> I donÕt.
>>I believe James currently admits that (a/7) is not
necessarily an
algebraic
>>integer for all integers x. His problem is that he has
proved by his own
>>ßawed methods that two of the aÕs are divisible by 7,
although you
will
>>never find out from him exactly what he means by this.
> ThatÕs all covered in my original post and yes, for those
who
> wondered, itÕs clear that some people donÕt 
bother to read
through my
> posts, as theyÕre just here to argue.
> For instance, I donÕt say divisible by 7 in the post. And
the
> accusation that my methods are ßawed is, youÕll 
notice,
given
> without any supporting mathematics.
> That arguing for the sake of arguing is the worst feature
of Usenet.
> If you read through my post youÕll find that I 
answered all
these
> issues, in detail.
Funny, I read through your post and found that you didnÕt
explain what
you were doing, where you were doing it, why you were doing
it, or how
to interpret it. Perhaps if you responded to a few of the
comments I
made it would clarify things.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: help
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4L76v27532;
My name is Gilbs, i just wanted ask if someone can help with
these 2
signal processing questions:
1. what is the analytic expression for the fourier transform
of x[n],
if x[n]=sin(pi*n/3)/(pi*n), n=...-1,0,1,2,...
===
Subject: Trigonometric proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB51PIN15309;
cos(x)> (1-x^2)/2
where 0<x
Here is one twist, you CANNOT use the MEAN VALUE THEOREM
Only trigonometry, no calculus.
===
Subject: Re: Trigonometric proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB5DCv408440;
>cos(x)> (1-x^2)/2
>where 0<x
>Here is one twist, you CANNOT use the MEAN VALUE THEOREM
>Only trigonometry, no calculus.
Even better: cos(x) > 1 - (x^2)/2. Hint:
(Linear) distance from P = (1, 0) to Q = (cos(x), sin(x))
is less than x = length of a circular trajectory from P to Q.
Todd Trimble
===
Subject: Re: Trigonometric proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB5DCwK08516;
>cos(x)> (1-x^2)/2
>where 0<x
>Here is one twist, you CANNOT use the MEAN VALUE THEOREM
>Only trigonometry, no calculus.
This is probably a typo and you want to show, using only
trigonometry,
that
cos(x) > 1 - x^2/2
where x is angle in radians. Since cos(x) >= -1 for all x, it
is
sufficient to show this for 1 - x^2/2 >= -1, i.e., x <= 2.
Since pi >
2, the angle x is in the 1st or 2nd quadrant. Let the unit
circle
(centered at the origin O and with radius r = 1) intersect
the x-axis
at the points A = [-1, 0] and B = [+1, 0]. Let the ray OC,
such that
the angle BOC = x < pi intersect the unit circle at the point
C. The
length of the arc BC of the unit circle is equal to x and the
chord BC
is shorter than the arc BC:
chord BC < x
Using Pythagorean theorem for the right angle triangle ABC
(with the
hypothenuse AB = 2r = 2),
x^2 > BC^2 = AB^2 - CA^2 = 4 - CA^2
CA^2 > 4 - x^2
Let D be the foot of the normal from the point C to the
line AB). Using Pythagorean theorem for the right angle
triangle ADC
with the hypothenuse CA,
AD^2 + DC^2 = CA^2
If 0 < x <= pi/2, OD = cos(x), DC = sin(x), and
AD = BO + OD = 1 + cos(x).
If pi/2 < x <= pi, OD = -cos(x), DC = sin(x), and
AD = BO - OD = 1 - (-cos(x)) = 1 + cos(x).
In both cases, AD = 1 + cos(x) and DC = sin(x).
[1 + cos(x)]^2 + sin^(x) = CA^2
1 + 2*cos(x) + cos^2(x) + 1 - cos^2(x) = CA^2
2 + 2*cos(x) = CA^2
Combining this with the inequality obtained previously,
2 + 2 cos(x) > 4 - x^2
cos(x) > 1 - x^2/2
and obviously (if your question does not have a typo),
cos(x) > 1 - x^2/2 = 1/2 + (1 - x^2)/2 > (1 - x^2)/2
===
Subject: Re: Trigonometric proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB5DCwo08530;
>cos(x)> (1-x^2)/2
>where 0<x
>Here is one twist, you CANNOT use the MEAN VALUE THEOREM
>Only trigonometry, no calculus.
Are we allowed to use the fact that:
cos(x) = 1 - (1/(2!))*x^2 + (1/(4!))*x^4 - (1/(6!))*x^6 + ...,
or must we confine this proof strictly to trigonometric
properties of
the cosine function?
===
Subject: Re: Trigonometric proof
>cos(x)> (1-x^2)/2
>where 0<x
>Here is one twist, you CANNOT use the MEAN VALUE THEOREM
>Only trigonometry, no calculus.
Clearly (1 - x^2)/2 <= 1/2 with equality at 0, and cos(x) >
1/2 for 0
<= x <= pi/3 with equality at pi/3, so cos(x) > 1/2 >= (1 -
x^2)/2 on
[0,pi/3]. On [pi/3, pi/2] cos(x) >= 0 with equality at pi/2.
But (1 -
x^2)/2 < 0 on this interval since 1 (radian) < pi/3. So on
that
interval cos(x) >= 0 > (1 - x^2)/2. That establishes the
inequality on
[0, pi/2]. That should get you started.
--Lynn
===
Subject: Calculus Help
Hi I am just starting out and I need some help with this
problem
Computer and simplify the following with the function f(a+h)
- f(a)
and the problem is 3x - 2
please show detailed instructions.
===
Subject: Re: Calculus Help
> Hi I am just starting out and I need some help with this
problem
> Computer and simplify the following with the function
f(a+h) - f(a)
> and the problem is 3x - 2
> please show detailed instructions.
f(x) = 3x - 2
f(a+h) = 3(a+h) - 2
f(a) = 3a - 2
f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
= 3a + 3h - 2 - 3a - 2
= 3h - 4
This goes against what I generally like to believe in (ie
have a crack
at it first, so we can help rather then do), but I have
noticed that
people on this board donÕt seem to mind.
I am also fairly new to higher level math, but have gotten so
much help
off this board, I just wanted to give back, so I await
someone more
expereienced to confirm this.
Cassandra
===
Subject: Re: Calculus Help
>> Hi I am just starting out and I need some help with this
problem
>> Computer and simplify the following with the function
f(a+h) - f(a)
>> and the problem is 3x - 2
>> please show detailed instructions.
> f(x) = 3x - 2
> f(a+h) = 3(a+h) - 2
> f(a) = 3a - 2
> f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
> = 3a + 3h - 2 - 3a - 2
> = 3h - 4
> This goes against what I generally like to believe in (ie
have a crack at
> it first, so we can help rather then do), but I have noticed
that people
> on this board donÕt seem to mind.
I mind. I believe that it is absolutely counterproductive to
just hand
solutions
to people. Especially individuals who simply post a question
without
giving
any kind of indication that they have thought at all about
what to do.
Have
they read the section? Identified and example in their book or
notes that
is
similar to the question? Notice how the question began: I am
just
starting out....
I often see this as code for I havenÕt really started and I
need to get
this done
for tomorrow.
DM
===
Subject: Re: Calculus Help
>> This goes against what I generally like to believe in (ie
have a crack at
>> it first, so we can help rather then do), but I have
noticed that people
>> on this board donÕt seem to mind.
>I mind. I believe that it is absolutely counterproductive to
just hand
>solutions to people. Especially individuals who simply post
a question
>without giving any kind of indication that they have thought
at all
>about what to do. Have they read the section? Identified an
example
>in their book or notes that is similar to the question?
Notice how the
>question began: I am just starting out.... I often see this
as code
>for I havenÕt really started and I need to get this done 
for
tomorrow.
Very well said!
ItÕs actually much easier just to solve problems for people,
but as
you say weÕre not doing them any good by doing that. Giving
useful
hints is hard, but worthwhile.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: Calculus Help
<TZWdnZsIYaNZhy7cRVn-iQ@comcast.com> Hi I am just starting
out and I need some help with this problem
>> Computer and simplify the following with the function
f(a+h) - f(a)
>> and the problem is 3x - 2
>> please show detailed instructions.
>f(x) = 3x - 2
IÕm amazed you made any sense of his utterances.
>f(a+h) = 3(a+h) - 2
>f(a) = 3a - 2
>f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
>= 3a + 3h - 2 - 3a - 2
>= 3h - 4
> I mind. I believe that it is absolutely counterproductive
to just hand
> solutions to people. Especially individuals who simply post
a question
> without giving any kind of indication that they have
thought at all
> about what to do. Have they read the section? Identified and
example
> in their book or notes that is similar to the question?
Notice how the
> question began: I am just starting out.... I often see this
as code
> for I havenÕt really started and I need to get this done 
for
tomorrow.
ItÕs worse than that. Calculus help will be of little use to
John R.
He needs help learning how to ask a coherent question. To
slow down
long enuf to speak in other than stream of conscienceness
thought.
To accept the woeful fact math isnÕt an instant easy grade.
===
Subject: Re: Calculus Help
>f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
>= 3a + 3h - 2 - 3a - 2
>= 3h - 4
f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
= 3a + 3h - 2 - 3a + 2
= 3h
In f(x) notation x is the input to the function while f(x) is
the output of the function!
Domain & Range!
--
Casey
===
Subject: Re: Calculus Help
>>f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
>>= 3a + 3h - 2 - 3a - 2
>>= 3h - 4
> f(a+h) - f(a) = (3(a+h) - 2) - (3a - 2)
> = 3a + 3h - 2 - 3a + 2
> = 3h
> In f(x) notation x is the input to the function while f(x)
is
> the output of the function!
> Domain & Range!
> --
> Casey
Grrrhhhh, that makes three silly errors in two days (at least
that I
have posted to a newsgroup).
===
Subject: Re: Mechanics problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB5DCuE08375;
Please donÕt call me dim...
I thought I might solve the following similar problem but 
IÕm
stuck
again:
Two rockets are fired vertically from launching pads side by
side.
The first rocket moves vertically upwards with an acceleration
of 7g
and the second with an acceleration of 9g. If the second
rocket is
fired 2 seconds after the first, find 
how long after its
launching the
second rocket overtakes the first.
--
Ok, so we have a similar problem and I should be able to
isolate t
somehow.
v = u + at
v = 0 + 68.6t
v = 0 + 88.2(t-2)
ThatÕs what I thought but it gives me a value of t=9 and the
bookÕs
answer is 14.94 secs.
So I went wrong somewhere along the line!
Can you help, please?
Jo
===
Subject: Re: Mechanics problem
> Please donÕt call me dim...
> I thought I might solve the following similar problem but
IÕm stuck
> again:
> Two rockets are fired vertically from launching pads side by
side.
> The first rocket moves vertically upwards with an
acceleration of 7g
> and the second with an acceleration of 9g. If the second
rocket is
> fired 2 seconds after the first, 
find how long after its
launching the
> second rocket overtakes the first.
> --
> Ok, so we have a similar problem and I should be able to
isolate t
> somehow.
> v = u + at
> v = 0 + 68.6t
> v = 0 + 88.2(t-2)
> ThatÕs what I thought but it gives me a value of t=9 and
the bookÕs
> answer is 14.94 secs.
Overtake does NOT mean the speeds are equal. It means the
distances
traveled are equal.
So write an equation for the distance travelled by the first
rocket as
a function of t (with t=0 being when the *first* rocket is
launched),
then write an eqaution for the distance travelled by the
second rocket
as a function of t (again, with t=0 being when the *first*
rocket is
launched -- do you see why?) and set those distances equal to
each
other.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: Mechanics problem
> Please donÕt call me dim...
> I thought I might solve the following similar problem but
IÕm stuck
> again:
> Two rockets are fired vertically from launching pads side by
side.
> The first rocket moves vertically upwards with an
acceleration of 7g
> and the second with an acceleration of 9g. If the second
rocket is
> fired 2 seconds after the first, 
find how long after its
launching the
> second rocket overtakes the first.
> --
> Ok, so we have a similar problem and I should be able to
isolate t
> somehow.
> v = u + at
> v = 0 + 68.6t
> v = 0 + 88.2(t-2)
> ThatÕs what I thought but it gives me a value of t=9 and
the bookÕs
> answer is 14.94 secs.
> So I went wrong somewhere along the line!
> Can you help, please?
> Jo
I think you need distance to be equal to each other - not
velocity. IÕm not
sure if the acceleration figures are correct or not.
Bill
===
Subject: Revolutionary Mathematics: Synopsis version
IÕm an amateur mathematician.
Essentially what I do is relate one polynomial to a family of
polynomials, using what I call a non-polynomial
factorization, which I
call that as the factoring of the primary polynomial is into
factors
that are themselves not polynomials.
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
where x is an integer
is related to the family of polynomials given by
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
where the roots of it are used in a factorization of P(x):
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
That equation defining the family of polynomials is found by
re-arranging of the original polynomial P(x), so that I go
from
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
to
P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
and the pattern I used might not quite be visible, so IÕll
make a
substitution using Y=5 and Z=7, to get
P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2 +
Z^3
so you can see how that can factor as
P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z).
The polynomial P(x) is a multiple of 49, as
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and the controversy arises when I divide 49 from the
factorization:
P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
as that happens without regard to the value of x, so I check
at x=0 to
see how the 49 divides off, which indicates that
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
which runs into a problem with what is called the ring of
algebraic
integers.
That problem is resolvable, but the resolution indicates a
problem in
algebraic number theory big enough that, for instance, it
indicates
that Wiles did not prove the Taniyama-Shimura Conjecture.
It impacts group theory and Galois Theory.
Not surpringly then, it is a controversial result.
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: Revolutionary Mathematics: Synopsis version
> IÕm an amateur mathematician.
> James Harris
> http://mathforprofit.blogspot.com/
Oh, yes, an amateur mathematician. That would explain why
line 2 of
your
website reads MY MATH DISCOVERIES, FOUND FOR PROFIT.
===
Subject: Re: Revolutionary Mathematics: Synopsis version
>> IÕm an amateur mathematician.
>> James Harris
>> http://mathforprofit.blogspot.com/
> Oh, yes, an amateur mathematician. That would explain why
line 2 of
your
> website reads MY MATH DISCOVERIES, FOUND FOR PROFIT.
...thus demonstrating that neither word in the phrase amateur
mathematician is true with reference to James. The phrase
aspiring
yet clueless gold-digger seems more appropriate.
--
Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you
improvise
fwbrown@bellsouth.net | if youÕre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: Revolutionary Mathematics: Synopsis version
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6DWRU30322;
>IÕm an amateur mathematician.
>Essentially what I do is relate one polynomial to a family of
>polynomials, using what I call a non-polynomial
factorization, which
I
>call that as the factoring of the primary polynomial is into
factors
>that are themselves not polynomials.
>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>where x is an integer
>is related to the family of polynomials given by
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>where the roots of it are used in a factorization of P(x):
>P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
>That equation defining the family of polynomials is found by
>re-arranging of the original polynomial P(x), so that I go
from
>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) +
7^3
>and the pattern I used might not quite be visible, so IÕll
make a
>substitution using Y=5 and Z=7, to get
>P(x) = 49(2401 x^3 - 147 x^2 + 3x) Y^3 - 3(-1 + 49 x ) YZ^2
+ Z^3
>so you can see how that can factor as
>P(x) = (Y a_1(x) + Z)(Y a_2(x)+ Z)(Y a_3(x) + Z).
>The polynomial P(x) is a multiple of 49, as
>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>and the controversy arises when I divide 49 from the
factorization:
>P(x)/49 = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)/49
>as that happens without regard to the value of x, so I check
at x=0
to
>see how the 49 divides off, which indicates that
>P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
This is a correct equation, but as you know, 5a_1(x)/7 cannot
be
an algebraic integer. You are dividing the factors by 49 in
the pattern (7, 7, 1). What you have shown here does not rule
out the possibility that there is another way to factor 49 out
of the three terms in parentheses, so that the divisors and
the
quotients are all algebraic integers.
>which runs into a problem with what is called the ring of
algebraic
>integers.
Some questions:
1. You define the constant term of a function h(x) to be h(0).
True or false?
2. If a function is a product of other functions, e.g.,
Q(x) = g_1(x) g_2(x) g_3(x),
then the product of the constant terms of g_1(x), etc.,
is the constant term of the product, Q(x).
That is, Q(0) = g_1(x) g_2(0) g_3(0).
True or false?
3. Elsewhere you have agreed that in general a_1(x), a_2(x),
and
a_3(x) are not coprime to 7 in the ring of algebraic integers.
True or false?
4. Moreover, elsewhere you have noted that there exist
algebraic
integers w_1(x), w_2(x), and w_3(x) such that:
a) Their product is 49
b) a_i(x)/w_i(x) is an algebraic integer for
i = 1, 2, and 3
c) 7/w_i(x) is also an algebraic integer for
i = 1, 2, and 3.
d) w_1(0) = w_2(0) = 7, and w_3(0) = 1.
e) For x > 0, the w_i(x) are not units in the ring
of algebraic integers.
True or false?
5. Therefore P(x)/49 can be factored as
P(x)/49
= (5 c_1(x) + d_1(x))(5 c_2(x) + d_2(x))(5 c_3(x) + d_3(x)),
where
c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x),
and these quotients are all algebraic integers.
True or false?
6. Therefore there is a way to factor 49 out of the
product
(5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
in such a way that in the quotient,
(5 c_1(x) + d_1(x))(5 c_2(x) + d_2(x))(5 c_3(x) + d_3(x))
all of the terms c_i(x) and d_i(x) are algebraic integers.
True or false ?
-------------------------------------------------------------
------------
Now, suppose you answer true to all those questions. (If you
do not, please explain why.)
Here is my understanding of your argument. Your claim is that
there is only one right way to divide 49 out of
(5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
because the constant term of P(x)/49 is P(0)/49 = 22, and the
constant terms of the individual factors are
(5 a_1(0) + 7) = 7,
(5 a_2(0) + 7) = 7, and
(5 a_3(0) + 7) = 22.
You note that the product
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 49 * 22 / 49 = 22.
This happens to be true for any x.
If you tried to divide, say, (5 a_3(x) + 7) by w_3(x), you
would
get
22/w_3(x)
in the place where the constant term was before the division.
Now you say, the constant terms by definition are constant, 
and
w_3(x) is a divisor of 7 and is therefore coprime to 22, so
it must
be the case that w_3(x) is a unit.
Right so far? Am I describing your argument correctly?
Here it is a little more succinctly. You say, because
of the fact that constant terms must be constant, that
the only way to divide 49 out of
(5a_1(x) + 7)(5a_2(x)+ 7)(5b_3(x) + 22)
is to factor 7 out of the first two terms and 1 out of
the last. Yet you know that a_1(x)/7 is not an algebraic
integer, so the only factorization that can work, doesnÕt.
Yet SOME factorization must work within the ring of
algebraic integers. Hence there is a problem with the ring
of algebraic integers. Right?
-------------------------------------------------------------
---------
There is just one little problem. The constant term of
(5 a_3(x) + 7)/w_3(x)
in general is not
22/w_3(x).
What is it?
By your own definition of constant term, it is
22/w_3(0).
Do you disagree with this? Or do you think it has to be
22/w_3(x)? Why?
If so, then the rest of what you claim does not follow:
22/w_3(x) is not a constant function (unless you assume what
you
want to prove). True, the PRODUCT of the wÕs is constant 
(and
equal to 49), but that does not imply that the individual
functions are constant. You have seen that before: the product
of nonconstant functions which take on values in the algebraic
integers can be constant.
Of course, since w_3(0) = 1, the constant term 22/w_3(0) = 22,
and from this you cannot conclude anything at all about w_3(x)
for x <> 0.
This is really the core of the error in your thinking.
In particular, you cannot conclude, for x nonzero, that w_3(x)
is a unit in the ring of algebraic integers. It is not true
that 22/w_3(x) is even an algebraic integer when x > 0.
It may be that you believe that when you divide the constant
term by w_3(x), you should always get an algebraic integer.
There
is no reason to think this. It is true if you divide 7 by
w_3(x),
but not if you divide 22 by w_3(x). If you disagree, please
explain why.
Of course w_3(x) is a unit in many larger rings. It is a unit
in the ring of algebraic numbers, for example. You can always
enlarge the ring of algebraic integers to obtain a larger ring
which is still not the whole field of algebraic numbers, and
in fact not a field at all, and in which w_3(x) is a unit. 
That
is *not* an interesting fact.
The bottom line is, you do NOT need a larger ring than the
algebraic integers to factor 49 out of
(5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7).
There is a perfectly good factorization where the divisors
and the
quotients are all algebraic integers. It is *not* the (7, 7,
1)
factorization. The fact that constant terms must be constant
is true, but it is not relevant. You think it is, but that is
because you are not applying the definition of constant term
correctly.
You do NOT need a larger ring to carry out a correct
factorization.
What you have done relies on the theorem quoted above, that
the
product of the constant terms equals the constant terms of the
product. A true theorem. But you do not apply it correctly.
You clearly believe, as you have indicated elsewhere, that
the constant term of (5 a_3(x) + 7) is 22/w_3(x). This is
incorrect. The correct constant term is 22/w_3(0). The latter
tells you nothing about w_3(x) for x <> 0.
>That problem is resolvable, but the resolution indicates a
problem in
>algebraic number theory big enough that, for instance, it
indicates
>that Wiles did not prove the Taniyama-Shimura Conjecture.
What you have done does not indicate any problem at all with
the
ring of algebraic integers. Your conclusion that there is some
kind of contradiction is based on an incorrect application of
your
own definition of constant term: a rookie error, to say the
least.
Nora B.
>It impacts group theory and Galois Theory.
>Not surpringly then, it is a controversial result.
>James Harris
href=http://mathforprofit.blogspot.com/>http://
mathforprofit.blogspot.co
m/</a>
===
Subject: Re: Revolutionary Mathematics: Synopsis version
Discussion, linux)
> 2. If a function is a product of other functions, e.g.,
> Q(x) = g_1(x) g_2(x) g_3(x),
> then the product of the constant terms of g_1(x), etc.,
> is the constant term of the product, Q(x).
> That is, Q(0) = g_1(x) g_2(0) g_3(0).
^ 0
> True or false?
--
Britney thought the idea of a pre-nup was vile, because she is
loved-up with Kevin and cannot envisage breaking up. However,
[...] no
one in Hollywood these days get married without brokering a
deal. [...] She had a long chat with Kevin and he was cool
about it.
===
Subject: Re: Revolutionary Mathematics: Synopsis version
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6KoGJ09574;
>> 2. If a function is a product of other functions, e.g.,
>> Q(x) = g_1(x) g_2(x) g_3(x),
>> then the product of the constant terms of g_1(x), etc.,
>> is the constant term of the product, Q(x).
>> That is, Q(0) = g_1(x) g_2(0) g_3(0).
> ^ 0
>> True or false?
>--
question. At least someone reads these things.
N.B.
>Britney thought the idea of a pre-nup was vile, because she
is
>loved-up with Kevin and cannot envisage breaking up.
However, [...]
no
>one in Hollywood these days get married without brokering a
>deal. [...] She had a long chat with Kevin and he was cool
about it.
===
Subject: Re: Revolutionary Mathematics: Synopsis version
> That problem is resolvable, but the resolution indicates a
problem in
> algebraic number theory big enough that, for instance, it
indicates
> that Wiles did not prove the Taniyama-Shimura Conjecture.
> It impacts group theory and Galois Theory.
How does it impact Galois theory? Even if what you say were
correct (which
it is not) then you would still have to explain how the
problem with
algebraic integers impacts Galois theory. Just making a claim
like that goes
nowhere.
===
Subject: Re: Revolutionary Mathematics: Synopsis version
> That problem is resolvable, but the resolution indicates a
problem in
> algebraic number theory big enough that, for instance, it
indicates
> that Wiles did not prove the Taniyama-Shimura Conjecture.
Just where did he go wrong? Can you state the Taniyama-Shimura
Conjecture for us?
===
Subject: maths problem
choose 4 of these digits. Each one must be different. Put one
digit in each
box
This makes two 2 digit numbers reading across and two digit
numbers reading
down. Add up all four of the numbers.
In this example the total is 100.
12 + 47 + 14 + 27 = 100
Heres it in the cube shape, roughly
1 - 2
4 - 7
you can use
1 2 3 4 5 6 7 8 9
got to make 200!!!!
===
Subject: Re: maths problem
> choose 4 of these digits. Each one must be different. Put
one digit in
> each
> box
> This makes two 2 digit numbers reading across and two digit
numbers
> reading
> down. Add up all four of the numbers.
> In this example the total is 100.
> 12 + 47 + 14 + 27 = 100
> Heres it in the cube shape, roughly
> 1 - 2
> 4 - 7
> you can use
> 1 2 3 4 5 6 7 8 9
> got to make 200!!!!
Well the easy answer is
6 - 2
4 - 7
This simply adds 50 to two of the numbers, thus giving a
total of 200
instead of 100.
===
Subject: Key point, non-polynomial factorization
IÕm in a weird situation where there are people quite
dedicated in
obscuring a result I have, so I want to be thorough in
highlighting
how it works.
As already IÕve given a long post in-depth, and a synopsis
post, IÕm
going to now isolate out a key point.
I relate one polynomial to a family of polynomials with a
factorization:
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
to
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
with the factorization
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
where the aÕs are the roots of the cubic family, for 
instance
-a_1^3 + 3(-1 + 49x)a_1^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0.
(To see details on the derivation go to my post Revolutionary
Mathematics: Non-polynomial Factorization.)
Now crucially you have that P(x) is a multiple of 49, as
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
and, it should be clear that you can divide off 49 from both
sides
without regard to the value of x.
ThatÕs basic, but itÕs the area where all the 
controversy
sets in, as
I say you can divide 49 off without regard to the value of x,
but
others have to agree for P(x), but step in and claim that is
not true
for the factorization
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
that the 49 divides off without regard to the value of x.
Essentially their arguments boil down to claiming that the
aÕs have
factors in common with 7 that must vary, and you know, 
thatÕs
kind of
a common-sense idea.
After all, before it would have seemed basic to believe that
how
factors of 7 distribute between the roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
is determined by just the expression itself!
What IÕm suggesting is counter-intuitive in that 
IÕm saying
that the
family of cubics is actually constrained by the factorization
of some
completely different polynomial.
Your gut instincts may keep telling you that thereÕs no way
that your
earlier instincts that a given cubic or family of cubics were
completely defined by just that cubics 
coefficients is wrong.
But, in science, and now in mathematics, human intuition, your
common-sense can fail you. In physics the best example
typically
given is quantum mechanics.
Logically, the relation between the cubic family and the
factorization
of the polynomial P(x) follows from
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7).
That is, counter to your intuition, mathematically whatÕs
important is
that the two can be related by that equation, and then
because it can
be related to the factorization, the cubic family is
constrained in a
way that goes counter to your intuition.
Your intuition is not mathematics. It can help you find
mathematics.
It can help you understand mathematics, or it can block you
if you
donÕt accept what follows mathematically.
Maybe mathematicians are spoiled. I was trained as a
physicist as I
have a B.Sc. in physics.
Now, 49 divides off of P(x), without regard to the value of
x, as
mathematically you have
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and thereÕs no trace of 49, left, and P(x)/49 might be
misleading in
that you can still see the 49, but thatÕs really just 
telling
you that
it is now removed.
If you accept that 49 divides off without regard to the value
of x,
then it doesnÕt matter what value of x I use next, and I 
pick
x=0.
At x=0, I get the cubic a^3 - 3a^2, which tells me that two
of the aÕs
equal 0, while one equals 3, plugging that into the
factorization
gives
P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = 22.
Now here is where a lot of people seem to get lost.
Logically, if terms are constant, then they donÕt vary with
x, but I
keep hearing from people who think that x=0 is a special
case, as if
when x changes then those terms constant with respect to x
will
change.
ThatÕs a logical contradiction.
If the terms are in fact independent of the value of x, then
they are
independent.
ItÕs a tautology.
If they are independent of x, then setting x=0 has NO EFFECT
on them,
but it does have the benefit of clearing out terms that ARE
dependent
on x.
Before these arguments about this that basic logic was not
argued
with, and IÕm sure lots of analysis is going on everyday,
where people
use it as itÕs been used for quite a while, without concern
that on
Usenet posters successfully question it day after day, month
after
month, and now year after year.
So why is there confusion? Why canÕt I just settle it, and
those
people just be the cranks?
Well I think for many of you your intuition gets in the way.
You look at
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and it just makes sense to you that it is itself independent
of some
freaky factorization of P(x). ItÕs almost as if your hearts
tell you
what must be, and when the mathematics says otherwise, you go
with
your hearts.
Mathematically though, the logic follows rather simply, which
is why
this is so frustrating for me, as I can trace out the logic,
repeatedly. Talk about constants being constant. Give actual
numbers
like 7 and 22, and it just not matter.
Mathematics hardly gets any simpler than this, but thereÕs a
fatal
ßaw if you donÕt understand that no matter how rational 
you
may
believe you are, or how objective you think you are, there
are places
and areas where your heart can get in the way, and your
emotions can
rule your mind.
IÕm asking you to be logical.
What happens then though? Well then you find out that 
IÕm
right and
yes, there is this major problem in algebraic number theory
that NEEDS
TO BE FIXED.
When I communicate with trained mathematicians, I donÕt get
objections
that come down to questioning whether or not constant terms
are
constant as they donÕt do the Usenet objections. They just
walk away.
So I have two different behaviors: irrational on Usenet, and
passive-aggressive off.
IÕm tired of it. I want you people to grow the hell up, and
start
using your minds as the problem does not go away just by
ignoring it.
Grow up.
James Harris
===
Subject: Re: Key point, non-polynomial factorization
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6DWP830196;
>IÕm in a weird situation where there are people quite
dedicated in
>obscuring a result I have, so I want to be thorough in
highlighting
>how it works.
>As already IÕve given a long post in-depth, and a synopsis
post, IÕm
>going to now isolate out a key point.
>I relate one polynomial to a family of polynomials with a
>factorization:
>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>with the factorization
>P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
>where the aÕs are the roots of the cubic family, for 
instance
>-a_1^3 + 3(-1 + 49x)a_1^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0.
>(To see details on the derivation go to my post Revolutionary
>Mathematics: Non-polynomial Factorization.)
>Now crucially you have that P(x) is a multiple of 49, as
>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
>and, it should be clear that you can divide off 49 from both
sides
>without regard to the value of x.
>ThatÕs basic, but itÕs the area where all the 
controversy
sets in, as
>I say you can divide 49 off without regard to the value of
x, but
>others have to agree for P(x), but step in and claim that is
not true
>for the factorization
>P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
>that the 49 divides off without regard to the value of x.
>Essentially their arguments boil down to claiming that the
aÕs have
>factors in common with 7 that must vary, and you know,
thatÕs kind of
>a common-sense idea.
>After all, before it would have seemed basic to believe that
how
>factors of 7 distribute between the roots of
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>is determined by just the expression itself!
Note that 49 is a factor of the constant term of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and of course, the roots are dependent on x. Thus
it would not be surprising to learn that whether the
roots are divisible by factors of 7 might be also be
dependent on x.
>What IÕm suggesting is counter-intuitive in that 
IÕm saying
that the
>family of cubics is actually constrained by the
factorization of some
>completely different polynomial.
>Your gut instincts may keep telling you that thereÕs no way
that your
>earlier instincts that a given cubic or family of cubics were
>completely defined by just that cubics 
coefficients is wrong.
>But, in science, and now in mathematics, human intuition,
your
>common-sense can fail you. In physics the best example
typically
>given is quantum mechanics.
>Logically, the relation between the cubic family and the
factorization
>of the polynomial P(x) follows from
>P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7).
>That is, counter to your intuition, mathematically whatÕs
important
is
>that the two can be related by that equation, and then
because it can
>be related to the factorization, the cubic family is
constrained in a
>way that goes counter to your intuition.
>Your intuition is not mathematics. It can help you find
mathematics.
>It can help you understand mathematics, or it can block you
if you
>donÕt accept what follows mathematically.
>Maybe mathematicians are spoiled. I was trained as a
physicist as I
>have a B.Sc. in physics.
>Now, 49 divides off of P(x), without regard to the value of
x, as
>mathematically you have
>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>and thereÕs no trace of 49, left,
No, you are making a mistake on this
point. There IS a trace of 49 left in this expression.
Remember how you are factoring P(x): as a polynomial in
the number 5. To be consistent with this, you must
rearrange P(x) as follows:
P(x) = 49(2401 x^3 - 147 x^2 + 3x) (5^3)
- 3(-1 + 49 x)(5)(7^2) + 7^3.
This is from one of your own posts of yesterday.
Now divide by 49: you get
P(x)/49 = (2401 x^3 - 147 x^2 + 3x)*5^3 - 3(-1 + 49 x)*5 + 7.
So where, on the right side, is the trace of 49 ?
ItÕs not in the term (2401 x^3 - 147 x^2 + 3x). For x not
equal to a multiple of 7, that term is coprime to 49.
ItÕs not in the term -3(-1 + 49 x). That is also coprime to
49.
ItÕs in the constant term, 7.
That feature of P(x) is there even after you have divided
by 49. ItÕs not actually the division by 49 which is
important.
Of course the whole expression is not divisible by 49
or any factor of 49. No one is claiming that. But that
7 in the constant term of P(x)/49 is actually part of
the explanation that a_1(x), a_2(x), and a_3(x) are not
coprime to 7 when x is nonzero. In fact, if that 7
were not there, you would not even be THINKING of factoring
P(x) in the form
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7).
>and P(x)/49 might be misleading in
>that you can still see the 49, but thatÕs really just
telling you
that
>it is now removed.
>If you accept that 49 divides off without regard to the
value of x,
>then it doesnÕt matter what value of x I use next, and I
pick x=0.
>At x=0, I get the cubic a^3 - 3a^2, which tells me that two
of the
aÕs
>equal 0, while one equals 3, plugging that into the
factorization
>gives
>P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = 22.
But if x > 0, the cubic is
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
So what?
This is a cubic polynomial in a. The constant term
with respect to a is -49(2401 x^3 - 147 x^2 + 3x).
Note that it has 49 as a factor.
Of course, when x = 0, this doesnÕt matter, because
the whole constant term -49(2401 x^3 - 147 x^2 + 3x)
is zero.
But when x > 0 and this polynomial in a is irreducible,
the constant term is nonzero and has 49 as a factor.
ThatÕs why this difference occurs. Since the polynomial
is irreducible, each of the roots a_1(x), a_2(x), and a_3(x)
must be non-coprime to 49.
>Now here is where a lot of people seem to get lost.
>Logically, if terms are constant, then they donÕt vary with
x, but I
>keep hearing from people who think that x=0 is a special
case, as if
>when x changes then those terms constant with respect to x
will
>change.
As can be seen VERY CLEARLY from the above, it IS a special
case.
>ThatÕs a logical contradiction.
ItÕs not a logical anything. No one is claiming that the
constant terms of P(x) or P(x)/49 or of any of constant terms
of the (5 a_i(x) + 7) change when x changes. The constant
terms are just the value of the functions when x = 0, from
your own definition.
>If the terms are in fact independent of the value of x, then
they are
>independent.
Clearly however the roots a_1, a_2, and a_3 are dependent on
x;
they are algebraic functions of x. Therefore their algebraic
properties are also dependent on x. Being coprime to 7
is an algebraic property. Therefore it is not a surprise
that their coprimeness to 7 is dependent on x.
>ItÕs a tautology.
ItÕs false.
>If they are independent of x,
TheyÕre not.
then setting x=0 has NO EFFECT on them,
>but it does have the benefit of clearing out terms that ARE
dependent
>on x.
They are only cleared out when x = 0. When x <> 0 they
are back in there.
>Before these arguments about this that basic logic was not
argued
>with, and IÕm sure lots of analysis is going on everyday,
where
people
>use it as itÕs been used for quite a while, without concern
that on
>Usenet posters successfully question it day after day, month
after
>month, and now year after year.
>So why is there confusion? Why canÕt I just settle it, and
those
>people just be the cranks?
>Well I think for many of you your intuition gets in the way.
>You look at
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>and it just makes sense to you that it is itself independent
of some
>freaky factorization of P(x). ItÕs almost as if your hearts
tell you
>what must be, and when the mathematics says otherwise, you
go with
>your hearts.
Look, wienie. I see a 49 in there as a factor
of the constant term. I see xÕs all over the place.
When you solve for the roots, they are going to be
variable functions of x. Clearly when x = 0, the
the cubic function defining the roots is
a^3 - 3*a^2,
and there are no factors of 7 or 49 anywhere in sight.
Why? Because the term in the cubic that has a factor of
49 is 0 when x = 0. Two of the roots are divisible by 7, but
that is only because they are 0; they are divisible by
ANYTHING.
The other root, 3, is coprime to 7: not a surprise.
The cubic in a in this case is REDUCIBLE.
But when x > 0, the cubic is
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and the term on the right is NONZERO in general and has
49 as a factor. This cubic in general is IRREDUCIBLE.
CLEARLY divisibility of the roots by factors of 7
could depend on x. It is NOT surprising that one of
the aÕs is coprime to 7 when x = 0, but none are coprime
to 7 when the polynomial is irreducible. CLEARLY there
is a mechanism which causes this to happen. Besides which,
it is a consequence of elementary Galois theory. Do you think
that Galois theory is wrong on this point ?
Not only that, but YOU YOURSELF in recent posts have
noted that, in general, a_1(x), a_2(x), and a_3(x) are
all non-coprime to 7 in the algebraic integers. Are
you now denying that ???
>Mathematically though, the logic follows rather simply,
which is why
>this is so frustrating for me, as I can trace out the logic,
>repeatedly. Talk about constants being constant. Give actual
numbers
>like 7 and 22, and it just not matter.
>Mathematics hardly gets any simpler than this, but thereÕs 
a
fatal
>ßaw if you donÕt understand that no matter how rational 
you
may
>believe you are, or how objective you think you are, there
are places
>and areas where your heart can get in the way, and your
emotions can
>rule your mind.
>IÕm asking you to be logical.
>What happens then though? Well then you find out that 
IÕm
right and
>yes, there is this major problem in algebraic number theory
that
NEEDS
>TO BE FIXED.
>When I communicate with trained mathematicians, I donÕt get
objections
>that come down to questioning whether or not constant terms
are
>constant as they donÕt do the Usenet objections. They just
walk
away.
Wonder why.
Nora B.
>So I have two different behaviors: irrational on Usenet, and
>passive-aggressive off.
>IÕm tired of it. I want you people to grow the hell up, and
start
>using your minds as the problem does not go away just by
ignoring it.
>Grow up.
>James Harris
===
Subject: Re: Key point, non-polynomial factorization
I think your entire post was spent arguing points that I
think most people
have no trouble with.
You have a polynomial:
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
It can be expressed as:
P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
Each of the a_*(x) may be expressed as:
a_*(x) = b_*(x) + const
where b_*(0)=0, and const =a_*(0)
The const terms are independant of x.
Its when you start trying to divide a_1(x) by 7, and then
making claims
about how it constitutes a crisis in mathematics that I thnk
people have
trouble understanding.
-Darren
> IÕm in a weird situation where there are people quite
dedicated in
> obscuring a result I have, so I want to be thorough in
highlighting
> how it works.
> As already IÕve given a long post in-depth, and a synopsis
post, IÕm
> going to now isolate out a key point.
> I relate one polynomial to a family of polynomials with a
> factorization:
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> to
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> with the factorization
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> where the aÕs are the roots of the cubic family, for
instance
> -a_1^3 + 3(-1 + 49x)a_1^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0.
> (To see details on the derivation go to my post
Revolutionary
> Mathematics: Non-polynomial Factorization.)
> Now crucially you have that P(x) is a multiple of 49, as
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> and, it should be clear that you can divide off 49 from
both sides
> without regard to the value of x.
> ThatÕs basic, but itÕs the area where all 
the controversy
sets in, as
> I say you can divide 49 off without regard to the value of
x, but
> others have to agree for P(x), but step in and claim that
is not true
> for the factorization
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7)
> that the 49 divides off without regard to the value of x.
> Essentially their arguments boil down to claiming that the
aÕs have
> factors in common with 7 that must vary, and you know,
thatÕs kind of
> a common-sense idea.
> After all, before it would have seemed basic to believe
that how
> factors of 7 distribute between the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> is determined by just the expression itself!
> What IÕm suggesting is counter-intuitive in that 
IÕm saying
that the
> family of cubics is actually constrained by the
factorization of some
> completely different polynomial.
> Your gut instincts may keep telling you that thereÕs no 
way
that your
> earlier instincts that a given cubic or family of cubics
were
> completely defined by just that cubics 
coefficients is wrong.
> But, in science, and now in mathematics, human intuition,
your
> common-sense can fail you. In physics the best example
typically
> given is quantum mechanics.
> Logically, the relation between the cubic family and the
factorization
> of the polynomial P(x) follows from
> P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)(5a_3(x) + 7).
> That is, counter to your intuition, mathematically whatÕs
important is
> that the two can be related by that equation, and then
because it can
> be related to the factorization, the cubic family is
constrained in a
> way that goes counter to your intuition.
> Your intuition is not mathematics. It can help you find
mathematics.
> It can help you understand mathematics, or it can block you
if you
> donÕt accept what follows mathematically.
> Maybe mathematicians are spoiled. I was trained as a
physicist as I
> have a B.Sc. in physics.
> Now, 49 divides off of P(x), without regard to the value of
x, as
> mathematically you have
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> and thereÕs no trace of 49, left, and P(x)/49 might be
misleading in
> that you can still see the 49, but thatÕs really just
telling you that
> it is now removed.
> If you accept that 49 divides off without regard to the
value of x,
> then it doesnÕt matter what value of x I use next, and I
pick x=0.
> At x=0, I get the cubic a^3 - 3a^2, which tells me that two
of the aÕs
> equal 0, while one equals 3, plugging that into the
factorization
> gives
> P(0)/49 = (5(0) + 7)(5(0)+ 7)(5(3) + 7)/49 = 22.
> Now here is where a lot of people seem to get lost.
> Logically, if terms are constant, then they donÕt vary 
with
x, but I
> keep hearing from people who think that x=0 is a special
case, as if
> when x changes then those terms constant with respect to x
will
> change.
> ThatÕs a logical contradiction.
> If the terms are in fact independent of the value of x,
then they are
> independent.
> ItÕs a tautology.
> If they are independent of x, then setting x=0 has NO
EFFECT on them,
> but it does have the benefit of clearing out terms that ARE
dependent
> on x.
> Before these arguments about this that basic logic was not
argued
> with, and IÕm sure lots of analysis is going on everyday,
where people
> use it as itÕs been used for quite a while, without 
concern
that on
> Usenet posters successfully question it day after day,
month after
> month, and now year after year.
> So why is there confusion? Why canÕt I just settle it, and
those
> people just be the cranks?
> Well I think for many of you your intuition gets in the way.
> You look at
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and it just makes sense to you that it is itself
independent of some
> freaky factorization of P(x). ItÕs almost as if your 
hearts
tell you
> what must be, and when the mathematics says otherwise, you
go with
> your hearts.
> Mathematically though, the logic follows rather simply,
which is why
> this is so frustrating for me, as I can trace out the logic,
> repeatedly. Talk about constants being constant. Give
actual numbers
> like 7 and 22, and it just not matter.
> Mathematics hardly gets any simpler than this, but thereÕs
a fatal
> ßaw if you donÕt understand that no matter how rational 
you
may
> believe you are, or how objective you think you are, there
are places
> and areas where your heart can get in the way, and your
emotions can
> rule your mind.
> IÕm asking you to be logical.
> What happens then though? Well then you find out that 
IÕm
right and
> yes, there is this major problem in algebraic number theory
that NEEDS
> TO BE FIXED.
> When I communicate with trained mathematicians, I donÕt 
get
objections
> that come down to questioning whether or not constant terms
are
> constant as they donÕt do the Usenet objections. They just
walk away.
> So I have two different behaviors: irrational on Usenet, and
> passive-aggressive off.
> IÕm tired of it. I want you people to grow the hell up, 
and
start
> using your minds as the problem does not go away just by
ignoring it.
> Grow up.
> James Harris
===
Subject: Re: Key point, non-polynomial factorization
IÕve been thinking about this some more, and it seems that
James has on one
side the polynomial
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
which he has shown may be expressible as:
P(x) = 7.Q(x) (where Q(x) evaluates to an algebraic integer)
Then he expresses the polynomial as a product of factors:
7.Q(x) = A(x)B(x)C(x)
Then he concludes that one of the factors on the right hand
side must have
7
as a factor.
But this jump is only valid if factorization is unique. (It
would certainly
follow in the ring of Integers...)
In the ring of algebraic integers, factorization is not
unique, therefore
it
does not follow that 7 is a factor of any of them.
Sure, he has shown that 7 is a factor when x=0, but this does
not imply
anything about the cases where x>0.
Is this what James is getting stuck on? Could it all be that
basic??
-Darren
===
Subject: Re: Key point, non-polynomial factorization
> IÕve been thinking about this some more, and it seems that
James has on
one
> side the polynomial
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> which he has shown may be expressible as:
> P(x) = 7.Q(x) (where Q(x) evaluates to an algebraic integer)
> Then he expresses the polynomial as a product of factors:
> 7.Q(x) = A(x)B(x)C(x)
> Then he concludes that one of the factors on the right hand
side must have
7
> as a factor.
> But this jump is only valid if factorization is unique. (It
would
certainly
> follow in the ring of Integers...)
> In the ring of algebraic integers, factorization is not
unique, therefore
it
> does not follow that 7 is a factor of any of them.
> Sure, he has shown that 7 is a factor when x=0, but this
does not imply
> anything about the cases where x>0.
> Is this what James is getting stuck on? Could it all be
that basic??
YouÕre close: in fact, factorization is unique (up to units)
in
polynomials as well. The problem is that his factors are *not*
polynomials, so they are not guaranteed to have any of the
nice
properties weÕre used to. His argument is similar to the
following:
(x+1)(x+2) is even for all integer values of x.
At x=0, we have 1*2, so the right-hand factor is divisible by
2 at x=0.
Therefor, for all integers x, x+2 is divisible by 2.
The problem is that he has factored his polynomial in a way
that does
not keep the 7 in a single factor for all values of x.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: algebra question
If n > 3, why is sqrt(-n) not prime in Z[ sqrt(-n) ] ?
===
Subject: Re: algebra question
>If n > 3, why is sqrt(-n) not prime in Z[ sqrt(-n) ] ?
For a hint, here is an exmple that might help you.
Consider Z[ sqrt(-5) ] and look at the element 2 + sqrt(-5).
Then,
9 = ( 2 + sqrt(-5) )( 2 - sqrt(-5) )
However, 9 = 3 * 3, so 3 * 3 is in the ideal generated by 2 +
sqrt(-5). That is 3 * 3 is in ( 2 + sqrt(-5) ). If 2 +
sqrt(-5) were
prime, then weÕd have 3 in ( 2 + sqrt(-5) ), but that is not
true.
So, 2 + sqrt(-5) is not prime. As a side note, this also
shows that
Z[ sqrt(-5) ] is not a unique factorization domain.
See if you can use this example to help with your problem.
Brian
===
Subject: Re: algebra question
days. My association with the Department is that of an
alumnus.
>>If n > 3, why is sqrt(-n) not prime in Z[ sqrt(-n) ] ?
>For a hint, here is an exmple that might help you.
>Consider Z[ sqrt(-5) ] and look at the element 2 + sqrt(-5).
>Then,
>9 = ( 2 + sqrt(-5) )( 2 - sqrt(-5) )
>However, 9 = 3 * 3, so 3 * 3 is in the ideal generated by 2 +
>sqrt(-5). That is 3 * 3 is in ( 2 + sqrt(-5) ). If 2 +
sqrt(-5) were
>prime, then weÕd have 3 in ( 2 + sqrt(-5) ), but that is 
not
true.
>So, 2 + sqrt(-5) is not prime. As a side note, this also
shows that
>Z[ sqrt(-5) ] is not a unique factorization domain.
>See if you can use this example to help with your problem.
I donÕt understand your example. Even in domains which are
not unique
factorization domains, there may be primes: primes are
non-units
which, when they divide ap roduct, will divide at least one
factor.
Suppose sqrt(-5) divides a + b*sqrt(-5), with a, b integers.
Then
sqrt(-5)*(x+y*sqrt(-5)) = a+b*sqrt(-5)
which means that -5y = a and x = b. Thus, sqrt(-5) divides
a+b*sqrt(-5) if and only if a is a multiple of 5.
Now assume that sqrt(-5) divides
(a+b*sqrt(-5))(x+y*sqrt(-5)). Then
5 must divide ax -5 by, which means that 5 must divide ax,
which means
that 5 must divide either a or x, which in turn means that
sqrt(-5)
divides either a+b*sqrt(-5) or x+y*sqrt(-5). Thus, sqrt(-5)
is a prime
in Z[sqrt(-5)].
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: algebra question
>If n > 3, why is sqrt(-n) not prime in Z[ sqrt(-n) ] ?
>>For a hint, here is an exmple that might help you.
>>Consider Z[ sqrt(-5) ] and look at the element 2 + sqrt(-5).
>>Then,
>>9 = ( 2 + sqrt(-5) )( 2 - sqrt(-5) )
>>However, 9 = 3 * 3, so 3 * 3 is in the ideal generated by 2
+
>>sqrt(-5). That is 3 * 3 is in ( 2 + sqrt(-5) ). If 2 +
sqrt(-5) were
>>prime, then weÕd have 3 in ( 2 + sqrt(-5) ), but that is
not true.
>>So, 2 + sqrt(-5) is not prime. As a side note, this also
shows that
>>Z[ sqrt(-5) ] is not a unique factorization domain.
>>See if you can use this example to help with your problem.
>I donÕt understand your example. Even in domains which are
not unique
>factorization domains, there may be primes: primes are
non-units
>which, when they divide ap roduct, will divide at least one
factor.
I was approaching the problem by using p to be prime if ( p )
was a
prime ideal. Thus, if xy were in ( p ), and p is a prime,
then either
x is in ( p ) or y is in ( p ).
In my example, we had 3*3 was in ( p ), but 3 was not in ( p
), so
( p ) was not a prime ideal, and p was not prime. Here,
p = 2 + sqrt( -5 ).
My note about the ring not being a UFD was simply a side note
that had
nothing to do with the problem.
>Suppose sqrt(-5) divides a + b*sqrt(-5), with a, b integers.
Then
>sqrt(-5)*(x+y*sqrt(-5)) = a+b*sqrt(-5)
>which means that -5y = a and x = b. Thus, sqrt(-5) divides
>a+b*sqrt(-5) if and only if a is a multiple of 5.
>Now assume that sqrt(-5) divides
(a+b*sqrt(-5))(x+y*sqrt(-5)). Then
>5 must divide ax -5 by, which means that 5 must divide ax,
which means
>that 5 must divide either a or x, which in turn means that
sqrt(-5)
>divides either a+b*sqrt(-5) or x+y*sqrt(-5). Thus, sqrt(-5)
is a prime
>in Z[sqrt(-5)].
According to what youÕve shown here, sqrt(-5) is a prime. 
The
OP must
have asked the question incorrectly. His question was to show
that
sqrt(-n) was not prime for n > 3. Either that, or they were
supposed
to come up with a counter-example. I hadnÕt thought of that
possibility, so I assumed the question was correct as asked
and I
threw up my example (without thinking very long). My hope was
to
inspire the OP towards a general solution.
Brian
===
Subject: Re: algebra question
days. My association with the Department is that of an
alumnus.
>>If n > 3, why is sqrt(-n) not prime in Z[ sqrt(-n) ] ?
>For a hint, here is an exmple that might help you.
>Consider Z[ sqrt(-5) ] and look at the element 2 + sqrt(-5).
>Then,
>9 = ( 2 + sqrt(-5) )( 2 - sqrt(-5) )
>However, 9 = 3 * 3, so 3 * 3 is in the ideal generated by 2 +
>sqrt(-5). That is 3 * 3 is in ( 2 + sqrt(-5) ). If 2 +
sqrt(-5) were
>prime, then weÕd have 3 in ( 2 + sqrt(-5) ), but that is 
not
true.
>So, 2 + sqrt(-5) is not prime. As a side note, this also
shows that
>Z[ sqrt(-5) ] is not a unique factorization domain.
>See if you can use this example to help with your problem.
>>I donÕt understand your example. Even in domains which are
not unique
>>factorization domains, there may be primes: primes are
non-units
>>which, when they divide ap roduct, will divide at least one
factor.
>I was approaching the problem by using p to be prime if ( p
) was a
>prime ideal. Thus, if xy were in ( p ), and p is a prime,
then either
>x is in ( p ) or y is in ( p ).
This is equivalent. I confess I didnÕt see your look at the
element
2+sqrt(-5), which would have explained the context.
>In my example, we had 3*3 was in ( p ), but 3 was not in ( p
), so
>( p ) was not a prime ideal, and p was not prime. Here,
>p = 2 + sqrt( -5 ).
>My note about the ring not being a UFD was simply a side
note that had
>nothing to do with the problem.
>>Suppose sqrt(-5) divides a + b*sqrt(-5), with a, b
integers. Then
>>sqrt(-5)*(x+y*sqrt(-5)) = a+b*sqrt(-5)
>>which means that -5y = a and x = b. Thus, sqrt(-5) divides
>>a+b*sqrt(-5) if and only if a is a multiple of 5.
>>Now assume that sqrt(-5) divides
(a+b*sqrt(-5))(x+y*sqrt(-5)). Then
>>5 must divide ax -5 by, which means that 5 must divide ax,
which means
>>that 5 must divide either a or x, which in turn means that
sqrt(-5)
>>divides either a+b*sqrt(-5) or x+y*sqrt(-5). Thus, sqrt(-5)
is a prime
>>in Z[sqrt(-5)].
>According to what youÕve shown here, sqrt(-5) is a prime.
In Z[sqrt(-5)]. Yes.
> The OP must
>have asked the question incorrectly. His question was to
show that
>sqrt(-n) was not prime for n > 3.
Yes, which I am sure is incorrect for n prime. But it need
not be
incorrect for composite squarefree n. For example, the
procedure above
with sqrt(-10) would lead to the conclusion that sqrt(-10)
divides
a+b*sqrt(-10) if and only if 10 divides a. Then sqrt(-10)
would divide
the product (a+b*sqrt(-10))(x+y*sqrt(-10)) if and only if 10
divides
ax-10by, if and only if 10 divides ax, but now it is easy to
come up
with examples where neither a nor x are divisible by 10 but
ax is.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Linear Algebra Help
Question: Show that the Gram-Schmidt process says that given
an
invertible nxn mx A, there is an upper triangular nxn mx B
such
that AB is an orthogonal mx.
Work done so far:
Taking the 2x2 case
If
A = [ a11 a12 ] and B = [ b11 b12 ]
[ a21 a22 ] [ 0 b22 ] ,
then direct computation shows that
AB = [ a11b11 a11b12 + a12b22 ]
[ a21b11 a21b12 + a22b22 ]
= [ b11*[a11] b12*[a11]+b22[a12] ]
[ [a12] [a22] [a22] ] .
That is, the first column of the product AB is just b11 times
the first column of A. The second column of the product is b12
times the first column of A, plus b22 times the second column
of
A.
What I havenÕt determined :
what matrix B will implement the Gram-Schmidt process, so that
the columns of the product are orthonormal vectors.
Any help would be greatly appreciated.
----------------------------------------------
* Binary Usenet Leeching Made Easy
* http://www.newsleecher.com/?usenet
----------------------------------------------
===
Subject: Re: Linear Algebra Help
> Question: Show that the Gram-Schmidt process says that
given an
> invertible nxn mx A, there is an upper triangular nxn mx B
such
> that AB is an orthogonal mx.
[...]
Check out
<http://www.csie.ncu.edu.tw/~chia/Course/LinearAlgebra/sec5-2
.pdf>
or Google QR factorization gram-schmidt and pick out
something that
works for you.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Natural Log Problem with Proportional Unknown
Exponents
The basic problem IÕm having is trying to solve for x.
300=12000e^(.025x)-12000e^(.02x)
I can do it with a graphing calculator but I would like to
know how it
could be done by hand. It doesnÕt seem to be covered by any
of the
Logarithmic Laws that work when there is only one term with
an unknown
exponent.
Tysen
===
Subject: Re: Natural Log Problem with Proportional Unknown
Exponents
> The basic problem IÕm having is trying to solve for x.
> 300=12000e^(.025x)-12000e^(.02x)
> I can do it with a graphing calculator but I would like to
know how it
> could be done by hand. It doesnÕt seem to be covered by 
any
of the
> Logarithmic Laws that work when there is only one term with
an unknown
> exponent.
Of course, anything that a graphing calculator can do can
also, at least in
principle, be done by hand. But I suppose that what you
really want is a
method, other than just numerical approximation, for solving
such an
equation. There is such a method, which expresses the
solution as a series.
<http://mathforum.org/discuss/sci.math/a/m/623863/623863>.
rewritten. There are many ways that this could be done. For
example,
300 = 12000e^(.025x) - 12000e^(.02x)
300 + 12000e^(.02x) = 12000e^(.025x)
.025e^(-.025x) + e^(-.005x) = 1
yields an equation in suitable form. Then (*), with A = 1/40,
a = -1/40,
B = 1 and b = -1/200, is the solution. If we truncate the
series, using,
say, just the first ten terms, to get an approximation, we 
find
x = 417090851369317809389830589/92343379137802193583984375
= 4.5167380191589199103...
A more accurate approximation of the solution (obtained using
MathematicaÕs
FindRoot) is
4.5167380191589199468,
showing that our approximation obtained by using ten terms of
(*) is
accurate through the sixteenth decimal place.
David Cantrell
===
Subject: Re: Natural Log Problem with Proportional Unknown
Exponents
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6DWOa30126;
>The basic problem IÕm having is trying to solve for x.
>300=12000e^(.025x)-12000e^(.02x)
If we let a = 0.025 and b = 0.02, the problem reduces to:
ln[a] - a*x = ln[ 1 - e^(-a*b*x) ].
Perhaps now it is more evident that there is no algebraic way
to solve
this problem.
Joseph A.
===
Subject: riemann sums.
Consider f(x) = (x^2)/2 - 8.
Interval [0,2]
Solve using limit as n approaches infinity, of (n variable)
(i=1)SIGMA, f(x
subI)(delta x)
Summation inside the brackets is Rn, which the Riemann sum
where the sample
points are chosen to be the right-hand endpoints of each
sub-interval.
Calculate Rn for f(x) on [0,2], and write your answer as a
function of n
without any summation signs. You will need the summation
formulas.
Rn =
lim n-->inf. Rn =.
IÕve been working on this goddamned hellish torture puzzle
for hours. IÕve
got something on the order of 30 pages of scrap paper used up
on this
problem. ItÕs due in 3 hours; I doubt anyone will respond in
time. I can
hope.
Anyway, simple stuff I doubt I can be wrong on:
deltaX = b-a / n = 2-0/ n = 2/n
XsubI = 0 +k(deltaX) = 0 + k(2/n).
And, when we get to it: k^2 = (n)(n+1)(2n+1)
lim n-->inf. sigma, [([(2k/n)^2] / 2) -8 ] (2/n)
...and I keep getting the wrong answer, so IÕll stop there.
Please, please
help me. IÕm ... so damned lost.
===
Subject: Re: riemann sums.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6DWO330145;
>Consider f(x) = (x^2)/2 - 8.
>Interval [0,2]
>Solve using limit as n approaches infinity, of (n variable)
(i=1)SIGMA, f(x
>subI)(delta x)
>Summation inside the brackets is Rn, which the Riemann sum
where the
sample
>points are chosen to be the right-hand endpoints of each
sub-interval.
>Calculate Rn for f(x) on [0,2], and write your answer as a
function
of n
>without any summation signs. You will need the summation
formulas.
>Rn =
>lim n-->inf. Rn =.
>deltaX = b-a / n = 2-0/ n = 2/n
>XsubI = 0 +k(deltaX) = 0 + k(2/n).
>And, when we get to it: k^2 = (n)(n+1)(2n+1)
>lim n-->inf. sigma, [([(2k/n)^2] / 2) -8 ] (2/n)
>...and I keep getting the wrong answer, so IÕll stop there.
Please,
please
>help me. IÕm ... so damned lost.
Okay, breathe. YouÕre almost there...
We have f(x) = (1/2)*x^2 - 8, for all x in [0,2].
For each number n, let P_n be the partition of [0,2] into n
equal
subintervals. Then for 0 < i < (n + 1), we have
x_i = (2*i)/n,
and
x_i - x_(i-1) = 2/n, (note: this is your delta x_i).
Using Riemann Sums, we can approximate the integral of f(x)
on the
given domain by
lim R(f, P_n) = R(f, P). (Note: R(f, P) is the integral of f)
n->oo
Observe that if we take the right-hand points as our Riemann
Sum
sample points, then
n
---
R(f, P) = lim [ f(x_i)*(x_i - x_(i-1)) ]
n->oo / [ ]
---
i=1
n
---
= lim [ f((2*i)/n)*(2/n) ]
n->oo / [ ]
---
i=1
n
---
= lim (2/n) [ (1/2)*((2*i)/n)^2 - 8 ]
n->oo / [ ]
---
i=1
n n
--- ---
= lim (2/n) [ (2*i^2)/n^2 - (2/n) [ 8 ]
n->oo / [ / [ ]
--- ---
i=1 i=1
n
---
= lim (4/n^3) [ i^2 ] - (2/n)*(8*n). (Eq.1)
n->oo / [ ]
---
i=1
Using the fact that the sum of the squares of the first m
integers can
be written as:
[m*(m + 1)*(2*m + 1)]/6,
Eq. 1 can be written as
R(f, P) = lim (4/n^3)*[(n*(n + 1)*(2*n + 1))/6] - 16
n->oo
= lim [ 8*n^3 + O(n^2) ]/[ 6*n^3 ] - 16
n->oo
= 8/6 - 16
= -44/3.
-----
You almost had it, but you were probably getting lost in your
(a)
NOTATION and (b) bad use of subscripts (c) incorrect formula
for the
sum of the squares of first m integers. Hope this helps.
Joseph A.
===
Subject: Re: riemann sums.
>Consider f(x) = (x^2)/2 - 8.
>>Interval [0,2]
>>Solve using limit as n approaches infinity, of (n variable)
> (i=1)SIGMA, f(x
>>subI)(delta x)
>>Summation inside the brackets is Rn, which the Riemann sum
where the
> sample
>>points are chosen to be the right-hand endpoints of each
> sub-interval.
>>Calculate Rn for f(x) on [0,2], and write your answer as a
function
> of n
>>without any summation signs. You will need the summation
formulas.
>>Rn =
>>lim n-->inf. Rn =.
>>deltaX = b-a / n = 2-0/ n = 2/n
>>XsubI = 0 +k(deltaX) = 0 + k(2/n).
>>And, when we get to it: k^2 = (n)(n+1)(2n+1)
>>lim n-->inf. sigma, [([(2k/n)^2] / 2) -8 ] (2/n)
>>...and I keep getting the wrong answer, so IÕll stop 
there.
Please,
> please
>>help me. IÕm ... so damned lost.
> Okay, breathe. YouÕre almost there...
> We have f(x) = (1/2)*x^2 - 8, for all x in [0,2].
> For each number n, let P_n be the partition of [0,2] into n
equal
> subintervals. Then for 0 < i < (n + 1), we have
> x_i = (2*i)/n,
> and
> x_i - x_(i-1) = 2/n, (note: this is your delta x_i).
> Using Riemann Sums, we can approximate the integral of f(x)
on the
> given domain by
> lim R(f, P_n) = R(f, P). (Note: R(f, P) is the integral of
f)
> n->oo
> Observe that if we take the right-hand points as our
Riemann Sum
> sample points, then
> n
> ---
> R(f, P) = lim [ f(x_i)*(x_i - x_(i-1)) ]
> n->oo / [ ]
> ---
> i=1
> n
> ---
> = lim [ f((2*i)/n)*(2/n) ]
> n->oo / [ ]
> ---
> i=1
> n
> ---
> = lim (2/n) [ (1/2)*((2*i)/n)^2 - 8 ]
> n->oo / [ ]
> ---
> i=1
> n n
> --- ---
> = lim (2/n) [ (2*i^2)/n^2 - (2/n) [ 8 ]
> n->oo / [ / [ ]
> --- ---
> i=1 i=1
> n
> ---
> = lim (4/n^3) [ i^2 ] - (2/n)*(8*n). (Eq.1)
> n->oo / [ ]
> ---
> i=1
> Using the fact that the sum of the squares of the first m
integers can
> be written as:
> [m*(m + 1)*(2*m + 1)]/6,
> Eq. 1 can be written as
> R(f, P) = lim (4/n^3)*[(n*(n + 1)*(2*n + 1))/6] - 16
> n->oo
> = lim [ 8*n^3 + O(n^2) ]/[ 6*n^3 ] - 16
> n->oo
> = 8/6 - 16
> = -44/3.
> -----
> You almost had it, but you were probably getting lost in
your (a)
> NOTATION and (b) bad use of subscripts (c) incorrect
formula for the
> sum of the squares of first m integers. Hope this helps.
notebook for future reference.
===
Subject: Re: riemann sums.
> deltaX = b-a / n = 2-0/ n = 2/n
OK.
> XsubI = 0 +k(deltaX) = 0 + k(2/n).
OK.
> And, when we get to it: k^2 = (n)(n+1)(2n+1)
Uh, no.
The sum of the first n squares is n(n+1)(2n+1)/6
Now weÕre ready...
You want to find
n
Sigma f(x_k) delta-x
k=0
f(x_k) = [(2k/n)^2]/2 - 8
= (4k^2/n^2)/2 - 8
= 2k^2/n^2 - 8
f(x_k) delta-x = [2k^2/n^2 - 8](2/n)
= 4k^3/n^3 - 16/n
So you want the sum
n n n
Sigma 4k^2/n^3 - 16/n = (4/n^3)Sigma k^2 - (16/n)Sigma 1
k=0 k=0 k=0
= (4/n^3)(n)(n+1)(2n+1)/6 - (16/n)(n+1)
= 4 (2n^3 + 2n^3 + n) n + 1
- ---------------- - 16 -----------
6 n^3 n
Now take the limit of that as n->+oo and you get
lim[above] = (4/6)(2) - 16(1)
= 4/3 - 16
= -44/3
Check against an integration.
Integral of (x^2)/2 - 8 is (x^3)/6 - 8x + C, so
the definite integral between 0 and 2 is:
(2^3)/6 - 8(2) + C - [0 - 0 + C]
= 8/6 - 16 = 4/3 - 16
= -44/3
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: riemann sums.
>> deltaX = b-a / n = 2-0/ n = 2/n
> OK.
>> XsubI = 0 +k(deltaX) = 0 + k(2/n).
> OK.
>> And, when we get to it: k^2 = (n)(n+1)(2n+1)
> Uh, no.
> The sum of the first n squares is n(n+1)(2n+1)/6
> Now weÕre ready...
> You want to find
> n
> Sigma f(x_k) delta-x
> k=0
> f(x_k) = [(2k/n)^2]/2 - 8
> = (4k^2/n^2)/2 - 8
> = 2k^2/n^2 - 8
> f(x_k) delta-x = [2k^2/n^2 - 8](2/n)
> = 4k^3/n^3 - 16/n
> So you want the sum
> n n n
> Sigma 4k^2/n^3 - 16/n = (4/n^3)Sigma k^2 - (16/n)Sigma 1
> k=0 k=0 k=0
> = (4/n^3)(n)(n+1)(2n+1)/6 - (16/n)(n+1)
> = 4 (2n^3 + 2n^3 + n) n + 1
> - ---------------- - 16 -----------
> 6 n^3 n
> Now take the limit of that as n->+oo and you get
> lim[above] = (4/6)(2) - 16(1)
> = 4/3 - 16
> = -44/3
Webworks (the online assignment program) accepts -44/3 as the
limit, but
doesnÕt accept your Rn. Am I misinterpreting what 
youÕve
written? I take it
that your Rn is (4/6) (2n^3 +2n^2 +n)/n^3 - (16/n)(n+1).
Yours originally said 2n^3 + 2n^3. I changed the second one
since I thought
you may have made a typo with the power. However, I tried
both 2n^3 + 2n^3
and 2n^3+2n^2, and webworks accepts neither.
===
Subject: Re: riemann sums.
> = 4 (2n^3 + 2n^3 + n) n + 1
> - ---------------- - 16 -----------
> 6 n^3 n
ThatÕs (2n^3 + 3n^2 + n). Of course, in the limit the
error doesnÕt change anything.
> Yours originally said 2n^3 + 2n^3. I changed the second one
since I
thought
> you may have made a typo with the power. However, I tried
both 2n^3 + 2n^3
> and 2n^3+2n^2, and webworks accepts neither.
You should have just gone back to n(n+1)(2n+1) and multiplied
it out
once you realized I had made a mistake whilst doing so.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: riemann sums.
>> = 4 (2n^3 + 2n^3 + n) n + 1
>> - ---------------- - 16 -----------
>> 6 n^3 n
> ThatÕs (2n^3 + 3n^2 + n). Of course, in the limit the
> error doesnÕt change anything.
>> Yours originally said 2n^3 + 2n^3. I changed the second
one since I
>> thought
>> you may have made a typo with the power. However, I tried
both 2n^3 +
>> 2n^3
>> and 2n^3+2n^2, and webworks accepts neither.
> You should have just gone back to n(n+1)(2n+1) and
multiplied it out
> once you realized I had made a mistake whilst doing so.
Webworks still says itÕs incorrect.
===
Subject: Solving Cubics Using A Linear Fractional
Transformation
Hello all,
If you are interested in an alternative way to solve the
cubic:
A New Way To Solve Cubics Using a Linear Fractional
Transformation
ABSTRACT: We provide a new method to solve the general cubic
equation
by
using a linear fractional transformation. This transformation,
sometimes
referred to as a Mobius transformation, can transform the
general
cubic
into the binomial form, though in a manner different from the
traditional
Tschirnhausen transformation that can also transform the
cubic into
the
binomial form. The resulting binomial is then simply solved
by the
extraction
of a single cube root.
Mathematics Subject Classification. Primary: 12E12; Secondary:
15A04
http://www.geocities.com/titus_piezas/cubics.html
Just click at the link to the pdf file.
P.S. ThereÕs also a new paper on sextics (less
paragraph-breaks
version). Just go to the index.
--Titus
===
Subject: Need some major help!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6FGea08281;
I have my final this week and my professor used the phrase if
you
know the review sheet you SHOULD be ok for the final. well a
lot of
the review sheet I cant get. I need some major help, anyone
that
could do some of my review questions for me so that would
show me how
to do them and be ready for my final this week. Email me ASAP
and
once you do I can email you back some of the questions for
you to
Josh
here is are a couple for example....
Show that F = zcosxzi + e^yj + xcosxzk is conservative and
find a
potential function for F.
Find the absolute maxima and minima and where they occur for
the
function f(x,y) = x^2+y^2+x^2y+4 on the square bounded by the
lines
x=1, x=-1, y=1, and y=-1.
===
Subject: Re: Need some major help!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6HrcW24333;
>Find the absolute maxima and minima and where they occur for
the
>function f(x,y) = x^2+y^2+x^2y+4 on the square bounded by
the lines
>x=1, x=-1, y=1, and y=-1.
If f(x,y) has a local minimum or maximum, then
df/dx = 2x(1 + y) = 0
df/dy = 2y + x^2 = 0
Solving for x,y:
y = -x^2/2
2x(1 - x^2/2) = 0
x = 0, sqrt(2), -sqrt(2)
if x = 0, then y = 0
if x = +- sqrt(2), then y = -1
Hence, you have 3 points to test for local minima or maxima:
[0,0]
[sqrt(2),-1]
[-sqrt(2),-1]
If f(x,y) has a local min or max, the jacobian of p = df/fx,
q = df/dy is greater or equal to 0. Otherwise, f(x,y) has a
saddle
point.
J(x,y) = d(p,q)/(x,y) = dp/dx*dq/dx - dp/dy*dq/dy =
= d^2f/dx^2*d^2f/dy^2 - [df^2/(dxdy)]^2
d^2f/dx^2 = 2(1 + y)
d^2f/dy^2 = 2
d^2f(dxdy) = 2x
J(x, y) = 2(1 + y)*2 - 4x^2 = 4 + 2y - 4x^2
J(0,0) = 4
J(sqrt(2),-1) = 4 - 2 - 8 = -6
So, the points [sqrt(2),-1] and [-sqrt(2),-1] are saddle
points, not
local min or max. f(0, 0) is a local min or max. If it is a
min, then
d^2f/dx^2 > 0 and d^2f/dy^2 > 0. If it is max, then d^2f/dx^2
< 0 and
d^2f/dy^2 < 0.
d^2f/dx^2(0, 0) = 2(1 + y) = 2
d^2f/dy^2(0, 0) = 2
Hence, the point [0,0] is a local minimum, f(0, 0) = 4. The
absolute
minima or maxima can also be on the boundary of the square.
If x = 1, then f( 1, y) = 1 + y^2 + y + 4 = y^2 + y + 5,
df/dy = 2y + 1, y = -1/2
If x = -1, then f(-1, y) = 1 + y^2 + y + 4 = y^2 + y + 5,
df/dy = 2y + 1, y = -1/2
If y = 1, then f(x, 1) = x^2 + 1 + x^2 + 4 = 2x^2 + 5,
df/dx = 4x, x = 0
If y = -1, then f(x, -1) = x^2 + 1 - x^2 + 4 = 5,
df/dx = 0, -1 <= x <= 1 arbitrary
The points on the boundary to test are [1, -1/2], [-1, -1/2],
[0, 1]:
f( 1,-1/2) = 1 + 1/4 - 1/2 + 4 = 4.75
f(-1,-1/2) = 1 + 1/4 - 1/2 + 4 = 4.75
f( 0, 1) = 1 + 4 = 5
Finally, testing the corners of the square:
f(1, 1) = 1 + 1 + 1 + 4 = 7
f(1,-1) = 1 + 1 - 1 + 4 = 5
f(-1,1) = 1 + 1 + 1 + 4 = 7
f(-1,-1) = 1 + 1 - 1 + 4 = 5
Clearly, the absolute naxima are the square corners [1, 1] and
[-1, 1], where f(1,1) = f(-1,1) = 7. The absolute minimum is
the local
minimum at the origin [0, 0], where f(0, 0) = 4.
===
Subject: Re: Need some major help!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6HF8a20810;
>Show that F = zcosxzi + e^yj + xcosxzk is conservative and
find a
>potential function for F.
F = z*cos(xz)*i + e^y*j + x*cos(xz)*k
F is conservative iff it can be expressed as the gradient of
a scalar
potential f(x,y,z):
F = grad f = df/dx*i + df/dy*j + df/dz*k
These are paretial derivatives. So, if F is conservative, then
df/dx = z*cos(xz)
df/dy = e^y
df/dz = x*cos(xz)
Pick one of these equations, say for df/dx, and integrate
with respect
to x, keeping y,z constant.
f(x,y,z) = z/z*sin(xz) + C(y,z) = sin(xz) + C(y,z)
Then
df/dz = x*cos(xz) = d/dz[sin(xz) + C(y,z)] = x*cos(xz) + dC/dz
Therefore, dC/dz = 0 and C = C(y), Finally,
df/dy = e^y = d/dy[sin(xz) + C(y)] = dC/dy
dC/dy = e^y
C(y) = e^y + K
f(x,y,z) = sin(xz) + C(y) = sin(xz) + e^y + K
Check:
df/dx = z*cos(xz)
df/dy = e^y
df/dz = x*cos(xz)
F = grad f as required.
===
Subject: Re: Need some major help!!!
> F = grad f as required.
Is that _grad_ as in vector operator, or _grade_ as in course
result?
===
Subject: Re: Need some major help!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8GXEb12528;
>> F = grad f as required.
>Is that _grad_ as in vector operator, or _grade_ as in
course result?
This is left to the reader as an exercise.
===
Subject: Re: Number Pattern Help
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6FjZ210717;
these are my numbers
0
7
26
63
?
find the next number
===
Subject: Re: Number Pattern Help
>these are my numbers
>find the next number
These arenÕt math questions they are psychology questions.
There are infinitely many answers --each as correct
mathematically as
any other. What weÕre being asked to do is to guess which 
one
the
questioner expects to see.
===
Subject: Re: Number Pattern Help
Just 1 of many possible solutions:
Table[n^3 - 1, {n, 5}]
{0, 7, 26, 63, 124}
HTH
--
Dana
> these are my numbers
> 26
> 63
> find the next number
===
Subject: Re: Number Pattern Help
posting-account=-m3APw0AAAA-5WXqbJX0X9WBnuKsH_Qg
116 ?
===
Subject: Re: Number Pattern Help
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> 116 ?
Surely 124?
===
Subject: Re: Number Pattern Help
>116 ?
> Surely 124?
116 is the answer if the rule generating the sequence is that
the
difference between entries n and n+1 is prime number 4n. That
rule is
almost as simple as the one giving 124.
--
Stefan Holm
ItÕs a pleasure to meet me. I hope you
never find a live turtle in your soup.
===
Subject: Re: Number Pattern Help
posting-account=-m3APw0AAAA-5WXqbJX0X9WBnuKsH_Qg
116 ?
> these are my numbers
> 26
> 63
> find the next number
===
Subject: Question w/o a deadline
So, anyway... that was for my sunday night webworks.
This one is just a problem IÕve been having issues with:
--
A toy manufactuer produces an inexpensive doll (Floppy) and
an expensive
doll (Mopsy) in units of x hundred and y hundred
respectively. Suppose it is
possible to produce dolls in such a way that y = (82 - 10x) /
(10-x), 0
less than or equal to X less than or equal to 8. The company
recieves twice
as much for selling a mopsy doll as for selling a Flopsy
doll. Find the
level of production for both x and y for which the totla
revenue derived
from selling these dolls is maximized. What vital assumption
must be made
about sales in this model?
--
First off, IÕm not sure where to start with the formulas 
here.
This much we have: y = (82 - 10x) / (10-x).
We have p(y) = 2x ( I think.)
Revenue(total) = Revenue - Cost
We want to max revenue, so we want to find R= 
CÕ.
IÕm do not know how to manipulate the given formula to 
find RÕ
and CÕ. I
looked through the bookÕs sample problems, but there are 
none
similar to
this - theyÕre all much more simplistic.
===
Subject: Re: Question w/o a deadline
> A toy manufactuer produces an inexpensive doll (Floppy) and
an expensive
> doll (Mopsy) in units of x hundred and y hundred
respectively. Suppose it
is
> possible to produce dolls in such a way that y = (82 - 10x)
/ (10-x), 0
> less than or equal to X less than or equal to 8. The
company recieves
twice
> as much for selling a mopsy doll as for selling a Flopsy
doll. Find the
> level of production for both x and y for which the totla
revenue derived
> from selling these dolls is maximized. What vital
assumption must be made
> about sales in this model?
The vital assumption is that you can sell everything that you
produce, so
that
revenue is 2 * y + x.
R = 2y + x = 2 (82 - 10x) / (10 - x) + x
R = (164 - 20x) / (10 - x) + x
R = (200 - 20x - 36) / (10 - x) + x
R = (20 * (10 - x) - 36) / (10 - x) + x
R = 20 + x - 36 / (10 - x)
R= 1 - 36 / (10 - x)^2 = 0
x1 = 16
x2 = 4
y1 = (82 - 160) / (10 - 16) = 78 / 6 = 13
y2 = (82 - 40) / (10 - 4) = 42 / 6 = 7
R1 = 42
R2 = 18
So, the max revenue is 42 at x = 16.
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Re: Mechanics problem - Any takers?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6IaOt28405;
Please...donÕt leave me behind...
===
Subject: phases
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6Jx8103644;
Question find vr
v1=6sin(theta+60) v2=10sin(theta-45)
I know the answer to r is 10.2 as i did it graphically but
when i come
to do the maths i donÕt get the same answer
Neil
===
Subject: Re: phases
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7DBKQ28148;
Ok here goes what i have done so far
Equation used sin(a+/-b)=Sina cosb+/-cosa sinb
V1=6sin(x+60) V2=10sin(x-45)
6(sinthetaCos60+costhetasin60)
6sinthetacos60+6costhetasin60
5.1961sintheta 3costheta
10(sinthetacos45-costhetasin-45)
10sinthetacos45-10costhetasin-45)
7.07sintheta 7.07costheta
5.1961sintheta 3costheta+7.07sintheta+7.07costheta
12.266sintheta 10.07costheta
Rsquared=asquared+bsquared
r=15.8
===
Subject: Re: phases
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7H1Pd19223;
>Ok here goes what i have done so far
>Equation used sin(a+/-b)=Sina cosb+/-cosa sinb
>V1=6sin(x+60) V2=10sin(x-45)
>6(sinthetaCos60+costhetasin60)
>6sinthetacos60+6costhetasin60
>5.1961sintheta 3costheta
>10(sinthetacos45-costhetasin-45)
>10sinthetacos45-10costhetasin-45)
>7.07sintheta 7.07costheta
>5.1961sintheta 3costheta+7.07sintheta+7.07costheta
>12.266sintheta 10.07costheta
>Rsquared=asquared+bsquared
>r=15.8
v1 = 6*(sin(theta)*cos(60) + cos(theta)* sin(60)) =
= 6*sin(theta)*cos(60) + 6*cos(theta)*sin(60) =
= 5.1961*sin(theta) + 3*cos(theta)
v2 = 10*(sin(theta)*cos(45) - cos(theta)*sin(-45))
= 10*sin(theta)*cos(45) - 10*cos(theta)*sin(-45)
= 7.07*sin(theta) + 7.07*cos(theta)
v1 + v2 =
= 5.1961*sin(theta) + 3*cos(theta) + 7.07*sin(theta) +
7.07*cos(theta)
= 12.266*sin(theta) + 10.07*cos(theta)
r^2 = a^2 + b^2
r = 15.8
Do you see that parenthesis might be helpful ?
You used cos(60) = sqrt(3)/2 ~ 0.866 and sin(60) = 1/2. This
is
incorrect, the correct values are cos(60) = 1/2 and sin(60) =
sqrt(3)/2. When in doubt about your memory of these values,
imagine an
equlateral triangle ABC with sides AB = BC = CA = 1 divided
in half by
one of its altitudes, say CO, where O is the foot of the
altitude on
the side AB. Clearly, angle A = 60 deg and AO = BO = AB/2 =
1/2. AO is
cos(60) and CO is sin(60). So cos(60) = 1/2 and by Pythagorean
theorem, sin(60) = sqrt(1^2 - (1/2)^2) = sqrt(3/4) =
sqrt(3)/2.
The second error is the double use of the minus sign in the
equation
for v2. You can write
sin(x - y) = sin(x)*cos(y) - cos(x)*sin(y)
or
sin(a - b) = sin(x)*cos(-y) + cos(x)*sin(-y)
and then use cos(-y) = cos(y) and sin(-y) = -sin(y).
Still, I do not know, what are you up to, you did not state
the
problem.
v1 = 6*(sin(theta)*cos(60) + cos(theta)* sin(60)) =
= 3*sin(theta) + 3*sqrt(3)*cos(theta)
v2 = 10*(sin(theta)*cos(45) - cos(theta)*sin(45))
= 5*sqrt(2)*sin(theta) - 5*sqrt(2)*cos(theta)
v1 + v2 =
= (3 + 5*sqrt(2))*sin(theta) + (3*sqrt(3) -
5*sqrt(2))*cos(theta) =
= a*sin(theta) + b*cos(theta)
r^2 = a^2 + b^2 =
= (3 + 5*sqrt(2))^2 + (3*sqrt(3) - 5*sqrt(2))^2 =
= 9 + 30*sqrt(2) + 50 + 27 - 30*sqrt(6) + 50 =
= 136 + 30*sqrt(2) - 30*sqrt(6) =
= 136 - 30*(sqrt(6) - sqrt(2)) ~
~ 136 - 30*1.0353 ~
~ 136 - 31.06 = 104.94
r ~ sqrt(104.94) ~ 10.244
===
Subject: Re: phases
>Question find vr
>v1=6sin(theta+60) v2=10sin(theta-45)
>I know the answer to r is 10.2 as i did it graphically but
when i come
>to do the maths i donÕt get the same answer
Good -- show us what you did and weÕll help you 
find your
mistake.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: help on C^2pi is dense in C[-pi,pi] with respect to
the L2-norm
Can somebody help me to prove that C^2pi is dense in
C[-pi,pi] with respect
to the L2-norm?
C^2pi is the set of the 2pi-periodic continuous functions
C[-pi,pi} is the set of the continuous functions in the
interval [-pi,pi]
the L2-norm^2 is defined as 1/(2pi) * integral
(abs(f(t))^2,t=[-pi,pi])
===
Subject: Re: help on C^2pi is dense in C[-pi,pi] with respect
to the
L2-norm
> Can somebody help me to prove that C^2pi is dense in
C[-pi,pi] with
respect
> to the L2-norm?
If f is in C[-pi,pi], let g = f except for small intervals
near the
endpoints, where g goes from the graph of f down to 0. Then g
is in C^2pi,
and the L^2 difference between f and g is small.
===
Subject: precalculus
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB6N3LD22206;
I have a ugly one that I just canÕt figure out 
where to begin.
Please
help.
Find an eighth degree polynomial P(x) that passes through the
following points:
(-4,-60314),(-3,-5545),(-2,-144),(-1,7),(0,2),(1,-9),(2,260),
(3,-6169),(4,-6
2658)
===
Subject: Re: precalculus
> Find an eighth degree polynomial P(x) that passes through
the
> following points:
>
(-4,-60314),(-3,-5545),(-2,-144),(-1,7),(0,2),(1,-9),(2,260),
(3,-6169),(4,-62
6
> 58)
LetÕs solve a simpler problem first:
Find an eighth degree polynomial P1(x) that passes through
the following
points:
(-4, 0)
(-3, 0)
(-2, 0)
(-1, 0)
(0, 0)
(1, 0)
(2, 0)
(3, 0)
(4, 1)
Because itÕs an 8th degree polynomial, and you know all its 
8
zeros, you
know
itÕs going to be of the form:
P1(x) = A (x + 4) (x + 3) (x + 2) (x + 1) x (x - 1) (x - 2)
(x - 3)
so the only question is what the value of A is. You get that
from the last
point
(4, 1):
P1(4) = A * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 1
A = 1/(1 * 2 * 3 * 4 * 5 * 6 * 7 * 8)
Why is this useful?
Now use a similar method to create P2(x) such that it passes
through:
(-4, 0), (-3, 0), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0),
(3, 1), and (4,
0)
Then find P3(x) that passes though ... you get the idea. Zeros
at all of -4
through 4, except for 2. Find P4(x) that has zeros at -4
through 4 except
for 1,
etc, until you have nine polynomials, P1 - P9, each of them
having zeros at
8 of
9 of the values of x you were given, and having value of 1 at
the remaining
one.
If you let R(x) = A1 * R1(x) + A2 * R2(x) for some A1 and A2,
what is the
value
of R(x) at -4, -3, -2, -1, 0, 1, 2, 3, and 4?
Now use that knowledge to combine P1 - P9 to form P.
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Re: precalculus
days. My association with the Department is that of an
alumnus.
>I have a ugly one that I just canÕt figure out 
where to
begin. Please
>help.
>Find an eighth degree polynomial P(x) that passes through the
>following points:
>(-4,-60314),(-3,-5545),(-2,-144),(-1,7),(0,2),(1,-9),(2,260)
,(3,-6169),(4,-
62658)
Say p(x) = a*x^8 + b*x^7 + c*x^6 + d*x^5 + e*x^4 + f*x^3 +
g*x^2 + h*x + m
Since the polynomial must go through (0,2), p(0) = m must be
equal to
2. So m=2.
Since it goes through (1,-9), you have that p(1)=-9, which
tells you
that
a + b + c + d + e + f + g + h + 2 = -9
or that
a + b + c + d + e + f + g + h = -11
Each point gives you an equation on a, b, c, d, e, f, g, h by
substituting. You will get 8 equations (having already used
(0,2) to
figure out the value of m), on these 8 unknowns. Solving the
system
will give you the values of a, b, c, d, e, f, g, and h.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Exponential and Logarithmic Functions (word problem
following the
exponential law) This is due tomorrow so I really need some
help
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB73FeU12152;
Ok this is one problem:
The population of a southern city follows the exponential
law. If the
population doubled in size over an 18-month period and the
current
population is 10,000, what will the population be 2 years
from now?
second problem:
The population of a midwestern city follows the exponential
law. If
the population decreased from 900,000 to 800,000 from 1993 to
1995,
what will the population be in 1997?
In my text book, it never actually calls any model the
exponential
growth law. IÕm thinking itÕs this model 
though:
A(t)= (initial number)e^kt, where A is original amount/initial
number, k is the growth rate, and t is time.... so for
example, if
A is 500, and k is .05: A(0)= 500e^.05t (when I write ^ I mean
the following numbers are exponents)
I donÕt know how to find the growth rate for 
those problems
when given
the time. IÕd really, greatly appreciate any help anyone can
give me.
===
Subject: Re: Exponential and Logarithmic Functions (word
problem following
the exponential law) This is due tomorrow so I really need
some help
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7DBLE28228;
> Ok this is one problem:
> The population of a southern city follows the exponential
law. If
the
>population doubled in size over an 18-month period and the
current
>population is 10,000, what will the population be 2 years
from now?
A(t) = I*e^(k*t), Eq.(1)
where we let t be expressed in MONTHS.
You know that current population is 10,000. Furthermore,
since the
current population is DOUBLE the population of 18-months ago,
we can
use the exponential growth formula to determine the rate, k,
of
population growth:
10000 = 5000*e^(18*k)
By basic algebra, the above can be written as:
2 = e^(18*k).
We must solve for k, so we take the natural logarithm of both
sides of
the above equation to obtain:
ln[2] = ln[ e^(18*k) ]
= 18*k.
Thus, we have our growth rate:
k = (1/18)*ln[2].
Now that we have our growth rate, Eq.(1) becomes:
A(t) = 10000*e^((1/18)*ln[2]*t).
To determine the population size 2 years from now, we must
first
convert time into months. Observe that
t = 2 years = 24 months.
Now we can plug into our growth rate formula. Thus,
A(24) = 10000*e^((1/18)*ln[2]*24)
= 10000*2^(4/3)
= 25198.4.
Therefore, we have shown that in two years, the population
size will
be roughly 25,200.
Carry on these same ideas to solve #2. Show us how far you
get with
it, using what youÕve just learned.
Joseph A.
===
Subject: Re: Exponential and Logarithmic Functions (word
problem following
the exponential law) This is due tomorrow so I really need
some help
Cc: hurleygurly916@aol.com
> I donÕt know how to find the growth rate for 
those problems
when given
> the time. IÕd really, greatly appreciate any help anyone
can give me.
Do it the way one generally does things. Plug in the known
numbers and solve for the unknowns.
If you have a function thatÕs known to be
of the form
p = Ae^(kt)
and you are told that at t=4, p=100 and at t=10, p=20000, you
know that
100 = Ae^(4k)
and
20000 = Ae^(10k)
What can you do to that pair of equations to
eliminate A and allow you to find k? Since once
you have k, you can immediately find A.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: even and odd functions.. general question..
can a function be both an even and an odd function, at the
same time? I
chose no.. but donÕt know how to back up my answer..
also..
Is it possible for a graph of a function to be symmetric with
respect to
explanations...
===
Subject: Re: even and odd functions.. general question..
Kenny,
The above posts have already answered your question, but you
may be
interested in this related fact.
Consider any function f whose domain is symmetric about 0
(this just
means that if the number p is in the domain, -p must be,
too). Then, we
can write f as a sum of two functions:
E(x)=(1/2)(f(x)+f(-x))
O(x)=(1/2)(f(x)-f(-x))
Then, E is an even function (because
E(-x)=(1/2)(f(-x)+f(x))=E(x)), and
O is an odd function (similarly).
But,
E(x)+O(x)=(1/2)(f(x)+f(-x))+(1/2)(f(x)-f(-x))
E(x)+O(x)=(1/2)f(x)+(1/2)f(-x)+(1/2)f(x)-(1/2)f(-x)=f(x)
So, any function with a symmetric domain in the real numbers
is the sum
of an even function and an odd function, and these two
functions are
uniquely defined.
Travis
> can a function be both an even and an odd function, at the
same time? I
> chose no.. but donÕt know how to back up my answer..
> also..
> Is it possible for a graph of a function to be symmetric
with respect to
> explanations...
===
Subject: Re: even and odd functions.. general question..
>can a function be both an even and an odd function, at the
same time? I
>chose no.. but donÕt know how to back up my answer..
ItÕs always a good idea to go to the 
definition.
An even function E is defined by E(-x) = E(x) for all x in the
domain.
An odd function )O is defined by O(-x) = -O(x) for all x in 
the
domain.
For a function to be both even and odd, you must have
f(-x) = f(x) and f(-x) = -f(x) for the entire domain;
in other words you need
f(x) = -f(x)
for the entire domain.
I may be missing something, but the only one I can see is the
trivial
constant function f(x) = 0.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: even and odd functions.. general question..
> can a function be both an even and an odd function, at the
same time? I
> chose no.. but donÕt know how to back up my answer..
Yes, but only the constant zero function, such that f(x) = 0
for all x.
> also..
> Is it possible for a graph of a function to be symmetric
with respect to
> explanations...
Yes, but only the constant zero function, such that f(x) = 0
for all x.
All even functions are line-symmetric about the y-axis.
All odd functions are point-symmetric about origin.
===
Subject: Re: even and odd functions.. general question..
posting-account=Tt3RvA0AAABur-q9NAlJK5fGGXUbcJWq
So what we have shown is that:
There exists a function f:D->R such that f(x) is even, odd,
and
symmetric about the x-axis for all x in D. Pretty good test
question,
IÕd say!
Joseph A.
>can a function be both an even and an odd function, at the
same
time? I
>chose no.. but donÕt know how to back up my answer..
> Yes, but only the constant zero function, such that f(x) =
0 for all
x.
>also..
>Is it possible for a graph of a function to be symmetric with
respect to
>explanations...
> Yes, but only the constant zero function, such that f(x) =
0 for all
x.
> All even functions are line-symmetric about the y-axis.
> All odd functions are point-symmetric about origin.
===
Subject: Re: even and odd functions.. general question..
Kenny Chunn Wrote:
> can a function be both an even and an odd function, at the
same time? I
> chose no.. but donÕt know how to back up my answer..
Well, letÕs see what happens:
(1) f(x) = f(-x)
(2) f(x) = -f(-x)
(2) implies that f(x) + f(-x) = 0.
substituting (1), we have: f(x) + f(x) = 0, implies f(x) = 0.
So it looks like the zero function has this property.
> also..
> Is it possible for a graph of a function to be symmetric
with respect to
> explanations...
If a function where symmetric with respect to the x axis, the
f(x) = y, and
f(x) = -y.
This is impossible since by defintion a function may only have
one value on
its domain.
Darren
===
Subject: Re: even and odd functions.. general question..
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7EToI02986;
>Kenny Chunn Wrote:
>> can a function be both an even and an odd function, at the
same
time? I
>> chose no.. but donÕt know how to back up my answer..
>Well, letÕs see what happens:
>(1) f(x) = f(-x)
>(2) f(x) = -f(-x)
>(2) implies that f(x) + f(-x) = 0.
>substituting (1), we have: f(x) + f(x) = 0, implies f(x) = 0.
>So it looks like the zero function has this property.
>> also..
>> Is it possible for a graph of a function to be symmetric
with
respect to
>> explanations...
>If a function where symmetric with respect to the x axis,
the f(x) =
y, and
>f(x) = -y.
>This is impossible since by defintion a function may only
have one
value on
>its domain.
Unless of course y= 0 for all x!
>Darren
===
Subject: Re: even and odd functions.. general question..
CouldnÕt the 0 function satisfy two as well? -0 = 0?
> Kenny Chunn Wrote:
>>can a function be both an even and an odd function, at the
same time? I
>>chose no.. but donÕt know how to back up my answer..
> Well, letÕs see what happens:
> (1) f(x) = f(-x)
> (2) f(x) = -f(-x)
> (2) implies that f(x) + f(-x) = 0.
> substituting (1), we have: f(x) + f(x) = 0, implies f(x) =
0.
> So it looks like the zero function has this property.
>>also..
>>Is it possible for a graph of a function to be symmetric
with respect to
>>explanations...
> If a function where symmetric with respect to the x axis,
the f(x) = y,
and
> f(x) = -y.
> This is impossible since by defintion a function may only
have one value
on
> its domain.
> Darren
===
Subject: Re: even and odd functions.. general question..
Yes. You are right!
> CouldnÕt the 0 function satisfy two as well? -0 = 0?
>Kenny Chunn Wrote:
>>can a function be both an even and an odd function, at the
same time?
I
>>chose no.. but donÕt know how to back up my answer..
>Well, letÕs see what happens:
>(1) f(x) = f(-x)
>(2) f(x) = -f(-x)
>(2) implies that f(x) + f(-x) = 0.
>substituting (1), we have: f(x) + f(x) = 0, implies f(x) = 0.
>So it looks like the zero function has this property.
>>also..
>>Is it possible for a graph of a function to be symmetric
with respect
to
>>explanations...
>If a function where symmetric with respect to the x axis,
the f(x) = y,
and
>f(x) = -y.
>This is impossible since by defintion a function may only
have one
value
on
>its domain.
>Darren
===
Subject: Why do two negatives equal a positive?
Why do two negatives equal a positive? How is it proven that
two negatives
equal a positive when multiplied together? I have always
known that a
negative times a negative equals a positive, but it has never
made sense to
me. Will someone please make sense of it?
===
Subject: Re: Why do two negatives equal a positive?
ThereÕs the old joke about an English professor telling the
class how
two negatives make a positive, but that two positives donÕt
make a
negative.
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
--
Oppie the Bear
aka TOJ (The Other John)
ÔRemoveMYWORRIES to email me!
When I die, I want to die like my uncle died;
peacefully sleeping like a baby!
Not screaming in fear and terror
like the passengers in his car.
===
Subject: Re: Why do two negatives equal a positive?
>Why do two negatives equal a positive? How is it proven that
two negatives
>equal a positive when multiplied together? I have always
known that a
>negative times a negative equals a positive, but it has
never made sense to
>me. Will someone please make sense of it?
Perhaps this will?
Google: two negatives positive .....8210 hits.
I was amazed that so many answered so quickly. Was my
question a common
question asked in math classes?
> Why do two negatives equal a positive? How is it proven
that two
> negatives equal a positive when multiplied together? I have
always known
> that a negative times a negative equals a positive, but it
has never made
> sense to me. Will someone please make sense of it?
> I was amazed that so many answered so quickly. Was my
question a common
> question asked in math classes?
ItÕs pretty common in some of my classes.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Why do two negatives equal a positive?
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
HI
In the real number system the following properties hold
1 Addition is associative
2 There is an additive identity
3 Every element has an additive inverse
4 Multiplication distributes over addition from the left and
right
(-a)(-b) = (-a)(-b) + 0 by 2
= (-a)(-b) + 0*b No matter how many times you add 0
the result is 0
= (-a)(-b) + (-a+a)*b by 3
= (-a)(-b) + ((-a)b + ab) by 4
= ((-a)(-b) + (-a)b) +ab by 1
= (-a)(-b + b) +ab by 4
= (-a)(0) +ab by 2
= ab
This may not make the kind of sense you want it to, however
if it
werenÕt true one of the conditions 1 to 4 would have to be
false and
that would make no sense.
===
Subject: Re: Why do two negatives equal a positive?
Well, itÕs certainly NOT UNtrue that two negatives make a
positive. I
would NOT say that this is NOT the case.
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
--
Oppie the Bear
aka TOJ (The Other John)
ÔRemoveMYWORRIES to email me!
When I die, I want to die like my uncle died;
peacefully sleeping like a baby!
Not screaming in fear and terror
like the passengers in his car.
===
Subject: Re: Why do two negatives equal a positive?
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
How about when you multiply a positive by a negative? Should
that be
negative? Then if two negatives didnÕt give a positive, 
youÕd
be stuck
in negative territory forever.
Of course that _is_ what happens when you multiply two
positives. It
doesnÕt seem quite fair...
--
john
===
Subject: Re: Why do two negatives equal a positive?
Picture the complex plane. Multiplying a number by -1 (= i^2)
rotates the
number 180 degrees counterclockwise, so multiplying by -1
twice .... . HTH.
Robert
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
===
Subject: Re: Why do two negatives equal a positive?
>Why do two negatives equal a positive? How is it proven that
two negatives
>equal a positive when multiplied together? I have always
known that a
>negative times a negative equals a positive, but it has
never made sense to
>me. Will someone please make sense of it?
How about this:
I am going to the store.
Means I *am* going to the store.
I am not going to the store.
Means I am *not* going to the store.
I am not not going to the store.
Hmm. That means I *am* going to the store since IÕm not 
*not*
going.
HTH.
CB
===
Subject: Re: Why do two negatives equal a positive?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7LduF11820;
>>Why do two negatives equal a positive? How is it proven
that two
negatives
>>equal a positive when multiplied together? I have always
known that
a
>>negative times a negative equals a positive, but it has
never made
sense to
>>me. Will someone please make sense of it?
>How about this:
> I am going to the store.
> Means I *am* going to the store.
> I am not going to the store.
> Means I am *not* going to the store.
> I am not not going to the store.
> Hmm. That means I *am* going to the store since IÕm not
*not*
going.
>HTH.
I assume that you are going to the store to get some bananas.
Could you get a gallon of milk while you are there?
- MO
===
Subject: Re: Why do two negatives equal a positive?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7Eta805872;
>Why do two negatives equal a positive? How is it proven that
two
negatives
>equal a positive when multiplied together? I have always
known that
a
>negative times a negative equals a positive, but it has
never made
sense to
>me. Will someone please make sense of it?
Consider the following sentences.
I have some banana. (same as I have N bananas, N > 0.)
I donÕt have any bananas. (same as I have zero bananas.)
I have no bananas. (same as I have zero bananas.)
I donÕt have no bananas. (same as I have some bananas.)
In the last sentence two negatives (donÕt and no) makes a
positive
(some).
Caution: The demonstration above should _not_ be construed to
mean
that two wrongs make a right. ;)
- MO
===
Subject: Re: Why do two negatives equal a positive?
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
First of all, the product of two negatives equals a positive,
but the
sum of two negatives is a negative. The operation matters.
As for why, consider the following:
2*3 = 6
2*2 = 4
2*1 = 2
2*0 = 0 Notice, dropping what 2 is multiplied by by 1 drops
the product
by 2, so...
2*-1 = -2
2*-2 = -4
2*-3 = -6 Justifying that a positive times negative is
negative.
Now:
-2*3 = -6
-2*2 = -4
-2*1 = -2
-2*0 = 0 Again, notice the pattern: drop by 1, increase
product by 2,
so...
-2*-1 = 2
-2*-2 = 4
-2*-3 = 6
ItÕs not a proof, but it also seems to be clearer than a
proof to most
people.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Why do two negatives equal a positive?
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
Do you want
-1 * X
to ever be anything apart from
0 - X
?
Phil
--
God was my co-pilot but we crashed in the mountains and I had
to eat him.
===
Subject: Re: Why do two negatives equal a positive?
> Why do two negatives equal a positive? How is it proven
that two
negatives
> equal a positive when multiplied together? I have always
known that a
> negative times a negative equals a positive, but it has
never made sense
to
> me. Will someone please make sense of it?
ItÕs like taking a video of someone walking backward, then
playing the
tape backward. YouÕll see them walk forward.
Mathematically it follows from the distributive law, as
others will post.
===
Subject: Can anyone help me tackle these problems? any help
is appreciated
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7DBLB28238;
A mountain has the equation
z= f(x,y)= H -(aplha)((x^2)+(mu)(y^2))
where H >0, alpha > 0, and mu(as in the greek letter) >1.
Starting at
the point (x-sub 0, y-sub 0, 0) at the mountain base, a
climber wants
to ascend the mountain by means of the steepest ascent curve.
1) Identify the level curves of the surface z=f(x,y) and
sketch a few
members of the family.
2) Calculate the gradient vector for the surface at an
arbitrary point
(x,y) in the domain of f and determine the differential eq.
for the
steepest ascewnt curve.
3) Solve the diff. euation in part 2 using the initial
condition
y(x-sub 0)= y- sub 0
4) Find parametric equations x= x(t), y= y(t) = z(t) for the
steepest
ascent curve on the mountain surface.
===
Subject: Re: Can anyone help me tackle these problems? any
help is
appreciated
>A mountain has the equation
>z= f(x,y)= H -(aplha)((x^2)+(mu)(y^2))
>where H >0, alpha > 0, and mu(as in the greek letter) >1.
Starting at
>the point (x-sub 0, y-sub 0, 0) at the mountain base, a
climber wants
>to ascend the mountain by means of the steepest ascent curve.
>1) Identify the level curves of the surface z=f(x,y) and
sketch a few
>members of the family.
Set f(x,y) = c. Because of the squared terms, certain values
of c
wonÕt work, but otherwise you should recognize the equation
of a
certain type of conic.
>2) Calculate the gradient vector for the surface at an
arbitrary point
>(x,y) in the domain of f and determine the differential eq.
for the
>steepest ascewnt curve.
Of course you know that grad f = < f_x, f_y> which you can
compute.
>3) Solve the diff. euation in part 2 using the initial
condition
>y(x-sub 0)= y- sub 0
see 4)
>4) Find parametric equations x= x(t), y= y(t) = z(t) for the
steepest
>ascent curve on the mountain surface.
First, you have a typo. If you get the parametric equations
for x and
y, you get z(t) = f(x(t), y(y)), not y(t) = z(t).
If R(t) = < x(t), y(t) > is the parametric equation of the
curve, then
it would be tangent to grad f. So xÕ(t) = f_x, 
yÕ(t) = f_y.
You can
use these to get dy/dx to use in 4) and you can solve them
directly to
get x(t) and y(t).
That should get you going.
--Lynn
===
Subject: ill love you forever if... STATISTICS QUESTION
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7DBJb28134;
question:
If Joey is signing up for classes and he has to take history
and
english... both are offered at 8am, 9am and 3 pm, If joey
randomly
picks classes what is the probablity that Joey gets history
at 3pm and
english at 8am?
===
Subject: Re: ill love you forever if... STATISTICS QUESTION
> question:
> If Joey is signing up for classes and he has to take
history and
> english... both are offered at 8am, 9am and 3 pm, If joey
randomly
> picks classes what is the probablity that Joey gets history
at 3pm and
> english at 8am?
With this particular question, because there are only three
time slots and
only two subjects, you can list all of the possibilities.
They are all
equally likely, and you are only interested in one of them,
so the
probability is one divided by the number of possibilities.
It would be more interesting to extend this to the general
case in which
there are m subjects, and n time slots.
===
Subject: Re: ill love you forever if... STATISTICS QUESTION
>> question:
>> If Joey is signing up for classes and he has to take
history and
>> english... both are offered at 8am, 9am and 3 pm, If joey
randomly
>> picks classes what is the probablity that Joey gets
history at 3pm and
>> english at 8am?
>With this particular question, because there are only three
time slots and
>only two subjects, you can list all of the possibilities.
They are all
>equally likely, and you are only interested in one of them,
so the
>probability is one divided by the number of possibilities.
>It would be more interesting to extend this to the general
case in which
>there are m subjects, and n time slots.
Say Joey used a spinner with outcomes 8, 9, and 3, and spins
it twice
to get the meeting times of his history and english requests.
Does
your sample space include (8,8), (9,9), and (3,3)? Or are you
assuming
he is not allowed those choices, or is smart enough to reject
them? It
matters.
--Lynn
===
Subject: Re: Why do two negatives equal a positive?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7DBOu28347;
By using the field axioms
(http://mathworld.wolfram.com/FieldAxioms.html)
first you can show (-1)(-1) = 1:
(-1)(-1) = (-1)(-1) + 0
= (-1)(-1) + (-1) + 1
= (-1)[(-1) + 1] + 1
= (-1) * 0 + 1
= 0 + 1 = 1
and that (-1)a = (-a):
(-1)a = (-1)a + 0
= (-1)a + a + (-a)
= a[(-1) + 1] + (-a)
= a * 0 + (-a)
= 0 + (-a) = (-a)
So (-a)(-b) = (-1)a * (-1)b
= (-1)(-1)ab
= 1 * ab = ab.
===
Subject: Re: Why do two negatives equal a positive?
>first you can show (-1)(-1) = 1:
>(-1)(-1) = (-1)(-1) + 0
> = (-1)(-1) + (-1) + 1
> = (-1)[(-1) + 1] + 1
> = (-1) * 0 + 1
> = 0 + 1 = 1
>and that (-1)a = (-a):
>(-1)a = (-1)a + 0
> = (-1)a + a + (-a)
> = a[(-1) + 1] + (-a)
> = a * 0 + (-a)
> = 0 + (-a) = (-a)
>So (-a)(-b) = (-1)a * (-1)b
> = (-1)(-1)ab
> = 1 * ab = ab.
===
Subject: Re: Why do two negatives equal a positive?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7Etaj05857;
well we have many smart mathematicians in this forum but I
bet that
they are getting you more confused and stressesed.
let me a 17 year old to tell you the real reason.
well it all dates when words meant more than numbers. to tell
you the
thruth I dont know who came up with that theory but it seems
to be
working pretty good.
well say that you have -1*-1= 1 right?
then we will rewrite the state ment
-1=1/-1 correct?
this means that -1=-1 correct?
well if we divide once again youÕll have
-1/-1 correct?
correct?
leaving with the answer 1/1( which is the negatives)*1/1( the
real
numbers)
leaving you with 1*1 which then you can assume that it surely
is one.
===
Subject: Re: Why do two negatives equal a positive?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBA3glo09952;
>well we have many smart mathematicians in this forum but I
bet that
>they are getting you more confused and stressesed.
>let me a 17 year old to tell you the real reason.
Go on...
>well it all dates when words meant more than numbers. to
tell you the
>thruth I dont know who came up with that theory but it seems
to be
>working pretty good.
>well say that you have -1*-1= 1 right?
Indeed, we must prove its truth using elementary mathematics.
This is
sort of a backwards derivation, where we begin with the
conjecture and
finally conclude with axiomatic truth. Now then, 
letÕs make the
statement more general. That is, for nonzero numbers a and b,
(-a)*(-b) = (a*b). (#)
>then we will rewrite the state ment
>-1=1/-1 correct?
Sure. Thus, we have
(-a) = (a*b)/(-a).
>this means that -1=-1 correct?
Well, wait a second. We have (-a) = (a*b)/(-a).
= (a*b)/(-1*a)
= (b)/(-1)
= (b)*(1/(-1))
= (b)*(1*(1/(-1)))
= (b)*(1*(1*(1/(-1))))
= (b)*(1*(1*(1*...*(1/(-1)...)))).
But how do we know
(b)*(1*(1*(1*...*(1/(-1)...)))) = (-b) ??
Remember, we must deduce truth WITHOUT using our conjecture,
otherwise
we are victim to circular logic.
Statement (#) can be even more generalized to
-(-a) = a.
The way we prove is by knowing that since -(-a) is a unique
number, it
has an additive inverse (-a). Thus,
-(-a) + (-a) = 0. (#2)
Here we impose the Field Axiomatic definition of difference.
That is,
we define the difference of two real numbers x and y, denoted
by x -
y, by
x - y = x + (-y).
Thus, eq.(#2) becomes:
-(-a) - a = 0.
Furthermore, by adding (a) to both sides of the equation we
obtain:
-(-a) = a.
Thus we have our groundwork for why a negative number
multiplied by a
negative number is positive.
Joseph A.
===
Subject: factors??
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7G5GV13273;
IÕm trying to figure this problem out and having 
trouble, can
anyone
help?
1. The factors of x4(power)-625 are?
my answer was x-5
===
Subject: Re: factors??
>IÕm trying to figure this problem out and 
having trouble, can
anyone
>help?
>1. The factors of x4(power)-625 are?
>my answer was x-5
ThatÕs _one_ of the factors.
Hint: how do you factor a^4-b^4? If you said (a-b)^4, you
gave the
same wrong answer as most students.
You need to know that p^2 - q^2 = (p+q)(p-q). (For some
reason, few
students seem to know this and fewer still can apply it when
needed.)
Look at a^4-b^4 as (a^2)^2 - (b^2)^2 and you see that it
factors as
(a^2+b^2)(a^2-b^2). Then of course you have to factor
a^2-b^2. (You
CANÕT factor a^2+b^2 over the reals, so donÕt 
try.)
Now apply that hint to your original problem.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: factors??
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7HZ1Q22112;
>IÕm trying to figure this problem out and 
having trouble, can
anyone
>help?
>1. The factors of x4(power)-625 are?
>my answer was x-5
This is only one factor, there are other factors:
x^4 - 625 = x^4 - 25^2 = x^4 - 5^4
x^4 - 5^4 = (x^2 - 5^2)*(x^2 + 5^5) =
= (x - 5)*(x + 5)*(x^2 + 5^2)
or including complex numbers
(x^2 + 5^2) = x^2 - 5^2*(-1) = x^2 - (5i)^2 = (x - 5i)*(x +
5i)
x^4 - 5^4 = (x - 5)*(x + 5)*(x - 5i)*(x + 5i)
===
Subject: Re: factors??
> IÕm trying to figure this problem out and 
having trouble,
can anyone
> help?
> 1. The factors of x4(power)-625 are?
> my answer was x-5
ThatÕs *one* of the factors of x^4 - 625.
Notice that x^4=(x^2)^2 and that 625=25^2
That should give you a leg up on further factoring x^4 - 625.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Anybody check these answers please?
I have a couple of homework questions, appreciate if someone
can take the
time to check my answers.
Q1. Given that voltages v1 = 12 sin 100 pi t and v2 = 20 sin
( 100 pi t +
(pi / 3) ). State the minimum value and the phase angle
(relative to v1) of
the resultant voltage v1 + v2 by writing the sum as a single
sinusoid.
My answer 28 sin ( 100 pi t + 0.667 )
Worked it out as follows:
sin(A+B) = sinAcosB+cosAsinB
to make it easier to write out I made 100pi = w
20sin(wt + (pi/3))
20 sin wt cos (pi/3) + 20 cos wt sin (pi/3)
= 20 sin wt * 0.5 + 20 cos wt * 0.866
= 10 sin wt + 18.32 cos wt
=> 12 sin wt + 10 sin wt + 18.32 cos wt
= 22 sin wt + 17.32 cos wt
I then put into R = wt + a (where a =alpha)
R^2 = SQRT 22^2 + 17.32^2
R = 28
tan a = 17.32/22
a = 0.667 radians
therefore v1 + v2 = 28 sin (wt + 0.667)
substituting 100pi back
v1 + v2 = 28 sin (100 pi t + 0.667)
Q2 In an ac circuit i = 100 sin 20 pi t amperes and v = 50
sin (20 pi t
-
( pi / 6 )) volts
Instantaneous power p, is given by p = vi watts
Use products-to-sums formulae to express p in a form
involving only one
sinusoid and hence state the maximum power.
My answer 2165 - 2500 cos ( 20 pi t - (pi / 6) )
Worked this one out as follows:
w = 20 pi
50 sin (wt - (pi/6)) * 100 sin wt
= 5000 sin (wt - (pi/6)) sin wt
sinA sinB = 0.5[cos (A - B) - cos (A + B)]
=> 5000 * 0.5[ cos (wt - (pi/6) - wt) - cos (wt - (pi/6) +
wt)]
= 2500[cos ( -pi/6) - cos (2wt - (pi/6))]
= 2500[0.866 - cos (2wt - (pi/6))]
= 2165 - 2500 cos (2wt - (pi/6))
Substituting 20 pi back for w
2165 - 2500 cos (20 pi t - (pi/6))
TIA
===
Subject: Re: Linear Algebra Problem, please help!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB7JNoo32219;
>Need to show that if ker A^r = ker A^(r+1) for some r, then
ker
>A^(r+1) = ker A^(r+2)
Just multiply both sides by A....I know that sounds stupid
and easy
but just take something in the kernel of A^r (also in kernel
of A^r+1)
and then consider what happens when you apply A again.
===
Subject: Re: Linear Algebra Problem, please help!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5ur25093;
Suppose that Ker(A^r)=Ker(A^[r+1]). The nontrivial part of
that
statement is that for any vector y, if A^[r+1](y)=0 then in
fact we
already have A^r(y)=0.
Now suppose that v is a vector for which A^[r+2](v)=0, then
A^[r+1]{A(v)}=0 and so from above A^r{A(v)}=0. But then
A^[r+1](v)=0
and so from above A^r(v)=0. That is, v is in Ker(A^r). In the
same
way we can extend this by induction to get
Ker(A^r)=Ker(A^[r+k]) for
all k>0.
-------------------------------------------------------------
---
>>Need to show that if ker A^r = ker A^(r+1) for some r, then
ker
>>A^(r+1) = ker A^(r+2)
>Just multiply both sides by A....I know that sounds stupid
and easy
>but just take something in the kernel of A^r (also in kernel
of
A^r+1)
>and then consider what happens when you apply A again.
===
Subject: question about a limit
IÕm trying to determine
lim f(n)
n->inf
where f(n)=(2n-1)!! / (2n)!!
ItÕs a strictly monotone (decreasing) and has 0 as a lower
bound. But what
is the limit?
I think it is zero (some quick and dirty numerical tests, and
mupad tells
me
so), but I canÕt prove it. Can anyone give me a hint?
===
Subject: Re: question about a limit
> IÕm trying to determine
> lim f(n)
> n->inf
> where f(n)=(2n-1)!! / (2n)!!
> ItÕs a strictly monotone (decreasing) and has 0 as a lower
bound. But
what
> is the limit?
> I think it is zero (some quick and dirty numerical tests,
and mupad tells
me
> so), but I canÕt prove it. Can anyone give me a hint?
f(n) = (1 - 1/(2n))*(1 - 1/(2n-2))*...*(3/4)*(1/2). So
log(f(n)) = sum (k=0,n-1) log[1 - 1/(2n-2k)].
Now use the fact that log(1+x) <= x for x > -1 (note y = x is
the tangent
line to y = log(1+x) at (0,0) and log is concave down), and
youÕll get what
you need (with apologies to the Rolling Stones).
===
Subject: Re: question about a limit
>IÕm trying to determine
>lim f(n)
>n->inf
>where f(n)=(2n-1)!! / (2n)!!
Are those really double factorials? If that is the case,
let u = (2n - 1)!, so we get
f(n) = u! / ( 2n * u )!
See if it helps to evaluate the limit in this construct.
Brian
>ItÕs a strictly monotone (decreasing) and has 0 as a lower
bound. But what
>is the limit?
>I think it is zero (some quick and dirty numerical tests,
and mupad tells
me
>so), but I canÕt prove it. Can anyone give me a hint?
===
Subject: Re: question about a limit
alt.math.undergrad:
>>IÕm trying to determine
>>lim f(n)
>>n->inf
>>where f(n)=(2n-1)!! / (2n)!!
> Are those really double factorials? If that is the case,
> let u = (2n - 1)!, so we get
> f(n) = u! / ( 2n * u )!
No, n!! = n(n-2)(n-4)...; e.g., 5!! = 5*3*1 = 15, and 6!! =
6*4*2 = 48. A more rigorous definition:
(-1)!! = 0!! = 1
n!! = n * (n - 2)!! for n > 0.
For Gunnar: (2n)! = (2n - 1)!! * 2^n * n! : just rearrange
the factors. Also, (2n)!! = 2^n * n! . Thus,
f(n) = (2n)!/(2^n * n!)^2 = C(2n, n)/4^n,
where C(p, q) is the binomial coefficient Ôp 
choose qÕ.
C_n, the n-th Catalan number, is C(2n, n)/(n+1), so f(n) =
(n+1)*C_n / 4^n. Asymptotically C_n ~
4^n / [sqrt(pi)*n^(3/2)] -- if memory serves, you can get
this easily from StirlingÕs approximation -- so f(n) does
indeed tend to 0.
Brian
===
Subject: Re: question about a limit
>>where f(n)=(2n-1)!! / (2n)!!
> Are those really double factorials? If that is the case,
Nope, semi-factorial
n!!=n(n-2)(n-4)...(2 if n even,1 if n odd)
8!!=8*6*4*2
7!!=7*5*3*1
===
Subject: Regression- probabilistic model
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB81wHJ01672;
How do i set up a probablistic model? Do i plug #Õs in or is
it just
abstract? i know hte model is Y=Bo+B1x1+b2x2+E, is that
simply what a
probabilistic model is? Im still confused as to what those
symbols
represent.
===
Subject: Help! Linear Algebra problem
Question: Let A be the 4x4 matrix :
0 0 0 0
0 1 sqrt(2) 0
0 sqrt(2) 1 0
-1 0 0 -1
Find an orthogonal mx U that so U(inverse)AU is a diagonal
matrix.
What IÕve done so far:
A-(lambda)I
, shufße columns to get
1-(lambda) sqrt(2) 0 0
sqrt(2) 1-(lambda) 0 0
0 0 -1 -1-x
0 0 -(lambda) -1
so ((1-lambda)^2-2)(lambda(1+lambda)-1)=0
so lambda = 1 + sqrt(2), 1-sqrt(2), (-1+sqrt(5))/2 and
(-1-sqrt
(5))/2
the rank of each is 3, I set the free variable to one... now
how
do I find the four independent eigenvalues? I 
canÕt seem to
----------------------------------------------
* Binary Usenet Leeching Made Easy
* http://www.newsleecher.com/?usenet
----------------------------------------------
===
Subject: Help! Linear Algebra problem(ignore first post)
Sorry, the first post had an incorrect matrix, I missed the 
-1.
Question: Let A be the 4x4 matrix :
0 0 0 -1
0 1 sqrt(2) 0
0 sqrt(2) 1 0
-1 0 0 -1
Find an orthogonal mx U that so U(inverse)AU is a diagonal
matrix.
What IÕve done so far:
A-(lambda)I
, shufße columns to get
1-(lambda) sqrt(2) 0 0
sqrt(2) 1-(lambda) 0 0
0 0 -1 -1-x
0 0 -(lambda) -1
so ((1-lambda)^2-2)(lambda(1+lambda)-1)=0
so lambda = 1 + sqrt(2), 1-sqrt(2), (-1+sqrt(5))/2 and
(-1-sqrt
(5))/2
the rank of each is 3, I set the free variable to one... now
how
do I find the four independent eigenvalues? I 
canÕt seem to
----------------------------------------------
* Binary Usenet Leeching Made Easy
* http://www.newsleecher.com/?usenet
----------------------------------------------
===
Subject: Re: Help! Linear Algebra problem(ignore first post)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB914lo31802;
>Sorry, the first post had an incorrect matrix, I missed the
-1.
>Question: Let A be the 4x4 matrix :
>0 0 0 -1
>0 1 sqrt(2) 0
>0 sqrt(2) 1 0
>-1 0 0 -1
>Find an orthogonal mx U that so U(inverse)AU is a diagonal
>matrix.
>What IÕve done so far:
>A-(lambda)I
>, shufße columns to get
>1-(lambda) sqrt(2) 0 0
>sqrt(2) 1-(lambda) 0 0
>0 0 -1 -1-x
>0 0 -(lambda) -1
>so ((1-lambda)^2-2)(lambda(1+lambda)-1)=0
>so lambda = 1 + sqrt(2), 1-sqrt(2), (-1+sqrt(5))/2 and
(-1-sqrt
>(5))/2
>the rank of each is 3, I set the free variable to one... now
how
>do I find the four independent eigenvalues? I 
canÕt seem to
| 0-k 0 0 -1 |
det(A - kI) = | 0 1-k sqrt(2) 0 | = 0
| 0 sqrt(2) 1-k 0 |
| -1 0 0 -1-k|
| 1-k sqrt(2) 0 |
det(A - kI) = (-k)*det|sqrt(2) 1-k 0 | -
| 0 0 -1-k|
| 0 1-k sqrt(2)|
- (-1)*det| 0 sqrt(2) 1-k | =
|-1 0 0 |
= k*[(1 - k)^2*(1 + k) - 2k*(1 + k)] - (1 - k)^2 + 2 =
= [(1 - k)^2 - 2]*[k*(1 + k) - 1] =
= (k^2 - 2k - 1)*(k^2 + k - 1) = 0
k = 1 +- sqrt(2), [-1 +-sqrt(5)]/2
~ 2.414, -0.414, 0.618, -1.618
You got the eigenvalues right, even though I do not see any
purpose in
your shufßing. Since A^T = A, A is symmetrical, all
eigenvalues are
real and the eigenvectors of different eigenvalues are
orthogonal. The
matrix U diagonalizing A (such that U^(-1)AU is diagonal) has
eigenvectors as columns and it is automatically orthogonal.
To get U,
you have to find the 4 eigenvectors
Xj = (u1j, u2j, u3j, u4j) corresponding to kj, j = 1, 2, 3, 4.
|a22-kj a23 a24 |
u1j = det(A - kI)11 = det| a32 a33-kj a34 | =
| a42 a43 a44-kj|
| 1-kj sqrt(2) 0 |
= det|sqrt(2) 1-kj 0 |
| 0 0 -1-kj |
| a11-kj a13 a14 |
u2j = det(A - kI)22 = det| a31 a33-kj a34 | =
| a41 a43 a44-kj|
| 0-kj 0 -1 |
= det| 0 1-kj 0 |
| -1 0 -1-kj |
|a11-kj a12 a14 |
u3j = det(A - kI)33 = det| a21 a22-kj a24| =
| a41 a42 a44-kj|
| 0-kj 0 -1 |
= det| 0 1-kj 0 |
| -1 0 -1-kj |
|a11-kj a12 a13 |
u4j = det(A - kI)44 = det| a21 a22-kj a23 | =
| a31 a32 a33-kj|
| 0-kj 0 0 |
= det| 0 1-kj sqrt(2)|
| 0 sqrt(2) 1-kj |
Eigenvector X1 for the eigenvalue k1 = 1 + sqrt(2):
| -sqrt(2) sqrt(2) 0 |
u11 = det| sqrt(2) -sqrt(2) 0 | = 0
| 0 0 -2-sqrt(2)|
|-1-sqrt(2) 0 -1 |
u21 = det| 0 -sqrt(2) 0 |
| -1 0 -2-sqrt(2)|
= -(1 + sqrt(2)*sqrt(2)*(2 + sqrt(2)) + sqrt(2) =
= -(2 + sqrt(2)^2 + sqrt(2) = -(6 + 3*sqrt(2))
|-1-sqrt(2) 0 -1 |
u31 = det| 0 -sqrt(2) 0 |
| -1 0 -2-sqrt(2)|
= -(1 + sqrt(2)*sqrt(2)*(2 + sqrt(2)) + sqrt(2)
= -(2 + sqrt(2)^2 + sqrt(2) = -(6 + 3*sqrt(2))
|-1-sqrt(2) 0 0 |
u41 = det| 0 -sqrt(2) sqrt(2)| = 0
| 0 sqrt(2) -sqrt(2)|
X1 ~ [0, -(6 + 3*sqrt(2)), -(6 + 3*sqrt(2)), 0] ~ [0, 1, 1,
0] should
be an eigenvector of A corresponding to the eigenvalue
k1 = 1 + sqrt(2). Check:
| 0 0 0 -1 | |0| | 0 |
A*X1 = | 0 1 sqrt(2) 0 |*|1| = |1 + sqrt(2)|
| 0 sqrt(2) 1 0 | |1| |sqrt(2) + 1|
| -1 0 0 -1 | |0| | 0 |
In a similar way, by plugging the remaining eigenvalues into
the above
equations, you can get the remaining eigenvectors:
X2 = [0, 1, -1, 0] corresponding to k2 = 1 - sqrt(2)
X3 = [1, 0, 0, -k3] corresponding to k3 = (-1 + sqrt(5))/2 > 0
X4 = [1, 0, 0, -k4] corresponding to k4 = (-1 - sqrt(5))/2 < 0
The matrix U is then
| 0 0 1 1 |
U = | 1 1 0 0 |
| 1 -1 0 0 |
| 0 0 (1 - sqrt(5))/2 (1 + sqrt(5))/2|
its inverse matrix is
| 0 1/2 1/2 0 |
U^(-1) = | 0 1/2 -1/2 0 |
|(5 + sqrt(5))/10 0 0 sqrt(5)/5 |
|(5 - sqrt(5))/10 0 0 sqrt(5)/5 |
and the matrix A is diagonalized as
|1+sqrt(2) 0 0 0 |
U^(-1)*A-*U = | 0 1-sqrt(2) 0 0 |
| 0 0 (-1+sqrt(5))/2 0 |
| 0 0 0 (-1-sqrt(5))/2|
===
Subject: Re: question on continuity of function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5ru24981;
My name is Aynalem Tekel. IÕm 19 years old. I live in
Ethiopia. IÕm
student at college in the 3rd year student. My study is
Secretarial
Science and Office Management... and in this study I educated
Maths
field and Please IÕm interested maths study by 
enternet...
please
write the answere..
Ayne
===
Subject: f(x) is integrable... is |f(x)| also? HELP!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5sZ24992;
im trying to prove or disprove that if f:[a,b]->R is
integrable, then
|f|:[a,b]->R is also integrable.
ive began the approach the proof as follows:
using the integrability criterion, we know that since f(x) in
integrable on [a,b], then for epsilon > 0 there exists some
partition
P such that
upper darboux sums - lower darboux sums < epsilon.
from this i do not know how to bring |f| into the picture.
HELP!!! i
have a test tomorrow and i have a feeling this will be on it!
jake
===
Subject: Re: f(x) is integrable... is |f(x)| also? HELP!!!
> im trying to prove or disprove that if f:[a,b]->R is
integrable, then
> |f|:[a,b]->R is also integrable.
> ive began the approach the proof as follows:
> using the integrability criterion, we know that since f(x)
in
> integrable on [a,b], then for epsilon > 0 there exists some
partition
> P such that
> upper darboux sums - lower darboux sums < epsilon.
> from this i do not know how to bring |f| into the picture.
Hint: ||x| - |y|| <= |x - y|.
> HELP!!! i
> have a test tomorrow and i have a feeling this will be on
it!
And we should care about your exam and the fact that youÕve
put off
studying for it why exactly?
===
Subject: Re: f(x) is integrable... is |f(x)| also? HELP!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB914mx31819;
>> im trying to prove or disprove that if f:[a,b]->R is
integrable,
then
>> |f|:[a,b]->R is also integrable.
>> ive began the approach the proof as follows:
>> using the integrability criterion, we know that since f(x)
in
>> integrable on [a,b], then for epsilon > 0 there exists some
partition
>> P such that
>> upper darboux sums - lower darboux sums < epsilon.
>> from this i do not know how to bring |f| into the picture.
>Hint: ||x| - |y|| <= |x - y|.
>> HELP!!! i
>> have a test tomorrow and i have a feeling this will be on
it!
>And we should care about your exam and the fact that youÕve
put off
>studying for it why exactly?
Who said anything about putting it off. I expect to earn the
highest
grade in course, as I always do. Some last minute help never
hurts...
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5u425106;
is n^3-n divisible by 24 the same proof
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBAFc6q07043;
>is n^3-n divisible by 24 the same proof
For n = 2, n^3 - n = 6, which is not divisible by 24.
For n = 4, n^3 - n = 60, which is not divisible by 24.
for n = 6, n^3 - n = 210, which is not divisible by 24.
for n = 10, n^3 - n = 990, which is not divisible by 24.
for n = 12, n^3 - n = 1716, which is not divisible by 24.
for n = 14, n^3 - n = 2730, which is not divisible by 24.
for n = 18, n^3 - n = 5814, which is not divisible by 24.
...
In fact, for any n, where n-1 and n+1 are twin primes,
n^3 - n = (n-1)n(n+1) cannot be divisible by 24.
Hey, is this a proof that the number surrounded by
twin primes is always divisible by 3?
- MO
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
> is n^3-n divisible by 24 the same proof
It does not require induction, only factoring and common
sense.
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECrg01154;
>> is n^3-n divisible by 24 the same proof
>It does not require induction, only factoring and common
sense.
Huh? 4^3 - 4 = 60, 6^3 - 6 = 210, etc.
Is true that 8 and 3 (hence 24) divide (2n + 1)^3 - (2n + 1).
Think modular.
Todd Trimble
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECqk01107;
>> is n^3-n divisible by 24 the same proof
>It does not require induction, only factoring and common
sense.
I know it does not require induction, but the OP asked where
did he go
wrong in an induction proof. BTW, I did not notice the date
of the OP.
Still, I have a problem to believe that n^3 - n is divisible
by 24
(haha, in the ring of integers).
===
Subject: Re: Induction: How to show n^3 - n is divisible by 24
>is n^3-n divisible by 24 the same proof
> It does not require induction, only factoring and common
sense.
Am I missing something? n = 4? n = 10? etc., etc.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Financial Equation: equation for this problem.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5tN25074;
the importance of equation to financial manager (5pages)
===
Subject: Mechanics - friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5tj25062;
I wonder if this is the right Forum, but here it goes anyway:
I am having a problem understanding the way some problems are
solved.
I can solve them but it is the basics that I donÕt 
understand.
Take the following problem:
It is being pulled up by a force of 500N at 10.bc upwards to
the
parallel of the plane. Find the acceleration when 
Ômewis
0.4.
Ok so the component down the plane is 245sin20 and the
perpendicular
is 245cos20.
Now when I look at the force being applied at 10.bc to the
plane, I
assume (wrongly!) that the component of the force parallel to
the
plane is 500sin10, but it turns out that this gives me the
force
perpendicular to the plane.
So why should 245sin20 act down the plane but to find the
force acting
up the plane we use cos rather than sin???
I hope somebody can give me a simple explanation that my
feeble brain
can cope qith!
===
Subject: Re: Mechanics - friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBAD3XQ25054;
>I wonder if this is the right Forum, but here it goes anyway:
>I am having a problem understanding the way some problems are
>solved. I can solve them but it is the basics that I donÕt
>understand.
>Take the following problem:
>It is being pulled up by a force of 500 N at 10.bc upwards
to the
>parallel of the plane. Find the acceleration when 
Ômewis
0.4.
>Ok so the component down the plane is 245sin20 and the
perpendicular
>is 245cos20.
>Now when I look at the force being applied at 10.bc to the
plane, I
>assume (wrongly!) that the component of the force parallel
to the
>plane is 500sin10, but it turns out that this gives me the
force
>perpendicular to the plane.
>So why should 245sin20 act down the plane but to find the
force
>acting up the plane we use cos rather than sin???
>I hope somebody can give me a simple explanation that my
feeble brain
>can cope with!
The body rests on an inclined plane. Under the specified 
forces
(gravity, pull, friction), it can move only up/down on this
plane. Not
all forces are parallel to the inclined plane (only the
friction force
is). A force has magnitude and direction, therefore it can be
taken as
a vector. Assume that the body moves length L, vector
parallel to the
inclined plane with the size |L|. In this, the resultant F of
all
forces (which does not have to be parallel to L) performs
work A
(scalar, does not have a direction)
A = F.L
which is a dot product of 2 vectors F and L. One way to
calculate the
dot product of 2 vectors is to multiply their sizes and the
cosine of
the angle a they form:
A = |F|*|L|*cos(a)
You project the force F to the direction of L. Or you can say
that you
find the component of the vector F in the direction of the
vector L.
This is always done with cosine of the angle between these 2
vectors.
rigidity
of the inclined plane, it does not cause any acceleration and
it does
not perform any work.
You have 3 forces acting on the body with the mass m = 25 kg:
1. Gravity F1 = m*g, |g| ~ 9.81 m/sec^2 (acceleration is also
a
vector). Its direction is vertical. The angle of the inclined
plane is
given as 20Á from the horizontal. This is not the angle
between F1 and
L (the inclined plane), so you have to calculate this angle:
a1 = 90Á + 20Á = 110Á
The component of the gravity force in the direction L (of the
inclined
plane) is
|F1|*cos(110Á) = -|F1|*cos(70Á) =
-25*9.8*cos(70Á) N
2. Pulling force F2, magnitude |F2| = 500 N, direction at the
angle a2
= 10Á to the inclined plane. The component in the direction
L is again
|F2|*cos(a2) = 500*cos(10Á) N
3. Friction force F3 = -k*|N|, where k is the coefficient of
friction
(I assume that this is k = 0.4). This force is parallel to the
inclined plane and I added the minus sign, because this force
acts
against the pull upwards. |N| is the force perpendicular to
the
surface, on which the body moves, i.e., to the inclined
plane. 2
forces have components in this direction, gravity and the
pull. So you
have to find the angles b1 and b2 of these forces with the
inclined
plane.
for gravity: angle b1 = 20Á,
force component |F1|*cos(b1) = |F1|*cos(20Á)
for the pull: angle = b2 = 90Á + 10Á =
100Á
force component |F2|*cos(100Á) =
-|F2|*cos(80Á)
|N| = |F1|*cos(20Á) - |F2|*cos(80Á)
|F3| = -k*|N|
Now you can add the components of the forces F1, F2, F3 in the
direction of the inclined plane. The resulting force F is
F = -|F1|*cos(70Á) + |F2|*cos(10Á) -
- k*[|F1|*cos(20Á) - |F2|*cos(80Á)] =
= -m*g*cos(70Á) + |F2|*cos(10Á) -
- k*[m*g*cos(20Á) - |F2|*cos(80Á)]
Substituting
m = 25 kg,
g = 9.81 m/sec^2,
|F2| = 500 N,
k = 0.4
cos(70Á) ~ 0.3420
cos(10Á) ~ 0.9848
cos(20Á) ~ 0.9397
cos(80Á) ~ 0.17365
F = -83.88 + 492.40 - 0.4*(230.46 - 86.82) =
= 408.52 - 57.46 = 351.06 N
The resulting acceleration a of the body with m = 25 kg is
a = F/m = 351.06/25 = 14.04 m/sec^2
===
Subject: Re: Mechanics - friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8GXuN12608;
<snipSo why should 245sin20 act down the plane but to find the
force
>acting up the plane we use cos rather than sin???
The plane is 20Á to the horizontal, which means that is
90Á-20Á = 70Á
to the vertical. The force of gravity m*g ~
25[kg]*9.8[m/sec^2] = 245
N is vertical, i.e., 70Á deg to the plane. The component of
the
gravity force parallel to the plane is 245[N]*cos(70Á).
The force of 500 N is pulling at 10Á to the plane. The
component of
this force parallel to the plane is 500[N]*cos(10Á).
You use cosine both times. But cos(70Á) =
sin(90Á-70Á) = sin(20Á).
===
Subject: Re: Mechanics - friction
> It is being pulled up by a force of 500N at 10.bc upwards
to the
> parallel of the plane. Find the acceleration when 
Ômewis
0.4.
> Ok so the component down the plane is 245sin20 and the
perpendicular
> is 245cos20.
> Now when I look at the force being applied at 10.bc to the
plane, I
> assume (wrongly!) that the component of the force parallel
to the
> plane is 500sin10, but it turns out that this gives me the
force
> perpendicular to the plane.
> So why should 245sin20 act down the plane but to find the
force acting
> up the plane we use cos rather than sin???
Draw two pictures.
In the first one, draw the inclined plane and a block sitting
on it. Draw
three
lines:
1) a vertical line through the center of the block (this is
the direction of
the
blockÕs weight)
2) a line perpendicular to the plane through the center of
the block, and
3) a line parallel to the plane through the center of the
block.
If the angle of the plane is 20 degrees, what is is the angle
between lines
1
and 2?
In the second picture, draw the inclined plane and a block
sitting on it,
and
the following three lines:
1) a line through the center of the block parallel to the
plane
2) a line through the center of the block perpendicular to
the plane
3) a line through the center of the block at 10 degrees to
the plane
What is the angle between lines 1 and 3?
Now the answer to your question should be clear if you
remember the
definition
of sine and cosine, but if itÕs not, then say so and 
weÕll
explain in more
detail.
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Re: Mechanics - friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8IuiO26465;
hello Miro
In example 1 you are drawing my attention to the fact that
the angle
between 1 and 2 is the same as the original 20.bc.
I know the definitions of sine and cosine but 
canÕt really see
where
youÕre heading to...
Sorry. Could you explain further?
>> It is being pulled up by a force of 500N at 10.bc upwards
to the
>> parallel of the plane. Find the acceleration when 
Ômewis
0.4.
>> Ok so the component down the plane is 245sin20 and the
perpendicular
>> is 245cos20.
>> Now when I look at the force being applied at 10.bc to the
plane, I
>> assume (wrongly!) that the component of the force parallel
to the
>> plane is 500sin10, but it turns out that this gives me the
force
>> perpendicular to the plane.
>> So why should 245sin20 act down the plane but to find the
force
acting
>> up the plane we use cos rather than sin???
>Draw two pictures.
>In the first one, draw the inclined plane and a block sitting
on it.
Draw three
>lines:
>1) a vertical line through the center of the block (this is
the
direction of the
>blockÕs weight)
>2) a line perpendicular to the plane through the center of
the block,
and
>3) a line parallel to the plane through the center of the
block.
>If the angle of the plane is 20 degrees, what is is the
angle between
lines 1
>and 2?
>In the second picture, draw the inclined plane and a block
sitting on
it, and
>the following three lines:
>1) a line through the center of the block parallel to the
plane
>2) a line through the center of the block perpendicular to
the plane
>3) a line through the center of the block at 10 degrees to
the plane
>What is the angle between lines 1 and 3?
>Now the answer to your question should be clear if you
remember the
definition
>of sine and cosine, but if itÕs not, then say so and 
weÕll
explain in
more
>detail.
>meeroh
>--
>If this message helped you, consider buying an item
>from my wish list: <<a
href=http://web.meeroh.org/wishlist>http://web.meeroh.org/
wishlist</a>>
===
Subject: Math Problem that I canÕt work out.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5uK25120;
The first number was simply 1.
The next, 11.
Then, 21, 1,211, and 111,221.
whatÕs the next number in the sequance?
===
Subject: Re: Math Problem that I canÕt work out.
http://dheera.net/sci/sequence_sol.php
> The first number was simply 1.
> The next, 11.
> Then, 21, 1,211, and 111,221.
> whatÕs the next number in the sequance?
===
Subject: Re: Math Problem that I canÕt work out.
> http://dheera.net/sci/sequence_sol.php
>The first number was simply 1.
>The next, 11.
>Then, 21, 1,211, and 111,221.
>whatÕs the next number in the sequance?
312211
check on run length encoding
===
Subject: working out angles??
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8GXEp12534;
hi there:
i am a self thought programmer working on some visual codes
and got
stuck in this simple prolem, i wonder if anyone can help.
given 3 points A(0,2), B(1.414,2) and C(1.414, 1.414), who
would i
work out in degrees of angle BAC? whatÕs the formula to work
out
angles in general?
CHUN
===
Subject: Re: working out angles??
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> hi there:
> i am a self thought programmer working on some visual codes
and got
> stuck in this simple prolem, i wonder if anyone can help.
> given 3 points A(0,2), B(1.414,2) and C(1.414, 1.414), who
would i
> work out in degrees of angle BAC? whatÕs the formula to
work out
> angles in general?
> CHUN
In the general case:
1. Find the lengths of the sides of triangle ABC (i.e. the
lengths AB,
AC and BC). Do this using Pythagorastheorem, which tells
that the
distance between the points (x1,y1) and (x2,y2) is
sqr[(x2-x1)^2 +
(y2-y1)^2]. Plug the coordinates into this formula to find the
lengths
of each of the sides.
2. Once you know the lengths of all three sides, you can find
any of
the angles using the cosine rule, which is explained at
http://mathworld.wolfram.com/LawofCosines.html. (The cosine
rule gives
you the cosine of the required angle, which will be a number
between -1
and 1. Once you have the cosine, use the inverse cosine
function to get
the angle, which will be between 0 and 180 degrees, or 0 to pi
radians).
However, in this specific case you have a right-angled
triangle, with
the right angle at B. The general method will still work, but
thereÕs
an easier way. Just use the basic properties of the
trigonometric
functions to find that Tan(BAC) = BC/AB. Having found
Tan(BAC), use the
inverse tangent function to find the angle. The lengths BC and
AB are
also particularly easy to work out in this example, because
of the
orientation of the triangle. If you draw it on a piece of
graph paper
youÕll see how.
===
Subject: Re: working out angles??
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8H8UJ16142;
>hi there:
>i am a self thought programmer working on some visual codes
and got
>stuck in this simple prolem, i wonder if anyone can help.
>given 3 points A(0,2), B(1.414,2) and C(1.414, 1.414), who
would i
>work out in degrees of angle BAC? whatÕs the formula to 
work
out
>angles in general?
>CHUN
Vector AB = (sqrt(2) - 0, 2 - 2) = (sqrt(2), 0)
Vector AC = (sqrt(2) - 0, 2 - sqrt(2)) = (sqrt(2), 2 -
sqrt(2))
|AB| = sqrt[sqrt(2)^2 + 0^2] = sqrt(2)
|AC| = sqrt[sqrt(2)^2 + (2 -sqrt(2))^2] = sqrt[2 + 2 -
4*sqrt(2) + 2]
= sqrt[8 - 4*sqrt(2)] = 2*sqrt[2 - sqrt(2)]
Scalar product:
AB.AC = sqrt(2)*sqrt(2) + 0*[2 - sqrt(2)] = 2
But AB.AC = |AB|*|AC|*cos(BAC)
cos(BAC) = 2/[sqrt(2)*2*sqrt(2 - sqrt(2))] =
= 1/sqrt[4 - 2*sqrt(2)]
===
Subject: Re: Induction: How to show n^3 - n is divisible by 3
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8GtJ514635;
>I donÕt know where IÕve gone wrong and 
IÕm hoping someone
can help me
>For n >= to 1 show n^3 - n is divisible by 3
>Tis is what IÕve done:
>Base Case: n = 1
>1^3 - 1 = 0 which is divisible by 3
>Inductive Step: Assume that n^3 - n is dibisble by 3 for n
>= 1.
>Prove that ( n + 1)^3 - ( n + 1) is divisible by 3.
>n^3 - n = n (n^2 -1 ) = n (n - 1) (n + 1)
>(n + 1)^3 - (n + 1) = (n + 1) [ (n + 1)^2 - 1]
> = (n + 1) [ (n + 1) - 1) ( ( n + 1) + 1) ]
> = ( n + 1) [ n ( n + 2) ]
> = n (n + 1) (n + 2)
>So where do I go from here?
>IÕd appreciate any help you can offer.
A. For n = 1, 1^3 - 1 = 0 is divisible by 3.
B. Assume that n^3 - n is divisible by 3 for some n >= 1,
i.e.,
n^3 - n = 3k, k >=0 is an integer (k >= 0 because n^3 >= n).
(n + 1)^3 - (n + 1) =
= n^3 + 3n^2 + 3n + 1 - n - 1 =
= n^3 - n + 3*(n^2 + n) =
= 3k + 3*(n^2 + n) =
= 3*(k + n^2 + n)
Therefore, (n + 1)^3 - (n + 1) is also divisible by 3.
===
Subject: Laplace Transforms
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8GtJv14621;
Can anyone get me started on the following Transform using
Convolution
Technique?
L^-1{ 8 / (s^2 + 1)^3}
Any help would be greatly appreciated.
NA
===
Subject: Re: Laplace Transforms
>Can anyone get me started on the following Transform using
Convolution
>Technique?
>L^-1{ 8 / (s^2 + 1)^3}
>Any help would be greatly appreciated.
Starting with f(t) and g(t), letÕs denote L(f) by F(s) and
L(g) by
G(s) and the convolution of f and g as f*g
So you know that L(f * g) = L(f) L(g) = F(s)G(s)
Stating this in inverse form it says that the inverse of the
product
of two transforms is the convolution of the inverses:
Linverse(F(s)G(s)) = f*g.
Getting to your problem, you could write
8 / (s^2 + 1)^3 = {8 / (s^2 + 1)} {1 / (s^2 + 1)^2} = F(s)G(s)
Presumably you know the inverse of F(s)=8 / (s^2 + 1), so if
you knew
the inverse of G(s) you could take their convolution for your
answer.
How do you find the inverse of G(s) = 1 / (s^2 + 1)^2 ? Well,
work it
first as the product of 1 / (s^2 + 1) with itself.
That should get you going. You are, of course, going to have
some
integrals to work out to simplify the convolution integrals.
:-)
--Lynn
===
Subject: Re: Laplace Transforms
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECnr00959;
>>Can anyone get me started on the following Transform using
Convolution
>>Technique?
>>L^-1{ 8 / (s^2 + 1)^3}
>>Any help would be greatly appreciated.
>>NA
>Starting with f(t) and g(t), letÕs denote L(f) by F(s) and
L(g) by
>G(s) and the convolution of f and g as f*g
>So you know that L(f * g) = L(f) L(g) = F(s)G(s)
>Stating this in inverse form it says that the inverse of the
product
>of two transforms is the convolution of the inverses:
>Linverse(F(s)G(s)) = f*g.
>Getting to your problem, you could write
>8 / (s^2 + 1)^3 = {8 / (s^2 + 1)} {1 / (s^2 + 1)^2} =
F(s)G(s)
>Presumably you know the inverse of F(s)=8 / (s^2 + 1), so if
you knew
>the inverse of G(s) you could take their convolution for
your answer.
>How do you find the inverse of G(s) = 1 / (s^2 + 1)^2 ? Well,
work it
>first as the product of 1 / (s^2 + 1) with itself.
>That should get you going. You are, of course, going to have
some
>integrals to work out to simplify the convolution integrals.
:-)
>--Lynn
sin(t)=> 1/(s^2+1),
and xÕÕ+ a^2.x = b*sin(t) => 
(s^2+a^2)X(s)=ab/(s^2+a^2)+
x0.s+x1 ...
X(s)=8 / (s^2 + 1)^3 is the operational solution of :
xÕÕÕÕ+2x[Capita
lOTilde]Õ+ x =8*sin(t) with x0=xÕ(0)= 0
the original x(t) = residu X(s)= at s = I ,
x(t)= re {d^2/ds^2 [ exp(st)/(s+I)^3 ]s=I
=(3-t^2)sin(t)-3tcos(t)
Friendly,Alain.
===
Subject: Re: Laplace Transforms
>sin(t)=> 1/(s^2+1),
>and xÕÕ+ a^2.x = b*sin(t) => 
(s^2+a^2)X(s)=ab/(s^2+a^2)+
x0.s+x1 ...
>X(s)=8 / (s^2 + 1)^3 is the operational solution of :
> 
xÕÕÕÕ+2x[Capital
OTilde]Õ+ x =8*sin(t) with x0=xÕ(0)= 0
>the original x(t) = residu X(s)= at s = I ,
> x(t)= re {d^2/ds^2 [ exp(st)/(s+I)^3 ]s=I
=(3-t^2)sin(t)-3tcos(t)
>Friendly,Alain.
Did you miss the part where the original poster specifically
asked how
to find the inverse **using the convolution theorem**?
--Lynn
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBAD3Xi25095;
===
Subject: Curl and Divergence
IÕm having trouble working out a proof for a formula. Since
IÕm not
really sure what the best way to represent it in plain text
would be,
IÕm going to write it in TeX format. If itÕs
incomprehensible, tell me
how better to represent it.
nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
Which I think is accurately represented as:
curl(curl(F) = grad(div(F)) - grad(grad(F))
IÕm very sorry about how awful that looks.
IÕm reasonably certain that I am supposed to take
F(x,y,z)=F_1(x,y,z)i + F_2(x,y,z)j + F_3(x,y,z)k
with i, j, and k the unit vectors.
Well, I worked out the left side to be a terribly ugly
looking thing
that IÕm pretty sure is right, by setting up a determinant 
to
work out
nabla times F and then again to get the final result.
I encounter trouble when I try to work with the right side of
the
equation. I compute first the divergence of F, and I get
div(F) = d/dx(F_1) + d/dy(F_2) + d/dz(F_3)
When I do grad(div(F)), I get a very long vector involving
several
second derivatives. Subtracting grad(grad(F)) gives me a
formula which
is unequal to what I got on the left of the equation, which
leads me to
believe I made a mistake in my method. I hope someone can
help to point
me in the right direction, since IÕm somewhat at a loss as 
to
what
mistake IÕve made.
IÕm sorry if what IÕve written is unclear. I 
can provide any
particular
step I took specifically, though I canÕt promise 
that it will
be
terribly easy to understand.
Tracy Poff
===
Subject: Re: Curl and Divergence
>IÕm having trouble working out a proof for a formula. Since
IÕm not
>really sure what the best way to represent it in plain text
would be,
>IÕm going to write it in TeX format. If itÕs
incomprehensible, tell me
>how better to represent it.
>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
>Which I think is accurately represented as:
>curl(curl(F) = grad(div(F)) - grad(grad(F))
>IÕm very sorry about how awful that looks.
>IÕm reasonably certain that I am supposed to take
>F(x,y,z)=F_1(x,y,z)i + F_2(x,y,z)j + F_3(x,y,z)k
>with i, j, and k the unit vectors.
>Well, I worked out the left side to be a terribly ugly
looking thing
>that IÕm pretty sure is right, by setting up a determinant
to work out
>nabla times F and then again to get the final result.
>I encounter trouble when I try to work with the right side
of the
>equation. I compute first the divergence of F, and I get
>div(F) = d/dx(F_1) + d/dy(F_2) + d/dz(F_3)
>When I do grad(div(F)), I get a very long vector involving
several
>second derivatives. Subtracting grad(grad(F)) gives me a
formula which
>is unequal to what I got on the left of the equation, which
leads me to
>believe I made a mistake in my method. I hope someone can
help to point
>me in the right direction, since IÕm somewhat at a loss as
to what
>mistake IÕve made.
>IÕm sorry if what IÕve written is unclear. I 
can provide any
particular
>step I took specifically, though I canÕt 
promise that it will
be
>terribly easy to understand.
>Tracy Poff
grad(grad(F), which might better be written as (del dot del)
F. I am
wondering if you expanded that correctly. For example, if:
F = < f, g, h> then (del dot del) F would be written as:
<f_xx + g_xx + h_xx, f_yy + g_yy + h_yy,f_zz + g_zz + h_zz>
Did you expand that term correctly?
--Lynn
===
Subject: Re: Curl and Divergence
>>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
> grad(grad(F), which might better be written as (del dot
del) F. I am
> wondering if you expanded that correctly. For example, if:
> F = < f, g, h> then (del dot del) F would be written as:
> <f_xx + g_xx + h_xx, f_yy + g_yy + h_yy,f_zz + g_zz + h_zz
Did you expand that term correctly?
> --Lynn
The term is certainly written nabla^2 F in my book, and
defined as:
f_xx + f_yy + f_zz
earlier in the chapter. It is given alternate representations
as
Delta F
or
nabla cdot (nabla F)
However, as I mentioned in a reply to another post, there is
a note
after this problem stating that ...a major part of the
problem is to
decipher an unfamiliar notation. This does not seem like a
very good
sort of problem to have, to me, but IÕm going to try 
assuming
that F is
meant to be a scalar valued function, as opposed to a vector
valued
function which it was defined to be earlier in this chapter of
my book,
and work with it that way.
Tracy Poff
===
Subject: Re: Curl and Divergence
>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
>> grad(grad(F), which might better be written as (del dot
del) F. I am
>> wondering if you expanded that correctly. For example, if:
>> F = < f, g, h> then (del dot del) F would be written as:
>> <f_xx + g_xx + h_xx, f_yy + g_yy + h_yy,f_zz + g_zz +
h_zz> Did you expand that term correctly?
>> --Lynn
>The term is certainly written nabla^2 F in my book, and
defined as:
>f_xx + f_yy + f_zz
>earlier in the chapter.
That is correct for scalar f, and not inconsistent with what
I am
saying. See below.
>It is given alternate representations as
>Delta F
>nabla cdot (nabla F)
>However, as I mentioned in a reply to another post, there is
a note
>after this problem stating that ...a major part of the
problem is to
>decipher an unfamiliar notation. This does not seem like a
very good
>sort of problem to have, to me, but IÕm going to try
assuming that F is
>meant to be a scalar valued function, as opposed to a vector
valued
>function which it was defined to be earlier in this chapter
of my book,
>and work with it that way.
>Tracy Poff
Tracy, the formula you are working is for a vector field. The
del^2
notation is sometimes used for (del dot del) as I suggested,
and it
must be interpreted as a scalar operator:
@^2/@x^2 + @^2/@y^2 + @^2/@z^2 where I use the @ for the
partial
derivative symbol (ugly, isnÕt it).
So your third term represents this scalar times the vector F,
but it
isnÕt really times; you have to perform the operation in 
each
component just as if you were multiplying a vector by a
scalar. If you
use the expression for (del dot del)F as I gave you above, the
identity you are trying to prove works.
Incidentally, the formula you are working on is reminiscent
of the
formula for three vectors:
A cross (B cross C) = (A dot C)B - (A dot B)C
where you take some care interpreting how the del is
interpreted.
--Lynn
===
Subject: Re: Curl and Divergence
> Tracy, the formula you are working is for a vector field.
The del^2
> notation is sometimes used for (del dot del) as I
suggested, and it
> must be interpreted as a scalar operator:
> @^2/@x^2 + @^2/@y^2 + @^2/@z^2 where I use the @ for the
partial
> derivative symbol (ugly, isnÕt it).
> So your third term represents this scalar times the vector
F, but it
> isnÕt really times; you have to perform the operation in
each
> component just as if you were multiplying a vector by a
scalar. If you
> use the expression for (del dot del)F as I gave you above,
the
> identity you are trying to prove works.
Since this seemed to match what I was trying to do already, I
went back
and very carefully verified all my work. After doing this
several times,
I found that I had mistaken an x for a y in my calculations
when
copying, and this was the reason why my answers had not been
matching.
encounter the need to trouble you again--IÕll watch my
writing more
carefully in the future.
Tracy Poff
===
Subject: Re: Curl and Divergence
> F = < f, g, h> then (del dot del) F would be written as:
<f_xx + g_xx + h_xx, f_yy + g_yy + h_yy,f_zz + g_zz + h_zz
Careless error here, it should of course be:
(del dot del)F
= < f_xx + f_yy + f_zz, g_xx + g_yy + g_zz, h_xx + h_yy +
h_zz >
--Lynn
===
Subject: Re: Curl and Divergence
days. My association with the Department is that of an
alumnus.
>IÕm having trouble working out a proof for a formula. Since
IÕm not
>really sure what the best way to represent it in plain text
would be,
>IÕm going to write it in TeX format. If itÕs
incomprehensible, tell me
>how better to represent it.
>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
>Which I think is accurately represented as:
>curl(curl(F) = grad(div(F)) - grad(grad(F))
Ehr... if F is a vector function, then grad(F) does not
really make
sense. Neither does nabla^2(F), as nabla^2 is usually
interpreted to
be the Laplacian operator, defined for scalar functions as
div(grad(f)).
[...]
>When I do grad(div(F)), I get a very long vector involving
several
>second derivatives. Subtracting grad(grad(F)) gives me a
formula which
>is unequal to what I got on the left of the equation, which
leads me to
>believe I made a mistake in my method. I hope someone can
help to point
>me in the right direction, since IÕm somewhat at a loss as
to what
>mistake IÕve made.
The method is fine, except for the grad(grad(F)) thing, which
does
not seem to make much sense to me.
[...]
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Curl and Divergence
>>IÕm having trouble working out a proof for a formula. 
Since
IÕm not
>>really sure what the best way to represent it in plain text
would be,
>>IÕm going to write it in TeX format. If 
itÕs
incomprehensible, tell me
>>how better to represent it.
>>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
>>Which I think is accurately represented as:
>>curl(curl(F) = grad(div(F)) - grad(grad(F))
> Ehr... if F is a vector function, then grad(F) does not
really make
> sense. Neither does nabla^2(F), as nabla^2 is usually
interpreted to
> be the Laplacian operator, defined for scalar functions as
div(grad(f)).
The specific directions are: Prove the vector formula. Looking
back in
the chapter in my book preceding this problem, it says that I
am to take
reproduces the equation as it is in my book, so if it does
not make
sense there is little I can do about it.
My book doesnÕt use curl, grad, div as ways of writing the
formulae; I
came across them on the internet and they seemed common, so I
tried to
reproduce it in that format for easier reading. If IÕve done
so
incorrectly, then please ignore them. The TeX code is
certainly correct.
>>When I do grad(div(F)), I get a very long vector involving
several
>>second derivatives. Subtracting grad(grad(F)) gives me a
formula which
>>is unequal to what I got on the left of the equation, which
leads me to
>>believe I made a mistake in my method. I hope someone can
help to point
>>me in the right direction, since IÕm somewhat at a loss as
to what
>>mistake IÕve made.
> The method is fine, except for the grad(grad(F)) thing,
which does
> not seem to make much sense to me.
Again, here I mean nabla^2 F, which is defined in my book as
the
Laplacian and does, indeed, appear to be used on a scalar
function.
I have just noticed a note in the book that says ...a major
part of the
problem is to decipher an unfamiliar notation. This does not
give me
high hopes as to my ability to properly solve this.
Should I, then, assume that the F here is a scalar function,
and that
the authors of the book were merely sadistic and gave me a
notation in
the exercises entirely different from that found in the text?
Tracy Poff
===
Subject: Re: Curl and Divergence
days. My association with the Department is that of an
alumnus.
>IÕm having trouble working out a proof for a formula. Since
IÕm not
>really sure what the best way to represent it in plain text
would be,
>IÕm going to write it in TeX format. If itÕs
incomprehensible, tell me
>how better to represent it.
>nabla times (nabla times F) = nabla (nabla cdot F) -
nabla^2 F
>Which I think is accurately represented as:
>curl(curl(F) = grad(div(F)) - grad(grad(F))
>> Ehr... if F is a vector function, then grad(F) does not
really make
>> sense. Neither does nabla^2(F), as nabla^2 is usually
interpreted to
>> be the Laplacian operator, defined for scalar functions as
div(grad(f)).
>The specific directions are: Prove the vector formula.
Looking back in
>the chapter in my book preceding this problem, it says that
I am to take
>reproduces the equation as it is in my book, so if it does
not make
>sense there is little I can do about it.
You can always write to the author and complain...
(IÕm going to switch notation to simplify the writing here 
in
ASCII. I
will write F = (G, H, K), where G, H, and K are scalar valued
functions; and I will use G_x to mean the partial of G with
respect to
x, H_yy for the second partial of H with respect to y, etc.)
I think I know what he means... If you think of nabla^2 as a
scalar,
then
nabla^2 F = ( nabla^2(G), nabla^2(H), nabla^2(K) )
and since G, H, K are scalar functions, this would make
sense. If you
do that, I believe you will find that, ->assuming that G, H,
and K
have continuous second partials<- (so that the mixed
derivatives are
equal, e.g., H_xz = H_zx, etc), you do indeed get equality.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: need a proof using Mean Value theorem
Can someone suggest an approach to this problem:
Using the Mean Value Theorem, given a function f that is
differentable on R
prove that if abs(f Ô(x))<1 for all x in R, then there is at
most one
solution for the equation f(x) = x.
===
Subject: Re: need a proof using Mean Value theorem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB914mo31835;
>Can someone suggest an approach to this problem:
>Using the Mean Value Theorem, given a function f that is
differentable on R
>prove that if abs(f Ô(x))<1 for all x in R, then there is at
most
>one solution for the equation f(x) = x.
Assume there are 2 different solutions x1 and x2, such that
x1 = f(x1), x2 = f(x2). By the mean value theorem, there
exist x0,
x1 < x0 < x2, such that
fÕ(x0)*(x2 - x1) = f(x2) - f(x1) = x2 - x1
Since x2 - x1 != 0, fÕ(x0) = 1, which contradicts the
assumption that
|fÕ(x)| < 1 everywhere.
===
Subject: Re: need a proof using Mean Value theorem
>Using the Mean Value Theorem, given a function f that is
> differentable on R
>prove that if abs(f Ô(x))<1 for all x in R, then there is at
most
>one solution for the equation f(x) = x.
> Assume there are 2 different solutions x1 and x2, such that
> x1 = f(x1), x2 = f(x2). By the mean value theorem, there
exist x0,
> x1 < x0 < x2, such that
Assuming x1 < x2.
> fÕ(x0)*(x2 - x1) = f(x2) - f(x1) = x2 - x1
> Since x2 - x1 != 0, fÕ(x0) = 1, which contradicts the
assumption that
> |fÕ(x)| < 1 everywhere.
The premise can be weakened to for all x, fÕ(x) /= 1
===
Subject: Optimization - Calculus question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECnF00978;
Here a third one that I canÕt seem to figure 
out. PLEASE help
:)
The production of blood cells plays an important role in
medical
research involving leukemia and other so-called dynamical
diseases. In
1977, a mathematical model was developed by A. Lasota that
involved
the cell production function
P(x)=Ax^s e^(-sx/r)
where A, s, and r are positive constants and x is the number
of
granulocytes (a type of white blood cell) present. Find the
granulocyte level x that maximizes the production function P.
How do
you know it is a maximum?
===
Subject: Re: Optimization - Calculus question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9IO0k25057;
>Here a third one that I canÕt seem to figure 
out. PLEASE help
:)
>The production of blood cells plays an important role in
medical
>research involving leukemia and other so-called dynamical
diseases.
In
>1977, a mathematical model was developed by A. Lasota that
involved
>the cell production function
>P(x)=Ax^s e^(-sx/r)
>where A, s, and r are positive constants and x is the number
of
>granulocytes (a type of white blood cell) present. Find the
>granulocyte level x that maximizes the production function
P. How do
>you know it is a maximum?
At a local maximum or minimum, PÕ(x) = 0:
PÕ(x) = A*s*x^(s-1)*e^(-sx/r) + A*x^s*(-s/r)*e^(-sx/r) =
= A*s*x^s*e^(-sx/r)*(1/x - 1/r) = 0
Divide the equation by A*s*x^s*exp(-sx/r) != 0:
1/x - 1/r = 0
x = r
This will be a local maximum, is P(x) < 0 at x = r.
P(x) = A*s^2*x^s*e^(-sx/r)*(1/x - 1/r)^2 +
+ A*s*x^s*e^(-sx/r)*(-1/x^2)
Plug x = r:
P(r) = -A*s*r^(s - 2)*e^(-s)
Obviously, P(r) < 0, i.e., P(x) has a local maximum at x = r.
===
Subject: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECrh01172;
how do you solve this question plz help!!! thx!!!
A current population is 2 000 000 and it increases 7%
annualy. However
each year a certain number of people die (this amount is
constant for
every year). How many people die each year if the population
is to
become 2 500 000 in 8 years?
plz list steps thx!!!
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC1vMw17522;
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy.
However
>each year a certain number of people die (this amount is
constant for
>every year). How many people die each year if the population
is to
>become 2 500 000 in 8 years?
>plz list steps thx!!!
year by simply plugging the values into the equation
t[n]=t[n-1](1.07)-x where t[0]=2,000,000 (all the values
within the []
are subscripts). Hence, this answer corresponds to that of
DanaÕs
method. However, my professor expects us to solve using
calculus, is
there a method to do this (that arrives at the same answer as
91,266
deaths)?
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBCLoD021812;
>>how do you solve this question plz help!!! thx!!!
>>A current population is 2 000 000 and it increases 7%
annualy.
>>However each year a certain number of people die (this
amount is
>>constant for every year). How many people die each year if
the
>>population is to become 2 500 000 in 8 years?
>>plz list steps thx!!!
>year by simply plugging the values into the equation
>t[n]=t[n-1](1.07)-x where t[0]=2,000,000 (all the values
within the
>[] are subscripts). Hence, this answer corresponds to that
of DanaÕs
>method. However, my professor expects us to solve using
calculus, is
>there a method to do this (that arrives at the same answer
as 91,266
>deaths)?
No, there is not. Calculus implies continuous changes. You
could
strictly require that 7 % of people living at the beginning
of the
year are born during the year. The rate m of population
increase
without deaths would have to be calculated:
dn = m*n*dt
n(t) = C*e^mt
Initial conditions:
n(0) = 2*10^6
n(1) = 1.07*2*10^6
1.07*2*10^6 = 2*10^6*e^m
m = ln(1.07)
m ~0.06766 = 6.766 %
With deaths, the differential equation is
dn = n*m*dt - k*dt
were k is the death rate (number of deths per unit time)
n(t) = -k*t + C*exp(0.07*t)
with the initial conditions
n(0) = 2*10^6
n(8) = 2.5*10^6
Hence C = 2*10^6 and
k = (2*e^(8m) - 2.5)*10^6 ~ 117047 deaths/year
which is still higher than 91266 deaths/year. The reason this
number
is higher is that deaths occur throughout the year, not
simultaneously
at the end of a year. Therefore, there are fewer people
during the
year to multiply to reach the increase 7 % anually. This
necessitates
a higher rate of population increase compared to the case,
when all
deaths occur simultaneously at the end of a year.
This increase, in turn, requires more deaths per year to
exactly meet
the goal of 2500000 population in 8 years.
===
Subject: Re: sequence problem: recursion formula most likely
involved
HereÕs my attempt. If you expand the equation for each year,
you can factor
the d (for deaths). this leaves the terms
d* (1 + r + r^2 + r^3 +r^4 + r^5 + r^6 + r^7)...
which can be written as below, where yr is the number of years
d*(-1 + r^yr)/(r - 1)
Solving for d, I get:
((r - 1)*(r^yr*start - end))/(r^yr - 1)
With:
{start -> 2000000, end -> 2500000, r -> 1.07, yr -> 8}
I get
deaths are
91266.1
per year
HTH
--
Dana
> how do you solve this question plz help!!! thx!!!
> A current population is 2 000 000 and it increases 7%
annualy. However
> each year a certain number of people die (this amount is
constant for
> every year). How many people die each year if the
population is to
> become 2 500 000 in 8 years?
> plz list steps thx!!!
===
Subject: Re: sequence problem: recursion formula most likely
involved
Your answer not correct because deaths are constant every
year. You can
find below to what IÕm referring. 
IÕd point out where, but as
you top
post, I do same which is not only a sloppy way to reply, 
itÕs
impolite.
> HereÕs my attempt. If you expand the equation for each
year, you can
factor
> the d (for deaths). this leaves the terms
> d* (1 + r + r^2 + r^3 +r^4 + r^5 + r^6 + r^7)...
> which can be written as below, where yr is the number of
years
> d*(-1 + r^yr)/(r - 1)
> Solving for d, I get:
> ((r - 1)*(r^yr*start - end))/(r^yr - 1)
> With:
> {start -> 2000000, end -> 2500000, r -> 1.07, yr -> 8}
> I get
> deaths are
> 91266.1
> per year
> HTH
> --
> Dana
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy. However
>each year a certain number of people die (this amount is
constant for
>every year). How many people die each year if the population
is to
>become 2 500 000 in 8 years?
>plz list steps thx!!!
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
> Your answer not correct because deaths are constant every
year.
DanaÕs answer *is* correct. ItÕs also 
essentially
equivalent to the one that I outlined in my first post in
this thread.
[...]
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
> alt.math.undergrad:
>Your answer not correct because deaths are constant every
year.
> DanaÕs answer *is* correct. ItÕs also 
essentially
equivalent to the one
> that I outlined in my first post in this thread.
Shucks another late night mishap. Strike it from the record!
But alas no, a privilege only for the politically anointed.
Do you recall the next line?
The moving pen, having write, moves on ...
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
>> alt.math.undergrad:
> Your answer not correct because deaths are constant every
year.
>> DanaÕs answer *is* correct. ItÕs also 
essentially
equivalent to the one
>> that I outlined in my first post in this thread.
> Shucks another late night mishap. Strike it from the record!
> But alas no, a privilege only for the politically anointed.
> Do you recall the next line?
> The moving pen, having write, moves on ...
Moves on: nor all your Piety nor Wit
Nor all your Tears wash out a Word of it,
Though X-No-Archive dims the Line a Bit.
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
>Shucks another late night mishap. Strike it from the record!
>But alas no, a privilege only for the politically anointed.
>Do you recall the next line?
>The moving pen, having write, moves on ...
> Moves on: nor all your Piety nor Wit
> Nor all your Tears wash out a Word of it,
by Omar Kaiam?
> Though X-No-Archive dims the Line a Bit.
To speak therein doesth lie a dim wit.
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
[...]
> Do you recall the next line?
> The moving pen, having write, moves on ...
>> Moves on: nor all your Piety nor Wit
>> Nor all your Tears wash out a Word of it,
> by Omar Kaiam?
By Omar Khayyam as translated by Edward Fitzgerald; as I
understand it, Fitzgerald tried very hard to retain the
spirit of the original but did not produce anything remotely
resembling a literal translation.
[...]
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
> how do you solve this question plz help!!! thx!!!
> A current population is 2 000 000 and it increases 7%
annually. However
> each year a certain number of people die (this amount is
constant for
> every year). How many people die each year if the
population is to
> become 2 500 000 in 8 years?
> plz list steps thx!!!
Sorry about the top post. I may be wrong, but I was just
using a growth of
7% each year followed by a decrease of a constant 91,266.1.
2,000,000 * 1.07-91266.1 -> 2,048,734
Then
2,048,734 * 1.07 - 91266.1 -> 2,100,879
etc...
After 8 years, I get 2,500,000
I used a math program to give it a quick check:
Population[n_] := n*1.07 - 91266.1
Round[NestList[Population, 2000000, 8]]
{
2000000,
2048734,
2100879,
2156675,
2216376,
2280256,
2348608,
2421744,
2500000
}
--
Dana
> Your answer not correct because deaths are constant every
year. You can
> find below to what IÕm referring. 
IÕd point out where, but
as you top
> post, I do same which is not only a sloppy way to reply,
itÕs impolite.
>> HereÕs my attempt. If you expand the equation for each
year, you can
>> factor
>> the d (for deaths). this leaves the terms
>> d* (1 + r + r^2 + r^3 +r^4 + r^5 + r^6 + r^7)...
>> which can be written as below, where yr is the number of
years
>> d*(-1 + r^yr)/(r - 1)
>> Solving for d, I get:
>> ((r - 1)*(r^yr*start - end))/(r^yr - 1)
>> With:
>> {start -> 2000000, end -> 2500000, r -> 1.07, yr -> 8}
>> I get
>> deaths are
>> 91266.1
>> per year
>> HTH
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
> how do you solve this question plz help!!! thx!!!
> A current population is 2 000 000 and it increases 7%
annualy. However
> each year a certain number of people die (this amount is
constant for
> every year). How many people die each year if the
population is to
> become 2 500 000 in 8 years?
Let p(n) be the population after n years. Of course p(0) =
2 000 000. Let d be the number who die each year. Then the
recurrence is
p(n+1) = 1.07*p(n) - d.
Have you solved recurrences of the type x(n+1) = ax(n) + b?
One fairly common elementary technique is to 
Ôunwrapit:
x(n) = ax(n-1) + b
= a[ax(n-2) + b] + b
= a^2 * x(n-2) + ab + b
= a^2 * [ax(n-3) + b] + ab + b
= a^3 * x(n-3) + a^2 * b + ab + b
= ...
= a^n * x(0) + a^(n-1)*b + a^(n-2)*b + ... + ab + b.
The terms after the first form a geometric series,
a^(n-1)*b + a^(n-2)*b + ... + ab + b,
whose sum you should know how to find in closed form. If
you apply this technique to your problem, youÕll get
p(n) = 1.07^n * p(0) + f(d),
where f(d) is some known function of d. Substitute n = 8
and solve for d.
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9G6VN12182;
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy.
However
>each year a certain number of people die (this amount is
constant for
>every year). How many people die each year if the population
is to
>become 2 500 000 in 8 years?
>plz list steps thx!!!
I am sure you can compute the future population without
deaths :
2 000 000 becoming in 8 years... ,
2 000 000*(1+7/100)^8 = G
G - 2 500 000 = deaths in 8 years ...
A constant amount for every year - as population increases
-means
a decreasing percentage of annual deaths :
Better health care and longevity inside your blessed country!
Alain.
===
Subject: Re: sequence problem: recursion formula most likely
involved
in alt.math.undergrad:
>>how do you solve this question plz help!!! thx!!!
>> A current population is 2 000 000 and it increases 7%
>> annualy. However each year a certain number of people
>> die (this amount is constant for every year). How many
>> people die each year if the population is to become 2
>> 500 000 in 8 years?
> I am sure you can compute the future population without
deaths :
> 2 000 000 becoming in 8 years... ,
> 2 000 000*(1+7/100)^8 = G
> G - 2 500 000 = deaths in 8 years ...
This approach will not work; it overestimates the correct
number. If d is the number of deaths per year, the d deaths
in the first year reduce the final population by 
1.07^7 * d
people, the d deaths in the second year reduce it by only
1.07^6 * d people, and so on. That is, the deaths in
different years have different effects on the final
population.
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBAD3WK25049;
>in alt.math.undergrad:
>how do you solve this question plz help!!! thx!!!
> A current population is 2 000 000 and it increases 7%
> annualy. However each year a certain number of people
> die (this amount is constant for every year). How many
> people die each year if the population is to become 2
> 500 000 in 8 years?
>> I am sure you can compute the future population without
deaths :
>> 2 000 000 becoming in 8 years... ,
>> 2 000 000*(1+7/100)^8 = G
>> G - 2 500 000 = deaths in 8 years ...
>This approach will not work; it overestimates the correct
>number. If d is the number of deaths per year, the d deaths
>in the first year reduce the final population by 
1.07^7 * d
>people, the d deaths in the second year reduce it by only
>1.07^6 * d people, and so on. That is, the deaths in
>different years have different effects on the final
>population.
>Brian
You are 100% right,
I should have proposed a step by step solution:
total deaths =d; population 2 000 000*(1+7/100)-d/10 first
year ...
Friendly,Alain.
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9G9Ab12348;
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy.
>However each year a certain number of people die (this
amount is
>constant for every year). How many people die each year if
the
>population is to become 2 500 000 in 8 years?
>plz list steps thx!!!
t - time in years
n(t) - current population
k - number of corpses/year
0.07 = 7%/year
Population change in time dt:
dn = 0.07*n*dt - k*dt
dn/dt = 0.07n - k
Solving the homogenous differential equation dn/dt = 0.07*n:
dn/n = 0.07*dt
ln|n| = 0.07*t + C1 = 0.07*t + ln(C2)
C1 is arbitrary constant, C2 = exp^C1 > 0
n = +- C2*exp(0.07*t) = C3*exp(0.07*t)
C3 is arbitrary constant
Assume C3 = C3(t) (variation of constants). Then
dn/dt = dC3/dt + 0.07*C3*exp(0.07*t) = dC3/dt + 0.07*n
Plug this into the differential equation with the right side:
dC3/dt + 0.07*n = 0.07*n - k
dC3/dt = -k
C3 = -k*t + C4
C4 is an arbitrary constant
n = (-k*t + C4)*exp(0.07*t)
Initial conditions:
at t = 0, n = 2*10^6
at t = 8, n = 2.5*10^6
Plug this into the solution n(t):
t = 0 => 2.5*10^6 = C4
t = 8 => 2.5*10^6 = (-8k + 2*10^6)*exp(0.07*8)
10^6*[2.5 - 2*exp(0.56)] = -8k*exp(0.56)
k = 10^6*[2*exp(0.56) - 2.5]/[8*exp(0.56)] =
= 10^6*[1/4 - 5/16*exp(-0.56)] =
= 10^6*[4 - 5*exp(-0.56)]/16 ~ 71497 ~ 71500 corpses/year
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
>>how do you solve this question plz help!!! thx!!!
>>A current population is 2 000 000 and it increases 7%
annualy.
>>However each year a certain number of people die (this
amount is
>>constant for every year). How many people die each year if
the
>>population is to become 2 500 000 in 8 years?
> t - time in years
> n(t) - current population
> k - number of corpses/year
> 0.07 = 7%/year
> Population change in time dt:
> dn = 0.07*n*dt - k*dt
> dn/dt = 0.07n - k
> Solving the homogenous differential equation dn/dt = 0.07*n:
[...]
> k = 10^6*[2*exp(0.56) - 2.5]/[8*exp(0.56)] =
> = 10^6*[1/4 - 5/16*exp(-0.56)] =
> = 10^6*[4 - 5*exp(-0.56)]/16 ~ 71497 ~ 71500 corpses/year
YouÕve used a continuous model to arrive at this 
figure, but
the problem is stated in terms of a discrete model. The
actual solution to the problem as stated is larger than your
figure. I wonÕt give it away completely, since 
IÕd prefer
for the original poster to arrive at it himself, but itÕs a
bit over 90 000.
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBA37Xc07177;
>alt.math.undergrad:
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy.
>However each year a certain number of people die (this
amount is
>constant for every year). How many people die each year if
the
>population is to become 2 500 000 in 8 years?
I was a little uneasy after hitting the <Send> button. Mainly
I made
arithmetic errors at the end (as usual):
>>Initial conditions:
>>at t = 0, n = 2*10^6
>>at t = 8, n = 2.5*10^6
>>Plug this into the solution n(t):
>>t = 0 => 2*10^6 = C4
>>t = 8 => 2.5*10^6 = -8k + 2*10^6*exp(0.07*8)
>>10^6*[2.5 - 2*exp(0.56)] = -8k*exp(0.56)
>>k = 10^6*[2*exp(0.56) - 2.5]/[8*exp(0.56)] =
>> = 10^6*[1/4 - 5/16*exp(-0.56)] =
>> = 10^6*[4 - 5*exp(-0.56)]/16 ~ 71497 ~ 71500 corpses/year
It should be like this:
Initial conditions:
at t = 0, n = 2*10^6
at t = 8, n = 2.5*10^6
Plug this into the solution n(t):
t = 0 => 2*10^6 = C4
t = 8 => 2.5*10^6 = -8k + 2*10^6*exp(0.07*8)
10^6*[2.5 - 2*exp(0.56)] = -8k
k = 10^6*[2*exp(0.56) - 2.5]/8 = 125168 corpses/year
year population
0 2000000
1 2019848
2 2050211
3 2091852
4 2145587
5 2212294
6 2292914
7 2388456
8 2500000
The annual 7% increase can be taken either as increase 7% in
one year
(like the normal interest, or discrete model) or an increase
rate per
unit time (like the compound interest, or continuous model).
Either
model is hardly true for human population growth, the correct
model
would probably use a differential equation with about 20 years
retardation.
===
Subject: Re: sequence problem: recursion formula most likely
involved
alt.math.undergrad:
>> alt.math.undergrad:
>>how do you solve this question plz help!!! thx!!!
>>A current population is 2 000 000 and it increases 7%
annualy.
>>However each year a certain number of people die (this
amount is
>>constant for every year). How many people die each year if
the
>>population is to become 2 500 000 in 8 years?
> I was a little uneasy after hitting the <Send> button.
Mainly I made
> arithmetic errors at the end (as usual):
>Initial conditions:
>at t = 0, n = 2*10^6
>at t = 8, n = 2.5*10^6
>Plug this into the solution n(t):
>t = 0 => 2*10^6 = C4
>t = 8 => 2.5*10^6 = -8k + 2*10^6*exp(0.07*8)
>10^6*[2.5 - 2*exp(0.56)] = -8k*exp(0.56)
>k = 10^6*[2*exp(0.56) - 2.5]/[8*exp(0.56)] =
> = 10^6*[1/4 - 5/16*exp(-0.56)] =
> = 10^6*[4 - 5*exp(-0.56)]/16 ~ 71497 ~ 71500 corpses/year
> It should be like this:
> Initial conditions:
> at t = 0, n = 2*10^6
> at t = 8, n = 2.5*10^6
> Plug this into the solution n(t):
> t = 0 => 2*10^6 = C4
> t = 8 => 2.5*10^6 = -8k + 2*10^6*exp(0.07*8)
> 10^6*[2.5 - 2*exp(0.56)] = -8k
> k = 10^6*[2*exp(0.56) - 2.5]/8 = 125168 corpses/year
> year population
> 0 2000000
> 1 2019848
> 2 2050211
> 3 2091852
> 4 2145587
> 5 2212294
> 6 2292914
> 7 2388456
> 8 2500000
> The annual 7% increase can be taken either as increase 7%
in one year
> (like the normal interest, or discrete model) or an
increase rate per
> unit time (like the compound interest, or continuous model).
True, and only knowledge of the course in which the problem
was assigned can tell us for sure which was intended. But
my experience suggests that the given wording is likelier to
go with the discrete interpretation than with the continuous
one.
Naja; whichever was actually wanted, the original poster now
has something to work from.
[...]
Brian
===
Subject: Re: sequence problem: recursion formula most likely
involved
> alt.math.undergrad:
>how do you solve this question plz help!!! thx!!!
>A current population is 2 000 000 and it increases 7%
annualy.
>However each year a certain number of people die (this
amount is
>constant for every year). How many people die each year if
the
>population is to become 2 500 000 in 8 years?
>> t - time in years
>> n(t) - current population
>> k - number of corpses/year
>> 0.07 = 7%/year
>> Population change in time dt:
>> dn = 0.07*n*dt - k*dt
>> dn/dt = 0.07n - k
>> Solving the homogenous differential equation dn/dt =
0.07*n:
> [...]
>> k = 10^6*[2*exp(0.56) - 2.5]/[8*exp(0.56)] =
>> = 10^6*[1/4 - 5/16*exp(-0.56)] =
>> = 10^6*[4 - 5*exp(-0.56)]/16 ~ 71497 ~ 71500 corpses/year
> YouÕve used a continuous model to arrive at this 
figure, but
> the problem is stated in terms of a discrete model. The
> actual solution to the problem as stated is larger than your
> figure. I wonÕt give it away completely, since 
IÕd prefer
> for the original poster to arrive at it himself, but itÕs 
a
> bit over 90 000.
> Brian
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
What indicates that the model must be discrete?
Aristotle Polonium
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+
===
Subject: Optimization question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECnh00945;
Hey guys, if you could help me solve this, IÕd really
appreciate it:
1. Aerobic rate is the rate at which the body consumes
oxygen. This
rate as a function of age(x) is sometimes modeled by:
A(x)=110[lnx-2/x]
for x>10. At what age is aerobic rate maximized?
===
Subject: Re: Optimization question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9Ghp116042;
>Hey guys, if you could help me solve this, IÕd really
appreciate it:
>1. Aerobic rate is the rate at which the body consumes
oxygen. This
>rate as a function of age(x) is sometimes modeled by:
>A(x)=110[lnx-2/x]
>for x>10. At what age is aerobic rate maximized?
A(x) is a sum of 2 increasing functions 110*ln(x) and -220/x,
the
first diverging and the second with lim(-220/x)->0 for x->inf.
Cannot
have a local maximum. The equation for A(x) appears to be
incorrect.
Maybe something like A(x) = 110*(ln(x) - x/20) would work.
===
Subject: Finding an explicit formula ... PLEASE HELP!!!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECqu01101;
Not sure how to work out these types of problems... any
assistance
would be great.
~Find an explicit formula for for the sequence if you know:
(1) Characteristic equation: (r-3)(r+4) = 0; 1st term = a(1)
= 3, 2nd
term = a(2) =0.
(2) Characteristic equation: (r-2)(r+3) = 0; a(1) = 3, a(2)
=0.
(3) Characteristic equation: (r+3)^2 = 0; a(1) = 3, a(2) =0.
(4) Characteristic equation: (r-2)^2 = 0; a(1) = -1, a(2) =1.
===
Subject: Re: Finding an explicit formula ... PLEASE HELP!!!
alt.math.undergrad:
> Not sure how to work out these types of problems... any
assistance
> would be great.
IÕd have expected the textbook for the course to have at
least a couple of examples of problems of this type, and the
procedure is pretty mechanical, so IÕm not sure where the
difficulty is.
> ~Find an explicit formula for for the sequence if you know:
> (1) Characteristic equation: (r-3)(r+4) = 0; 1st term =
a(1) = 3, 2nd
> term = a(2) =0.
The characteristic equation has roots 3 and -4; this tells
you that the general solution to the recurrence is
(*) a(n) = A*3^n + B*(-4)^n,
where A and B are constants. You need to choose A and B so
that this general solution gives you the right values of
a(1) and a(2). According to (*),
a(1) = 3A - 4B
a(2) = 9A + 16B.
(I simply substituted n = 1 to get a(1) and n = 2 to get
a(2).) On the other hand, you know that a(1) = 3 and a(2) =
0, so you can set up the following equations:
3 = 3A - 4B
0 = 9A + 16B
Solving this system will give you the desired values of A
and B. Subtracting 3 times the first equation from the
second gives me -9 = 28B, B = -9/28, and hence A = -(16/9)B
= 4/7, so the sequence is
a(n) = (4/7)*3^n - (9/28)*(-4)^n.
> (2) Characteristic equation: (r-2)(r+3) = 0; a(1) = 3, a(2)
=0.
This is of the same type as (1); you should be able to solve
it using the same technique.
> (3) Characteristic equation: (r+3)^2 = 0; a(1) = 3, a(2) =0.
> (4) Characteristic equation: (r-2)^2 = 0; a(1) = -1, a(2)
=1.
These are a little different, since the characteristic
equation has a double root, rather than two distinct roots,
as in (1) and (2). When thereÕs a double root at s, say,
the general solution is
(**) a(n) = A*s^n + B*n*s^n.
For instance, in (3) the general solution is
a(n) = A*(-3)^n + B*n*(-3)^n.
But the rest of the solution works the same as before:
substitute n = 1 and n = 2 into the general solution to see
what it gives for a(1) and a(2), then set these expressions
equal to the known values of a(1) and a(2) and solve the
resulting system for A and B.
Brian
===
Subject: Finding explicit formulas.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECp601090;
Need help on what to do for this types of problems:
(+) Find an explicit formula for for the sequence if you know:
(1) Characteristic equation: (r-3)(r+4) = 0; 1st term = a(1)
= 3, 2nd
term = a(2) =0.
(2) Characteristic equation: (r-2)(r+3) = 0; a(1) = 3, a(2)
=0.
(3) Characteristic equation: (r+3)^2 = 0; a(1) = 3, a(2) =0.
(4) Characteristic equation: (r-2)^2 = 0; a(1) = -1, a(2) =1.
===
Subject: Optimization - Calculus (last one)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECnR00982;
Ok, here is the last one... ANY help is appreciated:
In an experiment, a biologist introduces a toxin into a
bacterial
colony and then measures the effect on the population of the
colony.
Suppose that at time t (in minutes) the population is:
P(t)=5 + e^(-.04t)(t+1)
thousand. at what time will the population be the largest?
Find where
the graph of P has an inßection point, and interpret the
meaning of
this point in terms of the population.
THANK YOU!
===
Subject: Re: Optimization - Calculus (last one)
> Suppose that at time t (in minutes) the population is:
> P(t)=5 + e^(-.04t)(t+1)
> thousand. at what time will the population be the largest?
Find where
> the graph of P has an inßection point, and interpret the
meaning of
> this point in terms of the population.
Take the derivative and set it equal to zero to find potential
extrema. Perform the 2nd derivative test to see if each
extremum is a max or a min.
Take the 2nd derivative and set it equal to zero to
find potential inßection points, etc.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Optimization - one more
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9ECnn00969;
HereÕs another one, these questions have confused me for a
couple of
days, your help is appreciated:
Question:
Research indicates that the power P required by a bird to
maintain ßight is given by the formula
P(v)=(w^2/2pSv) + 1/2 pAv^3
where v is the relative speed of the bird, w is its weight, p
is the
density of the air, and S and A are constants associated with
birdÕs
size and shape. What speed will minimize the power expended
by the
bird? Assume that w, p, S and A are positive constants.
THANKS!
===
Subject: Re: Optimization - one more
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9IO1j25090;
>HereÕs another one, these questions have confused me for a
couple of
>days, your help is appreciated:
>Question:
>Research indicates that the power P required by a bird to
>maintain ßight is given by the formula
>P(v)=(w^2/2pSv) + 1/2 pAv^3
>where v is the relative speed of the bird, w is its weight,
p is the
>density of the air, and S and A are constants associated
with birdÕs
>size and shape. What speed will minimize the power expended
by the
>bird? Assume that w, p, S and A are positive constants.
>THANKS!
At a local minimum, PÕ(v) = 0 and P(v) > 0:
P(v) = w^2/(2pSv) + 1/2*pAv^3
PÕ(v)= -w^2/(2pSv^2) + 3/2*pAv^2 = 0
3/2*pAv^4 - w^2/(2pS) = 0
v^4 - 2/3*pAw^2/(2pS) = 0
v^4 - Aw^2/(3S) = 0
[v^2 + w*sqrt(A/(3S)]*[v^2 - w*sqrt(A/(3S)] = 0
The first factor is always positive, so it must be
v^2 - w*sqrt(A/(3S) = 0
v = +-sqrt(w)*(A/3S)^(1/4)
Since v must be positive, the negative root is not acceptable.
PÕ(v) = -w^2/(2pSv^2) + 3/2*pAv^2
P(v) = w^2/(pSv^3) + 3*pAv > 0
for any v > 0, including v = +sqrt(w)*(A/3S)^(1/4), P(v) has
a local
minimum at this speed. (It is not necessary to plug this v
into
P(v)).
===
Subject: probability question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9IO0625061;
was having a lot of trouble on this homework problem due
tommorow
five cards are selected from an ordinary deck determine the
probabilities
a prob (2pair|allface)
b. prob (one pair|all face)
c prob (four of a kind|all face)
d prob (full house|all face)
e. Five persons draw each a card from an ordinary deck what
is the
chance that no two cards are of equal value.
===
Subject: Re: probability question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBA4xsS15721;
>was having a lot of trouble on this homework problem due
tommorow
>five cards are selected from an ordinary deck determine the
>probabilities
>a prob (2pair|allface)
>b. prob (one pair|all face)
>c prob (four of a kind|all face)
>d prob (full house|all face)
>e. Five persons draw each a card from an ordinary deck what
is the
>chance that no two cards are of equal value.
12 faces in a deck
C(12,5) = 12!/(7!*5!) = (12*11*10*9*8)/(5*4*3*2*1) = 792 all
possibilities to choose 5 faces
a/ two pairs
1st pair 3*C(4,2) = 3*6 = 18 possibilities
2nd pair must be of different kind:
2*C(4,2) = 2*6 = 12 possibilities
order of the pairs is imaterial (divide by 2):
total number of 2 pair possibilities is 18*12/2 = 108
5th card must be different from the 2 pairs:
4 possibilities
total number of possibilities is 108*4 = 432
P(a) = 432/C(12,5) = 432/792 = 6/11
b/ one pair
1st pair 3*C(4,2) = 3*6 = 18 possibilities
the 3rd, 4th, 5th cards must be different from each other
and from the 1st pair:
0 possibilities
P(b) = 0
c/ four of a kind:
3 possibilities for the 1st 4 cards,
5th card can be any of the remaining 8 faces:
total number of possibilities is 3*8 = 24
P(c) = 24/C(12, 5) = 24/792 = 1/33
d/ full house:
3*C(4,3) = 3*4 = 12 possibilities for the 3 of a kind
pair cannot be of the same kind:
2*C(4,2) = 2*6 = 12 possibilities for the pair
total number of possibilities is 12*12 = 144
P(d) = 144/C(12, 5) = 144/792 = 2/11
e/ 3 of a kind, no pair:
3*C(4,3) = 3*4 = 12 possibilities for the 3 of a kind
4th card must be different from the 3 of a kind:
8 possibilities:
5th card must be different from the 3 of a kind and from the
4th:
4 possibilities
order of the 4th and 5th card is imaterial (divide by 2):
total number of differet 4th and 5th card: 8*4/2 = 16
total number of possibilities is 12*16 = 192
P(e) = 1/C(12, 5) = 192/792 = 8/33
Since faces have only 3 different values, you have to get
something
when drawing 5 cards. Hence, and it should be
P(a) + P(b) + P(c) + P(d) + P(e) = 1
6/11 + 0 + 1/33 + 2/11 + 8/33 = (18 + 1 + 6 + 8)/33 = 33/33 =
1
as required.
===
Subject: Re: probability question
>was having a lot of trouble on this homework problem due
tommorow
>five cards are selected from an ordinary deck determine the
>probabilities
>a prob (2pair|allface)
>b. prob (one pair|all face)
>c prob (four of a kind|all face)
>d prob (full house|all face)
>e. Five persons draw each a card from an ordinary deck what
is the
>chance that no two cards are of equal value.
> 12 faces in a deck
> C(12,5) = 12!/(7!*5!) = (12*11*10*9*8)/(5*4*3*2*1) = 792 all
> possibilities to choose 5 faces
> a/ two pairs
> 1st pair 3*C(4,2) = 3*6 = 18 possibilities
> 2nd pair must be of different kind:
> 2*C(4,2) = 2*6 = 12 possibilities
> order of the pairs is imaterial (divide by 2):
> total number of 2 pair possibilities is 18*12/2 = 108
> 5th card must be different from the 2 pairs:
> 4 possibilities
> total number of possibilities is 108*4 = 432
> P(a) = 432/C(12,5) = 432/792 = 6/11
> b/ one pair
> 1st pair 3*C(4,2) = 3*6 = 18 possibilities
> the 3rd, 4th, 5th cards must be different from each other
> and from the 1st pair:
> 0 possibilities
> P(b) = 0
> c/ four of a kind:
> 3 possibilities for the 1st 4 cards,
> 5th card can be any of the remaining 8 faces:
> total number of possibilities is 3*8 = 24
> P(c) = 24/C(12, 5) = 24/792 = 1/33
> d/ full house:
> 3*C(4,3) = 3*4 = 12 possibilities for the 3 of a kind
> pair cannot be of the same kind:
> 2*C(4,2) = 2*6 = 12 possibilities for the pair
> total number of possibilities is 12*12 = 144
> P(d) = 144/C(12, 5) = 144/792 = 2/11
> e/ 3 of a kind, no pair:
> 3*C(4,3) = 3*4 = 12 possibilities for the 3 of a kind
> 4th card must be different from the 3 of a kind:
> 8 possibilities:
> 5th card must be different from the 3 of a kind and from
the 4th:
> 4 possibilities
> order of the 4th and 5th card is imaterial (divide by 2):
> total number of differet 4th and 5th card: 8*4/2 = 16
> total number of possibilities is 12*16 = 192
> P(e) = 1/C(12, 5) = 192/792 = 8/33
> Since faces have only 3 different values, you have to get
something
> when drawing 5 cards. Hence, and it should be
> P(a) + P(b) + P(c) + P(d) + P(e) = 1
> 6/11 + 0 + 1/33 + 2/11 + 8/33 = (18 + 1 + 6 + 8)/33 = 33/33
= 1
> as required.
Interesting answers, but I donÕt think they respond to the
questions that
were asked.
===
Subject: Re: probability question
>>was having a lot of trouble on this homework problem due
tommorow
>>five cards are selected from an ordinary deck determine the
>>probabilities
>>a prob (2pair|allface)
>>b. prob (one pair|all face)
>>c prob (four of a kind|all face)
>>d prob (full house|all face)
>>e. Five persons draw each a card from an ordinary deck what
is the
>>chance that no two cards are of equal value.
<snip> Since faces have only 3 different values, you have to
get something
>> when drawing 5 cards. Hence, it should be
>> P(a) + P(b) + P(c) + P(d) + P(e) = 1
>> 6/11 + 0 + 1/33 + 2/11 + 8/33 = (18 + 1 + 6 + 8)/33 =
33/33 = 1
>> as required.
>Interesting answers, but I donÕt think they respond to the
questions
>that were asked.
I did not answer question e/, because it was a homework with
no work
shown. Still, I wanted to help a student in trouble. In order
to
perform a check of the answers a/ to d/, I added an answer to
a
different question:
e/ prob(3 of a kind, no pair|all faces)
How else do not the answers correspond to the questions asked?
Vladimir
===
Subject: Re: probability question
> Still, I wanted to help a student in trouble.
I donÕt really think you help a student by doing problems in
such detail.
ItÕs usually better for the student, and requires more
ingenuity on your
part, to give a well chosen hint or two.
===
Subject: Re: probability question
>was having a lot of trouble on this homework problem due
tommorow
Show us what you did; weÕll help you over the hump.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
I do not believe in a personal god and I have never denied
this but
have expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of
the world so far as our science can reveal it.
-- Einstein, in /The Human Side/, ed H Dukas and B Hoffman
===
Subject: Please donÕt forget Mechanics - Friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9N37L18623;
Please donÕt forget Mechanics - Friction
message from 08 Dec
===
Subject: Re: Please donÕt forget Mechanics - Friction
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Please donÕt forget Mechanics - Friction
> message from 08 Dec
Are you still stuck after having read the earlier replies, or
have you
got it sorted now?
===
Subject: Partitions and mod 4
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBA00kd23567;
I have a question thats stumping me:
Let A={x(member of)Z| x mod 4 = 1} and let
B={x member of) Z| x mod 4= 3} Is {A,B} a partition of the
set off
all odd integers?
===
Subject: Re: Partitions and mod 4
George,
Yes.
Suppose not. Then there would exist an odd integer k such that
k(mod 4)=0 or k(mod 4)=2.
If k(mod 4)=0, k=4n=2(2n) for some integer n.
If k(mod 4)=2, k=4n+2=2(2n+1) for some integer n.
In either case, 2 divides k, a contradiction.
Travis
> I have a question thats stumping me:
> Let A={x(member of)Z| x mod 4 = 1} and let
> B={x member of) Z| x mod 4= 3} Is {A,B} a partition of the
set off
> all odd integers?
===
Subject: Re: Partitions and mod 4
days. My association with the Department is that of an
alumnus.
[correcting top-posting]
>> I have a question thats stumping me:
>> Let A={x(member of)Z| x mod 4 = 1} and let
>> B={x member of) Z| x mod 4= 3} Is {A,B} a partition of the
set off
>> all odd integers?
>Yes.
>Suppose not. Then there would exist an odd integer k such
that
>k(mod 4)=0 or k(mod 4)=2.
>If k(mod 4)=0, k=4n=2(2n) for some integer n.
>If k(mod 4)=2, k=4n+2=2(2n+1) for some integer n.
>In either case, 2 divides k, a contradiction.
This proves the set of all odd integers is contained in the
union of A
and B. But there are two more things to show in order to show
that
{A,B} is indeed a partition:
(i) Every element of A and every element of B is an odd
integer;
(ii) A and B are disjoint.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Partitions and mod 4
Arturo,
YouÕre right, although these are even more trivial than the
below; I
intended originally to supply only the interesting part of
the proof.
(i) If k=1(mod 4), we can write k=4n+1 for some integer n.
Then, k(mod 2)=(4n)(mod 2)+(1)(mod 2)=1(mod 2). A similar
argument hods
for k=3(mod 4). Thus, every element of the union of A and B
is in the
odd integers.
(ii) mod is well-defined (if the intersection of A,B was
something other
than the empty set, mod would not be well-defined for some
integer).
Travis
> [correcting top-posting]
>I have a question thats stumping me:
>Let A={x(member of)Z| x mod 4 = 1} and let
>B={x member of) Z| x mod 4= 3} Is {A,B} a partition of the
set off
>all odd integers?
>>Yes.
>>Suppose not. Then there would exist an odd integer k such
that
>>k(mod 4)=0 or k(mod 4)=2.
>>If k(mod 4)=0, k=4n=2(2n) for some integer n.
>>If k(mod 4)=2, k=4n+2=2(2n+1) for some integer n.
>>In either case, 2 divides k, a contradiction.
> This proves the set of all odd integers is contained in the
union of A
> and B. But there are two more things to show in order to
show that
> {A,B} is indeed a partition:
> (i) Every element of A and every element of B is an odd
integer;
> (ii) A and B are disjoint.
===
Subject: Question about Graduate Study in Mathematics
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBA2AFd02286;
Next year I will be pursuing graduate studies in mathematics
with a
focus on financial engineering. IÕve been told I 
have strong
candidacy for programs like UC-Berkeley, Princeton, Carnegie
Mellon,
Georgia Tech, among others. I have indeed applied to these
schools,
but I have always wanted to study mathematics out of country.
I recently recieved a letter from International University
Bremen
(www.math.iu-bremen.du), a mathematics institution in Germany
that
emphasizes many areas of mathematical research. I have never
heard of
this school before (my ignorance) but would like to know how
highly it
is regarded in the mathematical community. Can someone
compare the
prestige of this school to perhaps that of one in USA. IÕve
read a
little about this school, and it was established in 1999.
Since that
time, does anyone know how its reputation has been? Advice or
information is greatly appreciated.
Joseph A.
===
Subject: Moments
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBALNZp07440;
Could you help me solve:
A large uniform plank of length 12 m and mass 40 Kg is held in
equilibrium by two small rollers P and Q ready to be pushed
into a
cutting machine.
The centres of the rollers are 1.5 m apart and the plank
presses up
against P and down against Q. The plank remains horizontal
and the
exerted forces are vertical.
There is a little diagram that shows P on top of the left end
of the
plank and Q under the plank 1.5 m along the plank. To the
right of Q
there is nothing for the remaining 10.5m.
Find the magnitude and direction of the forces exerted on the
plank by
P and Q separately.
---
I am at a total loss as to where to start!
I can only assume that you divide the mass into two areas:
40/10.5 and
40/1.5
The force on P could be 26+2/3 * 9.8 = 261.33
But thatÕs wrong and not likely...
Could you throw some light pls?
===
Subject: Re: Moments
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Could you help me solve:
> A large uniform plank of length 12 m and mass 40 Kg is held
in
> equilibrium by two small rollers P and Q ready to be pushed
into a
> cutting machine.
> The centres of the rollers are 1.5 m apart and the plank
presses up
> against P and down against Q. The plank remains horizontal
and the
> exerted forces are vertical.
> There is a little diagram that shows P on top of the left
end of the
> plank and Q under the plank 1.5 m along the plank. To the
right of Q
> there is nothing for the remaining 10.5m.
> Find the magnitude and direction of the forces exerted on
the plank
by
> P and Q separately.
> ---
Three forces act on the plank: P and Q (to be found), and the
force of
gravity (known). The plank is in equilibrium, so these three
forces
must exactly balance out. Here, balancing out means two
things:
(i) The net vertical force acting on the plank must be zero
(else the
plank would move up or down). This gives you one equation
involving the
forces at P and Q. (If there were any horizontal forces then
these
would have to balance too, but the problem states that there
are not.)
(ii) The net moment (turning force) on the plank must also be
zero
(else the plank would rotate). Choose any point on the plank,
and
calculate the net moment around that point (i.e. the sum of
the moments
exerted by all three forces). Give regard to the sign of the
turning
force: anticlockwise positive, clockwise negative (or vice
versa, it
matters not). This gives you a second equation involving the
forces at
P and/or Q.
Now combine the equations from (i) and (ii) to find the values
of the
forces at P and Q.
In (ii) you can choose any point on the plank around which to
calculate
moments. P or Q are the obvious choices, simply because it
makes the
algebra a bit easier. But to check your method you might like
to verify
that the choice of any other point gives exactly the same
answer.
===
Subject: Re: Moments
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBBDUmX21455;
Rich and Matt
The penny has dropped!
Jo
>> Could you help me solve:
>> A large uniform plank of length 12 m and mass 40 Kg is
held in
>> equilibrium by two small rollers P and Q ready to be
pushed into a
>> cutting machine.
>> The centres of the rollers are 1.5 m apart and the plank
presses up
>> against P and down against Q. The plank remains horizontal
and the
>> exerted forces are vertical.
>> There is a little diagram that shows P on top of the left
end of
the
>> plank and Q under the plank 1.5 m along the plank. To the
right of
Q
>> there is nothing for the remaining 10.5m.
>> Find the magnitude and direction of the forces exerted on
the plank
>> P and Q separately.
>> ---
>Three forces act on the plank: P and Q (to be found), and
the force
of
>gravity (known). The plank is in equilibrium, so these three
forces
>must exactly balance out. Here, balancing out means two
things:
>(i) The net vertical force acting on the plank must be zero
(else the
>plank would move up or down). This gives you one equation
involving
the
>forces at P and Q. (If there were any horizontal forces then
these
>would have to balance too, but the problem states that there
are
not.)
>(ii) The net moment (turning force) on the plank must also
be zero
>(else the plank would rotate). Choose any point on the
plank, and
>calculate the net moment around that point (i.e. the sum of
the
moments
>exerted by all three forces). Give regard to the sign of the
turning
>force: anticlockwise positive, clockwise negative (or vice
versa, it
>matters not). This gives you a second equation involving the
forces
at
>P and/or Q.
>Now combine the equations from (i) and (ii) to find the
values of the
>forces at P and Q.
>In (ii) you can choose any point on the plank around which to
calculate
>moments. P or Q are the obvious choices, simply because it
makes the
>algebra a bit easier. But to check your method you might
like to
verify
>that the choice of any other point gives exactly the same
answer.
===
Subject: Re: Moments
> Could you help me solve:
> A large uniform plank of length 12 m and mass 40 Kg is held
in
> equilibrium by two small rollers P and Q ready to be pushed
into a
> cutting machine.
> The centres of the rollers are 1.5 m apart and the plank
presses up
> against P and down against Q. The plank remains horizontal
and the
> exerted forces are vertical.
> There is a little diagram that shows P on top of the left
end of the
> plank and Q under the plank 1.5 m along the plank. To the
right of Q
> there is nothing for the remaining 10.5m.
> Find the magnitude and direction of the forces exerted on
the plank by
> P and Q separately.
First, draw a diagram
Fq
^
|
P | C
| Q |
| |
v v
Fp mg
P is at 0m, Q is at 1.5m.
mg is the force due to gravity on the board. You need to
remember
that it acts at the center of mass of the board, therefore C
is at 6m.
Now itÕs just a matter of imposing static equilibrium
conditions:
(a) Sum of forces is zero.
(b) Sum of torques is zero.
For (a), taking up as positive, we have:
-Fp + Fq - mg = 0
IÕll leave (b) to you. Remember that if the torques around
a point are zero, then the torques around any point are zero.
So you can choose to compute the torques around whichever
point you view as most convenient.
When all is said and done, conditions (a) and (b) will
provide you with two equations in two unknowns (Fp and Fq).
You use standard linear system solving procedures to find
Fp and Fq.
And remember to check your work.
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: PDE problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBB2Omb01518;
I needed some help with a PDE and stumbled upon
mathforum.org. Hope
some one can help me here.
(r*C)_t = (r*C)_rr --- (1)
where C = f(r,t)
_t specifys partial derivative wrt t
_rr specifys partial derivative twice wrt r
if we substitute
(r-phi)
n = -----
(1-phi)
where phi = f(t)
we get
(r*C)_t - {(1-n)/(1-phi)} * d(phi)/dt * (r*C)_n
= 1/(1-phi)^2 * (r*C)_nn
where _n specifys partial derivative wrt n
In the above equation I could not figure out how we got the
2nd term
in LHS.
===
Subject: Re: PDE problem
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> I needed some help with a PDE and stumbled upon
mathforum.org. Hope
> some one can help me here.
> (r*C)_t = (r*C)_rr --- (1)
> where C = f(r,t)
> _t specifys partial derivative wrt t
> _rr specifys partial derivative twice wrt r
> if we substitute
> (r-phi)
> n = -----
> (1-phi)
> where phi = f(t)
> we get
> (r*C)_t - {(1-n)/(1-phi)} * d(phi)/dt * (r*C)_n
> = 1/(1-phi)^2 * (r*C)_nn
> where _n specifys partial derivative wrt n
> In the above equation I could not figure out how we got the
2nd term
> in LHS.
My knowledge of solving PDEs is rudimentary to say the least,
but for
fun I had a look at this. DidnÕt get far!
I donÕt want in any w, s or f to hijack the original
question, but can
I ask a (possibly dumb) supplementary one to satisfy my own
curiosity?
Suppose that:
r and t are independent variables
C is a function of r and t
n is a function of r and t
C is not a function of n alone
Then what is the definition of C_n (the partial derivative of
C wrt
n)?
What I want to do is infinitesimally perturb n, and see how C
changes, in the usual way. However, I can infinitesimally
perturb n
in an infinite number of different ways depending on the
relative
proportions in which I change r and t, all of which seem to
give
different answers for C_n.
WhatÕs the way out?
===
Subject: Re: PDE problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBBEpUs28184;
>I needed some help with a PDE and stumbled upon
mathforum.org. Hope
>some one can help me here.
>(r*C)_t = (r*C)_rr --- (1)
>where C = f(r,t)
> _t specifys partial derivative wrt t
> _rr specifys partial derivative twice wrt r
>if we substitute
> (r-phi)
> n = -----
> (1-phi)
>where phi = f(t)
>we get
>(r*C)_t - {(1-n)/(1-phi)} * d(phi)/dt * (r*C)_n
> = 1/(1-phi)^2 * (r*C)_nn
>where _n specifys partial derivative wrt n
>In the above equation I could not figure out how we got the
2nd term
>in LHS.
Write g(r,t)=r.f(r,t) weÕve got d/dt g(r,t)= d^2/dr^2 g(r,t) 
,
we have a direct solution g(r,t)=k*exp(a^2*t+a*r)...
come on,
Alain.
===
Subject: Re: PDE problem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC1vIt17365;
no its a part of a bigger problem. With some BCÕs and 
ICÕs
along with
a moving boundary. Such that analytical solution does not
exist and
the substitution is made to immobilize the moving front
(phi)..
the diff. eqn is valid in the range phi < r < 1
and converting wrt n changes the range to 0 < n < 1
now we can make a grid and apply finite difference technique..
but all that information is not needed.. what is bugging me
is the
origin of 2nd term in LHS..
thx.
>>I needed some help with a PDE and stumbled upon
mathforum.org. Hope
>>some one can help me here.
>>(r*C)_t = (r*C)_rr --- (1)
>>where C = f(r,t)
>> _t specifys partial derivative wrt t
>> _rr specifys partial derivative twice wrt r
>>if we substitute
>> (r-phi)
>> n = -----
>> (1-phi)
>>where phi = f(t)
>>we get
>>(r*C)_t - {(1-n)/(1-phi)} * d(phi)/dt * (r*C)_n
>> = 1/(1-phi)^2 * (r*C)_nn
>>where _n specifys partial derivative wrt n
>>In the above equation I could not figure out how we got the
2nd term
>>in LHS.
>Write g(r,t)=r.f(r,t) weÕve got d/dt g(r,t)= d^2/dr^2 
g(r,t)
,
>we have a direct solution g(r,t)=k*exp(a^2*t+a*r)...
>come on,
>Alain.
===
Subject: JSH: Factoring Integers
posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
IÕm a working mathematician, which means that I keep at my
various
research interests, basically, well, on a continual basis.
For posters who try to pigeon-hole me it means that you have
specialists in terms of which posts they tend to reply to,
like there
will be some who will reply if I talk about my prime counting
formula,
and others who will reply if I talk about the problem with
the ring of
algebraic integers, and still others, when I talk about
factoring
integers.
Oh yeah, so I have multiple research interests in number
theory and now
a tendency to write papers and send them to journals.
Who out there thinks this situation is going to go on
indefinitely with
these losers tracking my posts talking down my work?
They have the tiger by the tale, and soon it will all be over.
It takes a special kind of stupidity to go after a working
mathematician, even an amateur one who has a history of major
errors.
As you never know whatÕs coming next as mathematics itself 
is
infinite
in extent.
ThereÕs always more to learn.
A real mathematician knows that, which is why after a while,
even if
youÕre dealing with weird crap like IÕve been 
dealing with,
you donÕt
worry, but keep working.
Like, hereÕs some more research, as IÕm 
looking at integer
factorization this time.
I focus on the relatively basic quadratic
yx^2 + Ax - k = T
with all integers, where T is the target, A is picked (more
later) and
the rest are to be determined.
Next I have
n = sqrt(A^2 + 4ky)
and
n + 2T = sqrt(A^2 + 4ky + 4yT)
and squaring both and subtracting the first from the second I
have
quite simply that
y = n + T
and substituting for y in the first equation, and solving for
n, gives
n = (4k +/- sqrt(16k^2 + 16kT + 4A^2))/2
which is
n = (2k +/- sqrt(4k^2 + 4kT + A^2))
and you can complete the square inside that square root to get
n = (2k +/- sqrt((2k + T)^2 + A^2 - T^2))
which gives you k, from the factorization of A^2 - T^2, which
gives you
n, and then you can work your way back to get all of the
variables.
Then you solve
yz^2 + Az - k = 0
for z, to get the factorization
(a_1 x + b_1)(a_2 x + b_2) = T
as solving for z you get, -b_1/a_1, and -b_2/a_2 as your
solutions,
where any factors in common between the numerator and
denominator are
divided out.
So does it work? I donÕt know. I do know that the algebra
seems
straightforward, but I also know that such approaches tend to
lead to a
dependency on the factorization of T, that simply may be
obscured by
the various algebraic gyrations.
Here thereÕs reason to think that maybe that 
didnÕt occur as
an
essential relation at the heart of this approach comes from
n = sqrt(A^2 + 4ky)
and
n + 2T = sqrt(A^2 + 4ky + 4yT)
as you have
sqrt(A^2 + 4ky) = sqrt(A^2 + 4ky + 4yT) mod T
and that may mean that a simple way to factor was sitting
right under
the noses of quite a few people for quite a long time.
But thatÕs the possibility to be determined, as if this
works, then
itÕs a very big deal. Even if it doesnÕt, the 
expressions are
beautiful enough for it to be a curious thing to figure out
WHY it
doesnÕt work!
Mathematical research is a joy. Luckily for you, I like to
share the
joy.
James Harris
===
Subject: Re: JSH: Factoring Integers
posting-account=jEbKGQ0AAADZF1UpkDsHa5gkWBqABnUE
[the same old song]
> James Harris
A mule is an animal with long funny ears, kicks up at
anything he hears
His back is brawny but his brain is weak, heÕs just plain
stupid with a
stubborn streak.
And by the way, if you hate to go to school, you may grow up
to be a
mule.
No watermelons in winter huh?
===
Subject: Re: Factoring Integers
> For posters who try to pigeon-hole me it means that ...
Who pigeon holes you. What does pigeon hole mean in this
context?
> Oh yeah, so I have multiple research interests in number
theory and now
> a tendency to write papers and send them to journals.
OK. Do you have any tendency to get those papers published?
> They have the tiger by the tale, and soon it will all be
over.
Yes... the tiger has a tail that looks like a hammer... nice
mixed
metaphor.
> As you never know whatÕs coming next as mathematics itself
is infinite
> in extent.
How to you estimate the extent of mathematics, and how do you
determine that
it is infinite? More o the point, what is your point with this
talk of the
extent of mathematics?
> ThereÕs always more to learn.
You should give that a try.
> yx^2 + Ax - k = T
Where did the motivation come for this? Or is it another
random sequence of
chicken scratches that goes nowhere, again?
> So does it work? I donÕt know.
Does it work at what. You never made a point that could be
taken as
something that works.
> Mathematical research is a joy. Luckily for you, I like to
share the
> joy.
===
Subject: Re: JSH: Factoring Integers
> They have the tiger by the tale, and soon it will all be
over.
Once your brilliance is recognized, the rest of the world
will stand in
humble awe.
> It takes a special kind of stupidity to go after a working
> mathematician, even an amateur one who has a history of
major errors.
> As you never know whatÕs coming next as mathematics itself
is infinite
> in extent.
ThatÕs the argument that an infinite number of 
monkeys would
eventually
come up with a proof of FLT.
> A real mathematician knows that, which is why after a
while, even if
> youÕre dealing with weird crap like IÕve 
been dealing with,
you donÕt
> worry, but keep working.
What, me worry?
> Mathematical research is a joy. Luckily for you, I like to
share the
> joy.
You certainly brightened up my otherwise dull morning.
===
Subject: Re: Solutions to GrimaldiÕs Discrete and
Combinatorial Mathematics
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC1vJe17407;
someone is selling it on ebay
www.ebay.com.tw
>Does anyone know where to find complete olutions to R.
GrimaldiÕs
>Discrete and Combinatorial Mathematics: An Applied
Introduction
>(Addison-Wesley)?
>I know there is answers and partial solutions in the last
pages of
the
>book, but are there any more complete solutions available
anywhere?
>Preferable a StudentÕs Solutions Manual, or if someone has
published
>something on the net.
===
Subject: Moments of despair!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC1vJW17381;
IÕm having a lot of trouble understanding a simple exercise:
A girder of negligible mass and length 5m is suspended in a
horizontal
position by vertical cables attached at points 0.6m and 3.6m
from one
other a mass of 12 tonnes. Find the tension in the cables.
I have the worked answer but canÕt understand it:
(call the cables A and B)
10000g * 0.6 + 3B = 4.4 * 12000g
The first part is fine (10000g * 0.6)
The second (3B) I can just about get assuming that B here
already
takes into account the moment from the 12000 Kg mass; and so
it means
the TWO moments -left (3m) and right (1.4 * 12000g). Please
correct
me if I am wrong.
What really defeats me is the equality: why should all this
be equal
to 4.4 * 12000g . I can see where the 4.4 comes from but
canÕt relate
one side of the equation to the other.
Could someone have a go at explaining it?
Jo
===
Subject: Re: Moments of despair!
> IÕm having a lot of trouble understanding a simple 
exercise:
> A girder of negligible mass and length 5m is suspended in a
horizontal
> position by vertical cables attached at points 0.6m and
3.6m from one
> other a mass of 12 tonnes. Find the tension in the cables.
Again, diagrams are your friend
Fb Fc
^ ^
| |
A | | D
| B C |
| |
v v
(10T)g (12T)g
and
A = 0m
B = 0.6m
C = 3.6m
D = 5m
Again, you have to impose the conditions that
the forces sum to zero and that the torques (or
moments, as they are also called) sum to zero.
For the forces, taking up as positive.
Fb + Fc - (10T)g - (12T)g = 0
Fb + Fc = (22T)g
As for the moments, itÕs clear from what you posted that
the worked answer chose to compute moments around B.
Which is fine, since one can choose to compute them
around any point for something assumed to be in equilibrium.
Remember, a moment is the force times the lever arm.
LetÕs take anti-clockwise to be positive and clockwise
to be negative. And remember, weÕre computing moments
around B, so the lever arm lengths are distances from B.
(10T)g * (0.6m) + Fb * (0m) + Fc * (3m) - (12T)g * (4.4m) = 0
or
(10T)g * 0.6m + Fc * 3m = (12T)g * 4.4m
The (Fc*3) is because Fc is applied 3m from the point around
which the moments are being computed (3.6 - 0.6 = 3). And
the (12T)g * 4.4m is because the force due to the 12T mass
is being applied 4.4m from the point around which the moments
are being computed (5 - 0.6 = 4.4).
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: Moments of despair!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBCDJRg09945;
Hello and thank you Rich.
As always, with your explanations everything is very clear.
There is just a subsidiary point that IÕm wondering about.
Fb = 152880N and Fc = 62720N
Common sense would say that the tension nearer the heavier
mass would
be greater if the distances were the same. As Fc is even
further away
from the end than Fb I would expect Fc to be greater than Fb.
But it isnÕt! I tried to work out what would happen if Fc
moved to 4
m away from Fb and the value decreases, seeming to prove that
the
closer you are to the edge the smaller the tension.
So whatÕs going on here?
Jo
>> IÕm having a lot of trouble understanding a simple
exercise:
>> A girder of negligible mass and length 5m is suspended in a
horizontal
>> position by vertical cables attached at points 0.6m and
3.6m from
one
>> other a mass of 12 tonnes. Find the tension in the cables.
>Again, diagrams are your friend
> Fb Fc
> ^ ^
> | |
> A | | D
> | B C |
> | |
> v v
> (10T)g (12T)g
>and
> A = 0m
> B = 0.6m
> C = 3.6m
> D = 5m
>Again, you have to impose the conditions that
>the forces sum to zero and that the torques (or
>moments, as they are also called) sum to zero.
>For the forces, taking up as positive.
> Fb + Fc - (10T)g - (12T)g = 0
> Fb + Fc = (22T)g
>As for the moments, itÕs clear from what you posted that
>the worked answer chose to compute moments around B.
>Which is fine, since one can choose to compute them
>around any point for something assumed to be in equilibrium.
>Remember, a moment is the force times the lever arm.
>LetÕs take anti-clockwise to be positive and clockwise
>to be negative. And remember, weÕre computing moments
>around B, so the lever arm lengths are distances from B.
> (10T)g * (0.6m) + Fb * (0m) + Fc * (3m) - (12T)g * (4.4m) =
0
> (10T)g * 0.6m + Fc * 3m = (12T)g * 4.4m
>The (Fc*3) is because Fc is applied 3m from the point around
>which the moments are being computed (3.6 - 0.6 = 3). And
>the (12T)g * 4.4m is because the force due to the 12T mass
>is being applied 4.4m from the point around which the moments
>are being computed (5 - 0.6 = 4.4).
>--
>Rich Carreiro
rlcarr@animato.arlington.ma.us
===
Subject: Re: Moments of despair!
<td0veg29qmg0@legacy>
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Hello and thank you Rich.
> As always, with your explanations everything is very clear.
> There is just a subsidiary point that IÕm wondering about.
> Fb = 152880N and Fc = 62720N
> Common sense would say that the tension nearer the heavier
mass would
> be greater if the distances were the same. As Fc is even
further
away
> from the end than Fb I would expect Fc to be greater than
Fb.
> But it isnÕt! I tried to work out what would happen if Fc
moved to 4
> m away from Fb and the value decreases, seeming to prove
that the
> closer you are to the edge the smaller the tension.
> So whatÕs going on here?
> Jo
I think you may have got Fb and Fc the wrong way round in
your answer.
Check again ... maybe the confusion stemmed from the fact
that you
started with the cables at A and B and then the labelling got
changed
to B and C...
===
Subject: Sketching a Surface Defined Parametrically
IÕm having quite a bit of trouble working out how to 
properly
sketch a
surface defined parametrically. I am given:
x=cos(u)*cos(v)
y=u
z=cos(u)*sin(v)
The format my book says to follow is to determine what shape
this is,
and then verify that it is correct (for example, verifying
that x^2 +
y^2 - z^2 = c for a hyperboloid of one sheet).
Unfortunately, I have no idea what the formula should be for
the
resulting shape, and I can find no useful relations between x,
y, and z.
If someone could give me an idea of how to proceed, IÕd be
very grateful.
Tracy Poff
===
Subject: Re: Sketching a Surface Defined Parametrically
> IÕm having quite a bit of trouble working out how to
properly sketch a
> surface defined parametrically. I am given:
> x=cos(u)*cos(v)
> y=u
> z=cos(u)*sin(v)
> The format my book says to follow is to determine what
shape this is,
> and then verify that it is correct (for example, verifying
that x^2 +
> y^2 - z^2 = c for a hyperboloid of one sheet).
> Unfortunately, I have no idea what the formula should be
for the
> resulting shape, and I can find no useful relations between
x, y, and z.
> If someone could give me an idea of how to proceed, IÕd be
very grateful.
> Tracy Poff
One helpful approach is to look at the curves you get by
holding one
parameter fixed while the other varies.
For example, letting v vary for fixed u gives circles about
the y-axis
in the plane y = u (parallel to the xz-plane) of radius
|cos(u)|
===
Subject: Re: Sketching a Surface Defined Parametrically
> IÕm having quite a bit of trouble working out how to
properly sketch a
> surface defined parametrically. I am given:
> x=cos(u)*cos(v)
> y=u
> z=cos(u)*sin(v)
> The format my book says to follow is to determine what
shape this is,
> and then verify that it is correct (for example, verifying
that x^2 +
> y^2 - z^2 = c for a hyperboloid of one sheet).
> Unfortunately, I have no idea what the formula should be
for the
> resulting shape, and I can find no useful relations between
x, y, and z.
> If someone could give me an idea of how to proceed, IÕd be
very grateful.
> Tracy Poff
Before I start scratching my head, are you sure there are
_two_
parameters? With just one, I think you can easily work it.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Sketching a Surface Defined Parametrically
>> IÕm having quite a bit of trouble working out how to
properly sketch a
>> surface defined parametrically. I am given:
>> x=cos(u)*cos(v)
>> y=u
>> z=cos(u)*sin(v)
> Before I start scratching my head, are you sure there are
_two_
>parameters? With just one, I think you can easily work it.
No, sheÕs correct. A non-degenerate surface needs two
parameters.
--Lynn
===
Subject: Re: Sketching a Surface Defined Parametrically
> IÕm having quite a bit of trouble working out how to
properly sketch a
> surface defined parametrically. I am given:
> x=cos(u)*cos(v)
> y=u
> z=cos(u)*sin(v)
> Unfortunately, I have no idea what the formula should be
for the
> resulting shape, and I can find no useful relations between
x, y, and
> z.
Well, since y=u, you can immediately eliminate u.
x = cos(y) cos(v)
z = cos(y) sin(v)
And whenever you see things in the form of Asin(w) and Acos(w)
like that, square and add is always a good trick to try.
x^2 = [cos(y)]^2 * [cos(v)]^2
z^2 = [cos(y)]^2 * [sin(v)]^2
so
x^2 + z^2 = [cos(y)]^2 [(cos(v))^2 + (sin(v))^2]
x^2 + z^2 = (cos(y))^2 * 1
x^2 + z^2 = (cos(y))^2
So thereÕs a relation between x, y, z. Unfortunately, I 
donÕt
know what to do with it, either :-)
--
Rich Carreiro rlcarr@animato.arlington.ma.us
===
Subject: Re: Sketching a Surface Defined Parametrically
> Well, since y=u, you can immediately eliminate u.
> x = cos(y) cos(v)
> z = cos(y) sin(v)
> And whenever you see things in the form of Asin(w) and
Acos(w)
> like that, square and add is always a good trick to try.
> x^2 = [cos(y)]^2 * [cos(v)]^2
> z^2 = [cos(y)]^2 * [sin(v)]^2
> so
> x^2 + z^2 = [cos(y)]^2 [(cos(v))^2 + (sin(v))^2]
> x^2 + z^2 = (cos(y))^2 * 1
> x^2 + z^2 = (cos(y))^2
> So thereÕs a relation between x, y, z. Unfortunately, I
donÕt
> know what to do with it, either :-)
Yes, I got to that point during one of my attempts, but since
I donÕt
know any figures with /quite/ that formula, I was forced to
abandon it.
Tracy Poff
===
Subject: Re: Sketching a Surface Defined Parametrically
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC5iUt03093;
>> Well, since y=u, you can immediately eliminate u.
>> x = cos(y) cos(v)
>> z = cos(y) sin(v)
>> And whenever you see things in the form of Asin(w) and
Acos(w)
>> like that, square and add is always a good trick to try.
>> x^2 = [cos(y)]^2 * [cos(v)]^2
>> z^2 = [cos(y)]^2 * [sin(v)]^2
>> so
>> x^2 + z^2 = [cos(y)]^2 [(cos(v))^2 + (sin(v))^2]
>> x^2 + z^2 = (cos(y))^2 * 1
>> x^2 + z^2 = (cos(y))^2
>> So thereÕs a relation between x, y, z. Unfortunately, I
donÕt
>> know what to do with it, either :-)
>Yes, I got to that point during one of my attempts, but
since I donÕt
>know any figures with /quite/ that formula, I was forced to
abandon
it.
>Tracy Poff
Go to the cylindrical coordinates with the y-axis as the axis
of
rotation:
x = r*cos(fi)
y = y
z = r*sin(fi)
x^2 + z^2 = (cos(y))^2
r^2 = cos^2(y)
This might be called a rotational coisinusoid, i.e., a surface
rotationally symmetrical with respect to the y-axis, kind of
an
infinite cylinder with variable radius equal to r = |cos(y)|.
===
Subject: Re: Sketching a Surface Defined Parametrically
> Go to the cylindrical coordinates with the y-axis as the
axis of
> rotation:
> x = r*cos(fi)
> y = y
> z = r*sin(fi)
> x^2 + z^2 = (cos(y))^2
> r^2 = cos^2(y)
> This might be called a rotational coisinusoid, i.e., a
surface
> rotationally symmetrical with respect to the y-axis, kind
of an
> infinite cylinder with variable radius equal to r = 
|cos(y)|.
Tracy Poff
===
Subject: Orlicz spaces
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBC4tMS31706;
I need help with a proof of a Orlicz spaces is a Banach
spaces with
respect to the luxembur norm , please
===
Subject: Re: Orlicz spaces
> I need help with a proof of a Orlicz spaces is a Banach
spaces with
> respect to the luxembur norm , please
Define your terms and state the question like someone who is
beyond a first
fingerpainting course.
===
Subject: mean value theorem
hi, i am given 5cos(.5x) and i need all x values which
satisfy the M.V.T.
from x = pi to x = 9pi /2.
first i found the slope of the secant line from pi to 9pi /2,
which turns
out to be .322
then i set the derivative of the function equal to the slope
and i solved
for x:
-2.5 sin (.5x) = .322
sin (.5x) = -.129
.5x = -.129
x = -.257
looking at the graph of the original function, i can see that
the x values
satisfying the M.V.T. are positive for the period pi to 9pi/2
and can even
estimate the values at roughly x = 6.5 and 12, but i canÕt
work it out
mathematically............please help.
===
Subject: Re: mean value theorem
>hi, i am given 5cos(.5x) and i need all x values which
satisfy the M.V.T.
>from x = pi to x = 9pi /2.
I just took about 20 minutes to type a response to you in
alt.math.
Now I see that it is already answered in this newsgroup.
DonÕt you
think it is a bit inconsiderate of you to post the question
separately
like that and waste our time?
--Lynn
===
Subject: Re: mean value theorem
days. My association with the Department is that of an
alumnus.
>hi, i am given 5cos(.5x) and i need all x values which
satisfy the M.V.T.
>from x = pi to x = 9pi /2.
>first i found the slope of the secant line from pi to 9pi /2,
which turns
>out to be .322
First recommendation is that you work with ->exact<- values,
not
approximations.
>then i set the derivative of the function equal to the slope
and i solved
>for x:
>-2.5 sin (.5x) = .322
>sin (.5x) = -.129
>.5x = -.129
>x = -.257
>looking at the graph of the original function, i can see
that the x values
>satisfying the M.V.T. are positive for the period pi to
9pi/2 and can even
>estimate the values at roughly x = 6.5 and 12, but i canÕt
work it out
>mathematically............please help.
The step from sin(.5x) = -.129 to .5x = -.129 was presumably
taken by
appealing to the arcsin function. What you must remember is
that
arcsin returns the only value between -pi/2 and pi/2 which
has sine
equal to the argument (in this case, -.129). But in each
period of
sin(x) there are other values that will have the same sine.
Since the function (5/2)sin(x/2) has period pi, every pi you
will get
another example of a value; that is, all values
x = -.257 + pi*k k=...., -2, -1, 0, 1, 2, 3, ...
will satisfy -2.5 sin(.5x) = .322.
Of these, only some are in your original interval; find them.
But you arenÕt done yet. If you look at the graph of sin(x),
you might
realize that you found a value for .5x which lies betwen
-pi/2 and 0;
but there is another value of .5x that will also have the
same sine:
it is symmetrically placed on the other side of -pi/2; that
is, it
will be at -pi + .129; so you also have
x = (-pi/2)+.129 + pi*k k = ..., -2, -1, 0, 1, 2, 3, ...
as solutions to -2.5 sin(.5x) = .322.
Some of these will be in your desired interval. Find them.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: mean value theorem
>>hi, i am given 5cos(.5x) and i need all x values which
satisfy the M.V.T.
>>from x = pi to x = 9pi /2.
>>first i found the slope of the secant line from pi to 9pi
/2, which turns
>>out to be .322
>First recommendation is that you work with ->exact<- values,
not
approximations.
>>then i set the derivative of the function equal to the
slope and i solved
>>for x:
>>-2.5 sin (.5x) = .322
>>sin (.5x) = -.129
>>.5x = -.129
>>x = -.257
>>looking at the graph of the original function, i can see
that the x
values
>>satisfying the M.V.T. are positive for the period pi to
9pi/2 and can
even
>>estimate the values at roughly x = 6.5 and 12, but i canÕt
work it out
>>mathematically............please help.
>The step from sin(.5x) = -.129 to .5x = -.129 was presumably
taken by
>appealing to the arcsin function. What you must remember is
that
>arcsin returns the only value between -pi/2 and pi/2 which
has sine
>equal to the argument (in this case, -.129). But in each
period of
>sin(x) there are other values that will have the same sine.
>Since the function (5/2)sin(x/2) has period pi, every pi you
will get
>another example of a value; that is, all values
sin(x/2) has period 4 pi, not 2 pi.
sin(2x) would have period pi.
Brian
>x = -.257 + pi*k k=...., -2, -1, 0, 1, 2, 3, ...
>will satisfy -2.5 sin(.5x) = .322.
>Of these, only some are in your original interval; find them.
>But you arenÕt done yet. If you look at the graph of 
sin(x),
you might
>realize that you found a value for .5x which lies betwen
-pi/2 and 0;
>but there is another value of .5x that will also have the
same sine:
>it is symmetrically placed on the other side of -pi/2; that
is, it
>will be at -pi + .129; so you also have
>x = (-pi/2)+.129 + pi*k k = ..., -2, -1, 0, 1, 2, 3, ...
>as solutions to -2.5 sin(.5x) = .322.
>Some of these will be in your desired interval. Find them.
===
Subject: Re: mean value theorem
days. My association with the Department is that of an
alumnus.
>hi, i am given 5cos(.5x) and i need all x values which
satisfy the
M.V.T.
>from x = pi to x = 9pi /2.
>first i found the slope of the secant line from pi to 9pi /2,
which
turns
>out to be .322
>>First recommendation is that you work with ->exact<-
values, not
approximations.
>then i set the derivative of the function equal to the slope
and i
solved
>for x:
>-2.5 sin (.5x) = .322
>sin (.5x) = -.129
>.5x = -.129
>x = -.257
>looking at the graph of the original function, i can see
that the x
values
>satisfying the M.V.T. are positive for the period pi to
9pi/2 and can
even
>estimate the values at roughly x = 6.5 and 12, but i canÕt
work it out
>mathematically............please help.
>>The step from sin(.5x) = -.129 to .5x = -.129 was
presumably taken by
>>appealing to the arcsin function. What you must remember is
that
>>arcsin returns the only value between -pi/2 and pi/2 which
has sine
>>equal to the argument (in this case, -.129). But in each
period of
>>sin(x) there are other values that will have the same sine.
>>Since the function (5/2)sin(x/2) has period pi, every pi
you will get
>>another example of a value; that is, all values
>sin(x/2) has period 4 pi, not 2 pi.
Sigh. And here I am asking similar questions in a final. You
are
So this:
>>x = -.257 + pi*k k=...., -2, -1, 0, 1, 2, 3, ...
should be
x = -.256 + 4pi*k, k=...., -2, -1, 0, 1, 2, 3, ...
and:
>>x = (-pi/2)+.129 + pi*k k = ..., -2, -1, 0, 1, 2, 3, ...
should be
x = (-pi/2)+.129 + 4*pi*k k = ..., -2, -1, 0, 1, 2, 3, ...
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: transversals
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBCLoEw21854;
I have a question.... could you recommend any sites wher ei
can find
problems on transversals. I looked everywhere. please help
ASAP. thanx
===
Subject: Re: transversals
> I have a question.... could you recommend any sites wher ei
can find
> problems on transversals. I looked everywhere. please help
ASAP. thanx
Searching on mathworld.com for transversal brings up a number
of diverse
results. I hope some are helpful.
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Chain Rule as re: Antiderivatives
Does it apply?
My problem is simple enough: eval. the integral x^5 *
((x^6)-9)^7.
Essentially, this becomes x^5 * u^7, where u = (x^6)-9.
IÕm just not sure if I need to find antid 
x^5*u^7, or if itÕs
going to
become something like x^5*u^7 * antid. u.
===
Subject: Re: Chain Rule as re: Antiderivatives
>Does it apply?
Yes. I have a similar illustration at
http://www.acad.sunytccc.edu/instruct/sbrown/calc/usubst.htm
which is probably easier to read than the plain text below.
>My problem is simple enough: eval. the integral x^5 *
((x^6)-9)^7.
>Essentially, this becomes x^5 * u^7, where u = (x^6)-9.
>IÕm just not sure if I need to find antid 
x^5*u^7, or if itÕs
going to
>become something like x^5*u^7 * antid. u.
You started with the right idea, but you need to apply
substitutions
consistently across the entire integrand.
YouÕre trying to integrate x^5 ((x^6-9)^7 dx, right?
Write u = x^6-9; then du = 6 x^5 dx.
In your integrand you have x^5 but you need 6x^5, so you
rewrite the
expression as
INTEGRAL (1/6) 6 x^5 ((x^6-9)^7 dx
(1/6)*6 = 1, so you havenÕt changed anything.
Move the constant multiplier 16 across the integral sign,
since
INTEGRAL[c f(x) dx] = c * INTEGRAL[f(x) dx]:
(1/6) INTEGRAL 6 x^5 ((x^6-9)^7 dx
Now the integrand can be rewritten in terms of u:
(1/6) INTEGRAL u^7 du
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
I do not believe in a personal god and I have never denied
this but
have expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of
the world so far as our science can reveal it.
-- Einstein, in /The Human Side/, ed H Dukas and B Hoffman
===
Subject: Re: Chain Rule as re: Antiderivatives
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Does it apply?
> My problem is simple enough: eval. the integral x^5 *
((x^6)-9)^7.
> Essentially, this becomes x^5 * u^7, where u = (x^6)-9.
> IÕm just not sure if I need to find antid 
x^5*u^7, or if
itÕs going
to
> become something like x^5*u^7 * antid. u.
Hope this doesnÕt confuse more than it helps, and that 
IÕm not
encouraging you to do something different from what youÕre
being
taught...
Once youÕve done a few of these specially constructed
integration
problems, and provided youÕre ßuent in differentiation, 
you
find you
can do them by inspection without the rigmarole of going
through the
substitutions.
The key thing to spot is that differentiating x^6-9 gives
6x^5, and x^5
is exactly the required other factor.
Therefore, you immediately know that differentiating
(x^6-9)^8 will
give you an expression of the correct form with regard to the
xÕs.
Specifically,
d/dx[(x^6-9)^8] = d/dx(x^6-9) * 8(x^6-9)^7
= 6x^5 * 8(x^6-9)^7
= 48x^5(x^6-9)^7
Then, adjusting for the constant 48 you see that
d/dx[(1/48)(x^6-9)^8] = x^5(x^6-9)^7
and that gives you the answer.
Like I say, this only comes right in special cases ... but
these
quite often appear in textbook integration problems.
===
Subject: Re: Chain Rule as re: Antiderivatives
> Does it apply?
> My problem is simple enough: eval. the integral x^5 *
((x^6)-9)^7.
> Essentially, this becomes x^5 * u^7, where u = (x^6)-9. 
IÕm
just not sure
> if I need to find antid x^5*u^7, or if itÕs 
going to become
something like
> x^5*u^7 * antid. u.
The u-substitution is the anti-chain rule.
u = x^6-9
du/dx = 6x^5
Since it looks like a fraction, let
du = (6x^5)dx.
int(x^5*(x^6 - 9)^7)dx
= (1/6)int(6x^5*(x^6 - 9)^7)dx
= (1/6)int[(x^6 - 9)^7(6x^5dx)
= (1/6)int[u^7du]
= (1/6) u^8 / 8 + C
= (1/48)(x^6-9)^8 + C,
which can be checked by taking its derivative.
===
Subject: Re: Chain Rule as re: Antiderivatives
James,
You have the right idea. (When you apply the technique to
integrals in
this way, itÕs called u-substitution.)
We have:
int[x^5*((x^6-9)^7) dx]
Now, set u=x^6-9.
YouÕre right in that we need to integrate x^5*u^7, but this
isnÕt the
form we need; we need everything to be written in terms of x
or u.
(That way, we can integrate with respect to one variable or
the other.)
Taking the derivative of u:
du/dx = 6x^5
We want to rearrange this so that an expression in terms of x
that looks
like something in our original integral is equal to an
expression in u.
Multiplying both sides by (dx/6), we have:
(x^5)dx=du/6
So, our integral is:
int[((x^6-9)^7)(x^5 dx)]
=int[u^7 (du/6)]
=(1/6)int[u^7 du]
=(1/6)*(u^8)/8
=(u^8)/48
WeÕve integrate, but weÕre not done: This 
formula doesnÕt
help anyone
who doesnÕt know how we defined u.
Substituting (using u=x^6-9),
(u^8)/8=((x^6-9)^8)/48
Travis
> Does it apply?
> My problem is simple enough: eval. the integral x^5 *
((x^6)-9)^7.
> Essentially, this becomes x^5 * u^7, where u = (x^6)-9.
> IÕm just not sure if I need to find antid 
x^5*u^7, or if
itÕs going to
> become something like x^5*u^7 * antid. u.
===
Subject: BOOK DIFFERENTIAL EQUATIONS ON SALE
Hi everybody,
I sell the book Elementary differential equations and
boundary value
problems 7th edition by Boyce and DiPrima (required for ODE
mat3270) for
150 HKD. Please contact me on my phone: 915 27 847
===
Subject: Can you help me solve?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBDELrv06852;
hi
Can you help me solve these three equations?
Find x
x^(1/3) + 2*x^(2/3)=3
Second one IÕm not shure how to proceed
sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of two
3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
8*sqrt(x+2)=4x+8+4-3x+2
8*sqrt(x+2)=x+14 /again on power of two
and now what?
( 2^0.5 + 0.125^(-1/3) )^(-1) * sqrt(3+2*sqrt(2))=
(2^(1/2) + 2 )^(-1) * ( 3+2*2^(1/2) )^(1/2)=
(2^(3/2))^(-1) * ( 3+2*2^(1/2) )^(1/2)=
(2^(-3/2))^2 * ( 3+2*2^(1/2) )^(1/2)=
sqrt( 2^(-3) * ( 3+2*2^(1/2) ))=...
I know IÕm doing something wrong
thank you
===
Subject: Re: Can you help me solve?
>sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of two
>3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
>8*sqrt(x+2)=4x+8+4-3x+2
>8*sqrt(x+2)=x+14 /again on power of two
> and now what?
Now square again.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
I do not believe in a personal god and I have never denied
this but
have expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of
the world so far as our science can reveal it.
-- Einstein, in /The Human Side/, ed H Dukas and B Hoffman
===
Subject: Re: Can you help me solve?
>x^(1/3) + 2*x^(2/3)=3
What do you notice about the two powers of x in that
equation? Does
that suggest a useful substitution?
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
I do not believe in a personal god and I have never denied
this but
have expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of
the world so far as our science can reveal it.
-- Einstein, in /The Human Side/, ed H Dukas and B Hoffman
===
Subject: Re: Can you help me solve?
> hi
> Can you help me solve these three equations?
> Find x
> x^(1/3) + 2*x^(2/3)=3
I assume you know what the solution of a general quadratic
is. Note
that this equation is quadratic in x^(1/3). Solve for x^(1/3)
and then
x.
> Second one IÕm not shure how to proceed
> sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of two
> 3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
> 8*sqrt(x+2)=4x+8+4-3x+2
> 8*sqrt(x+2)=x+14 /again on power of two
> and now what?
Why did you square the equation in the first place? 
ThatÕs
right to
get rid of the radical, so do it again. Now the equation is a
quadratic in x, solve for x.
> ( 2^0.5 + 0.125^(-1/3) )^(-1) * sqrt(3+2*sqrt(2))=
> (2^(1/2) + 2 )^(-1) * ( 3+2*2^(1/2) )^(1/2)=
> (2^(3/2))^(-1) * ( 3+2*2^(1/2) )^(1/2)=
a^k + a^j is not = a^(k+j)
(a^k)(a^j) = a^(k+j)
2^(1/2) + 2 is not = 2^(3/2)
> (2^(-3/2))^2 * ( 3+2*2^(1/2) )^(1/2)=
> sqrt( 2^(-3) * ( 3+2*2^(1/2) ))=...
> I know IÕm doing something wrong
> thank you
===
Subject: Re: Can you help me solve?
> hi
> Can you help me solve these three equations?
> Find x
> x^(1/3) + 2*x^(2/3)=3
Solve for quadratic in x^(1/3)
> Second one IÕm not shure how to proceed
> sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of two
> 3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
> 8*sqrt(x+2)=4x+8+4-3x+2
> 8*sqrt(x+2)=x+14 /again on power of two
> and now what?
Square both sides, solve the quadratic, and check for
extraneous
roots.
> ( 2^0.5 + 0.125^(-1/3) )^(-1) * sqrt(3+2*sqrt(2))=
> (2^(1/2) + 2 )^(-1) * ( 3+2*2^(1/2) )^(1/2)=
> (2^(3/2))^(-1) * ( 3+2*2^(1/2) )^(1/2)=
> (2^(-3/2))^2 * ( 3+2*2^(1/2) )^(1/2)=
> sqrt( 2^(-3) * ( 3+2*2^(1/2) ))=...
> I know IÕm doing something wrong
CanÕt tell what youÕre trying to do here.
> thank you
--
-- Geo. Michael Henry
No! Bad dog! I said sit! anonymous
===
Subject: Re: l_p norm-spaces
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBDGJW520287;
>I have the following problem: I need to prove that
>||x|| = (|x1|^p+|x2|^p+...)^1/p
>where x = (x1,x2,...) is a vector of the R^infty euclidean
space,
>defines a norm in this space (called the l_p space)
>N1: ||x||=0 <=> x =0
>N2: ||a*x|| = |a|*||x|| where a in R
>N3: ||x+y|| <= ||x|| + ||y||
>for any members of l_p.
>here p can attain any real values. The two first requirements
are not
>a problem, but the third (called the triangle inequality I
think) is,
>a BIG one. I just canÕt prove that the norm 
satisfies it. Any
help
>would be greatly appreciated.
Triangle inequality takes from classicalinequalities:
(SUM(Ai+Bi)^p)^1/p <= (SUMAi^p)^1/p+(SUMBi^p)^1/p (Minkowski
inequality), for all p>=1, so this is thw norm for p>=1
===
Subject: Re: Find the zeros of the function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBDK3P009243;
>Help!!!!
>g(x)=x^3 + 4x^2 +14x +20, zero: -1-3i
saves a lot of time by not doing the cubic formula for this
one,
because one zero is already given
(x-(-1+3i))(x-(-1-3i)(b)= x^3+4x^2+14x+20
and then you solve for b from there, you know that its
conjugate also
has to be a zero.
===
Subject: Re: Can you help me solve?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBDLiJl18018;
hi
> x^(1/3) + 2*x^(2/3)=3
>>Solve for quadratic in x^(1/3)
You mean I should try and factor it?
> sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of
two
> 3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
> 8*sqrt(x+2)=4x+8+4-3x+2
> 8*sqrt(x+2)=x+14 /again on power of two
> and now what?
> and now what?
>>Now square again.
If I square it again,I get
x^2+64x-68=0
I can only think of factoring in order to solve this,but I
donÕt think
it can be factored
> ( 2^0.5 + 0.125^(-1/3) )^(-1) * sqrt(3+2*sqrt(2))=
> (2^(1/2) + 2 )^(-1) * ( 3+2*2^(1/2) )^(1/2)=
> (2^(3/2))^(-1) * ( 3+2*2^(1/2) )^(1/2)=
> (2^(-3/2))^2 * ( 3+2*2^(1/2) )^(1/2)=
> sqrt( 2^(-3) * ( 3+2*2^(1/2) ))=...
> I know IÕm doing something wrong
>>CanÕt tell what youÕre trying to do here.
IÕm trying to compute this without using my calculator
thank you
===
Subject: Re: Can you help me solve?
Please fix your newsreader. It seems to be putting more >
characters before more recent material, which is exactly
backward
from the convention that has been accepted for over 15 years.
I have
fixed them for you, this time.
Oh yes, and please use a proper attribution line.
> x^(1/3) + 2*x^(2/3)=3
>>Solve for quadratic in x^(1/3)
>You mean I should try and factor it?
1. There is another common method for solving quadratics
besides
factoring, namely the formula.
2. What must you do to the equation before you can apply
_either_
method?
> 8*sqrt(x+2)=x+14
> and now what?
>>Now square again.
>If I square it again,I get x^2+64x-68=0
I donÕt believe so. I get x^2 - 36x + 196 = 0
>I can only think of factoring in order to solve this,but I
donÕt think
>it can be factored
The correct equation can be (if IÕve computed correctly), 
but
even if
it couldnÕt you have the formula.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
I do not believe in a personal god and I have never denied
this but
have expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of
the world so far as our science can reveal it.
-- Einstein, in /The Human Side/, ed H Dukas and B Hoffman
===
Subject: Re: Can you help me solve?
>> x^(1/3) + 2*x^(2/3)=3
>Solve for quadratic in x^(1/3)
>You mean I should try and factor it?
People have suggested that you substitute, say u = x^(1/3).
Doing that
is based on seeing that the second term with a variable has
it squared
relative to the first. That is, x^(2/3) is the square of
x^(1/3). Try
it!
>> sqrt(3x-2) =2*sqrt(x+2)-2 /I put both sides on power of
>two
>> 3x-2 =4*(x+2) - 8*sqrt(x+2) + 4
>> 8*sqrt(x+2)=4x+8+4-3x+2
>> 8*sqrt(x+2)=x+14 /again on power of two
>> and now what?
>> and now what?
>Now square again.
>If I square it again,I get
>x^2+64x-68=0
I donÕt think so. Try again. If you still get that, show us
how you
did the squaring, in detail.
>I can only think of factoring in order to solve this,but I
donÕt think
>it can be factored
Regardless of above... surely you know other ways to solve
quadratics
besides factoring. Factoring is the method least likely to
work. Never
hurts to use the quadratic formula.
>> ( 2^0.5 + 0.125^(-1/3) )^(-1) * sqrt(3+2*sqrt(2))=
>> (2^(1/2) + 2 )^(-1) * ( 3+2*2^(1/2) )^(1/2)=
hard to read, but ok, i think
>> (2^(3/2))
no. That was a + sign in there, not a *.
bob
>^(-1) * ( 3+2*2^(1/2) )^(1/2)=
>> (2^(-3/2))^2 * ( 3+2*2^(1/2) )^(1/2)=
>> sqrt( 2^(-3) * ( 3+2*2^(1/2) ))=...
>> I know IÕm doing something wrong
>CanÕt tell what youÕre trying to do here.
>IÕm trying to compute this without using my calculator
>thank you
===
Subject: Re: need an online course in discrete
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBDMhdK23289;
>I am searching for an online course for this summer in
discrete
>mathematics. It is the last course I need to fill math
requirements.
>Help!
Did you ever find a course? I am starting to look now.
===
Subject: Linear Algebra, Please help!
Question:
Given a symmetric nxn matrix A, show there is an orthogonal
mx U
of determinant 1 with U^(-1)AU= D a diagonal matrix.
This is what I have so far:
Suppose X1.....Xn is a basis of eignvalues of the nxn matrix A
of eigen values lambda....(lambda)n.
Set C= (X1.....Xn) (c is invertible)
Claim that if D is the diagonal (lambda)...(lambda)n then
CD=AC
Then the jth column of CD is CDj=C(0
.
(lambda)j
.
0)
= ((lambda)j)Xj=AXj
D=C^(-1)AC
if orthonormal, c is orthogonal. when c is orthogonal c
inverse
= C^(T)
so C^(T)AC if Ax1......Xn
C^(-1)AC=C^(T)AC
----------------------------------------------
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===
Subject: Quadratic Functions...
I need help determing the points to plot after solving
quadratic functions
that I must solve using the quadratic function, no other way
right now.
f(x)=x^2+2x+1
i use h=-b/2a to get -1
and plug f(-1) to get 0
and f(0)=1
i then use the quadratic equation to get x1x2=-1
I also note that a>0 , so the parabola opens up.
now my question is what points do I use to graph this thing?
KuLL@cox-internet.com
jkc014@latech.edu
Kennyz79@gmail.com
===
Subject: Re: Quadratic Functions...
Kenny,
Can you say more about what you mean by the plots to point?
One point thatÕs probably relevant here, and 
IÕm not sure
whether you
know it: Because (h,k) is the vertex of the graph, and f
relates a
horizontal coordinate to a vertical one, k=f(h).
You know that h=-1. So, k=f(h)=(-1)^2+2(-1)+1=0.
Travis
> I need help determing the points to plot after solving
quadratic
functions
> that I must solve using the quadratic function, no other
way right now.
> f(x)=x^2+2x+1
> i use h=-b/2a to get -1
> and plug f(-1) to get 0
> and f(0)=1
> i then use the quadratic equation to get x1x2=-1
> I also note that a>0 , so the parabola opens up.
> now my question is what points do I use to graph this thing?
> KuLL@cox-internet.com
> jkc014@latech.edu
> Kennyz79@gmail.com
===
Subject: Determine the equation of the quadratic function
from the
information given...
I need help with three problem types:
1) Vertex is at (1,1), y-intercept 3.
I have to put all of these in terms of f(x)=a(x-h)^2+k
2) y-intercept is 1, zeroes at plusminus2.
3) y-intercept is -2, goes through (1,-1) and (-1,-1).
if I have the vertex and 2points (x,y.) I just do not
understand the
variations.
jkc014@latech.edu
Kennyz79@gmail.com
KuLL@cox-internet.com
===
Subject: Re: Determine the equation of the quadratic function
from the
information
given...
Kenny,
IÕm not sure whether problem (1) is a question, or an
illustration of
what you already know, but just in case: You know that h=k=1.
Substituting, you get:
f(x)=a(x-1)^2+1
The y-intercept is 3; so, if x=0, f(x)=3. Substitute this in
the above
to find a.
You know how to solve them if you have a vertex and two
points. Your
problem (2) gives you exactly that: If your quadratic has
zeroes at 2,
-2, what two points must be on the graph? Alternately, you
know that a
quadratic must be symmetric about the line x=h. The points on
the graph
where y=0 must be mirror images of each other; you know what
these are,
so you can deduce h from this.
For problem (3), use the above symmetry argument again.
Travis
> I need help with three problem types:
> 1) Vertex is at (1,1), y-intercept 3.
> I have to put all of these in terms of f(x)=a(x-h)^2+k
> 2) y-intercept is 1, zeroes at plusminus2.
> 3) y-intercept is -2, goes through (1,-1) and (-1,-1).
> if I have the vertex and 2points (x,y.) I just do not
understand the
> variations.
> jkc014@latech.edu
> Kennyz79@gmail.com
> KuLL@cox-internet.com