mm-1259
>>If I were to give your two vectors, how would you go about
\[CapitalThorn]nding the
>>parallelogram formed by them?
>>Last night I thought that that was what the cross-product
formed. Now I
>>have no idea why I came to that conclusion.
>>...
>>And what of the cross-product. Is it just some point in
space that
>>doesnt even lie on the parallelogram? If so is its sole
relationship to
>>the parallelogram the fact that its magnitude (length) is
equal to the
>>area of the parallelogram (in square units)?
> Some questions to get your brain re-focussed:
> (1) What is the area of a triangle? Given the base and the
height?
> (2) What is the area of a triangle? Given two sides and the
included
angle?
> (3) Can you construct a parallelogram from, say, two
triangles?
> (4) What was your formula for ||w||?
> (5) Does the direction of w have anything to do with the
area of the
parallelogram? Its plane? etc.
> Tomasso.
Showing on a 3d graph that the formulae |u||v|sin& is the
area of the
paralleagram.
But your questions 4 & 5 are where I am stuck.
My formula for ||w|| is
|w| = a_1*b_1i + a_2*b_2j + a_3*b_3k
= -1j + 3j + 7k
Does the direction of w have anything to do with the area of
the
parallelogram, and its plane? I have no idea. I am assuming
by your
question that it does. So I will go look. But all incite
would be
appreciated, because I am really clueless on this one.
Cassandra
===
Subject: Re: Magnitude and area of a parallelogram
> But your questions 4 & 5 are where I am stuck.
> My formula for ||w|| is
> |w| = a 1*b 1i + a 2*b 2j + a 3*b 3k
> = -1j + 3j + 7k
Nearly. That is w (vector).
||w|| = area of parallelogram = sqrt (1+9+49) = your answer.
w is normal to the plan of the parallelogram.
Tomasso.
===
Subject: Re: Magnitude and area of a parallelogram
>>But your questions 4 & 5 are where I am stuck.
>>My formula for ||w|| is
>>|w| = a_1*b_1i + a_2*b_2j + a_3*b_3k
>> = -1j + 3j + 7k
> Nearly. That is w (vector).
> ||w|| = area of parallelogram = sqrt (1+9+49) = your answer.
> w is normal to the plan of the parallelogram.
> Tomasso.
had thought that area of the parallelogram to be sqrt(1 + 9 +
7^2). (But
wasnt sure, and I appreciate the \
con\[CapitalThorn]rmation).
doing the following:
sqrt((-1)^2 + 3^2 + 7^2) = sqrt(1 + 9 + 7) = sqrt(17).
I so much appreciated being able to show my results. Just to
be able to
check that I havent done any silly typos like the above.
Cassandra
===
Subject: Proving the cross product is orthogonal
I hope I dont seem like one of those louts that only turn \
up
at
assignment time!! However I am struggling with a proof. My
answer is not
coming out as expected, and I think I am making a silly
mistake
somewhere. If anyone could have a glance over it, that would
be
appreciated.
Q) a =(1, -2, 1) b =(3, 1, 0)
Find a x b and prove that your cross-product vector is
perpendicular to
each of the vectors a and b.
A) [I have worked out the cross-product, and I have proven
that a is
perpendicular to the cross-product, however I am struggling
to prove
that b is perpendicular to the cross-product]
a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
= -3i + 2j + 7k
Let w = -3i + 2j + 7k
Prove vector a is perpendicular to w
a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
cos&= (a.w)/(|a||w|) = 0
Prove vector a is perpendicular to w
b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
I know that is wrong, can anyone see what my mistkae is??
Cassandra
===
Subject: Re: Proving the cross product is orthogonal
> I hope I dont seem like one of those louts that only turn
up at
> assignment time!! However I am struggling with a proof. My
answer is not
> coming out as expected, and I think I am making a silly
mistake
> somewhere. If anyone could have a glance over it, that
would be
appreciated.
> Q) a =(1, -2, 1) b =(3, 1, 0)
> Find a x b and prove that your cross-product vector is
perpendicular to
> each of the vectors a and b.
> A) [I have worked out the cross-product, and I have proven
that a is
> perpendicular to the cross-product, however I am struggling
to prove
> that b is perpendicular to the cross-product]
Do you know about the dot product?
===
Subject: Re: Proving the cross product is orthogonal
>>I hope I dont seem like one of those louts that only turn
up at
>>assignment time!! However I am struggling with a proof. My
answer is not
>>coming out as expected, and I think I am making a silly
mistake
>>somewhere. If anyone could have a glance over it, that
would be
appreciated.
>>Q) a =(1, -2, 1) b =(3, 1, 0)
>>Find a x b and prove that your cross-product vector is
perpendicular to
>>each of the vectors a and b.
>>A) [I have worked out the cross-product, and I have proven
that a is
>>perpendicular to the cross-product, however I am struggling
to prove
>>that b is perpendicular to the cross-product]
> Do you know about the dot product?
Yes, I like the dot product...much easier then the cross
product I think!
It turned out my main mistake was silly mistakes in my
workings (ie 2*0
= 2 etc) I did this two places, I think it was only ßuke that
my
workings for a being perpendicular to a x b, after the silly
errors
where \[CapitalThorn]xed both proofs worked.
Cassandra
===
Subject: Re: Proving the cross product is orthogonal
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4GMN503869;
>I hope I dont seem like one of those louts that only turn
up at
>assignment time!! However I am struggling with a proof. My
answer is
not
>coming out as expected, and I think I am making a silly
mistake
>somewhere. If anyone could have a glance over it, that would
be
appreciated.
>Q) a =(1, -2, 1) b =(3, 1, 0)
>Find a x b and prove that your cross-product vector is
perpendicular
to
>each of the vectors a and b.
>A) [I have worked out the cross-product, and I have proven
that a is
>perpendicular to the cross-product, however I am struggling
to prove
>that b is perpendicular to the cross-product]
>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
> = -3i + 2j + 7k
Right heres your mistake. You seem to be thinking -2*0 = -2
and 1*0 = 1. Oops!
>Let w = -3i + 2j + 7k
>Prove vector a is perpendicular to w
>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>cos&= (a.w)/(|a||w|) = 0
>Prove vector a is perpendicular to w
>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
>I know that is wrong, can anyone see what my mistkae is??
Todd Trimble
===
Subject: Re: Proving the cross product is orthogonal
>>I hope I dont seem like one of those louts that only turn
up at
>>assignment time!! However I am struggling with a proof. My
answer is
> not
>>coming out as expected, and I think I am making a silly
mistake
>>somewhere. If anyone could have a glance over it, that
would be
> appreciated.
>>Q) a =(1, -2, 1) b =(3, 1, 0)
>>Find a x b and prove that your cross-product vector is
perpendicular
> to
>>each of the vectors a and b.
>>A) [I have worked out the cross-product, and I have proven
that a is
>>perpendicular to the cross-product, however I am struggling
to prove
>>that b is perpendicular to the cross-product]
>>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
>> = -3i + 2j + 7k
> Right heres your mistake. You seem to be thinking -2*0 = \
-2
> and 1*0 = 1. Oops!
>>Let w = -3i + 2j + 7k
>>Prove vector a is perpendicular to w
>>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
>>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
>>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>>cos&= (a.w)/(|a||w|) = 0
>>Prove vector a is perpendicular to w
>>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
>>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
>>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
>>I know that is wrong, can anyone see what my mistkae is??
> Todd Trimble
Big Ôoops alright, I checked that about 5 times \
and still
didnt see
it...*sigh*
===
Subject: Re: Proving the cross product is orthogonal
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4GMNR03877;
>I hope I dont seem like one of those louts that only turn
up at
>assignment time!! However I am struggling with a proof. My
answer is
not
>coming out as expected, and I think I am making a silly
mistake
>somewhere. If anyone could have a glance over it, that would
be
appreciated.
>Q) a =(1, -2, 1) b =(3, 1, 0)
>Find a x b and prove that your cross-product vector is
perpendicular
to
>each of the vectors a and b.
>A) [I have worked out the cross-product, and I have proven
that a is
>perpendicular to the cross-product, however I am struggling
to prove
>that b is perpendicular to the cross-product]
>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
> = -3i + 2j + 7k
>Let w = -3i + 2j + 7k
>Prove vector a is perpendicular to w
>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>cos&= (a.w)/(|a||w|) = 0
>Prove vector a is perpendicular to w
>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
>I know that is wrong, can anyone see what my mistkae is??
>Cassandra
Yes, cosine= (b.w)/(|b||w|)= 0 but did you notice that that
only
requires that b.w be 0? It is not necessary to \[CapitalThorn]nd |b| or \
|w|.
As to what you did wrong, I hate to say it but: Arithmetic!
you have:
a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
= -3i + 2j + 7k
Now what, exactly, is -2*0-1*1? 1*3- 1*0?
You seem to be under the impression that a*0= a.
Most people say a*0= 0.
===
Subject: Re: Proving the cross product is orthogonal
>>I hope I dont seem like one of those louts that only turn
up at
>>assignment time!! However I am struggling with a proof. My
answer is
> not
>>coming out as expected, and I think I am making a silly
mistake
>>somewhere. If anyone could have a glance over it, that
would be
> appreciated.
>>Q) a =(1, -2, 1) b =(3, 1, 0)
>>Find a x b and prove that your cross-product vector is
perpendicular
> to
>>each of the vectors a and b.
>>A) [I have worked out the cross-product, and I have proven
that a is
>>perpendicular to the cross-product, however I am struggling
to prove
>>that b is perpendicular to the cross-product]
>>a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
>> = -3i + 2j + 7k
>>Let w = -3i + 2j + 7k
>>Prove vector a is perpendicular to w
>>a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
>>|a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
>>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>>cos&= (a.w)/(|a||w|) = 0
>>Prove vector a is perpendicular to w
>>b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
>>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
>>|w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>>cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
>>I know that is wrong, can anyone see what my mistkae is??
>>Cassandra
> Yes, cosine= (b.w)/(|b||w|)= 0 but did you notice that that
only
> requires that b.w be 0? It is not necessary to \[CapitalThorn]nd |b| or
|w|.
> As to what you did wrong, I hate to say it but: Arithmetic!
> you have:
> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
> = -3i + 2j + 7k
> Now what, exactly, is -2*0-1*1? 1*3- 1*0?
> You seem to be under the impression that a*0= a.
> Most people say a*0= 0.
*grin* you are right, most people would say a*0=0. Normally I
say the
same....not sure what was up last night!
Cassie
===
Subject: Re: Proving the cross product is orthogonal - WRITE
WIDE AND LARGE
You simply overlooked the \[CapitalThorn]gures 0 in -2*0-1*1 and in
1*3-1*0. I guess
that you were writing too small print too tightly in the left
upper
corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) =
(-1, 3, 7);
etc., etc.
IHTH - Johan E. Mebius
> I hope I dont seem like one of those louts that only turn
up at
> assignment time!! However I am struggling with a proof. My
answer is
> not coming out as expected, and I think I am making a silly
mistake
> somewhere. If anyone could have a glance over it, that
would be
> appreciated.
> Q) a =(1, -2, 1) b =(3, 1, 0)
> Find a x b and prove that your cross-product vector is
perpendicular
> to each of the vectors a and b.
> A) [I have worked out the cross-product, and I have proven
that a is
> perpendicular to the cross-product, however I am struggling
to prove
> that b is perpendicular to the cross-product]
> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
> = -3i + 2j + 7k
> Let w = -3i + 2j + 7k
> Prove vector a is perpendicular to w
> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
> cos&= (a.w)/(|a||w|) = 0
> Prove vector a is perpendicular to w
> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)
> I know that is wrong, can anyone see what my mistkae is??
> Cassandra
===
Subject: Re: Proving the cross product is orthogonal - WRITE
WIDE AND LARGE
> You simply overlooked the \[CapitalThorn]gures 0 in -2*0-1*1 and in
1*3-1*0. I guess
> that you were writing too small print too tightly in the
left upper
> corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) =
(-1, 3, 7);
> etc., etc.
> IHTH - Johan E. Mebius
>> I hope I dont seem like one of those louts that only \
turn
up at
>> assignment time!! However I am struggling with a proof. My
answer is
>> not coming out as expected, and I think I am making a
silly mistake
>> somewhere. If anyone could have a glance over it, that
would be
>> appreciated.
>> Q) a =(1, -2, 1) b =(3, 1, 0)
>> Find a x b and prove that your cross-product vector is
perpendicular
>> to each of the vectors a and b.
>> A) [I have worked out the cross-product, and I have proven
that a is
>> perpendicular to the cross-product, however I am
struggling to prove
>> that b is perpendicular to the cross-product]
>> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k
>> = -3i + 2j + 7k
>> Let w = -3i + 2j + 7k
>> Prove vector a is perpendicular to w
>> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0
>> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)
>> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>> cos&= (a.w)/(|a||w|) = 0
>> Prove vector a is perpendicular to w
>> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7
>> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)
>> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)
>> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) =
-7/sqrt(620)
>> I know that is wrong, can anyone see what my mistkae is??
>> Cassandra
Much appreciated. I have been writing my equations in MS Word
using the
equation editor. Perhaps I should write them on paper Ôbig
and large
\[CapitalThorn]rst!!
Cassie
===
Subject: Re: Proving the cross product is orthogonal - WRITE
WIDE AND LARGE
> Much appreciated. I have been writing my equations in MS
Word using the
> equation editor. Perhaps I should write them on paper Ôbig
and large
> \[CapitalThorn]rst!!
With a pencil.
The trolls in this newsgroup post all kinds of nonsense,
sometimes on
purpose
and sometimes because they dont check their work.
===
Subject: Re: Proving the cross product is orthogonal - WRITE
WIDE AND LARGE
>>Much appreciated. I have been writing my equations in MS
Word using the
>>equation editor. Perhaps I should write them on paper Ôbig
and large
>>\[CapitalThorn]rst!!
> With a pencil.
> The trolls in this newsgroup post all kinds of nonsense,
sometimes on
purpose
> and sometimes because they dont check their work.
Yes, one of my biggest downfalls in life is that I make lots
of
mistakes. I once had to do a speed & accuracy test as part of
a pysch
test. I scored full marks (not sure how they calculated it),
but I
remember the pysch saying something about me getting alot
further
through the test then most, but making more mistakes then
most.
I used to have a turtle on my desk to remind myself to slow
down and
double check things. I \[CapitalThorn]nd it so dif\[CapitalThorn]cult, but \
it is
something I
should really work on. I have made two very elementary
mistakes in the
course of this thread...I do need to improve.
===
Subject: Just checking line slope
I hope this doesnt sound dumb, but is the following an
acceptable
formula for a straight line. I am sure it is, but just wanted
to check.
y = x/6
===
Subject: Re: Just checking line slope
> I hope this doesnt sound dumb, but is the following an
acceptable
> formula for a straight line. I am sure it is, but just
wanted to check.
> y = x/6
Add or subtract any number to your x,or y or 6 in this
equation and it
is still a straight line. I \[CapitalThorn]rst learnt it by visual
detection: If
the plot or graph is straight then it is linear, but if bent
it is not
so.There is only one way keeping it straight, so many ways to
bend it.
===
Subject: Re: Just checking line slope
<3Mcsd.58178$K7.34752@news-server.bigpond.net.au I hope this doesnt sound dumb, but is the following an
acceptable
> formula for a straight line. I am sure it is, but just
wanted to check.
> y = x/6
Yes. Checking your work is never dumb.
In general the equation of a straight line in the plane is
y = mx + b or x = a
Are you familiar with this general equation?
Your equation \[CapitalThorn]ts that general form for what values of m and
b?
===
Subject: Re: Just checking line slope
>>I hope this doesnt sound dumb, but is the following an
acceptable
>>formula for a straight line. I am sure it is, but just
wanted to check.
>>y = x/6
> Yes. Checking your work is never dumb.
> In general the equation of a straight line in the plane is
> y = mx + b or x = a
> Are you familiar with this general equation?
> Your equation \[CapitalThorn]ts that general form for what values of m
and b?
x/6 part. I have been struggling with some of those \[CapitalThorn]ner
rules for what
constitutes a linear equation.
My textbook says this...
The following *are* linear equations:
1) 3x + 2y = 7
2) x1-2x2 +10x3 + x4 = 0
3) 1/2x + y - (pi)z = sqrt(2)
4) (sin((pi)/2))x1 - 4x2 = e^2
The following *are not* linear equations:
5) xy + z = 2
6) e^x - 2y = 4
7) sin(x1) + 2x2 - 3x3 = 0
8) 1/x + 1/y = 4
Firstly my apologies for being unsure of ASCII syntax. Where
I have
written a number after the variable I am trying to indicate
the number
is a subscript, ie x1 is x subscript 1. What is the correct
syntax?
While I can understand why most of these are linear/not
linear. Some
have me confused.
Why is 4 okay, but not 7?
What is wrong with 8?
Cassie
===
Subject: Re: Just checking line slope
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4GMN803873;
>I hope this doesnt sound dumb, but is the following an
acceptable
>formula for a straight line. I am sure it is, but just
wanted to
check.
>y = x/6
>> Yes. Checking your work is never dumb.
>> In general the equation of a straight line in the plane is
>> y = mx + b or x = a
>> Are you familiar with this general equation?
>> Your equation \[CapitalThorn]ts that general form for what values of m
and b?
>x/6 part. I have been struggling with some of those \[CapitalThorn]ner
rules for
what
>constitutes a linear equation.
>My textbook says this...
>The following *are* linear equations:
>1) 3x + 2y = 7
>2) x1-2x2 +10x3 + x4 = 0
>3) 1/2x + y - (pi)z = sqrt(2)
>4) (sin((pi)/2))x1 - 4x2 = e^2
>The following *are not* linear equations:
>5) xy + z = 2
>6) e^x - 2y = 4
>7) sin(x1) + 2x2 - 3x3 = 0
>8) 1/x + 1/y = 4
>Firstly my apologies for being unsure of ASCII syntax. Where
I have
>written a number after the variable I am trying to indicate
the
number
>is a subscript, ie x1 is x subscript 1. What is the correct
syntax?
>While I can understand why most of these are linear/not
linear. Some
>have me confused.
>Why is 4 okay, but not 7?
>What is wrong with 8?
>Cassie
4 has (sin((pi)/2)) which is a number, a constant.
4 is exactly the same as Ax_1- 4x_2= B
7, on the other hand, has sin(x1) and x1 is a VARIABLE, not a
constant. It is functions of VARIABLES that count.
8 has 1/x and 1/y, which are quite different from just x and
y.
A linear function has the variables themselves, no powers or
functions, multiplied by numbers (not other variables) then
added or
subtracted.
===
Subject: Re: Just checking line slope
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB4GMPG03973;
>I hope this doesnt sound dumb, but is the following an
acceptable
>formula for a straight line. I am sure it is, but just
wanted to
check.
>y = x/6
>> Yes. Checking your work is never dumb.
>> In general the equation of a straight line in the plane is
>> y = mx + b or x = a
>> Are you familiar with this general equation?
>> Your equation \[CapitalThorn]ts that general form for what values of m
and b?
>x/6 part. I have been struggling with some of those \[CapitalThorn]ner
rules for
what
>constitutes a linear equation.
>My textbook says this...
>The following *are* linear equations:
>1) 3x + 2y = 7
>2) x1-2x2 +10x3 + x4 = 0
>3) 1/2x + y - (pi)z = sqrt(2)
>4) (sin((pi)/2))x1 - 4x2 = e^2
>The following *are not* linear equations:
>5) xy + z = 2
>6) e^x - 2y = 4
>7) sin(x1) + 2x2 - 3x3 = 0
>8) 1/x + 1/y = 4
>Firstly my apologies for being unsure of ASCII syntax. Where
I have
>written a number after the variable I am trying to indicate
the
number
>is a subscript, ie x1 is x subscript 1. What is the correct
syntax?
Many people use what you use (x1, etc.), but perhaps more
common is
x_1, or x_(1) or x_{1} when clarity is needed. The main thing
required is that the notation be uncluttered and unambiguous;
youre doing well on both counts.
>While I can understand why most of these are linear/not
linear. Some
>have me confused.
>Why is 4 okay, but not 7?
>What is wrong with 8?
(4) is ok because its in the form ax + by = c where a, b, c
are
*constants* [sin(pi/2) and e^2 are constants; remember that a
function applied to a constant yields another constant]. On
the
other hand, (7) is not linear because sin(x1) is not linear in
x1 [is not a constant times x1]. In (8), neither 1/x nor 1/y
is
linear (neither is a constant times the appropriate variable).
Todd Trimble
===
Subject: Re: Just checking line slope
<Shhsd.58594$K7.42034@news-server.bigpond.net.au My textbook says this...
> The following *are* linear equations:
> 1) 3x + 2y = 7
> 2) x1-2x2 +10x3 + x4 = 0
> 3) 1/2x + y - (pi)z = sqrt(2)
> 4) (sin((pi)/2))x1 - 4x2 = e^2
> The following *are not* linear equations:
> 5) xy + z = 2
> 6) e^x - 2y = 4
> 7) sin(x1) + 2x2 - 3x3 = 0
> 8) 1/x + 1/y = 4
> Firstly my apologies for being unsure of ASCII syntax.
Where I have
> written a number after the variable I am trying to indicate
the number
> is a subscript, ie x1 is x subscript 1. What is the correct
syntax?
x_1 and x^n for n-th power of x. Best readable ascii style is
to persist
like youre doing for the most part with adequate spaces \
like
x1 - 2x2 + 10x3 + x4 = 0
> While I can understand why most of these are linear/not
linear. Some
> have me confused.
> Why is 4 okay, but not 7?
sin pi/2 is a constant.
sin x_1 isnt ax_1 for some a in R.
> What is wrong with 8?
It doesnt have the form ax + by = c, the most general
equation for a
line, hence the expression linear, line like. Indeed, y + x =
4xy
Were you to graph the equations, what is linear and what is
not
would likely jump out at you for only lines are linear.
===
Subject: Re: Just checking line slope
> Were you to graph the equations, what is linear and what is
not
> would likely jump out at you for only lines are linear.
an equation looks linear on a graph, then it is linear.
Aah, I enjoy maths so much...but it is also very hard at
times...posibly
why I like it so much!
cassandra
===
Subject: Re: Vectors question 2
> I am working through a question for a university
assignment. I am not
> wanting to be given the answer, but am after some pointers.
> The question is:
> A train is travelling in a straight line at 24km/h. A movie
stunt-man
> moves at 4km/h straight across its roof, at right angles to
the
> direction of its motion. Draw a clear sketch of his
resultant movement
> relative to the ground, and \[CapitalThorn]nd the speed and direction of
that
movement.
> My take on this is as follows (way below is a very bad ASCII
> representation of the type of graph that is in my
head...de\[CapitalThorn]netly not
> to any sort of scale)
> I \[CapitalThorn]rst create a graph with x-axis distance travelled, and
y-axis time.
I would give up on attempting to depict time in this picture,
unless you
had some real attachment to drawing a 3-dimensional picture.
Why? Well,
the trains motion takes up one dimension, time a second \
one,
and the
stuntmans motion (which is not collinear with the \
trains,
if I read
the problem correctly) a third one.
Rather, imagine the displacements traversed over one unit of
time. That
unit can be of your choosing: one second, one hour, one
millenium, ...
you get to name it (I prefer a unit that makes calculations
trivial,
and that would suggest one hour; the problem is that you
would then
need to imagine a train that is 4 km wide. While the mental
image that
emerges is dif\[CapitalThorn]cult to believe in, the mental image is
absolutely
irrelevant; if the mental image is important, imagine the
time interval
to be 1 second or 1 millisecond). You can do this based on an
assumption
of constant velocities, and the simple (vector) relation s =
vt, where s
represents the (vector) displacment, v the (vector) velocity,
and t the
(scalar) time, which (by assumption) is a constant. If you
have a \[CapitalThorn]xed
time interval, the equation s = vt takes a velocity v,
multiplies it by
a \[CapitalThorn]xed time interval t, and obtains a vector that points from
the
starting position, to the position reached at the end, at
time t.
The problem is to deal with the vector addition (which is
what is meant
by the resultant) of velocities, and what you need to do is to
represent that. So, let the train be travelling along the +x
axis, and
the stuntman travelling along the +y direction. The
resultant, which
is the sum of the two vectors v_train and v_stuntman,
represents the
velocity vector of the stuntman with respect to the ground.
(It would
alternatively represent the velocity of the train with
respect to the
ground, if it were the stuntman running along the ground
carrying a
moving train. Ill ignore that interpretation.)
> I can then plot u = 24i + j to show the speed that the
train is
> travelling at.
> The line perpendicular to to u is the speed and direction
of the stunt
> man, the length of that line will be 4.
Note that you are devoting the vector i to displacement along
a single
(spatial) direction, and j to displacement along the time
direction.
This is incorrect. The vector u is \[CapitalThorn]ne. However, your vector
perpendicular to u will be (proportional to) either -i + 24 j
or
i - 24 j, since those are along the (only) direction
perpendicular
to u in the plane containing the directions i and j.
In the \[CapitalThorn]rst case, the stuntman is moving along the -i
direction. That
cant be right, since he must be moving perpendicular to the
trains
direction. In the second case, time is moving backwards. That
might be
nice for a sci-\fißick, but it doesnt \
really happen that way
in real
life.
You need to notice that the stuntmans motion is
perpendicular in
spatial terms: if the train moves east, he must be going
north or south.
Its the spatial content that you must model. Everyone and
everything
moves through time in exactly the same way (ignoring the
subtleties
that relativity forces us to deal with). Thus, the
description of the
motions of the train and the stuntman must show their motions
to be
perpendicular *in space*. Their behavior in time can be
reßected in
dealing with the magnitudes of their velocity vectors, and
those can
be made spatial in character by using the formula s = vt that
I
mentioned earlier.
The problem is that the i: space and j: time coordinate
system makes
for a single spatial dimension, and that doesnt allow for
two separate
spatial directions. Thats why I said earlier that this
approach would
require three dimensions to depict everything, and that it
would be best
to ignore devoting a dimension to the time variable.
> The line that forms the hyptonuse (v), represents the speed
(length),
> and direction (x,y) that the stuntman has travelled.
> Am I on the right track (no pun intended)? If not, could
you offer some
> suggestions.
Sorry, you have missed the train entirely (that pun was
irresistable).
Another will be along shortly, provided you follow the
instructions I
gave earlier.
> | 0
> | 0 0
> | 0 0
> | v 0 0 (24,1)
> | 0 0
> | 0 0 | 0 0 u
> | 0 0
> |0____________________________ Please note, this is an assignment, so I am not wanting
anyone to solve
> it for me. I do not want any reason to be accused of
wrongdoings.
> Cassandra
I hope the information I gave was both suf\[CapitalThorn]cient to get you
on your
way, and yet not too much.
Dale.
===
Subject: Re: Vectors question 2
>> I am working through a question for a university
assignment. I am not
>> wanting to be given the answer, but am after some pointers.
>> The question is:
>> A train is travelling in a straight line at 24km/h. A
movie stunt-man
>> moves at 4km/h straight across its roof, at right angles
to the
>> direction of its motion. Draw a clear sketch of his
resultant movement
>> relative to the ground, and \[CapitalThorn]nd the speed and direction of
that
>> movement.
> I hope the information I gave was both suf\[CapitalThorn]cient to get you
on your
> way, and yet not too much.
> Dale.
Dale,
written a few more times I think before I understand it all.
So after reading your response I began to think. Ignoring
vectors and i
& j etc. If I wanted to \[CapitalThorn]gure this out a few weeks ago (prior
to
learning this stuff) my method would have been as follows.
Put the direction of the train on the x-axis, the direction
of the
stuntman on the y-axis. (this is I think what you are trying
to say).
So now for the \[CapitalThorn]rst part of the question
--Sketch the resultant movement of the stuntman relative to
the ground--
The sketch would be the line y = x/6, for x >= 0.
The second part of the question
--\[CapitalThorn]nd the speed---
is answered by \[CapitalThorn]nding the length of the line.
speed = sqrt(4^2 + 24^2) = 4*sqrt(37) km/h
The \[CapitalThorn]nal part of the question then
--\[CapitalThorn]nd the...direction of that movement--
This would be where the vector math comes in.
tan& = b / a = 4 / 24 = 1 / 6
& = tan^-1(1/6) = 0.165
Phew, this seems alot nicer.
got me started at any rate. Any further comments?
Now for question number three...
Cassandra
===
Subject: Re: Vectors question 2
> So after reading your response I began to think. Ignoring
vectors and i
> & j etc. If I wanted to \[CapitalThorn]gure this out a few weeks ago
(prior to
> learning this stuff) my method would have been as follows.
> Put the direction of the train on the x-axis, the direction
of the
> stuntman on the y-axis. (this is I think what you are
trying to say).
That method is correct, and produces the correct answers...
so now my
question
is: what is it that you learned over the last few weeks that
confused you
into
trying to plot time as well?
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Re: Vectors question 2
>>So after reading your response I began to think. Ignoring
vectors and i
>>& j etc. If I wanted to \[CapitalThorn]gure this out a few weeks ago
(prior to
>>learning this stuff) my method would have been as follows.
>>Put the direction of the train on the x-axis, the direction
of the
>>stuntman on the y-axis. (this is I think what you are
trying to say).
> That method is correct, and produces the correct answers...
so now my
question
> is: what is it that you learned over the last few weeks
that confused you
into
> trying to plot time as well?
> meeroh
To answer your question, I overthought the question. We are
studying
vectors, including 3-dimensional vectors and projections.
Perhaps it was
because I had only just \[CapitalThorn]nished studying projections. Or
because I
thought that question must be more complex then it looked etc.
I really dont know why, however I am sort of glad because \
it
has
allowed me to gain alot better understanding into using
vectors, but
then making mistakes does usually have that effect.
cassandra
===
Subject: Re: Vectors question 2

>>So after reading your response I began to think. Ignoring
vectors and i
& j
>>etc. If I wanted to \[CapitalThorn]gure this out a few weeks ago (prior
to learning
this
>>stuff) my method would have been as follows.
>>Put the direction of the train on the x-axis, the direction
of the
stuntman
>>on the y-axis. (this is I think what you are trying to say).

> That method is correct, and produces the correct answers...
so now my
> question is: what is it that you learned over the last few
weeks that
> confused you into trying to plot time as well?
 To answer your question, I overthought the question. We are
studying
> vectors, including 3-dimensional vectors and projections.
Perhaps it was
> because I had only just \[CapitalThorn]nished studying projections. Or
because I
thought
> that question must be more complex then it looked etc.
> I really dont know why, however I am sort of glad because
it has allowed
me
> to gain alot better understanding into using vectors, but
then making
> mistakes does usually have that effect.
Well, if you want to consider this and include time as one of
your
dimensions,
then you have to realize that you have to project the
space-time coordinate
system into a space-only coordinate system before trying to
describe it in
terms
of visible effects.
For example, consider a single point traveling along a
straight line. Let
the
line of travel be the X axis and let time T be perpendicular
to it. Then, in
the
XT plane, the point will be describing some curve
(continuous, unless your
universe includes teleportation). For example, if the point
is oscillating
about
the origin, then the space-time curve describing its motion
would be a sine
wave
on the T axis. But a description of that motion would
probably not refer to
a
sine wave about the T axis, it would refer to an oscillatory
motion about
the
origin.
This seems like a trivial point until you start considering
more dimensions:
consider an XY plane and a time T axis perpendicular to it,
and two points
traveling in the plane -- one at a \[CapitalThorn]xed velocity along the X
axis, starting
at
the origin, and the other at the same velocity along the Y
axis, also
starting
at the origin. If you look at them in the XY plane, then
their trajectories
(namely, the X and Y axes) are perpendicular. However, if you
consider them
to
be points in XYT space, then the \[CapitalThorn]rst ones \
trajectory is a
straight line
in
the XT plane, and the second ones trajectory is a straight
line in the YT
plane, and those trajectories are not perpendicular! It is
only their
projections into the XY place (the space-only coordinate
system) that are
perpendicular.
Which brings me to your question; your mistake in considering
this with
distance
on the X axis and the time on the T axis was that when the
description of
physical motion of the stunt man was described as being
perpendicular to the
motion of the train, you drew the perpendicular in the XT
plane, which
doesnt
work.
What you need to do here is count the number of dimensions of
space
(clearly,
you need at least two, as the man the the train are moving at
right angles
to
each other. Then you add one for time, and you see that you
need a
three-dimensional system. In that system, the trajectory of
the train will
be a
straight line; if the train travels along the X axis, as it
does in your
original graph, then its XYT trajectory will be a straight
line in the XT
plane,
as it is in your original graph. However, the motion of the
man is,
according to
the problem statement, perpendicular to that of the train --
but the problem
statement refers to them in the space-only coordinate system,
which means
that
you have to have the man moving perpendicular not to the XYT
time-space
trajectory of the train, but to the XY spatial trajectory of
the train. If
you
do that, then everything comes out right, but its somewhat
complicated to
draw
that.
Now, there is no really good reason to involve time in this,
as the movement
of
the train (relative to the ground) and the man (relative to
the train) are
constant with time, so what you are asked for (movement of
the man relative
to
the ground) is also constant and you can draw it without
involving the T
axis at
all. However, I think you should know why your initial
solution didnt work
and
how you could have made it work.
meeroh
--
If this message helped you, consider buying an item
from my wish list: <http://web.meeroh.org/wishlist>
===
Subject: Re: Proving Trig Identities?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB3Frlo09977;
hello. i need some help proving some identities.
#1. cos40=1-8sin^2thetacos^2theta
#2. sin(pi/6-s)+cos(pi/3-s)=cos s
===
Subject: Re: Proving Trig Identities?
>hello. i need some help proving some identities.
>#1. cos40=1-8sin^2thetacos^2theta
I have no idea what this means. Please post again in some
notation
that we can understand.
>#2. sin(pi/6-s)+cos(pi/3-s)=cos s
There may be a more clever way, but you could use the
formulas for
sin(A-B) and cos(A-B) and simplify.
What have you tried?
--
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: Proving Trig Identities?
>hello. i need some help proving some identities.
>#1. cos40=1-8sin^2thetacos^2theta
Assuming thats 4theta on the left side. Think of \
cos(4theta)
as
cos(2*2theta) and use the double angle for cosine that
involves only
sines. That should get you started.
>#2. sin(pi/6-s)+cos(pi/3-s)=cos s
Use the addition formula for sin(a - b) and cos(a - b) and
the values
for the functions of pi/6 and pi/3.
--Lynn
===
Subject: TFC and contribution
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB3HG1V18324;
can you help me with this question guys?
The Selling price per unit (p) plus the Total Variable
Cost(xv) less
the Total Fixed Costs (TFC) per unit is equal to:
a. Contribution (C) to Total Fixed Costs (TFC)
b. Contribution (C) to loss, before the B.E.P.
c. Contribution (C) pro\[CapitalThorn]t, after the B.E.P.
d. Both a and c
e. none of the above
im just confused on this
===
Subject: Re: TFC and contribution
> can you help me with this question guys?
> The Selling price per unit (p) plus the Total Variable
Cost(xv) less
> the Total Fixed Costs (TFC) per unit is equal to:
> a. Contribution (C) to Total Fixed Costs (TFC)
> b. Contribution (C) to loss, before the B.E.P.
> c. Contribution (C) pro\[CapitalThorn]t, after the B.E.P.
> d. Both a and c
> e. none of the above
> im just confused on this
I have no idea. Whats a B.E.P.? Also, it would be helpful
for you to show
any
work/insights you have gained so far in solving the problem.
Bill
===
Subject: Re: JSH: Two sides
[snip delusion]
What does properly a unit mean?
Can you answer that simple question James Harris?
===
Subject: Re: JSH: Two sides
> [snip delusion]
> What does properly a unit mean?
> Can you answer that simple question James Harris?
JSH does not answer questions.
===
Subject: Animation on Mathematica or Maple
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB3LRdo09446;
Using Mathematica or Maple, I am trying to illustrate a
sphere S1
(stationary) with a much smaller sphere S2 (animated)
orbiting S1. I
would like to have the distance between S1 and S2 as close to
zero as
possible. One last thing Id like to do is be able to view \
the
animation from different angles (I know that Mathematica
allows you to
spin 3D objects in real time).
Any suggestions on how to make this happen. So far Ive only
got a
stationary preview of S1. Help would be greatly appreciated!!
Joseph A.
===
Subject: Re: Animation on Mathematica or Maple
>Using Mathematica or Maple, I am trying to illustrate a
sphere S1
>(stationary) with a much smaller sphere S2 (animated)
orbiting S1. I
>would like to have the distance between S1 and S2 as close
to zero as
>possible. One last thing Id like to do is be able to view
the
>animation from different angles (I know that Mathematica
allows you to
>spin 3D objects in real time).
>Any suggestions on how to make this happen. So far Ive \
only
got a
>stationary preview of S1. Help would be greatly appreciated!!
>Joseph A.
Ask and ye shall receive :-). I had a little time to kill this
evening, so here is some maple code -- it might not display
to well
here but, hey, its free and it works. I have put line \
spaces
between
each line of Maple instructions. R is the big sphere radius,
r the
small one, and d the separation between them. The number 80
in the
display statement can be raised for better resolution. Enjoy.
--Lynn
>restart;
>R:=10;r:=1;d:=0.2;
>big:=plot3d([R*sin(phi)*cos(theta),R*sin(phi)*sin(theta),R*
cos(phi)],theta=
0..2*Pi,phi=0..Pi,color=cyan):
>small:=t->plot3d([(R+r+d)*cos(t),(R+r+d)*sin(t),0]+[r*sin(
phi)*cos(theta),r
*sin(phi)*sin(theta),r*cos(phi)],
theta=0..2*Pi,phi=0..Pi,style=patchnogrid,color=brown):
>
orbit:=display(seq(small(k*2*Pi/80),k=0..80),insequence=true):
> display({orbit,big},scaling=constrained);
===
Subject: Re: Animation on Mathematica or Maple
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB42XU702002;
>Using Mathematica or Maple, I am trying to illustrate a
sphere S1
>(stationary) with a much smaller sphere S2 (animated)
orbiting S1. I
>would like to have the distance between S1 and S2 as close
to zero as
>possible. One last thing Id like to do is be able to view
the
>animation from different angles (I know that Mathematica
allows you
to
>spin 3D objects in real time).
>Any suggestions on how to make this happen. So far Ive \
only
got a
>stationary preview of S1. Help would be greatly appreciated!!
>Joseph A.
Here is the stationary view of the two spheres. I want S2 to
revolve
around S1. How can I do this using mathematica?
http://www.mindcraftstudios.com/mathematics/orbit.jpg
===
Subject: Re: Animation on Mathematica or Maple
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB42XTU01997;
Here is a stationary view of S1 (large sphere) and S2 (small
sphere).
What I want is S2 to revolve around S1. Also, I want to be
able to
see the rotation from any view.
Joseph A.
===
Subject: Re: Animation on Mathematica or Maple
>Here is a stationary view of S1 (large sphere) and S2 (small
sphere).
>What I want is S2 to revolve around S1. Also, I want to be
able to
>see the rotation from any view.
<< Graphics`Shapes`
s = Table[Show[
TranslateShape[Graphics3D[Sphere[0.1,12,12]],{1.5Cos[t],
1.5Sin[t],0}],
Graphics3D[Sphere[1, 12, 12]],
Boxed -> False],
{t, 0, 2Pi, Pi/24}];
Export[spheres.gif, s]
There are gif-speci\[CapitalThorn]c options that Export will accept to
modify
some of the presentation. There is also an option to Show or
maybe
Graphics3D, which I have forgotten and have spent ten minutes
unsuccessfully searching for, that will keep the spheres from
wobbling back and forth as the animation progresses. But maybe
this will give you an idea how to get started with this. If
you
get in a bind and this isnt enough to let you \
\[CapitalThorn]gure it out
then
throw mail at me and Ill \[CapitalThorn]ddle with this a \
little more.
There are some bugs in Exported animated gifs, \
Ive reported
the
ones that Ive found and havent heard whether \
they have been
\[CapitalThorn]xed in the latest version or not.
I just tried rendering this with the Windows XP SP2 Picture
and Fax
Viewer and it fails to display this. But IrFanView is happy
with
the result. You will likely \[CapitalThorn]nd that if you push the edges of
the
implementations of animated gif viewers that you will turn up
bugs
and features that you didnt expect.
===
Subject: JSH: James, you be the judge...
Here is how James Harris operates:
1. Start with some goofy polynomial that was a leftover from
one of his many
failed FLT proofs. No explanation of motivation or reasoning
provided for
choosing that particular polynomial or explanation of why it
might be
meaningful in his new context.
2. Avoid being speci\[CapitalThorn]c about such important details as which
ring he
intends to work in.
3. Dream up a bunch of weird non-standard terminology like
properly
unit,
dividing off, etc. Use terms incorrectly, such as referring to
multiple
when he should use the term factor. Also, make use of standard
terminology
like distributive property and constant term without actually
understanding it or using it properly. Also, add lots of
vague distractions
such as the equation has no memory, you can see the 7s in
there,
cant
you? to act as a smoke screen.
4. Apply a bunch of algebraic manipulations, some of which
are just wrong.
Make inappropriate generalizations from speci\[CapitalThorn]c cases to
general cases,
etc.
5. Get a result that seems contradictory, and assume that the
error was with
the core of algebra rather than the much more likely
explanation that he
himself might have made an error.
6. Through out a bunch of paranoid ad hominem attacks on
critics. Then
hypocritically claim that all his critics resort to social
crap and personal
attacks rather than sticking to mathematics, logic, and
reasoning.
We are supposed to believe that his arbitrarily chosen goofy
polynomial just
happens to one that, when probed by the genius of James
Harris, give
results that shake mathematics to its core!
James, you be the judge. Are you a pile of crap?
===
Subject: JSH: Simply fascinating
The math here is so readily understandable that its \
actually
almost
as interesting watching how people react to it, as anything
else.
For instance, Ive 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 its 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
Ive 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.
Its 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, its simple to the point of trivial.
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 whats 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
doesnt 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.
In terms of dif\[CapitalThorn]culty, my proof is about as easy as it gets \
in
algebraic number theory, in terms of the actual mathematics.
But the concepts are where there is a problem, and the social
hang-up
is in accepting that theres this simple technique that \
shows
a BIG
problem, which can invalidate claims of proof for, well, over
a
hundred plus years.
So the mathemtics is EASY for a trained mathematician to
follow. The
social implications are hard, if social stuff is important to
you, and
clearly from what Ive seen it is to many of you.
For instance, at this point Ive removed all objections
raised in
detail. Like I can explain supposed counter examples to my
work. I
can give an actual example where you can see the
factorization play
out--just as the theory shows it must. And you probably know
that my
research is the work that can be said to have gone to a
journal which
at least claims it does formal peer review.
They thanked me for the paper said the reviewers liked it,
and then
some sci.mathers got together--actually literally \
conspiring
online
in posts on sci.math--sent them emails and the editors yanked
my
papers THE NEXT DAY.
They had it for nine months. Id corresponded with them for \
a
while,
even corrected them when they called me Dr. Harris as I \
dont
have a
Ph.d, and I told them I was an independent researcher with
concerns
about how my work would be handled. They kept saying no
problem, ok,
all that matters is whats correct.
Then they yanked my paper after sci.mathers emailed them:
All the pertinent facts are in my favor.
So whats the hold-up?
My research shows that some mistakes were made over a hundred
years
ago, and a lot of people missed them, and gave proofs which
were
not, and are not proofs.
Mathematics is unforgiving. It doesnt care about the social
implications of the truth. So it doesnt matter
mathematically that a
LOT of people out there are terribly dependent on the false
beliefs
and incorrect results, but it DOES matter a lot to those
people!
I call their behavior passive-aggessive, as by dragging their
feet,
taking as long as possible before acknowledging my research,
or worse,
hoping to NEVER acknowledge it at all, they are passively
hoping to
escape mathematical truth, in what amounts to a very
aggressive way.
I liken their behavior to judges at a race, who watch a
runner break a
world record, and then lie about his time, refuse to admit he
even
\[CapitalThorn]nished the race, and some even call him names!!!
Theyre turning the way its supposed to work \
with a major
discovery,
upside down.
And its silly behavior as eventually the truth will come
out, and you
know what Ill do then?
Probably go to the beach. Ill also hang out in some bars.
Yup, Ill
de\[CapitalThorn]nitely hang out in some bars, preferably near a beach.
Yup, you guessed it, Ill do my best to forget about them, \
as
why
bother worrying about silly people who do silly things.
Lifes too short.
James Harris
http://mathforpro\[CapitalThorn]t.blogspot.com/
===
Subject: Re: JSH: Simply fascinating
> Ive 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).
In general thats not true at all. For example, consider:
(x+1)*(x+2)*(x+3)
If you expand it out to get a polynomial in the usual form,
you see
that the constant term is 6. But 6 is not a factor of any of
the three
polymomial factors, in fact neither 2 nor 3 is a factor of
any of them
say what you really meant to say? Or maybe you were totally
mistaken
and need to just retract what you said.
> My research shows that some mistakes were made over a
hundred years
> ago, and a lot of people missed them, and gave proofs which
were
> not, and are not proofs.
Its true that some mistakes were made long ago, the worst \
in
my
opinion being the \[CapitalThorn]rst developers of set theory who concocted
a set
of all sets, which was subsequently proven not to exist. But
Ive seen
no evidence that *you* have ever done any research that
showed any such
past mistakes of others. Instead all I see is *you* making
mistakes
yourself and not admitting them.
Do you know how to factor 7 in the ring of integers with
sqrt(7)
adjoined? Thats easy, of course you do, right? But do you
know how to
factor 7 in the ring of integers with sqrt(2) adjoined? Now
if you can
\[CapitalThorn]gure out that simple arithmetic problem, \
heres something
more
interesting: Find all primes p (other than 2 and 7 which \
Ive
already
shown you) such that 7 can be nontrivially factored in the
ring of
integers with sqrt(p) adjoined. Finally, \[CapitalThorn]nd all \
\[CapitalThorn]nite sets
of two or
more primes p1,p2,...,pn such that 7 is composite in the ring
you get
when you adjoin all the sqrts of those primes but its prime
in the
ring you get when you adjoin all but one of those sqrts. Same
question
if you allow adjoining sqrt(-1) in addition to adjoining sqrt
of a
prime. (Note: When I refer to a prime p or primes p1,p2..., I
mean
prime in the ring of integers. When I refer to 7 being prime
or
composite, I mean prime or composite in the ring of integers
adjoined
with the various sqrts.)
===
Subject: Re: JSH: Simply fascinating
> Ive 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).
> In general thats not true at all. For example, consider:
> (x+1)*(x+2)*(x+3)
But he said, e.g. in your example, having factored your
polynomial into
(x+1), (x+2), and (x+3), 1, 2, and 3 must be factors of 6,
the constant
term of your expanded polynomial.
> If you expand it out to get a polynomial in the usual form,
you see
> that the constant term is 6. But 6 is not a factor of any
of the three
> polymomial factors, in fact neither 2 nor 3 is a factor of
any of them
He didnt say, with respect to your example, that 6 must be \
a
factor of 1,
2, and 3. Rather that 1, 2, and 3 must be factors of 6.
KeithK
> say what you really meant to say? Or maybe you were totally
mistaken
> and need to just retract what you said.
> My research shows that some mistakes were made over a
hundred years
> ago, and a lot of people missed them, and gave proofs which
were
> not, and are not proofs.
> Its true that some mistakes were made long ago, the worst
in my
> opinion being the \[CapitalThorn]rst developers of set theory who
concocted a set
> of all sets, which was subsequently proven not to exist.
But Ive seen
> no evidence that *you* have ever done any research that
showed any such
> past mistakes of others. Instead all I see is *you* making
mistakes
> yourself and not admitting them.
> Do you know how to factor 7 in the ring of integers with
sqrt(7)
> adjoined? Thats easy, of course you do, right? But do you
know how to
> factor 7 in the ring of integers with sqrt(2) adjoined? Now
if you can
> \[CapitalThorn]gure out that simple arithmetic problem, \
heres something
more
> interesting: Find all primes p (other than 2 and 7 which
Ive already
> shown you) such that 7 can be nontrivially factored in the
ring of
> integers with sqrt(p) adjoined. Finally, \[CapitalThorn]nd all \
\[CapitalThorn]nite sets
of two or
> more primes p1,p2,...,pn such that 7 is composite in the
ring you get
> when you adjoin all the sqrts of those primes but its
prime in the
> ring you get when you adjoin all but one of those sqrts.
Same question
> if you allow adjoining sqrt(-1) in addition to adjoining
sqrt of a
> prime. (Note: When I refer to a prime p or primes p1,p2...,
I mean
> prime in the ring of integers. When I refer to 7 being
prime or
> composite, I mean prime or composite in the ring of
integers adjoined
> with the various sqrts.)
===
Subject: Re: JSH: Simply fascinating
> 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 Ôx, why did you have \
to set Ôx
to zero to
uncover them? Evaluating a univariate polynomial at any
numeric value of
the variable produces a numeric result. Are all these results
Ôconstants?
> The logic is inescapable.
> Lifes too short.
I think yours is too long.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Simply fascinating
> The math here is so readily understandable that its
actually almost
> as interesting watching how people react to it, as anything
else.
> For instance, Ive 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 its 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
> Ive 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.
> Its 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, its 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 whats 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
doesnt 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, youre 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 cant you see it?
> The logic is inescapable.
True enough.
> In terms of dif\[CapitalThorn]culty, 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.
> Lifes too short.
> James Harris
> http://mathforpro\[CapitalThorn]t.blogspot.com/
===
Subject: Re: JSH: Simply fascinating
> The math here is so readily understandable that its
actually almost
> as interesting watching how people react to it, as anything
else.
 For instance, Ive 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 its 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
> Ive 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.
 Its 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, its 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, Im going to answer the Nora Baron poster yet again.
And I want readers to understand that Ive replied to this
poster who
you know lies anyway as its a guy posting as a woman using \
a
name
thats 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 Id said.
Just deleted out everything, and you know what? I STILL so
people
replying about how supposedly I dont answer the objections
from Nora
Baron.
Its partly a game for some people on sci.math, \
Im sure, and
partly a
case where many readers are hoodwinked as *they* dont know
its 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 Im 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 hasnt happened yet from what Ive \
seen, but I keep
hoping.
Ok, so whats wrong with the posters \
assertion?
Well, the ws the posters gives are factors of 49. But 49
comes into
the picture because 49 is a multiple of the polynomial. But
the ws
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
dont seem to get it.)
Notice the posters 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 cant 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 its sad as youre 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 its
actually
almost
>> as interesting watching how people react to it, as
anything else.
>> For instance, Ive 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 its 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
>> Ive 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.
>> Its 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, its 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, Im going to answer the Nora Baron poster yet again.
>And I want readers to understand that Ive replied to this
poster who
>you know lies anyway as its a guy posting as a woman using
a name
>thats 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 Id 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 dont answer the objections
from
Nora
>Baron.
You STILL so people replying ??? Really!
>Its partly a game for some people on sci.math, \
Im sure,
and partly
a
>case where many readers are hoodwinked as *they* dont know
its 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 Im 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 hasnt happened yet from what Ive \
seen, but I keep
hoping.
Whine, whine, whine. Why not go straight to the math rather
and
skip the propaganda ?
>Ok, so whats wrong with the posters \
assertion?
>Well, the ws the posters gives are factors of 49. But 49
comes into
>the picture because 49 is a multiple of the polynomial. But
the ws
>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 ws is 49. You
divide P(x) by 49. You divide the product of the gs by
the product of the ws. 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 gs by the product of the
ws,
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 ds 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 ds 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
de\[CapitalThorn]nition 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
Harriss own de\[CapitalThorn]nition of constant term. Why is \
he so
reluctant to apply that de\[CapitalThorn]nition? 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?
Wheres the trick?
Everything works out just as it should: IF you correctly
identify the constant terms.
Wheres the trick, Mr. Harris ?
>(Notice Nora Baron gives you clues, but somehow sci.math
readers
>dont 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 posters 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 dont, then you get the
wrong answer. This is Mr. Harriss mistake.
This is not rocket science. This is not Galois theory, or
algebraic number theory, or \[CapitalThorn]eld 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 \
de\[CapitalThorn]nition.
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 its
actually almost
>> as interesting watching how people react to it, as
anything else.
>> For instance, Ive 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 its 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
>> Ive 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.
>> Its 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, its 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, Im going to answer the Nora Baron poster yet again.
>And I want readers to understand that Ive replied to this
poster who
>you know lies anyway as its a guy posting as a woman using
a name
>thats 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 Id 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 dont answer the objections
from Nora
>Baron.
You STILL so people replying ??? Really!
>Its partly a game for some people on sci.math, \
Im sure,
and partly a
>case where many readers are hoodwinked as *they* dont know
its 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 Im 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 hasnt happened yet from what Ive \
seen, but I keep
hoping.
Whine, whine, whine. Why not go straight to the math rather
and
skip the propaganda ?
>Ok, so whats wrong with the posters \
assertion?
>Well, the ws the posters gives are factors of 49. But 49
comes into
>the picture because 49 is a multiple of the polynomial. But
the ws
>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 ws is 49. You
divide P(x) by 49. You divide the product of the gs by
the product of the ws. 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 gs by the product of the
ws,
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 ds 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 ds 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
de\[CapitalThorn]nition of constant term of a function, but he \
doesnt 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. Harriss own de\[CapitalThorn]nition of \
constant term. Why
is he so reluctant to apply that de\[CapitalThorn]nition? 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?
Wheres the trick?
Everything works out just as it should: IF you correctly
identify the constant terms.
Wheres the trick, Mr. Harris ?
>(Notice Nora Baron gives you clues, but somehow sci.math
readers
>dont 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 posters 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 dont, then you get the
wrong answer. This is Mr. Harriss mistake.
This is not rocket science. This is not Galois theory, or
algebraic number theory, or \[CapitalThorn]eld 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 \
de\[CapitalThorn]nition.
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 its
actually almost
> as interesting watching how people react to it, as anything
else.
 For instance, Ive 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 its 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
> Ive 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.
 Its 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, its 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, Im going to answer the Nora Baron poster yet again.
> And I want readers to understand that Ive replied to this
poster who
> you know lies anyway as its a guy posting as a woman \
using
a name
> thats 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 de\[CapitalThorn]nition of the word \
Ômultiple.
> 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 Noras 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 Noras 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 Noras 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. Ive argued with this poster for
some
time and noted the lying, and the gender lying is just
another example
of a trend.
Some people dont 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 wouldnt
disagree with
me, and there wouldnt 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 dont 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 whats 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 inde\[CapitalThorn]nitely by rather basic tactics.
I say, the strategy is about to die, and the passive-aggessive
strategy of hoping that inde\[CapitalThorn]nitely the public will be fooled
will be
shown to be one of professional suicide.
James Harris
===
Subject: Re: JSH: Simply fascinating
> What does Noras 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. Ive 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
Harris math 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 dont 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 Ôlies are \
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 whats your problem? If well-trained mathematicians have
discussed
your work with you, surely you dont 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.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Simply fascinating
> The math here is so readily understandable that its
actually almost
> as interesting watching how people react to it, as anything
else.
> For instance, Ive 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 its 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
> Ive 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.
> Its 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, its 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 whats 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
doesnt 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 theres 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 Occams Razor here! Just a
thought.
===
Subject: Graphing functions using rules for shifting
stretching and
reßecting
f(x)= 1 / x+2
is the \[CapitalThorn]rst 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 didnt 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 \[CapitalThorn]rst 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 didnt 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. Hes 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 Im 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 *in\[CapitalThorn]nity* 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 thats obvious with polynomial factors but some
sci.mathers
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 as and bs 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 doesnt matter how complicated that \
as and bs
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
dont 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 its
divided off, but STILL people have argued with me quite
successfully,
and I \[CapitalThorn]gure many of you, despite what I say here, remain
unconvinced
that Im right.
And you are no different than Creationists arguing with
scientists
against evolution. Or people who dont believe man landed on
the
moon.
You are no different, and in fact worse, as here its
mathematics,
with absolute proof.
You people who cant accept mathematics are no different \
from
those
other people who cant 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
dont like.
I understand the *desire* to have certain things be true, but
in
mathematics that doesnt matter. At the end of the day, you
put away
your emotions, and go with whats true--if you truly value
mathematics.
James Harris
http://mathforpro\[CapitalThorn]t.blogspot.com/
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that Im 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 *in\[CapitalThorn]nity* 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 thats obvious with polynomial factors but some
sci.mathers
> 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 as and bs 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 doesnt matter how complicated that \
as and
bs 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
> dont believe in evolution, but worse.
Here yet again, lacking a proof, you try to convince
people with an oversimpli\[CapitalThorn]ed 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 as 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 as 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 as
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 \[CapitalThorn]rst 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 \[CapitalThorn]rst term
by w_1(1) = a_1(1) and the second term by
w_2(1) = a_2(1) !
Does this work? Lets see ---
The \[CapitalThorn]rst 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! Thats
good!
But I know from the de\[CapitalThorn]nition 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 whats 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 \[CapitalThorn]rst and second terms respectively
by 3 and 1. When x = 1, you divide the \[CapitalThorn]rst 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 Im 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 *in\[CapitalThorn]nity* 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:
isnt 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 thats obvious with polynomial factors but some
sci.mathers
> 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 as and bs 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 doesnt matter how complicated that \
as and
bs 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
> dont 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 its
> divided off, but STILL people have argued with me quite
successfully,
> and I \[CapitalThorn]gure many of you, despite what I say here, remain
unconvinced
> that Im right.
> And you are no different than Creationists arguing with
scientists
> against evolution. Or people who dont believe man landed
on the
> moon.
> You are no different, and in fact worse, as here its
mathematics,
> with absolute proof.
> You people who cant accept mathematics are no different
from those
> other people who cant 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
> dont like.
> I understand the *desire* to have certain things be true,
but in
> mathematics that doesnt matter. At the end of the day, \
you
put away
> your emotions, and go with whats true--if you truly value
> mathematics.
> James Harris
> http://mathforpro\[CapitalThorn]t.blogspot.com/
===
Subject: Re: JSH: Without a trace
> Some of you may have noticed that Im talking about
abstractions now
> at higher and higher levels, which I think should help.
Im 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 \[CapitalThorn]nger prints,
DNA or other forensic evidence.
Its 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 perpetrators blood and other bodily
ßuids can be left at the scene. Lets 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 Im 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.
Cant 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 its 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, youd see where your
errors are.
Youre just too pompous to see that.
Dave
===
Subject: Re: JSH: Without a trace
> If you really understood mathematics, youd see where your
errors are.
> Youre just too pompous to see that.
> Dave
Yet Ive 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 its a sad lie for some of them to just repeat that they
have or
that I dont answer objections as Ive \
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 its a sad lie for some of them to just repeat that
they have or
> that I dont answer objections as Ive \
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 Ôp equals 1/2, \
you take the
q, where Ôq
equals cake.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
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 cant 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 \[CapitalThorn]rst stone is thrown,
h= -4.9 t^2+ 20t for the \[CapitalThorn]rst 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 hs 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.
Thats actually
h = -4.9(t - 1)^2 + 25(t - 1)
> When the stones collide, both those hs will be the same
for the
> same t. Set them equal and solve for t.
And when youll do that, youll 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 books answer (20.4 m) nor does it agree
with Richs
coment about getting two answers (probably by using the
quadratic
formula.
So Im 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
Thats the correct answer --- for the TIME of the collision.
> Which is not the books 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 Richs coment about getting two
answers
I was wrong about that. I dont 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, \[CapitalThorn]ddling 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 \[CapitalThorn]nd 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
de\[CapitalThorn]nition 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, \[CapitalThorn]ddling 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 \[CapitalThorn]nd
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, \[CapitalThorn]ddling 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, \[CapitalThorn]ddling 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
Im an amateur mathematician who has found that using some
simple
ideas I can show a remarkable error in thinking which
unfortunately
underpins much of whats 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 dont 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 lets 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 its 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 Ill relate it to a family of polynomials, where that
multiple 49
is very important.
Now heres 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 its 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, youll 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
Ill 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 as 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 as, 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 its not a multiple of the cubic, as you have \
the
term
3(-1 + 49x)
which is of course, coprime to 49.
Thats important, as somehow with this technique of \
factoring
P(x) in
a special way, Ive related that factorization of a
polynomial that
has 49 as a multiple, to a family of polynomials that do not.
Its 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, lets stop to consider what \
Ive
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 de\[CapitalThorn]nes the
as. 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 in\[CapitalThorn]nity 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 \[CapitalThorn]nd out is to check directly. However,
the as
are rather complex being de\[CapitalThorn]ned by the cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and it might seem extraordinarily dif\[CapitalThorn]cult to \
\[CapitalThorn]nd 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 as equal 0, at x=0, while one equals 3.
So I need to pick the as 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, arent I just checking at a *single* value, for
something
thats 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
thats 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 doesnt vary.
Checking at x=0 clears out those terms that vary, leaving
those that
do not, revealing that for two of the terms, whats left
over, is 7,
while for one, whats left over is 22.
Thats just a fact. Its such a simple fact \
that one of the
more
remarkable things over the years Ive 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
thatd be
\[CapitalThorn]ne, 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 \[CapitalThorn]rst two
as 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 Ive shown a result that indicates that two of the
as have 7
as a factor, without regard to the value of x.
Thats HUGE as it turns out that a long time ago (like over \
a
hundred
years ago) msthematicians decided that you couldnt make \
such
a
determination if the roots of a polynomial were all
irrational and
didnt 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.
Youve 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 theres another
problem.
Over a hundred years ago, back in the late 1800s
mathematicians
studied polynomials special in that they had a leading
coef\[CapitalThorn]cient of
1 or -1, and integer coef\[CapitalThorn]cients. Polynomials with a leading
coef\[CapitalThorn]cient of 1 or -1 are called monic, so more technically,
they
studied monic polynomials with integer coef\[CapitalThorn]cients.
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 cant \[CapitalThorn]nd 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 \[CapitalThorn]nd algebraic integers w_1, w_2, and w_3,
such that
w_1 w_2 w_3 = 49
where the ws are the respective factors of
(5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
when I just said that you cant 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 theres 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, its different, but all of the mathematical
operations are
valid ones.
So what gives?
Lets 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 cant 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 as 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, heres where its rather strange, as u_1 \
and u_2 are
algebraic
integers i.e. roots of a monic polynomial with integer
coef\[CapitalThorn]cients,
while u_3 is not.
So, for that reason, u_1 and u_2 are NOT units in the ring of
algebraic integers.
Heres 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
S(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 its 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 cant 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
coef\[CapitalThorn]cients 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, Ill say, quirky.
Actually its worse than quirky, as not understanding the
weirdness
that can arise from focusing on the roots of monic
polynomials with
integer coef\[CapitalThorn]cients you can prove lots of things with the
ring of
algebraic integers that are in fact mathematically false.
Its 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 Fermats Last Theorem, \
dont work.
They dont work because theyre based on an \
improper
understanding of
the ring of algebraic integers. Works like Wiless 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://mathforpro\[CapitalThorn]t.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 Im \[CapitalThorn]xing 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 thats 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 theyre \
\[CapitalThorn]ghting to
defend,
which is why they have all that obsessive energy. And youll
understand why Im working so hard to get this work past the
people
trying to block it.
Im an amateur mathematician who has found that using some
simple
ideas I can show a remarkable error in thinking which
unfortunately
underpins much of whats 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 dont 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 lets 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 its 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 Ill relate it to a family of polynomials, where that
multiple 49
is very important.
Now heres 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 its 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, youll 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
Ill 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 as 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 as, 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 its not a multiple of the cubic, as you have \
the
term
3(-1 + 49x)
which is of course, coprime to 49.
Thats important, as somehow with this technique of \
factoring
P(x) in
a special way, Ive related that factorization of a
polynomial that
has 49 as a multiple, to a family of polynomials that do not.
Its 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, lets stop to consider what \
Ive
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 de\[CapitalThorn]nes the
as. 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 in\[CapitalThorn]nity 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 \[CapitalThorn]nd out is to check directly. However,
the as
are rather complex being de\[CapitalThorn]ned by the cubic
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and it might seem extraordinarily dif\[CapitalThorn]cult to \
\[CapitalThorn]nd 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 as equal 0, at x=0, while one equals 3.
So I need to pick the as 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, arent I just checking at a *single* value, for
something
thats 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
thats 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 doesnt vary.
Checking at x=0 clears out those terms that vary, leaving
those that
do not, revealing that for two of the terms, whats left
over, is 7,
while for one, whats left over is 22.
Thats just a fact. Its such a simple fact \
that one of the
more
remarkable things over the years Ive 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
thatd be
\[CapitalThorn]ne, 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 \[CapitalThorn]rst two
as 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 Ive shown a result that indicates that two of the
as have 7
as a factor, without regard to the value of x.
Thats HUGE as it turns out that a long time ago (like over \
a
hundred
years ago) msthematicians decided that you couldnt make \
such
a
determination if the roots of a polynomial were all
irrational and
didnt 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.
Youve 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 theres another
problem.
Over a hundred years ago, back in the late 1800s
mathematicians
studied polynomials special in that they had a leading
coef\[CapitalThorn]cient of
1 or -1, and integer coef\[CapitalThorn]cients. Polynomials with a leading
coef\[CapitalThorn]cient of 1 or -1 are called monic, so more technically,
they
studied monic polynomials with integer coef\[CapitalThorn]cients.
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 cant \[CapitalThorn]nd 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 \[CapitalThorn]nd algebraic integers w_1, w_2, and w_3,
such that
w_1 w_2 w_3 = 49
where the ws are the respective factors of
(5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
when I just said that you cant 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 theres 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, its different, but all of the mathematical
operations are
valid ones.
So what gives?
Lets 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 cant 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 as 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, heres where its rather strange, as u_1 \
and u_2 are not
algebraic integers i.e. roots of a monic polynomial with
integer
coef\[CapitalThorn]cients, 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.
Heres 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
S(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 its 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 cant 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
coef\[CapitalThorn]cients 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, Ill say, quirky.
Actually its worse than quirky, as not understanding the
weirdness
that can arise from focusing on the roots of monic
polynomials with
integer coef\[CapitalThorn]cients you can prove lots of things with the
ring of
algebraic integers that are in fact mathematically false.
Its 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 Fermats Last Theorem, \
dont work.
They dont work because theyre based on an \
improper
understanding of
the ring of algebraic integers. Works like Wiless 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://mathforpro\[CapitalThorn]t.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 Im \[CapitalThorn]xing here.
OOPS! The original was correct, so my correction was wrong.
Thats the second time Ive done something \
like this which
is, yes,
worrisome.
I guess Im so used to making mistakes that I now \
\[CapitalThorn]nd
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 Im \[CapitalThorn]xing 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 thats 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 theyre \
\[CapitalThorn]ghting
to defend,
> which is why they have all that obsessive energy. And \
youll
> understand why Im working so hard to get this work past
the people
> trying to block it.
Better question: why wasnt 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 youve 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 Im \[CapitalThorn]xing 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 yesterdays post, why \
it
was
wrong and what is correct.
Heres the only differences between the two posts:
,----[ Yesterday ]
| But, heres where its rather strange, as \
u_1 and u_2 are
algebraic
| integers i.e. roots of a monic polynomial with integer
coef\[CapitalThorn]cients,
| 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, heres where its rather strange, as \
u_1 and u_2 are
not
| ^^^
| algebraic integers i.e. roots of a monic polynomial with
integer
| coef\[CapitalThorn]cients, 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, \[CapitalThorn]ghting for what
should be
mine, having to get past weak-minded academics who are
\[CapitalThorn]ghting to
block my success. But I shall prevail!!! -- James S. Harris
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> ... ad in\[CapitalThorn]nitum ...
What I want to know is if its 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)
ÔRemove MYWORRIES to email me!
Calulus and alcohol dont mix.
So remember ...
Dont drink and derive!
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
Discussion, linux)
>> ... ad in\[CapitalThorn]nitum ...
[...]
> I should introduce you to my sister.
Youre weird.
--
Jesse F. Hughes
Leaving things always seems to \[CapitalThorn]x me,
Running seems to ease my worried mind.
-- Bad Livers, Honey, Ive Found a Brand New Way
===
Subject: Re: Revolutionary Mathematics: Non-polynomial
Factorization
> Im an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whats 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 dont 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 lets 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 its distinctive
in an
> important way, as it has 49 as a multiple
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
FACTOR FACTOR
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FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
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FACTOR FACTOR
FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
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FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR FACTOR
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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
> Im an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whats 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 lets 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 its 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 Ill relate it to a family of polynomials, where that
multiple 49
> is very important.
Ok, so it can be factored in Z[x].
> Now heres 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 its 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, youll \
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
Ill 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 as 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 as, 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 as 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 its 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?
> Thats important, as somehow with this technique of
factoring P(x) in
> a special way, Ive 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.
> Its 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
havent stated what properties the a_i(x) have, so \
its
dif\[CapitalThorn]cult 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, lets stop to consider what
Ive 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 its
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 de\[CapitalThorn]nes the
as. 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 de\[CapitalThorn]nition 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 in\[CapitalThorn]nity 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 \[CapitalThorn]nd a factorization of P(x)/49 that is somehow related to
the
peculiar way you chose to factor P(x).
> The simplest way to \[CapitalThorn]nd out is to check directly. However,
the as
> are rather complex being de\[CapitalThorn]ned by the cubic
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and it might seem extraordinarily dif\[CapitalThorn]cult to \
\[CapitalThorn]nd out
anything about
> them, so is the cause lost?
Back up. The fact that they are roots of a cubic is, in
itself,
something weve found out about them quite simply. The
dif\[CapitalThorn]culty 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)s are not polynomials, they do not
necessarily
have constant terms. Also, since we have no stated goal yet,
it is
dif\[CapitalThorn]cult 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 as equal 0, at x=0, while one equals 3.
> So I need to pick the as 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, arent I just checking at a *single* value, for
something
> thats 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
> thats 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 doesnt vary.
Pretty much by de\[CapitalThorn]nition. 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, whats left
over, is 7,
> while for one, whats left over is 22.
> Thats just a fact. Its such a simple fact \
that one of the
more
> remarkable things over the years Ive found is the ability
of some
> people to argue around it.
Its 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.
Heres 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
thatd be
> \[CapitalThorn]ne, but with that value, the terms with x get in the way,
but at
> x=0, they do not.
Heres 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 \[CapitalThorn]rst two
as 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 Ive shown a result that indicates that two of the
as have 7
> as a factor, without regard to the value of x.
How do you \[CapitalThorn]gure that? You have shown that two of the \
as are
0 when
x=0, but that is far from showing the as have 7 as a factor
in their
ring. Granted, the most obvious ring to me for the as to
exist in is
the ring of algebraic functions, in which *every* function
has 7 as a
factor. Did you want the as to be in a more restricted \
ring?
If so,
which?
> Thats HUGE as it turns out that a long time ago (like \
over
a hundred
> years ago) msthematicians decided that you couldnt make
such a
> determination if the roots of a polynomial were all
irrational and
> didnt 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, wouldnt it
be useful to
refer to it a little earlier in your discussion?
> Youve 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, youre 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 theres another
problem.
And perhaps they do. It depends on what ring you believe
these roots
live in. If the as 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 1800s
mathematicians
> studied polynomials special in that they had a leading
coef\[CapitalThorn]cient of
> 1 or -1, and integer coef\[CapitalThorn]cients. Polynomials with a \
leading
> coef\[CapitalThorn]cient of 1 or -1 are called monic, so more
technically, they
> studied monic polynomials with integer coef\[CapitalThorn]cients.
> 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 cant \[CapitalThorn]nd any that \
when divided
by 7, give
> an algebraic integer.
This suggests that your earlier work failed to take this
possible goal
into account. Thats ok, it had no stated goal and no \
special
efforts
were taken to insure any properties were preserved. Given
that, its
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 \[CapitalThorn]nd algebraic integers w_1, w_2, and w_3,
such that
> w_1 w_2 w_3 = 49
> where the ws are the respective factors of
> (5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
> when I just said that you cant 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 as \
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 theres
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, its
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, its different, but all of the mathematical
operations are
> valid ones.
> So what gives?
Perhaps the constant terms didnt 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?
> Lets 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 cant work with these numbers called algebraic
integers!
Well, since you introduced non-algebraic integers with those
1/7s, 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: whats it all that works now? I dont \
know what the
us 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 ws above.
> That is, for some reason, while two of the as 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, heres where its rather strange, as \
u_1 and u_2 are
algebraic
> integers i.e. roots of a monic polynomial with integer
coef\[CapitalThorn]cients,
> 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 dont even appear to be numbers. They appear \
to
be
functions. Did you mean for them to be numbers?
> Heres 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
> S(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 its 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 cant 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 de\[CapitalThorn]ned.
> So why all this talk about algebraic integers? They seem
kind of
> messy at this point.
No they dont, 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
> coef\[CapitalThorn]cients 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, Ill say, quirky.
> Actually its worse than quirky, as not understanding the
weirdness
> that can arise from focusing on the roots of monic
polynomials with
> integer coef\[CapitalThorn]cients 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
thats not
a big deal.
> Its a nasty little bug.
No, its 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 Fermats Last Theorem, \
dont
work.
Huh? Care to \[CapitalThorn]ll in a detail or three?
> They dont work because theyre based on an \
improper
understanding of
> the ring of algebraic integers. Works like Wiless 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 youve 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 didnt take long to read over. The implications are less
stunning
than you think. Heres 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 coef\[CapitalThorn]ecients that it solves. If it really is as
trivial as
you
say, then why dont you simply tell us what value of x to
use, and which
polynomial (a/7) is a root of.
Im not looking at this with any bias or slant. I (like you)
would \[CapitalThorn]nd it
very interesting if you really found something important, but
I just dont
get your argument, and no amount of hand-waving about
constants has yet
convinced me otherwise.
-Darren
> Im an amateur mathematician who has found that using some
simple
> ideas I can show a remarkable error in thinking which
unfortunately
> underpins much of whats 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 dont 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 lets 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 its 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 Ill relate it to a family of polynomials, where that
multiple 49
> is very important.
> Now heres 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 its 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, youll \
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
Ill 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 as 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 as, 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 its not a multiple of the cubic, as you have
the term
> 3(-1 + 49x)
> which is of course, coprime to 49.
> Thats important, as somehow with this technique of
factoring P(x) in
> a special way, Ive related that factorization of a
polynomial that
> has 49 as a multiple, to a family of polynomials that do
not.
> Its 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, lets stop to consider what
Ive 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 de\[CapitalThorn]nes the
as. 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 in\[CapitalThorn]nity 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 \[CapitalThorn]nd out is to check directly. However,
the as
> are rather complex being de\[CapitalThorn]ned by the cubic
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and it might seem extraordinarily dif\[CapitalThorn]cult to \
\[CapitalThorn]nd 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 as equal 0, at x=0, while one equals 3.
> So I need to pick the as 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, arent I just checking at a *single* value, for
something
> thats 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
> thats 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 doesnt vary.
> Checking at x=0 clears out those terms that vary, leaving
those that
> do not, revealing that for two of the terms, whats left
over, is 7,
> while for one, whats left over is 22.
> Thats just a fact. Its such a simple fact \
that one of the
more
> remarkable things over the years Ive 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
thatd be
> \[CapitalThorn]ne, 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 \[CapitalThorn]rst two
as 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 Ive shown a result that indicates that two of the
as have 7
> as a factor, without regard to the value of x.
> Thats HUGE as it turns out that a long time ago (like \
over
a hundred
> years ago) msthematicians decided that you couldnt make
such a
> determination if the roots of a polynomial were all
irrational and
> didnt 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.
> Youve 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 theres another
problem.
> Over a hundred years ago, back in the late 1800s
mathematicians
> studied polynomials special in that they had a leading
coef\[CapitalThorn]cient of
> 1 or -1, and integer coef\[CapitalThorn]cients. Polynomials with a \
leading
> coef\[CapitalThorn]cient of 1 or -1 are called monic, so more
technically, they
> studied monic polynomials with integer coef\[CapitalThorn]cients.
> 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 cant \[CapitalThorn]nd 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 \[CapitalThorn]nd algebraic integers w_1, w_2, and w_3,
such that
> w_1 w_2 w_3 = 49
> where the ws are the respective factors of
> (5a_1(x) + 7), (5a_2(x)+ 7), and (5a_3(x) + 7)
> when I just said that you cant 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 theres
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, its different, but all of the mathematical
operations are
> valid ones.
> So what gives?
> Lets 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 cant 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 as 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, heres where its rather strange, as \
u_1 and u_2 are
algebraic
> integers i.e. roots of a monic polynomial with integer
coef\[CapitalThorn]cients,
> while u_3 is not.
> So, for that reason, u_1 and u_2 are NOT units in the ring
of
> algebraic integers.
> Heres 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
> S(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 its 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 cant 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
> coef\[CapitalThorn]cients 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, Ill say, quirky.
> Actually its worse than quirky, as not understanding the
weirdness
> that can arise from focusing on the roots of monic
polynomials with
> integer coef\[CapitalThorn]cients you can prove lots of things with the
ring of
> algebraic integers that are in fact mathematically false.
> Its 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 Fermats Last Theorem, \
dont
work.
> They dont work because theyre based on an \
improper
understanding of
> the ring of algebraic integers. Works like Wiless 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://mathforpro\[CapitalThorn]t.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 as are divisible by 7,
although you
will
never \[CapitalThorn]nd out from him exactly what he means by this.
Thats 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
as are
divisible by 7. He might say they are properly divisible by 7
or
something, although I dont 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 as
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 de\[CapitalThorn]ne these terms. When people
ask speci\[CapitalThorn]c
questions he answers by simply reposting his original proof
and later
claims Ive 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 coef\[CapitalThorn]ecients that it solves. If it really is
as trivial as
you
> say, then why dont you simply tell us what value of x to
use, and which
> polynomial (a/7) is a root of.
> Im not looking at this with any bias or slant. I (like
you) would \[CapitalThorn]nd
it
> very interesting if you really found something important,
but I just
dont
> 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 isnt
entirely logical, but for James sake they go along with the
following
precise de\[CapitalThorn]nition:
DEFINITION: the constant term of f(x) is f(0).
Nora and others even believe this is James own \
de\[CapitalThorn]nition,
and for sure he
may even have said this at some point, but I dont believe
this is whats
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 dont 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
coef\[CapitalThorn]cients, so Im hoping this will help you \
understand what
Im
saying.
It will be a real test of you as a mathematician, but Im
hoping youll
come
through and be able to follow - its not \
dif\[CapitalThorn]cult - in the
end its 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 dont.
> 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 as are divisible by 7,
although you
will
> never \[CapitalThorn]nd out from him exactly what he means by this.
Thats all covered in my original post and yes, for those \
who
wondered, its clear that some people dont \
bother to read
through my
posts, as theyre just here to argue.
For instance, I dont say divisible by 7 in the post. And \
the
accusation that my methods are ßawed is, youll 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 youll \[CapitalThorn]nd 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 dont.
>>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 as are divisible by 7,
although you
will
>>never \[CapitalThorn]nd out from him exactly what he means by this.
> Thats all covered in my original post and yes, for those
who
> wondered, its clear that some people dont \
bother to read
through my
> posts, as theyre just here to argue.
> For instance, I dont say divisible by 7 in the post. And
the
> accusation that my methods are ßawed is, youll \
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 youll \[CapitalThorn]nd that I \
answered all
these
> issues, in detail.
Funny, I read through your post and found that you didnt
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
suf\[CapitalThorn]cient 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 con\[CapitalThorn]ne 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 \[CapitalThorn]rst, so we can help rather then do), but I have
noticed that
people on this board dont 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 con\[CapitalThorn]rm 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 \[CapitalThorn]rst, so we can help rather then do), but I have noticed
that people
> on this board dont 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? Identi\[CapitalThorn]ed 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 havent 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 \[CapitalThorn]rst, so we can help rather then do), but I have
noticed that people
>> on this board dont 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? Identi\[CapitalThorn]ed 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 havent really started and I need to get this done \
for
tomorrow.
Very well said!
Its actually much easier just to solve problems for people,
but as
you say were 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
Im 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? Identi\[CapitalThorn]ed 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 havent really started and I need to get this done \
for
tomorrow.
Its 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 isnt 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 dont call me dim...
I thought I might solve the following similar problem but \
Im
stuck
again:
Two rockets are \[CapitalThorn]red vertically from launching pads side by
side.
The \[CapitalThorn]rst rocket moves vertically upwards with an acceleration
of 7g
and the second with an acceleration of 9g. If the second
rocket is
\[CapitalThorn]red 2 seconds after the \[CapitalThorn]rst, \[CapitalThorn]nd \
how long after its
launching the
second rocket overtakes the \[CapitalThorn]rst.
--
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)
Thats what I thought but it gives me a value of t=9 and the
books
answer is 14.94 secs.
So I went wrong somewhere along the line!
Can you help, please?
Jo
===
Subject: Re: Mechanics problem
> Please dont call me dim...
> I thought I might solve the following similar problem but
Im stuck
> again:
> Two rockets are \[CapitalThorn]red vertically from launching pads side by
side.
> The \[CapitalThorn]rst rocket moves vertically upwards with an
acceleration of 7g
> and the second with an acceleration of 9g. If the second
rocket is
> \[CapitalThorn]red 2 seconds after the \[CapitalThorn]rst, \
\[CapitalThorn]nd how long after its
launching the
> second rocket overtakes the \[CapitalThorn]rst.
> --
> 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)
> Thats what I thought but it gives me a value of t=9 and
the books
> 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 \[CapitalThorn]rst
rocket as
a function of t (with t=0 being when the *\[CapitalThorn]rst* 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 *\[CapitalThorn]rst*
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 dont call me dim...
> I thought I might solve the following similar problem but
Im stuck
> again:
> Two rockets are \[CapitalThorn]red vertically from launching pads side by
side.
> The \[CapitalThorn]rst rocket moves vertically upwards with an
acceleration of 7g
> and the second with an acceleration of 9g. If the second
rocket is
> \[CapitalThorn]red 2 seconds after the \[CapitalThorn]rst, \
\[CapitalThorn]nd how long after its
launching the
> second rocket overtakes the \[CapitalThorn]rst.
> --
> 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)
> Thats what I thought but it gives me a value of t=9 and
the books
> 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. Im not
sure if the acceleration \[CapitalThorn]gures are correct or not.
Bill
===
Subject: Revolutionary Mathematics: Synopsis version
Im 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 de\[CapitalThorn]ning 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 Ill
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://mathforpro\[CapitalThorn]t.blogspot.com/
===
Subject: Re: Revolutionary Mathematics: Synopsis version
> Im an amateur mathematician.
> James Harris
> http://mathforpro\[CapitalThorn]t.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
>> Im an amateur mathematician.
>> James Harris
>> http://mathforpro\[CapitalThorn]t.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 tails in a crack, you
improvise
fwbrown@bellsouth.net | if youre 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;
>Im 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 de\[CapitalThorn]ning 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 Ill
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 de\[CapitalThorn]ne 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 de\[CapitalThorn]nition 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 \[CapitalThorn]rst 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, doesnt.
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 de\[CapitalThorn]nition 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 ws 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 \[CapitalThorn]eld of algebraic numbers, and
in fact not a \[CapitalThorn]eld 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 de\[CapitalThorn]nition 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 de\[CapitalThorn]nition 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://mathforpro\[CapitalThorn]t.blogspot.com/>http://
mathforpro\[CapitalThorn]t.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
Im 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 Ive given a long post in-depth, and a synopsis
post, Im
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 as 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.
Thats basic, but its 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
as have
factors in common with 7 that must vary, and you know, \
thats
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 Im suggesting is counter-intuitive in that \
Im 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 theres no way
that your
earlier instincts that a given cubic or family of cubics were
completely de\[CapitalThorn]ned by just that cubics \
coef\[CapitalThorn]cients 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 whats
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 \[CapitalThorn]nd
mathematics.
It can help you understand mathematics, or it can block you
if you
dont 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 theres no trace of 49, left, and P(x)/49 might be
misleading in
that you can still see the 49, but thats 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 doesnt 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 as
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 dont 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.
Thats a logical contradiction.
If the terms are in fact independent of the value of x, then
they are
independent.
Its a tautology.
If they are independent of x, then setting x=0 has NO EFFECT
on them,
but it does have the bene\[CapitalThorn]t of clearing out terms that ARE
dependent
on x.
Before these arguments about this that basic logic was not
argued
with, and Im sure lots of analysis is going on everyday,
where people
use it as its 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 cant 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). Its 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 theres a
fatal
ßaw if you dont 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.
Im asking you to be logical.
What happens then though? Well then you \[CapitalThorn]nd out that \
Im
right and
yes, there is this major problem in algebraic number theory
that NEEDS
TO BE FIXED.
When I communicate with trained mathematicians, I dont get
objections
that come down to questioning whether or not constant terms
are
constant as they dont do the Usenet objections. They just
walk away.
So I have two different behaviors: irrational on Usenet, and
passive-aggressive off.
Im 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;
>Im 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 Ive given a long post in-depth, and a synopsis
post, Im
>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 as 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.
>Thats basic, but its 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
as have
>factors in common with 7 that must vary, and you know,
thats 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 Im suggesting is counter-intuitive in that \
Im 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 theres no way
that your
>earlier instincts that a given cubic or family of cubics were
>completely de\[CapitalThorn]ned by just that cubics \
coef\[CapitalThorn]cients 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 whats
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 \[CapitalThorn]nd
mathematics.
>It can help you understand mathematics, or it can block you
if you
>dont 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 theres 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 ?
Its 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.
Its not in the term -3(-1 + 49 x). That is also coprime to
49.
Its in the constant term, 7.
That feature of P(x) is there even after you have divided
by 49. Its 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 thats 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 doesnt 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
as
>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 doesnt 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.
Thats 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 dont 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.
>Thats a logical contradiction.
Its 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 de\[CapitalThorn]nition.
>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.
>Its a tautology.
Its false.
>If they are independent of x,
Theyre not.
then setting x=0 has NO EFFECT on them,
>but it does have the bene\[CapitalThorn]t 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 Im sure lots of analysis is going on everyday,
where
people
>use it as its 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 cant 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). Its 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 xs 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 de\[CapitalThorn]ning 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 as 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 theres \
a
fatal
>ßaw if you dont 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.
>Im asking you to be logical.
>What happens then though? Well then you \[CapitalThorn]nd out that \
Im
right and
>yes, there is this major problem in algebraic number theory
that
NEEDS
>TO BE FIXED.
>When I communicate with trained mathematicians, I dont get
objections
>that come down to questioning whether or not constant terms
are
>constant as they dont 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.
>Im 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
> Im 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 Ive given a long post in-depth, and a synopsis
post, Im
> 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 as 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.
> Thats basic, but its 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
as have
> factors in common with 7 that must vary, and you know,
thats 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 Im suggesting is counter-intuitive in that \
Im 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 theres no \
way
that your
> earlier instincts that a given cubic or family of cubics
were
> completely de\[CapitalThorn]ned by just that cubics \
coef\[CapitalThorn]cients 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 whats
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 \[CapitalThorn]nd
mathematics.
> It can help you understand mathematics, or it can block you
if you
> dont 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 theres no trace of 49, left, and P(x)/49 might be
misleading in
> that you can still see the 49, but thats 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 doesnt 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 as
> 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 dont 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.
> Thats a logical contradiction.
> If the terms are in fact independent of the value of x,
then they are
> independent.
> Its a tautology.
> If they are independent of x, then setting x=0 has NO
EFFECT on them,
> but it does have the bene\[CapitalThorn]t of clearing out terms that ARE
dependent
> on x.
> Before these arguments about this that basic logic was not
argued
> with, and Im sure lots of analysis is going on everyday,
where people
> use it as its 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 cant 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). Its 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 theres
a fatal
> ßaw if you dont 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.
> Im asking you to be logical.
> What happens then though? Well then you \[CapitalThorn]nd out that \
Im
right and
> yes, there is this major problem in algebraic number theory
that NEEDS
> TO BE FIXED.
> When I communicate with trained mathematicians, I dont \
get
objections
> that come down to questioning whether or not constant terms
are
> constant as they dont do the Usenet objections. They just
walk away.
> So I have two different behaviors: irrational on Usenet, and
> passive-aggressive off.
> Im 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
Ive 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
> Ive 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??
Youre 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 were 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 wed 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 wed 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 dont 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)].
--
Its not denial. Im 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 wed 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 dont 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 youve 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 hadnt 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 wed 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 dont 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 didnt 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 youve 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.
--
Its not denial. Im 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 \[CapitalThorn]rst column of the product AB is just b11 times
the \[CapitalThorn]rst column of A. The second column of the product is b12
times the \[CapitalThorn]rst column of A, plus b22 times the second column
of
A.
What I havent 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 Im 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 doesnt 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 Im 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 doesnt 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 \[CapitalThorn]rst ten terms, to get an approximation, we \
\[CapitalThorn]nd
x = 417090851369317809389830589/92343379137802193583984375
= 4.5167380191589199103...
A more accurate approximation of the solution (obtained using
Mathematicas
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 Im 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 in\[CapitalThorn]nity, 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 =.
Ive been working on this goddamned hellish torture puzzle
for hours. Ive
got something on the order of 30 pages of scrap paper used up
on this
problem. Its 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 Ill stop there.
Please, please
help me. Im ... 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 in\[CapitalThorn]nity, 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 Ill stop there.
Please,
please
>help me. Im ... so damned lost.
Okay, breathe. Youre 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 \[CapitalThorn]rst 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 \[CapitalThorn]rst 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 in\[CapitalThorn]nity, 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 Ill stop \
there.
Please,
> please
>>help me. Im ... so damned lost.
> Okay, breathe. Youre 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 \[CapitalThorn]rst 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 \[CapitalThorn]rst 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 \[CapitalThorn]rst n squares is n(n+1)(2n+1)/6
Now were ready...
You want to \[CapitalThorn]nd
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 de\[CapitalThorn]nite 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 \[CapitalThorn]rst n squares is n(n+1)(2n+1)/6
> Now were ready...
> You want to \[CapitalThorn]nd
> 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
doesnt accept your Rn. Am I misinterpreting what \
youve
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
Thats (2n^3 + 3n^2 + n). Of course, in the limit the
error doesnt 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
> Thats (2n^3 + 3n^2 + n). Of course, in the limit the
> error doesnt 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 its 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 Classi\[CapitalThorn]cation. Primary: 12E12; Secondary:
15A04
http://www.geocities.com/titus_piezas/cubics.html
Just click at the link to the pdf \[CapitalThorn]le.
P.S. Theres 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 \[CapitalThorn]nal this week and my professor used the phrase if
you
know the review sheet you SHOULD be ok for the \[CapitalThorn]nal. 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 \[CapitalThorn]nal 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
\[CapitalThorn]nd 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
\[CapitalThorn]nd 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
?
\[CapitalThorn]nd the next number
===
Subject: Re: Number Pattern Help
>these are my numbers
>\[CapitalThorn]nd the next number
These arent math questions they are psychology questions.
There are in\[CapitalThorn]nitely many answers --each as correct
mathematically as
any other. What were 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
> \[CapitalThorn]nd 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
Its a pleasure to meet me. I hope you
never \[CapitalThorn]nd a live turtle in your soup.
===
Subject: Re: Number Pattern Help
posting-account=-m3APw0AAAA-5WXqbJX0X9WBnuKsH_Qg
116 ?
> these are my numbers
> 26
> 63
> \[CapitalThorn]nd the next number
===
Subject: Question w/o a deadline
So, anyway... that was for my sunday night webworks.
This one is just a problem Ive 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, Im 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 \[CapitalThorn]nd R = \
C.
Im do not know how to manipulate the given formula to \
\[CapitalThorn]nd R
and C. I
looked through the books sample problems, but there are \
none
similar to
this - theyre 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...dont 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 \[CapitalThorn]nd 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 dont 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 \[CapitalThorn]nd 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 dont get the same answer
Good -- show us what you did and well help you \
\[CapitalThorn]nd 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 de\[CapitalThorn]ned 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 cant \[CapitalThorn]gure 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)
Lets solve a simpler problem \[CapitalThorn]rst:
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 its an 8th degree polynomial, and you know all its \
8
zeros, you
know
its 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 \[CapitalThorn]nd 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 cant \[CapitalThorn]gure 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
\[CapitalThorn]gure 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.
--
Its not denial. Im 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. Im thinking its 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 dont know how to \[CapitalThorn]nd the growth rate for \
those problems
when given
the time. Id 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
\[CapitalThorn]rst
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 youve 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 dont know how to \[CapitalThorn]nd the growth rate for \
those problems
when given
> the time. Id 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 thats 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 \[CapitalThorn]nd k? Since once
you have k, you can immediately \[CapitalThorn]nd 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 dont 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 de\[CapitalThorn]ned.
Travis
> can a function be both an even and an odd function, at the
same time? I
> chose no.. but dont 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 dont know how to back up my answer..
Its always a good idea to go to the \
de\[CapitalThorn]nition.
An even function E is de\[CapitalThorn]ned by E(-x) = E(x) for all x in the
domain.
An odd function )O is de\[CapitalThorn]ned 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 dont 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,
Id say!
Joseph A.
> can a function be both an even and an odd function, at the
same
time? I
> chose no.. but dont 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 dont know how to back up my answer..
Well, lets 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 de\[CapitalThorn]ntion 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 dont know how to back up my answer..
>Well, lets 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 de\[CapitalThorn]ntion 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..
Couldnt 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 dont know how to back up my answer..
> Well, lets 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 de\[CapitalThorn]ntion a function may only
have one value
on
> its domain.
> Darren
===
Subject: Re: even and odd functions.. general question..
Yes. You are right!
> Couldnt 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 dont know how to back up my answer..
> Well, lets 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 de\[CapitalThorn]ntion 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?
> 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?
Well since -1 is the additive inverse of +1 it follows that
-1 +1 =0
Now multipy both sides by -1
-1(-1) +-1(1) = -1(0) =0
so we know (-1)(-1) +(-1)(+1) = 0 now subtract (-1)(+1) from
both sides
we have (-1)(-1) = -(-1)(+1) or
(-1)(-1) = (+1)(+1) = +1
That is a minus times a minus equals a plus. You can use this
result to
show that it works for any other set of integers as well
===
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 iBKFrbj26194;
i have answered this question using the properties of the
number line.
unfortunately, i cannot post this here since it contains
graphics.
hence, i have posted it on my webpage at
if you have anything to say about it, please mail to
abhishek@deydas.com.
thank you.
>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?
Im supposed to pay a bunch of $10 parking tickets. The
effect on my
income is -$10 for each. Three of the tickets are thrown out
of court
(-3). In effect, I made an income of $30. (-10)(-3) = +30.
Mike Lepore - email delete the 5 - http://www.crimsonbird.com/
===
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 iBGLfP111198;
Each day you borow 5 bucks for 5 days.
therefore: -5 -5 -5 -5 -5 -5=5*(-5)=-25
Therefore:
- ( -5 -5 -5 -5 -5)= 5+5+5+5+5=-[5*(-5)]=-5*(-5)=25
Happy?
>> 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
>> me. Will someone please make sense of it?
>Im supposed to pay a bunch of $10 parking tickets. The
effect on my
>income is -$10 for each. Three of the tickets are thrown out
of
court
>(-3). In effect, I made an income of $30. (-10)(-3) = +30.
>Mike Lepore - email delete the 5 - <a
href=http://www.crimsonbird.com/>http://www.crimsonbird.com/<
/a>
===
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?
> Im supposed to pay a bunch of $10 parking tickets. The
effect on my
> income is -$10 for each. Three of the tickets are thrown
out of court
> (-3). In effect, I made an income of $30. (-10)(-3) = +30.
Now that you have made that public, the IRS expects its share
(assuming youre in the US).
Bill
Swap \[CapitalThorn]rst and last parts of username and ISP for address.
===
Subject: Re: Why do two negatives equal a positive?
Theres the old joke about an English professor telling the
class how
two negatives make a positive, but that two positives dont
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)
ÔRemove MYWORRIES 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?
Its 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
werent 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, its 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)
ÔRemove MYWORRIES 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 didnt give a positive, \
youd
be stuck
in negative territory forever.
Of course that _is_ what happens when you multiply two
positives. It
doesnt 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?
> Picture the complex plane. Multiplying a number by -1 (=
i^2) rotates
the
> number 180 degrees counterclockwise, so multiplying by -1
twice .... .
HTH.
That (-a)(-b) = ab is fact about rings. It should not be
necessary to
invoke the complex \[CapitalThorn]eld to prove it.
===
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.
> That (-a)(-b) = ab is fact about rings. It should not be
necessary to
> invoke the complex \[CapitalThorn]eld to prove it.
No, its not necessary, but I think that answers to this
question can be
not just mathematical, but also psychological/pedagogical.
Given a
mathematical notion (such as multiplying negative numbers),
how can I
most easily visualize it, so that I can remember it easily?
Or, how can
I convey my idea to someone else? When you devise a proof of
some
conjecture (and thereby transform it into a theorem), I
imagine you
dont throw at it all the de\[CapitalThorn]nitions you ever \
knew, but
rather have a
pretty good (internal, non-verbal) idea of what you want to
show, and of
how to reach the goal, and then you plug in the de\[CapitalThorn]nitions
and logical
steps as needed to make the argument solid and easily
understandable.
Imagination and the ability to visualize are important, and
different
people do this in different ways. An argument that appeals to
one
person may puzzle another.
-- Vincent Johns <vjohns@alumni.caltech.edu>
Please feel free to quote anything I say here.
===
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 Im 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 Im 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 dont have any bananas. (same as I have zero bananas.)
I have no bananas. (same as I have zero bananas.)
I dont have no bananas. (same as I have some bananas.)
In the last sentence two negatives (dont 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
Its 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?
Its like taking a video of someone walking backward, then
playing the
tape backward. Youll 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
wont 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 x(t) = f_x, \
y(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 \[CapitalThorn]eld axioms
(http://mathworld.wolfram.com/FieldAxioms.html)
\[CapitalThorn]rst 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 iBKDNsc11202;
using the \[CapitalThorn]eld axioms is perfect except for a major ßaw \
which
(atleast in my eyes) has rendered it useless. we are trying
to prove
here that when a negetive and a positive integer is
multiplied the
result is positive. why? but in your solution, you have used
the same
property in the third line without proving it \[CapitalThorn]rst.
>By using the \[CapitalThorn]eld axioms
>(<a
href=http://mathworld.wolfram.com/FieldAxioms.html>http://
mathworld.wolf
ram.com/FieldAxioms.html</a>)
>\[CapitalThorn]rst 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?
> using the \[CapitalThorn]eld axioms is perfect except for a major ßaw
which
> (atleast in my eyes) has rendered it useless. we are trying
to prove
> here that when a negetive and a positive integer is
multiplied the
> result is positive.
No, we are trying to show that when two negative integers are
multiplied
the
result is positive. Multiplying a positive and a negative
will give you a
negative...
> why? but in your solution, you have used the same
> property in the third line without proving it \[CapitalThorn]rst.
Here are the second and third lines, if I understand you
correctly:
So where exactly has Matt used the same property without
proof?
(Answer: he hasnt!)
And BTW many people consider top-posting as you have done to
be rude (just
to warn you...)
Mike.
>By using the \[CapitalThorn]eld axioms
>(<a
href=http://mathworld.wolfram.com/FieldAxioms.html>http://
mathworld.wolf
ra
m.com/FieldAxioms.html</a>)
>\[CapitalThorn]rst 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?
>\[CapitalThorn]rst 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 youll 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
\[CapitalThorn]nally conclude with axiomatic truth. Now then, \
lets 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 de\[CapitalThorn]nition of difference.
That is,
we de\[CapitalThorn]ne 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;
Im trying to \[CapitalThorn]gure 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??
>Im trying to \[CapitalThorn]gure this problem out and \
having trouble, can
anyone
>help?
>1. The factors of x4(power)-625 are?
>my answer was x-5
Thats _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
CANT factor a^2+b^2 over the reals, so dont \
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;
>Im trying to \[CapitalThorn]gure 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??
> Im trying to \[CapitalThorn]gure this problem out and \
having trouble,
can anyone
> help?
> 1. The factors of x4(power)-625 are?
> my answer was x-5
Thats *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
Im trying to determine
lim f(n)
n->inf
where f(n)=(2n-1)!! / (2n)!!
Its 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 cant prove it. Can anyone give me a hint?
===
Subject: Re: question about a limit
> Im trying to determine
> lim f(n)
> n->inf
> where f(n)=(2n-1)!! / (2n)!!
> Its 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 cant 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
youll get what
you need (with apologies to the Rolling Stones).
===
Subject: Re: question about a limit
>Im 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
>Its 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 cant prove it. Can anyone give me a hint?
===
Subject: Re: question about a limit
alt.math.undergrad:
>>Im 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 de\[CapitalThorn]nition:
(-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 coef\[CapitalThorn]cient Ôp \
choose q.
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 Stirlings 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 #s 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 Ive 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 \[CapitalThorn]nd the four independent eigenvalues? I \
cant seem to
----------------------------------------------
* Binary Usenet Leeching Made Easy
* http://www.newsleecher.com/?usenet
----------------------------------------------
===
Subject: Help! Linear Algebra problem(ignore \[CapitalThorn]rst post)
Sorry, the \[CapitalThorn]rst 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 Ive 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 \[CapitalThorn]nd the four independent eigenvalues? I \
cant seem to
----------------------------------------------
* Binary Usenet Leeching Made Easy
* http://www.newsleecher.com/?usenet
----------------------------------------------
===
Subject: Re: Help! Linear Algebra problem(ignore \[CapitalThorn]rst post)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB914lo31802;
>Sorry, the \[CapitalThorn]rst 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 Ive 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 \[CapitalThorn]nd the four independent eigenvalues? I \
cant 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 \[CapitalThorn]nd 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. Im 19 years old. I live in
Ethiopia. Im
student at college in the 3rd year student. My study is
Secretarial
Science and Of\[CapitalThorn]ce Management... and in this study I educated
Maths
\[CapitalThorn]eld and Please Im 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 youve
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 youve
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 \[CapitalThorn]nancial 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 dont \
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 \
Ômew is
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 \[CapitalThorn]nd 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 dont
>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 \
Ômew is
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 \[CapitalThorn]nd 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 speci\[CapitalThorn]ed \
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
\[CapitalThorn]nd 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 20 from the horizontal. This is not the angle
between F1 and
L (the inclined plane), so you have to calculate this angle:
a1 = 90 + 20 = 110
The component of the gravity force in the direction L (of the
inclined
plane) is
|F1|*cos(110) = -|F1|*cos(70) =
-25*9.8*cos(70) N
2. Pulling force F2, magnitude |F2| = 500 N, direction at the
angle a2
= 10 to the inclined plane. The component in the direction
L is again
|F2|*cos(a2) = 500*cos(10) N
3. Friction force F3 = -k*|N|, where k is the coef\[CapitalThorn]cient 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 \[CapitalThorn]nd the angles b1 and b2 of these forces with the
inclined
plane.
for gravity: angle b1 = 20,
force component |F1|*cos(b1) = |F1|*cos(20)
for the pull: angle = b2 = 90 + 10 =
100
force component |F2|*cos(100) =
-|F2|*cos(80)
|N| = |F1|*cos(20) - |F2|*cos(80)
|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(70) + |F2|*cos(10) -
- k*[|F1|*cos(20) - |F2|*cos(80)] =
= -m*g*cos(70) + |F2|*cos(10) -
- k*[m*g*cos(20) - |F2|*cos(80)]
Substituting
m = 25 kg,
g = 9.81 m/sec^2,
|F2| = 500 N,
k = 0.4
cos(70) ~ 0.3420
cos(10) ~ 0.9848
cos(20) ~ 0.9397
cos(80) ~ 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 \[CapitalThorn]nd the
force
>acting up the plane we use cos rather than sin???
The plane is 20 to the horizontal, which means that is
90-20 = 70
to the vertical. The force of gravity m*g ~
25[kg]*9.8[m/sec^2] = 245
N is vertical, i.e., 70 deg to the plane. The component of
the
gravity force parallel to the plane is 245[N]*cos(70).
The force of 500 N is pulling at 10 to the plane. The
component of
this force parallel to the plane is 500[N]*cos(10).
You use cosine both times. But cos(70) =
sin(90-70) = sin(20).
===
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 \
Ômew is
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 \[CapitalThorn]nd the
force acting
> up the plane we use cos rather than sin???
Draw two pictures.
In the \[CapitalThorn]rst 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
blocks 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
de\[CapitalThorn]nition
of sine and cosine, but if its not, then say so and \
well
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 de\[CapitalThorn]nitions of sine and cosine but \
cant really see
where
youre 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 \
Ômew is
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 \[CapitalThorn]nd the
force
acting
>> up the plane we use cos rather than sin???
>Draw two pictures.
>In the \[CapitalThorn]rst 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
>blocks 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
de\[CapitalThorn]nition
>of sine and cosine, but if its not, then say so and \
well
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 cant work out.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB8D5uK25120;
The \[CapitalThorn]rst number was simply 1.
The next, 11.
Then, 21, 1,211, and 111,221.
whats the next number in the sequance?
===
Subject: Re: Math Problem that I cant work out.
http://dheera.net/sci/sequence_sol.php
> The \[CapitalThorn]rst number was simply 1.
> The next, 11.
> Then, 21, 1,211, and 111,221.
> whats the next number in the sequance?
===
Subject: Re: Math Problem that I cant work out.
> http://dheera.net/sci/sequence_sol.php
> The \[CapitalThorn]rst number was simply 1.
> The next, 11.
> Then, 21, 1,211, and 111,221.
> whats 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? whats 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? whats 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 Pythagoras theorem, 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 \[CapitalThorn]nd the
lengths
of each of the sides.
2. Once you know the lengths of all three sides, you can \[CapitalThorn]nd
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 speci\[CapitalThorn]c case you have a right-angled
triangle, with
the right angle at B. The general method will still work, but
theres
an easier way. Just use the basic properties of the
trigonometric
functions to \[CapitalThorn]nd that Tan(BAC) = BC/AB. Having found
Tan(BAC), use the
inverse tangent function to \[CapitalThorn]nd 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
youll 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? whats 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 dont know where Ive gone wrong and \
Im hoping someone
can help me
>For n >= to 1 show n^3 - n is divisible by 3
>Tis is what Ive 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?
>Id 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), lets 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 \[CapitalThorn]nd the inverse of G(s) = 1 / (s^2 + 1)^2 ? Well,
work it
\[CapitalThorn]rst 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), lets 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 \[CapitalThorn]nd the inverse of G(s) = 1 / (s^2 + 1)^2 ? Well,
work it
>\[CapitalThorn]rst 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 x+ 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 :
x+2x\[Capita\
lOTilde]+ x =8*sin(t) with x0=x(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 x+ 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 :
> \
x+2x\[Capital\
OTilde]+ x =8*sin(t) with x0=x(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 speci\[CapitalThorn]cally
asked how
to \[CapitalThorn]nd 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
Im having trouble working out a proof for a formula. Since
Im not
really sure what the best way to represent it in plain text
would be,
Im going to write it in TeX format. If its
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))
Im very sorry about how awful that looks.
Im 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 Im pretty sure is right, by setting up a determinant \
to
work out
nabla times F and then again to get the \[CapitalThorn]nal result.
I encounter trouble when I try to work with the right side of
the
equation. I compute \[CapitalThorn]rst 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 Im somewhat at a loss as \
to
what
mistake Ive made.
Im sorry if what Ive written is unclear. I \
can provide any
particular
step I took speci\[CapitalThorn]cally, though I cant promise \
that it will
be
terribly easy to understand.
Tracy Poff
===
Subject: Re: Curl and Divergence
>Im having trouble working out a proof for a formula. Since
Im not
>really sure what the best way to represent it in plain text
would be,
>Im going to write it in TeX format. If its
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))
>Im very sorry about how awful that looks.
>Im 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 Im pretty sure is right, by setting up a determinant
to work out
>nabla times F and then again to get the \[CapitalThorn]nal result.
>I encounter trouble when I try to work with the right side
of the
>equation. I compute \[CapitalThorn]rst 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 Im somewhat at a loss as
to what
>mistake Ive made.
>Im sorry if what Ive written is unclear. I \
can provide any
particular
>step I took speci\[CapitalThorn]cally, though I cant \
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
de\[CapitalThorn]ned 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 Im going to try \
assuming
that F is
meant to be a scalar valued function, as opposed to a vector
valued
function which it was de\[CapitalThorn]ned 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
de\[CapitalThorn]ned 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 Im going to try
assuming that F is
>meant to be a scalar valued function, as opposed to a vector
valued
>function which it was de\[CapitalThorn]ned 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 \[CapitalThorn]eld. 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, isnt it).
So your third term represents this scalar times the vector F,
but it
isnt 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 \[CapitalThorn]eld.
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, isnt it).
> So your third term represents this scalar times the vector
F, but it
> isnt 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 veri\[CapitalThorn]ed 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--Ill 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.
>Im having trouble working out a proof for a formula. Since
Im not
>really sure what the best way to represent it in plain text
would be,
>Im going to write it in TeX format. If its
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, de\[CapitalThorn]ned 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 Im somewhat at a loss as
to what
>mistake Ive made.
The method is \[CapitalThorn]ne, except for the grad(grad(F)) thing, which
does
not seem to make much sense to me.
[...]
--
Its not denial. Im just very selective \
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Curl and Divergence
>>Im having trouble working out a proof for a formula. \
Since
Im not
>>really sure what the best way to represent it in plain text
would be,
>>Im going to write it in TeX format. If \
its
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, de\[CapitalThorn]ned for scalar functions as
div(grad(f)).
The speci\[CapitalThorn]c 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 doesnt 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 Ive 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 Im somewhat at a loss as
to what
>>mistake Ive made.
> The method is \[CapitalThorn]ne, 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 de\[CapitalThorn]ned 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.
>Im having trouble working out a proof for a formula. Since
Im not
>really sure what the best way to represent it in plain text
would be,
>Im going to write it in TeX format. If its
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, de\[CapitalThorn]ned for scalar functions as
div(grad(f)).
>The speci\[CapitalThorn]c 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...
(Im 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 \[CapitalThorn]nd 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.
--
Its not denial. Im 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
f(x0)*(x2 - x1) = f(x2) - f(x1) = x2 - x1
Since x2 - x1 != 0, f(x0) = 1, which contradicts the
assumption that
|f(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.
> f(x0)*(x2 - x1) = f(x2) - f(x1) = x2 - x1
> Since x2 - x1 != 0, f(x0) = 1, which contradicts the
assumption that
> |f(x)| < 1 everywhere.
The premise can be weakened to for all x, f(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 cant seem to \[CapitalThorn]gure \
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 cant seem to \[CapitalThorn]gure \
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, P(x) = 0:
P(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
Danas
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 Danas
>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
Heres 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
\[CapitalThorn]nd below to what Im referring. \
Id point out where, but as
you top
post, I do same which is not only a sloppy way to reply, \
its
impolite.
> Heres 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.
Danas answer *is* correct. Its also \
essentially
equivalent to the one that I outlined in my \[CapitalThorn]rst 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.
> Danas answer *is* correct. Its also \
essentially
equivalent to the one
> that I outlined in my \[CapitalThorn]rst 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.
>> Danas answer *is* correct. Its also \
essentially
equivalent to the one
>> that I outlined in my \[CapitalThorn]rst 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
> \[CapitalThorn]nd below to what Im referring. \
Id point out where, but
as you top
> post, I do same which is not only a sloppy way to reply,
its impolite.
>> Heres 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 \
Ôunwrap it:
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 \[CapitalThorn]rst form a geometric series,
a^(n-1)*b + a^(n-2)*b + ... + ab + b,
whose sum you should know how to \[CapitalThorn]nd in closed form. If
you apply this technique to your problem, youll 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 \[CapitalThorn]rst year reduce the \[CapitalThorn]nal 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 \[CapitalThorn]nal
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 \[CapitalThorn]rst year reduce the \[CapitalThorn]nal 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 \[CapitalThorn]nal
>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 \[CapitalThorn]rst
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
Youve used a continuous model to arrive at this \
\[CapitalThorn]gure, but
the problem is stated in terms of a discrete model. The
actual solution to the problem as stated is larger than your
\[CapitalThorn]gure. I wont give it away completely, since \
Id prefer
for the original poster to arrive at it himself, but its 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
> Youve used a continuous model to arrive at this \
\[CapitalThorn]gure, but
> the problem is stated in terms of a discrete model. The
> actual solution to the problem as stated is larger than your
> \[CapitalThorn]gure. I wont give it away completely, since \
Id prefer
> for the original poster to arrive at it himself, but its \
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, Id 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, Id 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
\[CapitalThorn]rst 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.
Id 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 Im not sure where the
dif\[CapitalThorn]culty 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 \[CapitalThorn]rst 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 theres 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 \[CapitalThorn]nd 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
\[CapitalThorn]nd 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;
Heres 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
birds
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;
>Heres 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 birds
>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, P(v) = 0 and P(v) > 0:
P(v) = w^2/(2pSv) + 1/2*pAv^3
P(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 \[CapitalThorn]rst 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.
P(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
\[CapitalThorn]ve 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
>\[CapitalThorn]ve 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
>\[CapitalThorn]ve 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 dont 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
>>\[CapitalThorn]ve 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 dont 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 dont really think you help a student by doing problems in
such detail.
Its 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; well 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 dont forget Mechanics - Friction
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB9N37L18623;
Please dont forget Mechanics - Friction
message from 08 Dec
===
Subject: Re: Please dont forget Mechanics - Friction
posting-account=hXoQ7w0AAADSdZxlCHUb3O5k5oe4oxj9
> Please dont 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.
--
Its not denial. Im 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,
Youre 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-de\[CapitalThorn]ned (if the intersection of A,B was
something other
than the empty set, mod would not be well-de\[CapitalThorn]ned 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 \[CapitalThorn]nancial engineering. Ive 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. Ive
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 thats 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 \[CapitalThorn]nd 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.