mm-217
2inators to get the sum. But ts is not an off the> wall
technique. Although these students don't know what 
they're
doing there> is a scenario when ts technique is quite valid.
For example if today I> answer 12 math questions correctly out
of 14- so my score is 12/14- and> tomorrow when I answer 5
questions correctly out of 6 I can safely say that> I answered
17 out of 20 for a score of 80%. That is, (12 / 14) # ( 5 / 6)
=> 17 / 20 = 80% where # is some new operation. My question is
===
Re: running count not popular (execpt in card counting)-why
not?> Some elementary math students add fractions incorrectly
by simply adding the> numerators and the denominators to get
the sum. But ts is not an off the> wall technique. Although
these students don't know what they're doing 
there> is a
scenario when ts technique is quite valid. For example if
today I> answer 12 math questions correctly out of 14- so my
score is 12/14- and> tomorrow when I answer 5 questions
correctly out of 6 I can safely say that> I answered 17 out of
20 for a score of 80%. That is, (12 / 14) # ( 5 / 6) => 17 / 20
= 80% where # is some new operation. My question is why do you
tnk> that ts operation # is not popular?It *is* popular. Add
the components. Vector addition.It is not an operation on
rational numbers [ (12 / 14) # ( 5 / 6) is not (6 / 7) # ( 5 /
===
popular (execpt in card counting)-why
not?http://www.giganews.com/info/dmca.html>Some elementary
math students add fractions incorrectly by simply adding
the>numerators and the denominators to get the sum. But ts is
not an off the>wall technique. Although these students don't
know what they're doing there>is a scenario when ts 
technique
is quite valid. For example if today I>answer 12 math
questions correctly out of 14- so my score is 12/14-
and>tomorrow when I answer 5 questions correctly out of 6 I
can safely say that>I answered 17 out of 20 for a score of
80%. That is, (12 / 14) # ( 5 / 6) =>17 / 20 = 80% where # is
some new operation. My question is why do you tnk>that ts
operation # is not popular?What do you mean, not popular? 
It's
not used when the sum of two fractions is needed, because it
doesn't give the sum of two fractions.In situations like you
describe, where # is what's needed, # is 
what'sused. (I
sometimes get an evil thrill from performing # but writing
itas + in adding up partial credit on students' papers...)As
has been pointed out, you should note that # is really not
anoperation on _fractions_ per se: 2/4 and 1/2 are the same
===
Re: running count not popular (execpt in card counting)-why
not?> Some elementary math students add fractions incorrectly
by simply adding> the numerators and the denominators to get
the sum. But ts is not an off> the wall technique. Although
these students don't know what they're doing> 
there is a
scenario when ts technique is quite valid. For example if>
today I answer 12 math questions correctly out of 14- so my
score is> 12/14- and> tomorrow when I answer 5 questions
correctly out of 6 I can safely say> that I answered 17 out of
20 for a score of 80%. That is, (12 / 14) # ( 5> / 6) = 17 / 20
= 80% where # is some new operation. My question is why do> you
tnk that ts operation # is not popular?I tnk that operation is
call the harmonic mean. The operation is a weighted average
where the weight for each data point is the number of
available points for that assignment. However, you can't
perform the operation on just the fractions, you need the
weight too. In other words, The result will change if you
change the representations of some of the fractions. For
example, 1/2 # 1/3 = 2/5 and 1/2 # 1/3 = 2/4 # 1/3 = 3/7. So
it's not well-defined as an operation on 
fractions. It's only
defined when you have a weight along with each fraction.Have a
===
been getting some feedback in sci.math for the best shape for a
===
spiders I've been doing some research on some formations on
Mars called Martianspiders:http://www.martianspiders.com/These
seem to form their branches almost invariably in Fibonacci
numbers.According to geologists I have asked there seems to be
no geologicalprocesses that form Fibonacci brancng. Does anyone
know other processesthat might form them and if there is some
way to make a statistical proof asto whether they are
Fibonacci branches or not. For
example:http://www.msss.com/moc_gallery/e13_e18/images/E13/
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id h9SDNDm11470;.....My name is I'm
from Indonesia, a universty student with major computer
science. I have a project that i have to explain and made an
example programabout wigner ville distribution. I don't
understand about the wigner ville distribution (WVD). how
tocalculate the WVD, how to use the formula.I hope you can
help me, give the explanation. The major problem I have is I
can not calculate the WVD,I dont know the algorithm. I hope
you really can help me. In here I don't have books or source
===
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id h9SDNAt11463;The original and
exceedingly more difficult problem is similar to that stated
below; however, one ball is slightly different weight than the
other 11 balls (either heavier or lighter), but we're not 
sure
wch. The solution must a) identify the odd ball, AND, b)
determine if the ball is heavier OR lighter. Again, only 3
comparison weighs using the jeweler's scale.Solving ts 
problem
has stumped many a genius!h><pre>Does anybody know of an
elegant solution to the twelve billiard
ball>>problem?>>(Twelve apparently identical billiard balls,
one of wch is slightly>>heavier than the rest, determine wch
is the heavier one making only three>>weigngs.)> Step 1:> Put
four balls on each side of the balance. If one side is>
heavier, use those balls in the next step. If neither side is>
heavier, use the four balls that are not on the balance.> Step
2:> Put two balls on each side of the balance. Use the>
heavier side in the next step. > Step 3:> Put one ball on each
side of the balance. One will be> heavier.> Alternate Step 2:>
Put one ball on each side of the balance. If one side is>
heavier, you are done. If not, use the left-over balls> in the
===
friend about a problem he saw in a>>newspaper. The problem
involves basic algebra and the question is>>what is the order
of power of exp, parens, +, -, * and /. I say the>>problem
5(9+3)-3*4 is not properly defined since it could be either>I
am assumming 5(9+3) is 5*(9+3).>(5*12) - (3*4)>60-12 =
48>There was a rather large thread on ts subject a few weeks
multiply/divide and add/subtract>being equal. For equal
operations work from left to right. Exponents>and parenthesis
===
Re: Pretty Calculus I Problem by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9SDNHD11475;(retry: since the post sent yesterday seems to
have been lost)>I'm trying to create a function in ts
form.>f(x) = ax^3 + bx^2 + cx + d>These are the
prerequisites:>a, b, c, d are all integers.>If ax^3 + bx^2 +
cx + d = 0, then the factors are also all integers.>In other
words, each of the x-intercepts of the functions consists>out
of integers.>When the derivative 3ax^2 + 2bx + c = 0, the
solution to x is also an>integer.>When I set the second
derivative 6ax + 2b to zero, the solution of x>is also an
integer.>In other words, ts is a pretty Calculus 1 problem.>Is
there a way outside of the trial and error method that I can do
ts?>>Suppose that the three roots of the cubic are s, t and
0.Then the cubic is:x^3 - (s+t) x^2 + s t x = 0and its
derivative is:3 x^2 - 2 (s+t) x + s t = 0The roots of the
derivative are:x = [ (s+t) +- sqrt( (s+t)^2 - 3 s t ) ] / 3So
the problem reduces to finding s and t such that(s^2 - s t +
t^2) is a square,or finding a triangle with a 60 degree angle
and integer sides.Such solutions can be found by squaring
Eisenstein integers.(2 + 3 w)^2 = (4 + 12 w + 9 w^2) = (-8 - 3
w^2)since w^3 = 1, and 1 + w + w^2 = 0.Using 3 and 8 for s and
t produces 7^2, and 4/3 or 6 for x.Triple to get integers:
s=9, t=24, x1=4, x2=18, inßection at 11.x^3 - 33 x^2 + 216 x =
0Another example:(5 + w)^2 = (25 + 10 w + w^2) = (13 - 2 w - 11
w^2)(x + 13)(x - 2)(x - 11) = x^3 - 147 x + 286wch is Robert
===
output by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id h9SDOXl11645;I got a
problem.Let u(t) be a function on time, and theta be a fixed
threshold.u(t) starts from zero and increases to some value
with time t.t is considered in a small window. A threshold
function f is defined such that f(u(t))=t when u(t) exceeds
theta, otherwise 0.I have to get a linear/non linear function
wch can approximate the output of the above function.Could
anybody please suggest me some books or anytng relevant to get
===
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9SDNwc11572;><pre>>-Matthew, making full disclosure: 0 _is_ a
natural number, and it is also>a positive real number,>>I
presume ts is a joke. :-) Otherwise, you will need to be
prepared to say>>that 0 is *also* a negative real number! >I
believe the Bourbaki way is to say that 0 is a positive
number, but>it is not *strictly* positive. Likewise, 0 is a
negative number, but>it is not *strictly* negative.>--
===
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9SK0Iu07061;I'm looking for information on what 
Riemann's
zeta function says about the distribution of the primes
(assuming the conjecture is true about the zero's all having
real part 1/2). Specifically, could we start with the
implications for the primes, prove them, and then work
backwards to prove Riemann's hypothesis? I'm 
starting on my
master's in math, but some of the math involved in studying
the zeta function seems a little opaque to me still. Anyway,
if you could steer me in the right direction, I'd appreciate
===
information on what Riemann's zeta function says about the >
distribution of the primes (assuming the conjecture is true
about the zero's > all having real part 1/2). 
Specifically,
could we start with the > implications for the primes, prove
them, and then work backwards to prove > Riemann's 
hypothesis?
In theory, perhaps. In practice, not very likely. In general
the way we prove tngs about primes is by first proving tngs
about zeta. E.g., the way it was proved that the number of
primes up to x is asymptotic to x / log x was by proving that
zeta has no zeros on the line real part 1. To put it another
way, if we could prove stuff about primes without using zeta,
no one would be interested in proving the Riemann
===
another way, if we could prove stuff about primes without
>using zeta, no one would be interested in proving the Riemann
>hypothesis.I would, since there is an equivalent problem
involving cosine transforms, inwch I have more than passing
===
for information on what Riemann's zeta function says about 
the
>> distribution of the primes (assuming the conjecture is true
about the>zero's >> all having real part 1/2). 
Specifically,
could we start with the >> implications for the primes, prove
them, and then work backwards to prove >> Riemann's
hypothesis? >In theory, perhaps. In practice, not very likely.
In general the way >we prove tngs about primes is by first
proving tngs about zeta. >E.g., the way it was proved that the
number of primes up to x is >asymptotic to x / log x was by
proving that zeta has no zeros >on the line real part 1. >To
put it another way, if we could prove stuff about primes
without >using zeta, no one would be interested in proving the
===
Convergence on a space with no topology by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
===
$Revision: 1.9 primary) id h9SJxME06936;>Subject: Was:
Convergence on a space with no topologyConvergence on a space
with no topology got cluttered relativelyfast. In my opinion,
it was getting hard for anyone reading themessage for the 
first
time to follow. Therefore, I try to give themain points of what
was said some sort of chronological order.>With all that
feedback, you need present revision.>Skip the blah blah and
just get to the punch line.>Are you reinventing the Hausdorff
metric for sets?>Do you know what the Hausdorff metric for
sets is?Let (X, d) be a space whose metric is induced by a
norm.Call (A_n) subset B* subset P(x), the set of all subsets
of X,a *Cauchy sequence if (R = reals, N = natural numbers)For
all p > 0, p in R, exists q in N for all n,m > q for alla in
A_n exists b in A_m: d(a,b) < p.>Toss B* out, it isn't being
used nor is it defined.If (r_n) subset R is a Cauchy-sequence
and and r_n !=0 for all n,then (B(x ,r_n)) is a
*Cauchy-sequence for all x in X, whereB is the open ball of
radius r_n at x (we could presumablyjust as easily take closed
balls of radius r_n at x)>That's r_n > 0 for all n > 0 for 
the
following interesting example:>Let the metric space be {0,1}
subset Reals, x = 0 and>r_n = 1 + 1/n for even n, and r_n = 1
- 1/n for odd n.>Let p = 1/2. Pick any q in N. 2q, 2q+1 > q>1
in B(0, 1 + 1/2q) and B(0, 1 - 1/(2q+1)) = {0}.>So no b in
B(0, 1 - 1/(2q+1)) with d(1,b) < 1/2.>Proposition is false.But
why?I repeat from above,Let (X, d) be a space whose metric is
induced by a norm.Call (A_n) subset B* subset P(x), the set of
all subsets of X,a *Cauchy sequence if (R = reals, N = natural
numbers)...If (X,d) is a space whose metric is induced by a
norm,then X must be a vector space- since a norm is only
definedon a vector space (nevertheless, sorry for not
explicitly stating ts).But your counterexample space {0,1} is
not a vector space.Indeed, if you look at my proof of the
proposition, youwill see that it relies on (t_n a) also being
in the space wheret_n > 0 is a scalar... ,Ok? My explicit
definition>-- Define C-seq as> for all eps > 0, 
some n in N
with>for all j,k > n, a in A_j, b in A_k, d(a,b) <
eqs>Theorem>0 < r_n -> 0 implies B(x,r_n) is
===
C-sequence.Subject: Re: Was: Convergence on a space with no
sets?Do you know what the Hausdorff metric for sets is?How
about studying up on the Hausdorff metric? It's a metric for
P(S),based upon the metric for S. Much like what your
attempting. So getsmart, instead of just stumbling around, 
find
out what others have alreadycomplished decades ago that have
bearing on your attempts.>>Let (X, d) be a space whose metric
is induced by a norm.>>Call (A_n) subset B* subset P(x), the
set of all subsets of X,>>a *Cauchy sequence if (R = reals, N
= natural numbers)>>For all p > 0, p in R, exists q in N for
all n,m > q for all>>a in A_n exists b in A_m: d(a,b) < p.Toss
B* out, it isn't being used nor is it 
defined.>>If (r_n) subset
R is a Cauchy-sequence and and r_n !=0 for all n,>>then (B(x
,r_n)) is a *Cauchy-sequence for all x in X, where>>B is the
open ball of radius r_n at x (we could presumably>>just as
easily take closed balls of radius r_n at x)That's r_n > 0 
for
all n > 0> for the following interesting example:Let the metric
space be {0,1} subset Reals, x = 0 andr_n = 1 + 1/n for even n,
and r_n = 1 - 1/n for odd n.Let p = 1/2. Pick any q in N. 2q,
2q+1 > q1 in B(0, 1 + 1/2q) and B(0, 1 - 1/(2q+1)) = {0}.So no
b in B(0, 1 - 1/(2q+1)) with d(1,b) < 1/2.Proposition is
false.> But why?'Cause I've given a counter 
example.> I repeat
from above,> Let (X, d) be a space whose metric is induced by a
norm.> Call (A_n) subset B* subset P(x), the set of all subsets
of X,> a *Cauchy sequence if (R = reals, N = natural
numbers)...Explain what that useless B* is about.> If (X,d) is
a space whose metric is induced by a norm,> then X must be a
vector space- since a norm is only defined> on a vector space
(nevertheless, sorry for not explicitly stating ts).Baloney,
you said metric space and a norm can be defined for a
groupmaking it a topological group or conversely certain
metrics for atopological group can induce or be given norms.
So if you want vectorspace, say you want a vector space, a
normed vector space or even an innerproduct space.> But your
counterexample space {0,1} is not a vector space.> Indeed, if
you look at my proof of the proposition, you> will see that it
relies on (t_n a) also being in the space where> t_n > 0 is a
scalar...It's not? It's a one dimensional 
vector space over
Z_2, that is withscalars integers modulus 2. Yes, Z_2 is a
field, a finite field. As ithas the 
discrete topology, it can be
given a discrete metric such as d(0,1) = d(1,0) = 1, d(0,0) =
d(1,1) = 0.For a norm, |0| = 0, |1| = 1 will do.> Ok? My
explicit definitionNok, back to the drawing board with
it.Perhaps you should study up. See what else has been
done.Would you like a definition of the Hausdorff
metric?William's Metatheorem: Whatever math I dream up is
already old hat.-- Define C-seq as> for all eps > 0, some n in
N withfor all j,k > n, a in A_j, b in A_k, d(a,b) <
eqsTheorem0 < r_n -> 0 implies B(x,r_n) is
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id h9SK0mH07164;I dunno if it's just
me, but a lot of the proofs in (undergraduate real) analysis
(presented in the book and/or homework probems) I've seen
always seems to involve some clever trick that I would never
have thought of. A lot of the tricks seem pretty unmotivated,
too, but maybe that's just the books I'm 
using. Is ts how tngs
are gonna be, or are there general guidelines (i.e. good/bad
tngs to try/not try) when tackling these problems? Maybe I'm
===
problemshttp://www.giganews.com/info/dmca.html>I dunno if 
it's
just me, but a lot of the proofs in (undergraduate real)
analysis (presented in the book and/or homework probems) 
I've
seen always seems to involve some clever trick that I would
never have thought of. A lot of the tricks seem pretty
unmotivated, too, but maybe that's just the books 
I'm using.
Is ts how tngs are gonna be, or are there general guidelines
(i.e. good/bad tngs to try/not try) when tackling these
problems? Maybe I'm just being an idiot...You should 
find the
Enter key on your keyboard, by the way...Anyway: Yes, a lot of
those proofs can seem like tricks you never woulda thought of.
You're studying a subject that's been 
developedover the course
of centuries - every once in a wle someone smart_tnks_ of
sometng that nobody ever woulda thought of.Some of the tricks
are just tricks. But a lot of them are tricks youwill see
again and again - the second time you see a given trickused it
will make more sense, because you'll say aha, someonemight 
tnk
of ts because he recalls the same trick being usedin that
other problem. After you've seen a trick used three orfour
times it becomes a standard technique, not a trick, andyou tnk
===
Re: Is Nature a Logical System? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9SK2si07529;>Is Nature a Logical System?A Laconic answer is
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id h9T31qx04985;>> How can you tell if
a set of simultaneous equations has solutions or not?>> e.g.
12x - 4y = 8, -9x + 3y = 6>> or>> x^3 - y^2 = 0, x = y>> As
far as I can tell, you can work out a value for x and a value
for y >> every>> time...? So what defines whether there are
solutions or not?>I may be missing sometng in the question,
but if the equations are >linear, then, generally, a pair of
simultaneous equations a1x + b1y = c1, >a2x + b2y = c2 have a
solution if the determinant>|a1 b1|>|a2 b2|>evaluates to
non-zero. For larger numbers of variables the determinant
>gets more complicated (bigger).>In your example 12*3 -
(-4)*(-9) = 0, therefore no solution.>-- >Using M2, Opera's
revolutionary e-mail client: <a
href=http://www.opera.com/m2/>http://www.opera.com/m2/</a>
That's not QUITE correct. A linear system of equation has a
UNIQUE SOLUTION if and only if the determinant of the
coefficients is non-zero. If it is 0 it may have no solution 
or
it may have an infinite number of solutions. The example 12x -
4y = 8, -9x + 3y = 6 has no solution but 12x - 4y = 8, -9x +
===
Simultaneous Equation without Solutions> How can you tell if a
set of simultaneous equations has solutions or > not? e.g. 12x
- 4y = 8, -9x + 3y = 6 or x^3 - y^2 = 0, x = y As far as I can
tell, you can work out a value for x and a value for y> every>
time...? So what defines whether there are solutions or not?>>
I may be missing sometng in the question, but if the equations
are>> linear, then, generally, a pair of simultaneous equations
a1x + b1y = >> c1,>> a2x + b2y = c2 have a solution if the
determinant>> |a1 b1|>> |a2 b2|>> evaluates to non-zero. For
larger numbers of variables the determinant>> gets more
complicated (bigger).>> In your example 12*3 - (-4)*(-9) = 0,
therefore no solution.>> -->> Using M2, Opera's 
revolutionary
e-mail client: <a >>
href=http://www.opera.com/m2/>http://www.opera.com/m2/</a>>
That's not QUITE correct. A linear system of equation has a
UNIQUE > SOLUTION if and only if the determinant of the
coefficients is > non-zero. If it is 0 it may have no solution
or it may have an infinite > number of solutions.> The example
12x - 4y = 8, -9x + 3y = 6 has no solution but> 12x - 4y = 8,
-9x + 3y =-6 has an infinite number of > 
solutions.You're quite
right, Yogi.That's what comes of composing replies after
midnight!-- Using M2, Opera's revolutionary e-mail client:
===
C^2=A^2+B^2 by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id h9T31wF05023;>>Hey
there..ive been doing a lot of mechanics, and i tnk ive
forgotten, but>>why are perpendiculars always the shortest
distance? if i need to elobrate>>on ts..please say>>btw i am
dealing with triangles ...have searched on the interent, but
have>>not reached a satisfactory answer... for your help>>>
The shortest distance from a point to a line is along the
>perpendicular from the point to the line.> The reason is
that, by the Pythagorean theorem >c^2= a^2+ b^2, the
hypotenuse is always the longest side of a right >triangle.
Given any non-perpendicular line, it is the hypotenuse of> the
right triangle formed by the perpendicular and so is
longer.>^2YOUR INFORMATION HELPED, , BUT ONLY A LITTLE. I NEED
A LITTLE MORE INFORMATION ON THE PYTHAGOREAN
===
there..ive been doing a lot of mechanics, and i tnk ive
forgotten, butwhy are perpendiculars always the shortest
distance? if i need to elobrateon ts..please saybtw i am
dealing with triangles ...have searched on the interent, but
havenot reached a satisfactory answer... for your help> The
shortest distance from a point to a line is along the
perpendicular from the point to the line.> The reason is that,
by the Pythagorean theorem c^2= a^2+ b^2, the hypotenuse is
always the longest side of a right triangle. Given any
non-perpendicular line, it is the hypotenuse of>the right
triangle formed by the perpendicular and so is longer.^2> YOUR
INFORMATION HELPED, , BUT ONLY A LITTLE. I NEED A LITTLE MORE >
INFORMATION ON THE PYTHAGOREAN THEOREM.Google shows more thatn
75000 ts on Pythagorean Theorem, of wch one is
===
Re: Pythagorean theorem. C^2=A^2+B^2RyahenhHey there..ive been
doing a lot of mechanics, and i tnk ive forgotten,butwhy are
perpendiculars always the shortest distance? if i need
toelobrateon ts..please saybtw i am dealing with triangles
...have searched on the interent, buthavenot reached a
satisfactory answer... for your help> The shortest distance
from a point to a line is along theperpendicular from the
point to the line.> The reason is that, by the Pythagorean
theoremc^2= a^2+ b^2, the hypotenuse is always the longest
side of a righttriangle. Given any non-perpendicular line, it
is the hypotenuse of>the right triangle formed by the
perpendicular and so is longer.^2> YOUR INFORMATION HELPED, ,
BUT ONLY A LITTLE. I NEED A LITTLE MOREINFORMATION ON THE
PYTHAGOREAN
THEOREM.http://www.cut-the-knot.org/geometry.shtmland click
scroll down until you see Pythagoras. Forty-odd proofs,
===
C^2=A^2+B^2simplest constructible proof:inscribe a rigth
trigon into a circle (on a diameter;you have to construct the
diameter, firstly), andconstruct semicircles on the two
legs;use the simple decomposition to show the similarities.ts
is known as the lunes proof.(NB: the Eisntein proof is
simpler, butit's merely diagramatic, and amounts to the same
numberof compass constructions. more to the point,he'd never
have made that proof, unless he haddone the constructions,
ipso-facto .-)so, I also call it, the Ienstien Proof. >> The
reason is that, by the Pythagorean theorem>>c^2= a^2+ b^2, the
hypotenuse is always the longest side of a right>>triangle.
Given any non-perpendicular line, it is the hypotenuse of>>
the right triangle formed by the perpendicular and so is
longer.>>^2 > http://www.cut-the-knot.org/geometry.shtml> and
click scroll down until you see Pythagoras. Forty-odd proofs,
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id h9T324405057;What does the graph of
sinx^2 (x squared) look like? I was also wondering what the
===
squared> What does the graph of sinx^2 (x squared) look like?
I was also wondering what the graph of cosx^2 looked like.I
answered a different question. I thought you were askingabout
(sin x)^2 and (cos x)^2.Virgil answered the correct question.
There is a symmetrybecause the functions are even. The
oscillations will getcloser and closer together as you get
farther from theorigin. Since sin(y) is 0 at y = 0, pi, 2*pi,
3*pi, ... n*pi,...Then sin(x^2) will be 0 at x^2 = 0, pi,
2*pi, ... n*pi...That is, at x = 0, pi, pi*sqrt(2),
pi*sqrt(3),..., pi*sqrt(n).That should help you plot it.
===
sinx^2 (x squared) look like? I was also wondering what the
graph of cosx^2 looked like.Since cos(x) and sin(x) are
horizontally-sfted versions ofthe same curve, so are their
squares.There are a couple identities that are very useful
incalculus, but also useful for answering your question:
cos^2(x) = (1 + cos(2x))/2 sin^2(x) = (1 - cos(2x))/2cos(2x)
varies between -1 and +1, with a period of... well,you should
be able to figure that out.(1 + cos(2x)) sfts it upward by 1,
so it oscillatesbetween 0 and 2.(1 + cos(2x)) scales it by a
factor of 2, so it oscillatesbetween 0 and 1.(1 - cos(2x))
also oscillates between 0 (when the cosine is +1)and 2 (when
the cosine is -1), but at different places than(1 +
cos(2x)).So they both look like sine waves, scaled and sfted
===
x squared> What does the graph of sinx^2 (x squared) look like?
I was also wondering > what the graph of cosx^2 looked
like.Your notation would be clearer if you used more
parentheses:sin(x^2) versus (sin(x))^2, for example.Both
sin(x^2) and cos(x^2) are even functions so the graphs are
symmetric about the y-axis.As x starts from 0 and increases ,
sin(x^2) first looks a bit like sin(x), oscillating between y 
=
-1 and y = 1, but with the oscillations getting closer together
as x increases, instead of being the same distance across.
Similarly for cos(x^2) being a bit like cos(x) for small x,
===
logistic function with 2 parameters experts,since i don't 
have
a very good background in maths(have leftghschool for ages), it
would be thankful if any maths/stat expertwould help me out.i
need to implement ts function in my software for
curvingfitting/tracing purpose:Y= Ymin + (Ymax - Ymin) / (1 +
exp(-(X - A) / B))i've got a set of sample points:X Y-----
-----0 0.25 (=Ymin)1 0.3752 0.53 0.6254 0.755 0.8756 17 1.6258
2.259 3.62510 511 6.37512 7.7513 8.37514 915 9.12516 9.2517
9.37518 9.519 9.62520 9.75 (=Ymax)the curve looks like ts: * *
* * * * **my question is here: what's the algorithm of 
working
out A and B from thegivenpoints, so that i can trace the
===
function with 2 parameters> experts,> since i don't have a
very good background in maths(have left> ghschool for ages),
it would be thankful if any maths/stat expert> would help me
out.> i need to implement ts function in my software for
curving> fitting/tracing purpose:> Y= Ymin + (Ymax - Ymin) / 
(1
+ exp(-(X - A) / B))Here's a simple method: Rewrite the 
above
to the following equivalentequation:log[(Ymax-Y)/(Y-Ymin)] =
alpha + beta * X, where alpha = A/B and beta= -1/B.The left
hand side contains known values, and the right hand side is
linearin X. So here you can use standard linear regression to
find the two unknownPlease don't panic over the 
fact that the
left hand side is undefined forX=Xmin and X=Xmax, just skip
over those points.Ts was the quick-and-dirty method. You may
refine your parameters bytweaking slightly on Ymin and Ymax.
Presumably, Ymin should be slightlysmaller and Ymax should be
slightly larger. A simple method to obtain Yminand Ymax is to
use your newly calculated values for A and B and insert
theminto your original equation, regarding them now as fixed
constants. Insteadtreat now Ymin and Ymax as unknown
parameters and use linear regression todetermine them. These
values should be more accurate, but don't forget touse 
common
sense!You can then repeat the entire procedure, find better
values for A and B,then find better values for Ymin and Ymax,
then find even better values forA and B, etc.Remember to plug
in all the parameters into your equation and plot theentire
curve and compare with your parameters. Common
===
michael, that's clear enough!yan.> experts,>since i 
don't have
a very good background in maths(have left>ghschool for ages),
it would be thankful if any maths/stat expert>would help me
out.>i need to implement ts function in my software for
curving>fitting/tracing purpose:>Y= Ymin + (Ymax - Ymin) / (1 
+
exp(-(X - A) / B))> Here's a simple method: Rewrite the 
above
to the following equivalent> equation:> log[(Ymax-Y)/(Y-Ymin)]
= alpha + beta * X, where alpha = A/B and beta> = -1/B.> The
left hand side contains known values, and the right hand side
islinear> in X. So here you can use standard linear regression
to find the twounknown> Please don't panic over 
the fact that
the left hand side is undefined for> X=Xmin and X=Xmax, just
skip over those points.> Ts was the quick-and-dirty method.
You may refine your parameters by> tweaking slightly on Ymin
and Ymax. Presumably, Ymin should be slightly> smaller and
Ymax should be slightly larger. A simple method to obtain
Ymin> and Ymax is to use your newly calculated values for A
and B and insertthem> into your original equation, regarding
them now as fixed constants.Instead> treat now Ymin and Ymax 
as
unknown parameters and use linear regression to> determine
them. These values should be more accurate, but don't forget
to> use common sense!> You can then repeat the entire
procedure, find better values for A and B,> then 
find better
values for Ymin and Ymax, then find even better valuesfor> A
and B, etc.> Remember to plug in all the parameters into your
equation and plot the> entire curve and compare with your
previous post is whether a batter such as>B. Bonds can t a
homerun from simply just tossing the ball in the>air from
homeplate and then seeing whether a homerun can be t.
Those>who have played baseball know ts test in that they do
fielding>practice in many instances where a batter with bat in
one hand and>with ball in another tosses the ball into the air
and tries to t it.>One can say the ball has 0 speed from the
pitcher's mound direction.>And I really do not know what the
slowest speed attainable from the>pitcher in major league
sports to carry the ball over the homeplate???>pitcher pitches
a 90 mph ball.>Ts test that I speak of is designed to see
whether a genuine slugger>of baseball can t a homerun from a 0
speed ball from homeplate.>I heard from the announcers that the
field park in Florida for the>Marlins is vastly deeper than 
the
park in New York. But I am wondering>of a Barry Bonds test of
a 0 speed ball and whether Bonds can t that>ball a homerun??
If he can t a 0 speed ball from Yankee stadium as a>homerun
then my opinion is that baseball stadiums are
grossly>underdistanced and that they need to be altered such
that no tter in>baseball is able to t a homerun from a 0 speed
ball.>In fact, my opinion is that the slowest speed pitch
possible, what is>it??? Is it 10 mph that a pitcher in the
major leagues can get across>the plate?? My opinion is that
all pitches of ts slowest speed>should not be able to be t as
a homerun in any major league baseball>park and that these
parks should be re landscaped such that their>distances make
it impossible for any slugger to t a 10 mph pitch as>a
homerun.>Now I understand that aluminum bats are not allowed
in major leagues.>I feel they should be. And do the above
testing with aluminum bats. I>would love to hear the sound of
a pitch t from a aluminum bat. But>if these bats can cause
death and injury to players then keep the ban>on aluminum
bats.>If a Barry Bonds is unable to t a 0 speed ball as a
homerun then>that indicates to me that the Optimal Strategy
Pitcng in Baseball is>not in the direction of fast pitcng but
in the direction of slow>pitcng.>I need a test of the average
slow speed of a baseball that gets it>across the plate as a
strike. I am simply guessing of 10 mph but need>an accurate
number.>And the reason that no pitchers in modern baseball do
a slow pitch is>because of social antagonism that they demand
and expect all pitchers>to deliver fast pitches. But if the
rules in Baseball allow slow>pitches then that will be the
future trend.>And the distance of all baseball parks outfield
should be a>physics-science determination. Obviously a Barry
Bonds is able to t>homeruns from 90 mph pitches in virtually
every baseball park. But is>he able to t homeruns in every
park from a 30 mph pitch?? So the>distance of every major
league baseball park should be determined by>the speed of
pitch and those parks that are inadequate should
be>relandscaped and designed better. So how can you justify to
any>baseball fan that a Barry Bonds can t a homerun in their
stadium>from a simple hand tossed 0 pitch speed?? You cannot
justify that and>that the field distance needs to be 
fixed.
Almost as if a Grandmother>can go out to that field and t a
homerun.>So that in the future, when a Barry Bonds comes to
the plate and>scared of m tting a homerun, then a stream of
slow pitches is>rendered instead of a forced walk to the base.
In fact, if a pitcher>in the future perfects a slow pitch with
a spin on the ball that>almost every batter will have an
impossible time of tting a homerun.>But whether the scores in
baseball will become larger because so many>singles or doubles
are t remains a question for slow pitcng.>The trouble with
baseball at the moment is that science and physics>have never
rally given the sport any inputs and that the rules and>agenda
of baseball as a sport has been governed more by the
tillation>effect on fans. Most every fan of baseball is
unaware at the moment of>these writings that the pitcher is
the dominate feature of baseball>where most fans immediately
tnk of the batting slugger. And what if>a macne, a pitcng
macne that is anchored on the pitcng mound>designed to deliver
a fast ball at a specified speed in every pitch.>The fans 
would
balk and baseball lose much of its appeal. And in
that>circumstance we can truly say that the slugger in
baseball is of all>importance>and not the pitcher.>So I have
raised many important questions above. (1) what is the>slowest
speed to>deliver in major league pitches (2) can the present
day sluggers t a>homerun in the stadiums from a 0 speed toss
up (3) are most stadiums>inadequate in distance of homerun (4)
link and connect speed of pitch>to outfield distance 
(5)research
the advantages of slow speed pitcng>in baseball>Arcmedes
Plutonium>whole entire Universe is just one big atom where
dots>of the electron-dot-cloud are galaxiesActually, my gut
tells me that it would be easier to t a home runfrom just
throwing the ball up into the air and tting it
because,everytng else being equal, it would simply require
less impulse toget it moving at the same speed. Given roughly
the same amount oftime of contact between bat and ball, ts
would require less force. Remember that impulse is the change
in momentum. First of all, thebatter needs to overcome the
momentum of the ball in s owndirection, reverse it, and
provide sufficient momentum to get it overthe wall. By tossing
it up, he's essentially starting the ball fromrest and 
won't
need to overcome the initial momentum of the ball. Soit
requires less
alluded to in my previous post is whether a batter such asB.
Bonds can t a homerun from simply just tossing the ball in
theair from homeplate and then seeing whether a homerun can be
t. Thosewho have played baseball know ts test in that they do
fieldingpractice in many instances where a batter with bat in
one hand andwith ball in another tosses the ball into the air
and tries to t it.One can say the ball has 0 speed from the
pitcher's mound direction.And I really do not know what the
slowest speed attainable from thepitcher in major league
sports to carry the ball over the homeplate???pitcher pitches
a 90 mph ball.Ts test that I speak of is designed to see
whether a genuine sluggerof baseball can t a homerun from a 0
speed ball from homeplate.I heard from the announcers that the
field park in Florida for theMarlins is vastly deeper than the
park in New York. But I am wonderingof a Barry Bonds test of a
0 speed ball and whether Bonds can t thatball a homerun?? If he
can t a 0 speed ball from Yankee stadium as ahomerun then my
opinion is that baseball stadiums are grosslyunderdistanced
and that they need to be altered such that no tter inbaseball
is able to t a homerun from a 0 speed ball.In fact, my opinion
is that the slowest speed pitch possible, what isit??? Is it 10
mph that a pitcher in the major leagues can get acrossthe
plate?? My opinion is that all pitches of ts slowest
speedshould not be able to be t as a homerun in any major
league baseballpark and that these parks should be re
landscaped such that theirdistances make it impossible for any
slugger to t a 10 mph pitch asa homerun.Now I understand that
aluminum bats are not allowed in major leagues.I feel they
should be. And do the above testing with aluminum bats. Iwould
love to hear the sound of a pitch t from a aluminum bat. Butif
these bats can cause death and injury to players then keep the
banon aluminum bats.If a Barry Bonds is unable to t a 0 speed
ball as a homerun thenthat indicates to me that the Optimal
Strategy Pitcng in Baseball isnot in the direction of fast
pitcng but in the direction of slowpitcng.I need a test of the
average slow speed of a baseball that gets itacross the plate
as a strike. I am simply guessing of 10 mph but needan
accurate number.And the reason that no pitchers in modern
baseball do a slow pitch isbecause of social antagonism that
they demand and expect all pitchersto deliver fast pitches.
But if the rules in Baseball allow slowpitches then that will
be the future trend.And the distance of all baseball parks
outfield should be aphysics-science determination. Obviously a
Barry Bonds is able to thomeruns from 90 mph pitches in
virtually every baseball park. But ishe able to t homeruns in
every park from a 30 mph pitch?? So thedistance of every major
league baseball park should be determined bythe speed of pitch
and those parks that are inadequate should berelandscaped and
designed better. So how can you justify to anybaseball fan
that a Barry Bonds can t a homerun in their stadiumfrom a
simple hand tossed 0 pitch speed?? You cannot justify that
andthat the field distance needs to be fixed. 
Almost as if a
Grandmothercan go out to that field and t a homerun.So that in
the future, when a Barry Bonds comes to the plate andscared of
m tting a homerun, then a stream of slow pitches isrendered
instead of a forced walk to the base. In fact, if a pitcherin
the future perfects a slow pitch with a spin on the ball
thatalmost every batter will have an impossible time of tting
a homerun.But whether the scores in baseball will become
larger because so manysingles or doubles are t remains a
question for slow pitcng.The trouble with baseball at the
moment is that science and physicshave never rally given the
sport any inputs and that the rules andagenda of baseball as a
sport has been governed more by the tillationeffect on fans.
Most every fan of baseball is unaware at the moment ofthese
writings that the pitcher is the dominate feature of
baseballwhere most fans immediately tnk of the batting
slugger. And what ifa macne, a pitcng macne that is anchored
on the pitcng mounddesigned to deliver a fast ball at a
specified speed in every pitch.The fans would balk and 
baseball
lose much of its appeal. And in thatcircumstance we can truly
say that the slugger in baseball is of allimportanceand not
the pitcher.So I have raised many important questions above.
(1) what is theslowest speed todeliver in major league pitches
(2) can the present day sluggers t ahomerun in the stadiums
from a 0 speed toss up (3) are most stadiumsinadequate in
distance of homerun (4) link and connect speed of pitchto
outfield distance (5)research the advantages of slow speed
pitcngin baseballArcmedes Plutoniumwhole entire Universe is
just one big atom where dotsof the electron-dot-cloud are
galaxies> Actually, my gut tells me that it would be easier to
t a home run> from just throwing the ball up into the air and
tting it because,> everytng else being equal, it would simply
require less impulse to> get it moving at the same speed.
Given roughly the same amount of> time of contact between bat
and ball, ts would require less force.> Remember that impulse
is the change in momentum. First of all, the> batter needs to
overcome the momentum of the ball in s own> direction, reverse
it, and provide sufficient momentum to get it over> the wall. 
By
tossing it up, he's essentially starting the ball from> rest
and won't need to overcome the initial momentum of the ball.
So> it requires less force.Nope.. The energy imparted by the
pitch will be transferred to thebat and redirected back into
imparted inertial force to the ball..A bat swung with all else
being equal will drive the ball furtherif is traveling at say
100mph and a shorter distance if it weretraveling at say
80mph.Starting from a standing stop you only have the imparted
force of theenergy of the bat ( converted kinetic form derived
from the transfer ofthe applied energy of the swinger) where
as in a pitched ball you havethe same applied force plus the
redirected vector force of the pitchersimparted
energy.-----Some where witn the Quantum
StateHttp://Paul.Mays.Com/story.htmlhttp://paul.mays.com/
tells me that it would be easier to t a home run>from just
throwing the ball up into the air and tting it
because,>everytng else being equal, it would simply require
less impulse to>get it moving at the same speed. Given roughly
the same amount of>time of contact between bat and ball, ts
would require less force.>Remember that impulse is the change
in momentum. First of all, the>batter needs to overcome the
momentum of the ball in s own>direction, reverse it, and
provide sufficient momentum to get it over>the wall. By 
tossing
it up, he's essentially starting the ball from>rest and 
won't
need to overcome the initial momentum of the ball. So>it
requires less force.> Nope.. The energy imparted by the pitch
will be transferred to the> bat and redirected back into
imparted inertial force to the ball..> A bat swung with all
else being equal will drive the ball further> if is traveling
at say 100mph and a shorter distance if it were> traveling at
say 80mph.> Starting from a standing stop you only have the
imparted force of the> energy of the bat ( converted kinetic
form derived from the transfer of> the applied energy of the
swinger) where as in a pitched ball you have> the same applied
force plus the redirected vector force of the pitchers>
imparted energy.> Indeed, physics tells us that the 100mph
pitch momentum is added onto the t ball for extra long
distance.What we need now for baseball is for the famous
sluggers such as Bonds,Sosa, et al, to run the test. Can any
one of them t a homerun in anymajor league baseball stadium
from a 0 speed self pitched ball from thehomeplate? If Barry
Bonds is able to t such a self pitched 0 speedball in any
stadium of baseball, then that stadium is utterly inadequateas
far as distance is concerned and needs to be re-designed for it
is acheating baseball field.And also, the best pitches in
baseball, because the physics says so orproves it, are slow
pitches that have a wicked spin on the ball.Physics says that
our modern day baseball games are being played to
adumb-strategy and not the optimal strategy. In that a pitcher
with fastballsmaybe able to get a few strikeouts but also is
increasing the chances verygreatly of pitcng homeruns.So that
if a Barry Bonds or Sammy Sosa (excuse the spelling) were to
cometo the plate, then the best pitches to them would be the
slowest possiblepitch with a wicked spin on the ball.In
practice season ts year for all major league baseball teams,
should try out what I am saying above and see if slow pitcng
can retire theirbatters faster and better than fastball
pitcng.Arcmedes Plutoniumwhole entire Universe is just one big
atom where dotsof the electron-dot-cloud are galaxies
===
electron-dot-cloud are galaxiesSubject: Equalizing Golf to
Baseball Re: B. Bonds test on distance of playing field Re:
for baseball is for the famous sluggers such as Bonds,> Sosa,
et al, to run the test. Can any one of them t a homerun in
any> major league baseball stadium from a 0 speed self pitched
ball from the> homeplate? If Barry Bonds is able to t such a
self pitched 0 speed> ball in any stadium of baseball, then
that stadium is utterly inadequate> as far as distance is
concerned and needs to be re-designed for it is a> cheating
baseball field.Now suppose a Bonds or Sosa and all the other
sluggers of baseball are reluctantto do the test above.
Suppose they refuse.Well, modern physics is so very good that
we can compute to some accuracy asto how Bonds or Sosa would
fare in such a test, even if they refuse to actuallydo
it.Because Golf is similar to Baseball in that the ball tee-ed
off is a ball at zerospeed and would simulate a self pitched
baseball wch is nearly zero speed.So, all I would have to do
is compute a factor of a golf club compared to abaseball bat.
Compute another factor of the aerodynamics of a golf
ballcompared to a major league baseball. Then one more factor
of the musclesused in baseball compared to the muscles used in
golf.Plugging in all those factors in an equation.Finding out
what the average long drive in golf is and using that number
inthe equation. When all is said and done I should be able to
give a fairly accurate number forthe longest drive that a
Barry Bonds or Sammy Sosa can musterfor a zero pitched
baseball.Easier yet, if Bonds or Sosa or some other notorious
slugger plays golf and hasrecorded s longest golf fairway
drive. I can use that number into the above equation andhave
an accurate distance for s maximum t into outfield of azero
pitch ball.Is it not wonderful how modern science can answer
tngs and then have the people go out andfill in and 
confirm the
details. Yes, truly wonderful.Arcmedes Plutonium,
a_plutonium@hotmail.comwhole entire Universe is just one big
Actually, my gut tells me that it would be easier to t a home
run> from just throwing the ball up into the air and tting it
because,> everytng else being equal, it would simply require
less impulse to> get it moving at the same speed. Given
roughly the same amount of> time of contact between bat and
ball, ts would require less force. Remember that impulse is
the change in momentum. First of all, the> batter needs to
overcome the momentum of the ball in s own> direction, reverse
it, and provide sufficient momentum to get it over> the wall. 
By
tossing it up, he's essentially starting the ball from> rest
and won't need to overcome the initial momentum of the ball.
So> it requires less force.Nope.. The energy imparted by the
pitch will be transferred to the>> bat and redirected back
into imparted inertial force to the ball..A bat swung with all
else being equal will drive the ball further>> if is traveling
at say 100mph and a shorter distance if it were>> traveling at
say 80mph.Starting from a standing stop you only have the
imparted force of the>> energy of the bat ( converted kinetic
form derived from the transfer of>> the applied energy of the
swinger) where as in a pitched ball you have>> the same
applied force plus the redirected vector force of the
pitchers>> imparted energy.>Indeed, physics tells us that the
100mph pitch momentum is added onto the >t ball for extra long
distance.I don't know what physics text you found that one 
in.
How can anincrease in the initial momentum lead to an increase
in the finalmomentum, when the two are pointed in different
directions? I'mtrying to keep the analysis simple by only
looking at linear momentum.In truth, to examine ts problem
correctly, you need to treat it aselastic collision between
bat and ball, conserving both scalar kineticenergy and vector
momentum.After that analysis, we obtain ts single equation
(wch can befound in most elementary physics text involving
elastic collisions):( M + m ) Vbf = ( m - M) Vbi + 2 M
VBiwhereM = mass of batm = mass of ballVbi = initial speed of
ballVbf = final speed of ballVBi = initial speed of CM of 
batIf
the ball is just tossed up into the air, Vbi is essentially
zerowhen the ball is t, so ( M + m ) Vbf = 2 M VBiIf the ball
is pitched from sixty feet away with an initial speed Vbi,(M +
m ) Vbf = ( m - M) Vbi + 2 M VBiSince usually M > m, ts makes
the final speed of in the ball in thelatter case less than in
Actually, my gut tells me that it would be easier to t a home
run>> from just throwing the ball up into the air and tting it
because,>> everytng else being equal, it would simply require
less impulse to>> get it moving at the same speed. Given
roughly the same amount of>> time of contact between bat and
ball, ts would require less force.>> Remember that impulse is
the change in momentum. First of all, the>> batter needs to
overcome the momentum of the ball in s own>> direction,
reverse it, and provide sufficient momentum to get it over>>
the wall. By tossing it up, he's essentially starting the 
ball
from>> rest and won't need to overcome the initial momentum 
of
the ball. So>> it requires less force.>>>Nope.. The energy
imparted by the pitch will be transferred to the>bat and
redirected back into imparted inertial force to the ball..>A
bat swung with all else being equal will drive the ball
further>if is traveling at say 100mph and a shorter distance
if it were>traveling at say 80mph.>Starting from a standing
stop you only have the imparted force of the>energy of the bat
( converted kinetic form derived from the transfer of>the
applied energy of the swinger) where as in a pitched ball you
have>the same applied force plus the redirected vector force
of the pitchers>imparted energy.>> Indeed, physics tells us
that the 100mph pitch momentum is added onto the> t ball for
extra long distance.> What we need now for baseball is for the
famous sluggers such as Bonds,> Sosa, et al, to run the test.
Can any one of them t a homerun in any> major league baseball
stadium from a 0 speed self pitched ball from the> homeplate?
If Barry Bonds is able to t such a self pitched 0 speed> ball
in any stadium of baseball, then that stadium is utterly
inadequate> as far as distance is concerned and needs to be
re-designed for it is a> cheating baseball field.> And also,
the best pitches in baseball, because the physics says so or>
proves it, are slow pitches that have a wicked spin on the
ball.> Physics says that our modern day baseball games are
being played to a> dumb-strategy and not the optimal strategy.
In that a pitcher withfastballs> maybe able to get a few
strikeouts but also is increasing the chances very> greatly of
pitcng homeruns.> So that if a Barry Bonds or Sammy Sosa
(excuse the spelling) were to come> to the plate, then the
best pitches to them would be the slowest possible> pitch with
a wicked spin on the ball.> In practice season ts year for all
major league baseball teams, should> try out what I am saying
above and see if slow pitcng can retire their> batters faster
and better than fastball pitcng.-----Some where witn the
Quantum
StateHttp://Paul.Mays.Com/story.htmlhttp://paul.mays.com/
mayday.htmlhttp://paul.mays.com/rainy.htmlScience tries to
answer the question: How? How docells act in the body? How do
you design an airplanethat will ßy faster than sound? How is a
molecule ofinsulin constructed? Religion, by contrast, tries
toanswer the question: Why? Why was man created?Why ought I to
tell the truth? Why must there be sorrowor pain or death?
Science attempts to analyze howtngs and people and animals
behave; it has no concernwhether ts behavior is good or bad,
is purposeful ornot. But religion is precisely the quest for
suchanswers: whether an act is right or wrong, good or bad,and
===
the angle : you are> measuring an angle for only a subset of
all tngs for wch> one could measure angles. :-)When I look for
a competing non-conventional (non-conventional) logic andwork
with coordinates x, y, z, a1 , I have a horizontal angle in
the xyplane, a distance in the xy plane, a total distance from
the origin to theendpoint, and a vertical angle based on the
total distance and the net of zand a1...So, an inverse
produces the x coordinate, the y coordinate, and thenet of z
and a1 . Two distances and two angles required and only a
partialinverse is possible.Compare to the KBH development wch
requires an angle off the x axis to theendpoint, an angle off
the y axis to the endpoint, and the total distancefrom origin
to endpoint. Then the inverse is the x coordinate, the
ycoordinate, and the distance from the intermediate xy
location to theendpoint...and again only a partial inverse
operation.I assume the conventional logic in any case is a
geometry where theoperations apply up the the z coordinate,
then the origin is translated, andthe operations repeated for
the next three points. Ts is very good buthere I doubt the
significance of a total distance from origin 
toendpoint....when
beyond three dimensions.If the distance from origin to endpoint
is significant when beyond threedimensions then I wonder about
an angle from origin to endpoint...even ifthe angle is as
===
= 0, and then use the 3D formula .2D:Let z = 0 and use the 3D
formula .3D:Distance = Sqrt(x^2 + y^2 + z^2)X-dangle =
ArcTan(Sqrt(y^2 + z^2) / x)Y-dangle = ArcTan(Sqrt(x^2 + z^2) /
y)Note that all three paramters are required when usingdangle
terminology.Beyond 3D:Distance = Sqrt(x^2 + y^2 + z^2 + a1^2 +
... + an^2)X-dangle = ArcTan(Sqrt(y^2 + z^2 + a1^2 + ... +
an^2) / x)Y-dangle = ArcTan(Sqrt(x^2 + z^2 + a1^2 + ... +
an^2) / y)Properties:A 2D mapping output of the distance and
output of the distance and Y-dangle producesthe original y
coordinate. Next, the y coordinate of 2D mapping 
Ôa' can
beoperated with the y coordinate of 2D mapping 
Ôb'to produce
the distance from the intermediate x-y location to the
endpointas:Lump = Sqrt(ya^2 - yb^2)where ya >= yb .Note that
when working in 3D, Lump will equal to the z
===
y = 0, z = 0, and then use the 3D formula .> 2D:> Let z = 0 and
use the 3D formula .> 3D:> Distance = Sqrt(x^2 + y^2 + z^2)>
ArcTan(Sqrt(x^2 + z^2) / y)> Note that all three paramters are
required when using> dangle terminology.> Beyond 3D:> Distance
= Sqrt(x^2 + y^2 + z^2 + a1^2 + ... + an^2)> X-dangle =
ArcTan(Sqrt(y^2 + z^2 + a1^2 + ... + an^2) / x)> Y-dangle =
ArcTan(Sqrt(x^2 + z^2 + a1^2 + ... + an^2) / y)> Properties:>
A 2D mapping output of the distance and X-dangle produces the
original x> coordinate. Then a 2D mapping output of the
distance and Y-dangle produces> the original y coordinate.
Next, the y coordinate of 2D mapping Ôa' can be> 
operated with
the y coordinate of 2D mapping Ôb'> to produce 
the distance
from the intermediate x-y location to the endpoint> as:> Lump
= Sqrt(ya^2 - yb^2)> where ya >= yb .> Note that when working
in 3D, Lump will equal to the z coordinate.> -Development of
KBHGosh.The Dangle system is implemented as a quadrant system
where (x, y) isquadrant 1, (x, -y) is quadrant 2, (-x, -y) is
quadrant 3, and (-x, y) isquadrant 4 . Next, a z-coordinate
less than zero outputs a negative signX-dangle and a negative
===
solution toy'' = x*exp(x)*ysomeone suggested 
it may be related
to airy functions, and i can see 
howy''=exp(x)y might be...but
i cant seem to solve for the above equation. anyideas? anyone
have maple, or mathematica, or sometng else they can bungts in
===
equation?>need a solution to>y'' = 
x*exp(x)*yMaple 9 can't find
a closed-form solution.>someone suggested it may be related to
airy functions, and i can see how>y''=exp(x)y 
might be...but i
cant seem to solve for the above equation. any>ideas? anyone
have maple, or mathematica, or sometng else they can bung>ts
in and see what happens.Actually the solutions for 
y'' =
exp(x) y are in terms of Bessel functions: y(x) = _C1
===
Re: difficult riccati equation?> ,> need a solution to> 
y'' =
x*exp(x)*yThe standard trick is to reduce the problem to a
first-order ODE, byintroducing the new dependent variable z(x)
= y'(x) / y(x). The down-side isyou get a non-linear ODE, 
but
===
problem finding a solution to an equation:Some info 
first: S, A,
and B are known numbers (and if ts helps, S>= 0).Solve for X: S
= 0.1*[(0.001+X)^(A*X^B) - 1]I have come to a step where:
[(S/0.1)+1]^(1/A) = (0.001+X)^(X^B)and cannot continue. I
tried with raising e to the both sides andtaking the natural
log after in order to remove that X^B somehow. I'mstuck. in
===
Solutions>As far as I can tell, you can work out a value for x
and a value for y every>time...? So what defines whether there
are solutions or not?Here's one that doesn't 
have a solution -
see if you can see the differencewhen you try to solve for x
===
Equation without SolutionsHow can you tell if a set of
simultaneous equations has solutions or not?e.g. 12x - 4y = 8,
-9x + 3y = 6orx^3 - y^2 = 0, x = y> e.g. for your second set of
equations, x = 0, y = 0 is the only> solution. And here I could
have sworn that x = y = 1 was another solution to the system
===
ts problem?Suppose M^m is a smooth submanifold of R^n. The
Gauss map of M is themap r: M -> Gr(m, n) wch assigns to each
p in M the pointcorresponding to TM_m. Show that r is a smooth
===
sehr intersesting, der Matter ist almost gesettled !
===
Fermat (among others) are acknowledged for theircontributions
to the foundations of calculus, but only the beginning of
whatNewton & Leibniz took much farther. for the input!! Can
anyone help me with ts question? Why is Newton more
widelyaccepted as the man who developed calculus, instead of
Leibniz?> And why isn't proper credit given to earlier work
by, e.g.,> Arcmedes, Fermat?> FWIW, our current notation stems
from both Liebniz and Newton.> Physicist are more prone to use
Newton's x dot notation, but both> mathematicians and
physicists use both when appropriate. The notation> used for
partial derivatives also stems from Liebniz's notation.> -- 
>
> Unsolicited bulk E-mail will be subject to legal action. I
reserve> the right to publicly post or ridicule any abusive
E-mail.> Reply to domain Patriot dot net user shmuel+news to
contact me. Do> not reply to
===
have cosidered a software project as a vector in the given
space.Soevery project can be divided into sub-modules. Every
sub-module can bedivided into packages and every package can,
in turn, be subdividedinto classes. Each class can be divided
into a class variable, aconstructor, and a method. Thus, the
space in wch the project isrepresented can be understood as
consisting of the following: Modules = M Package = P Classes =
C Class Variable = v Class Constructor = c Class Method = m.The
magnitude of these elements together will constitute the state
ofthe project or will represent the project as a vector in the
space.To analyze a project at a given stage, it is necessary to
maintain theprevious state of the project.Let us consider that
Delivery1 for a given project consists of thefollowing:5
modules, 30 different packages, 100 classes, 200
variables(irrespective of the encapsulation), 125
constructors, and 500 methodsIn ts case, the project can be
represented as: Delivery1 = 5M + 30 P + 100 C + 200v + 125c +
500mHere what you tnk Module,package and Classes are othogonal
===
inbehaviour or not???Please I need a urgent help on ts.Subject:
project as a vector in the given space.So> every project can
be divided into sub-modules. Every sub-module can be> divided
into packages and every package can, in turn, be subdivided>
into classes. Each class can be divided into a class variable,
a> constructor, and a method. Thus, the space in wch the
project is> represented can be understood as consisting of the
following:> Modules = M> Package = P > Classes = C> Class
Variable = v> Class Constructor = c > Class Method = m.Project
= (M1,M2,M3,...,Mn)M1 = (P1,P2,P3,...,Po)...P1 = (...)......>
The magnitude of these elements together will constitute the
state of> the project or will represent the project as a
vector in the space.Huh?It's just the magnitude of the
variables, constructors, and methods, from what I can tell.>
To analyze a project at a given stage, it is necessary to
maintain the> previous state of the project.Huh?> Let us
consider that Delivery1 for a given project consists of the>
following:> 5 modules, 30 different packages, 100 classes, 200
variables> (irrespective of the encapsulation), 125
constructors, and 500 methods> In ts case, the project can be
represented as: > Delivery1 = 5M + 30 P + 100 C + 200v + 125c
+ 500m> Here what you tnk Module,package and Classes are
othogonal in> behaviour or not???Orthogonal cannot begin to
apply to the construction you've offered. It sounds like 
your
model of the project is severely ßawed with respect to
whatever you're trying to do.A package would be 
(effectively)
a *subspace* of a module. The same with classes to packages.
That is, if the notion of tnking of ts as a vector space is
even useful, wch I doubt. Do you understand what a vector
space is?What is ts supposed to let you accomplish, aside from
===
cosidered a software project as a vector in the given
space.So>every project can be divided into sub-modules. Every
sub-module can be>divided into packages and every package can,
in turn, be subdivided>into classes. Each class can be divided
into a class variable, a>constructor, and a method. Thus, the
space in wch the project is>represented can be understood as
consisting of the following:> Modules = M> Package = P >
Classes = C> Class Variable = v> Class Constructor = c > Class
Method = m.> Project = (M1,M2,M3,...,Mn)> M1 =
(P1,P2,P3,...,Po)> ...> P1 = (...)> ...> ...The magnitude of
these elements together will constitute the state of>the
project or will represent the project as a vector in the
space.> Huh?> It's just the magnitude of the variables,
constructors, and methods, > from what I can tell.To analyze a
project at a given stage, it is necessary to maintain
the>previous state of the project.> Huh?Let us consider that
Delivery1 for a given project consists of the>following:>5
modules, 30 different packages, 100 classes, 200
variables>(irrespective of the encapsulation), 125
constructors, and 500 methodsIn ts case, the project can be
represented as: >Delivery1 = 5M + 30 P + 100 C + 200v + 125c +
500mHere what you tnk Module,package and Classes are othogonal
in>behaviour or not???> Orthogonal cannot begin to apply to
the construction you've offered. It > sounds like your model
of the project is severely ßawed with respect to > whatever
you're trying to do.> A package would be (effectively) a
*subspace* of a module. The same > with classes to packages.
That is, if the notion of tnking of ts as > a vector space is
even useful, wch I doubt. Do you understand what a > vector
space is?> What is ts supposed to let you accomplish, aside
from delay coding?You are right will do some more work and get
===
software project as a vector in the given space.So> every
project can be divided into sub-modules. Every sub-module can
be> divided into packages and every package can, in turn, be
subdivided> into classes. Each class can be divided into a
class variable, a> constructor, and a method. Thus, the space
in wch the project is> represented can be understood as
consisting of the following:> Modules = M> Package = P >
Classes = C> Class Variable = v> Class Constructor = c > Class
Method = m.> The magnitude of these elements together will
constitute the state of> the project or will represent the
project as a vector in the space.> To analyze a project at a
given stage, it is necessary to maintain the> previous state
of the project.> Let us consider that Delivery1 for a given
project consists of the> following:> 5 modules, 30 different
packages, 100 classes, 200 variables> (irrespective of the
encapsulation), 125 constructors, and 500 methods> In ts case,
the project can be represented as: > Delivery1 = 5M + 30 P +
100 C + 200v + 125c + 500m> Here what you tnk Module,package
and Classes are othogonal in> behaviour or not???> Please I
need a urgent help on ts.As a working software engineer I
don't tnk any of ts is valid. If you're saying 
that your
project consists of 50 modules and you've coded 25 and
therefore you're 50% done, believe me, that is wrong. When 
all
your coding is complete, you're still barely 50% done once 
you
start testing. Likewise, what sense does it make to count
variables as a deliverable? My project has 100 variables and
I've defined 50 of them so I'm 
half done? How many variables
===
required>I have cosidered a software project as a vector in
the given space.So>every project can be divided into
sub-modules. Every sub-module can be>divided into packages and
every package can, in turn, be subdivided>into classes. Each
class can be divided into a class variable, a>constructor, and
a method. Thus, the space in wch the project is>represented can
be understood as consisting of the following:> Modules = M>
Package = P > Classes = C> Class Variable = v> Class
Constructor = c > Class Method = m.The magnitude of these
elements together will constitute the state of>the project or
will represent the project as a vector in the space.To analyze
a project at a given stage, it is necessary to maintain
the>previous state of the project.Let us consider that
Delivery1 for a given project consists of the>following:>5
modules, 30 different packages, 100 classes, 200
variables>(irrespective of the encapsulation), 125
constructors, and 500 methodsIn ts case, the project can be
represented as: >Delivery1 = 5M + 30 P + 100 C + 200v + 125c +
500mHere what you tnk Module,package and Classes are othogonal
in>behaviour or not???Please I need a urgent help on ts.> As a
working software engineer I don't tnk any of ts is valid. If 
>
you're saying that your project consists of 50 modules and
you've coded > 25 and therefore you're 50% 
done, believe me,
that is wrong. The variable are taken after a phase of a
project is over,to study theexecution cycle i.e.if there is
any delay is it due to impropersoftware design or due to
someother reason.>When all your coding is complete, you're
still barely 50% done onceyou start> testing. Likewise, what
sense does it make to count variables as a > deliverable? My
project has 100 variables and I've defined 50 of 
them > so I'm
half done? How many variables you define is irrelevant to >
anytng.You are right, If you have 100 variable and you define
50 it is notthat you have completed 50% of work.But
identifying 100 variable isthe challenge and then defining
properly is another designingpart.Work is constituted
logically during development.When i say delivery It means it
is tested delivery.My intension was to define Project so
movement or drift can beindentify easily and become a good
===
urgent help ???>I have cosidered a software project as a
vector in the given space.So>every project can be divided into
sub-modules. Every sub-module can be>divided into packages and
every package can, in turn, be subdivided>into classes. Each
class can be divided into a class variable, a>constructor, and
a method. Thus, the space in wch the project is>represented can
be understood as consisting of the following:> Modules = M>
Package = P > Classes = C> Class Variable = v> Class
Constructor = c > Class Method = m.>The magnitude of these
elements together will constitute the state of>the project or
will represent the project as a vector in the space.>To
analyze a project at a given stage, it is necessary to
maintain the>previous state of the project.>Let us consider
that Delivery1 for a given project consists of
the>following:>5 modules, 30 different packages, 100 classes,
200 variables>(irrespective of the encapsulation), 125
constructors, and 500 methods>In ts case, the project can be
represented as: >Delivery1 = 5M + 30 P + 100 C + 200v + 125c +
500m>Here what you tnk Module,package and Classes are othogonal
in>behaviour or not???>Please I need a urgent help on
===
help ???I have cosidered a software project as a vector in the
given space.Soevery project can be divided into sub-modules.
Every sub-module can bedivided into packages and every package
can, in turn, be subdividedinto classes. Each class can be
divided into a class variable, aconstructor, and a method.
Thus, the space in wch the project isrepresented can be
understood as consisting of the following:> Modules = M>
Package = P > Classes = C> Class Variable = v> Class
Constructor = c > Class Method = m.The magnitude of these
elements together will constitute the state ofthe project or
will represent the project as a vector in the space.To analyze
a project at a given stage, it is necessary to maintain
theprevious state of the project.Let us consider that
Delivery1 for a given project consists of thefollowing:5
modules, 30 different packages, 100 classes, 200
variables(irrespective of the encapsulation), 125
constructors, and 500 methodsIn ts case, the project can be
represented as: Delivery1 = 5M + 30 P + 100 C + 200v + 125c +
500mHere what you tnk Module,package and Classes are othogonal
inbehaviour or not???Please I need a urgent help on
===
dependency logic REVISEDIn sci.physics, Brian Quincy
Hutcngs<QncyMI@netscape.net><bde404c9.0310281352.16bc64c3@
ghostly entities? :-)> have you gotten recruits from the Sokal
School of Numbertheory,> or what?> anyway, you don't tnk 
that
Einstein made any mistakes> in s math?... how about s
hypotheses?> and what about Ienstien?Hmm...maybe we can get
Sir Francis Bacon in there somehow.As well as Elvis Presley.
:-)> >> And now readers should notice the psych out game.>
--Dec.2000 ÔWAND' Chairman Paul 
O'Neill, reelected to Board.
Newsish? >
http://www.rand.org/publications/randreview/issues/rr.12.00/ >
===
http://members.tripod.com/~american_almanac-- Subject: Re: Math
happen when you deal with a> discoverer is that you find that
tngs you thought were simple,> suddenly seem complicated and
iffy. So I'm now facing a problem where> people are doubting
algebra, wch leads to the fun question of, just> what is the
logic here?> The issue has to do with a factor like x+2 and
how you know that 2 is> constant.Simple, 2 is constant.> If
the logic is expanded upon, my hope is that many of you will>
realize what is happening in my core error proof, and quit
doubting,> if you are doubting, as if you're not, and
understand it, then that's> another problem entirely.> Ok, 
so
how do you know that with even a polynomial factor, like x+2,>
that 2 is *constant* and not related to x?right.> What's the
mathematical logic?> I've talked about setting x=0, wch 
would
reveal the constant factor> wch you can look at, and I tnk
that for some of you, the ability> to *physically* see
constants in polynomial factors, has left you> without ever
knowing the *why* bend how it works.And, when x = 0, x+2 is
divisible by 2.> How about ts?> There are an infinity of 
number
and letter combinations, of wch x+2> is just one of that
infinity. If you set x=0 though, you just have 2,> so how do
you know from whence it came?You don't.> I mean, with just 
2,
well, it's just 2, so there's no reference 
back> to x, or y,
or any other of that infinity of possibilities.Correct.> The
answer is that 2 is just a number that has no other
information> except itself.Correct.> Logically, you can say
that 2 is *independent* of any particular> variable, even if
it's associated with it, like in an expression like>
x+2.Correct.> Now, as 0 clears out variables set to 0, you can
show independent> terms by setting a particular value to 0, as
that reveals everytng> that can be associated with an infinity
of other tngs.> It's like how it could be y+2, or z+2, or
alpha + 2, as seeing just 2,> doesn't tell you anytng about 
an
association.> Possibly the problem for some of you is that you
*remember* the> previous association. That is, you know that
you had x+2, then you> set x=0, so in your mind you see a
phantom x that holds the> association.Here's the problem: 
your
argument boils down to:If x=0, x+2 is divisible by 2. Therefor,
x+2 is divisible by 2 for all integer values of x.When someone
points out that when x=1, x+2 is *not* divisible by 2, we get
insulted.> That's ok, but it doesn't have a 
*mathematical*
reality because your> memory is irrelevant.> The reality is
that as far as the math is concerned 2 is a number that> can
be by itself or associated with x or an infinity of other
tngs,> and the math isn't going to pick or choose based on
what YOU remember.Correct.> It handles all possibilities at
once.> And with *just* a number, it sees independence.2 is
independent of x. Got it.> That's why setting a variable to 
0
is such a useful and powerful tool,> as it reveals infinite
possibilities by removing a particular> association.You are
trying to make it do more than it can.> My argument works on
that principle, as by setting x or m, depending> on what I'm
calling the variable, I find that I have 7, 7 and 22 as>
*constant* terms for factors of a polynomial.Agreed. 7,7,22
are the constant terms.> I also know that polynomial has a
factor that is constant as that> factor is 49.Agreed.> When I
divide the polynomial by 49 I have constant terms of 1, 1,
and> 22, wch tells me sometng.Here's the problem. 49 
doesn't
go into *just* the constant terms. It must *also* go into the
variable terms. See the above example with x+2 being divisible
by 2.You have come to the (incorrect) conclusion that you can
ignore the behavior of the variable terms and focus only on
the behavior of the constant terms. Ts would be nice, but it
is not true. You are focusing on the constant terms, but they
tell you *notng* about the variable terms. Those terms are
also relevant to what is happening in your factorization and
cannot be ignored.> Now those numbers don't care about x or 
m.
They don't *know* about x> or m, but you may remember them, 
so
you tnk your memory has some> effect, but the math sees
*infinite* possibility, wle you may fixate> on an 
association
you remember.Your conclusion reintroduces x or m. Therefore,
we *must* consider them from the beginning, or be very careful
when reintroducing them.> In the realm of infinite 
possibility,
it's not possible for 7, 7, and> 22 to go to 1, 1, and 22
based on the value of x, where they check it> first to decide
how they'll behave, because they're just 
NUMBERS, and> they
don't have enough data associated with them to be making
those> kinds of decisions.But you *did* base it on the value
of x. You set x=0. That is a particular value of x. More
importantly, you are ignoring the variable terms wch you are
making your conclusion about.> It's like, if human beings on
earth constrain 7, 7 and 22 so that they> have to worry about
the value of x and m, then, wow, we've affected> all reality
in all dimensions and all possibilities, just with our> MEMORY
of there being an x or m associated in some interesting>
expression.> But, of course, your thoughts don't constrain
those numbers as they> keep operating quite well throughout
Totality, despite what you may> believe, or remember.> Just
remember that when you're considering the issue of constants
in a> math argument.Done. Now please remember what variable
terms do when considering the issues of *variables* in a math
===
of the more odd tngs that can happen when you deal with
a>discoverer is that you find that tngs you thought were
simple,>suddenly seem complicated and iffy. So I'm now 
facing
a problem where>people are doubting algebra, wch leads to the
fun question of, just>what is the logic here?The issue has to
do with a factor like x+2 and how you know that 2
is>constant.> Simple, 2 is constant.Hold that thought.Um,
there's one other tng first as the poster C. 
Bond pointed outin
a post in another thread I've had a sign mistake, as 
I've
beengivingP(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078but
the correct polynomial isP(x) = 14706125 x^3 - 900375 x^2 -
17640 x + 1078.So with the correction, here's the math once
again.Notice how I'll be strongly emphasizing constant terms
all the waydown.P(x) = 14706125 x^3 - 900375 x^2 - 17640 x +
1078wch has a constant term that is 1078.Well P(x) can also be
written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 +
49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5
a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 +
49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that
the constant terms for the three factors areall 7, wch can't 
be
right, as the constant term of P(x) is 1078, sosetting x=0,
revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the
cubic defining the a's with x=0 isa^3 - 3a^2, 
wch has roots, 0,
0 and 3, and I've picked a_1 and a_2for 0, so that leaves 
a_3
with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep
indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5
b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and
now my constant terms work out correctly.But P(x) has 49 as a
factor as every term in P(x) = 14706125 x^3 - 900375 x^2 -
17640 x + 1078has 49 as a factor, so I can divide by 49, and
dividing 1078 by 49gives me 22, as the new constant term.Well
that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 +
22)is the only way that the constant terms keep
===
the more odd tngs that can happen when you deal with
a>discoverer is that you find that tngs you thought were
simple,>suddenly seem complicated and iffy. So I'm now 
facing
a problem where>people are doubting algebra, wch leads to the
fun question of, just>what is the logic here?The issue has to
do with a factor like x+2 and how you know that 2
is>constant.> Simple, 2 is constant.Ok, if it's that simple
then you have no problem with what follows.Readers should note
that I'm hoping that getting people to admit theobvious will
stop them from lying about the argument that follows.Please
read carefully with the poster's admission in mind.Notice 
how
I'll be strongly emphasizing constant terms all the
waydown.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078wch
has a constant term that is 1078.Well P(x) can also be written
out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2
+ 7)(5 a_3 + 7)where the a's are roots ofa^3 + 3(-1 + 
49x)a^2
- 49(2401 x^3 - 147 x^2 + 3x).Notice it *appears* that the
constant terms for the three factors areall 7, wch can't be
right, as the constant term of P(x) is 1078, sosetting x=0,
revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the
cubic defining the a's with x=0 isa^3 - 3a^2, 
wch has roots, 0,
0 and 3, and I've picked a_1 and a_2for 0, so that leaves 
a_3
with a value of 3 when x=0.So let a_3 = b_3 + 3, where I keep
indices matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5
b_3 + 5(3) + 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and
now my constant terms work out correctly.But P(x) has 49 as a
factor as every term in P(x) = 14706125 x^3 - 900375 x^2 +
17640 x + 1078has 49 as a factor, so I can divide by 49, and
dividing 1078 by 49gives me 22, as the new constant term.Well
that means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 +
22)is the only way that the constant terms keep
===
the more odd tngs that can happen when you deal with
a>discoverer is that you find that tngs you thought were
simple,>suddenly seem complicated and iffy. So I'm now 
facing
a problem where>people are doubting algebra, wch leads to the
fun question of, just>what is the logic here?The issue has to
do with a factor like x+2 and how you know that 2
is>constant.>>Simple, 2 is constant.> Ok, if it's that 
simple
then you have no problem with what follows.James, you
obviously didn't finish reading my response. The 
key phrase is
variable terms. Go back and try again. With your second
attempt, please respond to all my points, not just the first
===
Math dependency logic REVISED>One of the more odd tngs that can
happen when you deal with adiscoverer is that you find that 
tngs
you thought were simple,suddenly seem complicated and iffy. So
I'm now facing a problem wherepeople are doubting algebra, 
wch
leads to the fun question of, justwhat is the logic here?The
issue has to do with a factor like x+2 and how you know that 2
isconstant.Simple, 2 is constant.>Ok, if it's that simple 
then
you have no problem with what follows.> James, you obviously
didn't finish reading my response. The key 
phrase > is variable
terms. Go back and try again. With your second attempt, >
please respond to all my points, not just the first one.I
*created* ts thread. Reasonable people might assume I created
itfor a reason, and in fact my point is to show the rather
bizarreinsanity of raising questions about a proof that relies
on 7 and othernumbers being CONSTANT, so when you said that it
was simple, I wentfrom there.Now there isn't any other point
to be made, and I should know as Icreated the thread to make
the point that it is SIMPLE.Now readers can see the actual
argument to understand what I mean.Readers should note that
I'm hoping that getting people to admit theobvious will stop
them from lying about the argument that follows.Please read
carefully with the poster's admission in mind.Notice how 
I'll
be strongly emphasizing constant terms all the waydown.P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078wch has a constant
term that is 1078.Well P(x) can also be written out asP(x)=
7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 +
7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).Notice it *appears* that the constant terms
for the three factors areall 7, wch can't be right, as the
constant term of P(x) is 1078, sosetting x=0, revealsP(0) =
(5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining
the a's with x=0 isa^3 - 3a^2, wch has roots, 0, 0 and 3, 
and
I've picked a_1 and a_2for 0, so that leaves a_3 with a 
value
of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices
matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3)
+ 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my
constant terms work out correctly.But P(x) has 49 as a factor
as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x +
1078has 49 as a factor, so I can divide by 49, and dividing
1078 by 49gives me 22, as the new constant term.Well that
means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is
===
Re: Math dependency logic REVISED... > Notice how I'll be
strongly emphasizing constant terms all the way > down. P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078Works the same
argument with the polynomial Q(x) = 6125 x^3 + 6125 x^2 - 6370
x + 1078? If not, what step fails? wch has a constant term that
is 1078. Well P(x) can also be written out as P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I
can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)
where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x). Notice it *appears* that the constant terms
for the three factors are > all 7, wch can't be right, as 
the
constant term of P(x) is 1078, so > setting x=0, reveals P(0)
= (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic
defining the a's with x=0 is a^3 - 3a^2, wch has 
roots, 0, 0
and 3, and I've picked a_1 and a_2 > for 0, so that leaves 
a_3
with a value of 3 when x=0. So let a_3 = b_3 + 3, where I keep
indices matched. Then I have P(x) = (5 a_1 + 7)(5 a_2 + 7)(5
b_3 + 5(3) + 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) and
now my constant terms work out correctly. But P(x) has 49 as a
factor as every term in P(x) = 14706125 x^3 - 900375 x^2 -
17640 x + 1078 has 49 as a factor, so I can divide by 49, and
dividing 1078 by 49 > gives me 22, as the new constant term.
Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5
b_3 + 22) is the only way that the constant terms keep matcng.
===
it. don't fall into Will Twentyman's trap! 
<deletives
expleted> > is variable terms. Go back and try again. With
your second attempt, > please respond to all my points, not
===
just the first one.Subject: Re: Math dependency logic
deal with a>discoverer is that you find that tngs you thought
were simple,>suddenly seem complicated and iffy. So I'm now
facing a problem where>people are doubting algebra, wch leads
to the fun question of, just>what is the logic here?The issue
has to do with a factor like x+2 and how you know that 2
is>constant.>>>James, revising ts post is not going to help.
What on earth do you >>mean how do you know 2 is constant? Do
you mean that sometimes it's 3? >>Technical folk sometimes
make a joke like 2+1 = 4, for suitable values >>of 2. But 
it's
a joke, not a plosopcal argument. >>What are you talking
about?> Mathematicians have been fighting me on the issue of
constant terms> being constant. No they haven't. You simply
haven't eithera) read the counter-arguments, orb) understood
the counter-arguments >It's crazy, it's wacky, 
and I find it
infuriating, so> I set up a stage for context for readers on
the other newsgroups. > Notice the replies I got, and now pay
attention carefully to the> mathematics these people are
fighting.You are mis-representing our objections. Is ts
deliberate deceit?> Notice how I'll be strongly emphasizing
constant terms all the way> down.> P(x) = 14706125 x^3 -
900375 x^2 + 17640 x + 1078> wch has a constant term that is
1078.> Well P(x) can also be written out as> P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> so I
can factor to get> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)>
where the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).> Notice it *appears* that the constant terms
for the three factors are> all 7, wch can't be right, as the
constant term of P(x) is 1078, so> setting x=0, reveals> P(0)
= (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic
defining the a's with x=0 is> a^3 - 3a^2, wch 
has roots, 0, 0
and 3, and I've picked a_1 and a_2> for 0, so that leaves 
a_3
with a value of 3 when x=0.> So let a_3 = b_3 + 3, where I
keep indices matched. Then I have> P(x) = (5 a_1 + 7)(5 a_2 +
7)(5 b_3 + 5(3) + 7)> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 +
22)> and now my constant terms work out correctly.> But P(x)
has 49 as a factor as every term in > P(x) = 14706125 x^3 -
900375 x^2 + 17640 x + 1078> has 49 as a factor, so I can
divide by 49, and dividing 1078 by 49> gives me 22, as the new
constant term.> Well that means that> P(x)/49 = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 b_3 + 22)> is the only way that the constant terms
keep matcng.*Ts* is the step that is objected to, because of
how you are dividing 7 into the *non-constant* terms a_1 and
a_2. The constant terms are not the issue.> And if you let
mathematicians debate that conclusion or cast doubt on> it,
wle calling me names and acting like I'm the one 
who's crazy,>
then you piss on the intellectual basis for civilization
itself.> Why let mathematicians get away with it? What's in 
it
for you?> Why let them trample on civilization, as if honesty
and integrity in a> discipline don't mean anytng if enough
people in that discipline> don't like the truth?> 
What's wrong
with you people?> Don't ANY of you believe in anytng?I 
believe
in the fact that you are wrong. You are not correctly
representing our arguments, and your logic is ßawed. Until you
can correctly *state* the counter-arguments, you will never
make any kind of headway in your effort to convince us that
===
REVISED>>One of the interesting tngs I've noticed about
mathematics is that the>>truth or falsity of a mathematical
statement is not affected by the>>number of times it is
repeated.> You can say that again.>>One of the interesting
tngs I've noticed about mathematics is that thetruth or
falsity of a mathematical statement is not affected by
===
dependency logic REVISED> I'm *tired* of the games.
Mathematicians are running and there are> posters trying to
help them run, but I'm giving notice, I'll 
keep> putting the
information out there.> Notice how I'll be strongly
emphasizing constant terms all the way> down.> P(x) = 14706125
x^3 - 900375 x^2 + 17640 x + 1078> wch has a constant term that
is 1078.> Well P(x) can also be written out as> P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3No, it
can't. Did you ever try expanding that last version of P(x) 
to
see if it actually matchedthe original version? Not likely. The
original version is:P(x) = 14706125 x^3 - 900375 x^2 + 17640 x
+ 1078Your factored version expands to:P(x) = 14706125 x^3 -
900375 x^2 - 17640 x + 1078See the sign error? Run, James,
run.The correct factorization is:P(x)= 7^2(2401 x^3 - 147 x^2
- 3x) (5^3) + 3(1 + 49 x )(5)(7^2) + 7^3and is certainly not
what you posted.Beep! You are wrong again. Your subsequent
argument is fatally broken.To the question of whether I
believe in anytng: I believe the correction I provided
abovecorresponds to the truth and your post is wrong.Wacky,
isn't it? But, hey, it's just basic math. Yup, 
yup,
yup!----Democracy: The triumph of popularity over
===
dependency logic REVISED>One of the more odd tngs that can
happen when you deal with a>discoverer is that you find that
tngs you thought were simple,>suddenly seem complicated and
iffy. So I'm now facing a problem where>people are doubting
algebra, wch leads to the fun question of, just>what is the
logic here?> No one is doubting algebra. We're just certain
that you aren't using> it properly.Math society is in 
crisis.
I've found sometng that mathematicianshad just decided 
wasn't
possible--an error in core--and rather thandeal with it,
they're trying to run away.So I get to chase them 
down.Here's
the math once again.Notice how I'll be strongly emphasizing
constant terms all the waydown.P(x) = 14706125 x^3 - 900375
x^2 + 17640 x + 1078wch has a constant term that is 1078.Well
P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 +
3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to
getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are
roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 +
3x).Notice it *appears* that the constant terms for the three
factors areall 7, wch can't be right, as the constant term 
of
P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) +
7)(5(3) + 7) = 7(7)(22)as the cubic defining the 
a's with x=0
isa^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked a_1
and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So
let a_3 = b_3 + 3, where I keep indices matched. Then I
haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5
a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work
out correctly.But P(x) has 49 as a factor as every term in
P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078has 49 as a
factor, so I can divide by 49, and dividing 1078 by 49gives me
22, as the new constant term.Well that means thatP(x)/49 = (5
a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the
===
logic REVISED> Notice how I'll be strongly emphasizing
constant terms all the way> down.> P(x) = 14706125 x^3 -
900375 x^2 + 17640 x + 1078I still would like to see a
demonstration using P(x)=x^3-x+6> wch has a constant term that
is 1078.> Well P(x) can also be written out as> P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> so I
can factor to get> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)>
where the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).> Notice it *appears* that the constant terms
for the three factors are> all 7, wch can't be right, as the
constant term of P(x) is 1078, so> setting x=0, reveals> P(0)
= (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic
defining the a's with x=0 is> a^3 - 3a^2, wch 
has roots, 0, 0
and 3, and I've picked a_1 and a_2> for 0, so that leaves 
a_3
with a value of 3 when x=0.> So let a_3 = b_3 + 3, where I
keep indices matched. Then I have> P(x) = (5 a_1 + 7)(5 a_2 +
7)(5 b_3 + 5(3) + 7)> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 +
22)> and now my constant terms work out correctly.> But P(x)
has 49 as a factor as every term in > P(x) = 14706125 x^3 -
900375 x^2 + 17640 x + 1078> has 49 as a factor, so I can
divide by 49, and dividing 1078 by 49> gives me 22, as the new
constant term.> Well that means that> P(x)/49 = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 b_3 + 22)> is the only way that the constant terms
===
Notice how I'll be strongly emphasizing constant terms all 
the
way>down.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078>> I
still would like to see a demonstration using P(x)=x^3-x+6How
about P(x) = 11x + 123?P(x) = 11x + 123 = 11^2 + 11x + 2 gives
the non-polynomial factorization (11 + (x+sqrt(x^2 - 8))/2)(11
+ (x-sqrt(x^2 - 8))/2) = 11x + 123where the constant term is
*still* given by setting x=0, wch givesP(0) = (11 +
===
Math dependency logic REVISED> >> Notice how I'll be 
strongly
emphasizing constant terms all the way>> down.P(x) = 14706125
x^3 - 900375 x^2 + 17640 x + 1078>> I still would like to see
a demonstration using P(x)=x^3-x+6> How about P(x) = 11x +
123?> P(x) = 11x + 123 = 11^2 + 11x + 2 > gives the
non-polynomial factorization > (11 + (x+sqrt(x^2 - 8))/2)(11 +
(x-sqrt(x^2 - 8))/2) = 11x + 123Interesting, but pointless.
Using the same form, I can factor aCONSTANT into
polynomials:Q(x) = 314 = (17 + 7x - sqrt (49x^2+238x-25))*(17
+ 7x + sqrt(49x^2+238x-25))Not only Q(0) gives the constant
314, but every value of x does so.The two factors when
evaluated at 0 gives a factorization of 314 overthe set of
complex integers. Q(0) = (17 - 5i)*(17 + 5i) = 314Similarily,
evaluating the factors in ANY of YOUR factorizations atx=n
just gives a factorization of P(n) over complex numbers. You
havenot discovered anytng groundbreaking. All you did is show
that realnumbers can have complex factors.> where the constant
term is *still* given by setting x=0, wch gives> P(0) = (11 +
===
Re: Math dependency logic REVISEDbut he *did* dyscover
sometng, even if it took youto point out what it is. isn't 
it
always for the best,to recreate the original dyscoveries in
mathematical physics?(of course, James is just dealing with
dry, pure math .-) >P(x) = 11x + 123 = 11^2 + 11x + 2 gives
the non-polynomial factorization (11 + (x+sqrt(x^2 - 8))/2)(11
+ (x-sqrt(x^2 - 8))/2) = 11x + 123>> Interesting, but
pointless. Using the same form, I can factor a> CONSTANT into
polynomials:> Q(x) = 314 = (17 + 7x - sqrt
(49x^2+238x-25))*(17 + 7x + sqrt> (49x^2+238x-25))> Not only
Q(0) gives the constant 314, but every value of x does so.>
The two factors when evaluated at 0 gives a factorization of
314 over> the set of complex integers. Q(0) = (17 - 5i)*(17 +
5i) = 314> Similarily, evaluating the factors in ANY of YOUR
factorizations at> x=n just gives a factorization of P(n) over
complex numbers. You have> not discovered anytng
groundbreaking. All you did is show that real> numbers can
have complex
factors.http://www.cecaust.com.au/http://members.tripod.com/~
===
REVISED>> Notice how I'll be strongly emphasizing constant
terms all the way> down.P(x) = 14706125 x^3 - 900375 x^2 +
17640 x + 1078> I still would like to see a demonstration
using P(x)=x^3-x+6How about P(x) = 11x + 123?P(x) = 11x + 123
= 11^2 + 11x + 2 gives the non-polynomial factorization (11 +
(x+sqrt(x^2 - 8))/2)(11 + (x-sqrt(x^2 - 8))/2) = 11x + 123>>
Interesting, but pointless. Using the same form, I can factor
a> CONSTANT into polynomials:> Q(x) = 314 = (17 + 7x - sqrt
(49x^2+238x-25))*(17 + 7x + sqrt> (49x^2+238x-25))Yeah, yeah,
that's easy. You're not doing anytng special, 
and mypoint is
actually that there's notng *wrong* with what 
I'm doing.> Not
only Q(0) gives the constant 314, but every value of x does
so.Wch isn't refuting my point. After all, 
it's STILL
constant.It amazes me that so many of you keep arguing around
ts point, wchis such an incredible waste of time. Numbers like
7 or 314 areCONSTANTS, and setting a variable equal to 0
reveals constant terms.If the newsgroup would like to accept
at ts time that constants areconstant, then much progress will
have been made. > The two factors when evaluated at 0 gives a
factorization of 314 over> the set of complex integers. Q(0) =
(17 - 5i)*(17 + 5i) = 314> Similarily, evaluating the factors
in ANY of YOUR factorizations at> x=n just gives a
factorization of P(n) over complex numbers. You have> not
discovered anytng groundbreaking. All you did is show that
real> numbers can have complex factors.What? Did you just pull
that out of your butt Brian Smith? You askedfor an example,
indicating lack of understanding, so I *gave* you anexample,
and now you try to tell me what I'm doing, completely out
ofcontext?Are you sane?My point remains that the argument 
I've
shown is correct.That's the only real point.And here it is 
for
those who wonder what it is.Notice how I'll be strongly
emphasizing constant terms all the waydown.P(x) = 14706125 x^3
- 900375 x^2 - 17640 x + 1078wch has a constant term that is
1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 -
147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can
factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the
a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 
+
3x).Notice it *appears* that the constant terms for the three
factors areall 7, wch can't be right, as the constant term 
of
P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) +
7)(5(3) + 7) = 7(7)(22)as the cubic defining the 
a's with x=0
isa^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked a_1
and a_2for 0, so that leaves a_3 with a value of 3 when x=0.So
let a_3 = b_3 + 3, where I keep indices matched. Then I
haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)P(x) = (5
a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my constant terms work
out correctly.But P(x) has 49 as a factor as every term in
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078has 49 as a
factor, so I can divide by 49, and dividing 1078 by 49gives me
22, as the new constant term.Well that means thatP(x)/49 = (5
a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is the only way that the
===
logic REVISED>> Notice how I'll be strongly emphasizing
constant terms all the way> down.P(x) = 14706125 x^3 - 900375
x^2 + 17640 x + 1078> I still would like to see a
demonstration using P(x)=x^3-x+6How about P(x) = 11x +
123?P(x) = 11x + 123 = 11^2 + 11x + 2 gives the non-polynomial
factorization (11 + (x+sqrt(x^2 - 8))/2)(11 + (x-sqrt(x^2 -
8))/2) = 11x + 123>***CORRECTION*** Interesting, but
pointless. Using the same form, I can factor a CONSTANT into
non-polynomials: Q(x) = 314 = (17 + 7x - sqrt
(49x^2+238x-25))*(17 + 7x + sqrt (49x^2+238x-25)) Not only
Q(0) gives the constant 314, but every value of x does so. The
two factors when evaluated at 0 gives a factorization of 314
over the set of complex integers. Q(0) = (17 - 5i)*(17 + 5i) =
314 Similarily, evaluating the factors in ANY of YOUR
factorizations at x=n just gives a factorization of P(n) over
complex numbers. You have not discovered anytng
groundbreaking. All you did is show that real numbers can have
complex factors. >where the constant term is *still* given by
setting x=0, wch givesP(0) = (11 + sqrt(-8)/2)(11 -
===
logic REVISED> >> Notice how I'll be strongly emphasizing
constant terms all the way>> down.P(x) = 14706125 x^3 - 900375
x^2 + 17640 x + 1078>> I still would like to see a
demonstration using P(x)=x^3-x+6> How about P(x) = 11x + 123?>
P(x) = 11x + 123 = 11^2 + 11x + 2 > gives the non-polynomial
factorization > (11 + (x+sqrt(x^2 - 8))/2)(11 + (x-sqrt(x^2 -
8))/2) = 11x + 123> where the constant term is *still* given
by setting x=0, wch gives> P(0) = (11 + sqrt(-8)/2)(11 -
sqrt(-8)/2) = 123So would you write a1 = (x + sqrt(x^2-8))/2
a2 = (x - sqrt(x^2-8))/2in analogy to your cubic
factorization?Whatever you would write for a1 and a2, what are
theconstant terms of a1 and a2 and what is the
analogousargument about their factorability?Because to me it
doesn't look like I can writea1(x) = g(x) + a1(0) in any way
===
Re: Math dependency logic REVISEDI take it back,about JSH not
using formulae with smaller coefficients;, dude! >> I still
would like to see a demonstration using P(x)=x^3-x+6How about
P(x) = 11x + 123?P(x) = 11x + 123 = 11^2 + 11x + 2 gives the
non-polynomial factorization (11 + (x+sqrt(x^2 - 8))/2)(11 +
(x-sqrt(x^2 - 8))/2) = 11x + 123where the constant term is
*still* given by setting x=0, wch givesP(0) = (11 +
sqrt(-8)/2)(11 - sqrt(-8)/2) = 123> So would you write > a1 =
(x + sqrt(x^2-8))/2> a2 = (x - sqrt(x^2-8))/2> in analogy to
your cubic factorization?> Whatever you would write for a1 and
a2, what are the> constant terms of a1 and a2 and what is the
===
Math dependency logic REVISED... > Math society is in crisis.
I've found sometng that mathematicians > had just decided
wasn't possible--an error in core--and rather than > deal 
with
it, they're trying to run away.I tnk it is you who is in a
crisis. > Here's the math once again.Yes, repeat upon
repeat... Notice how I'll be strongly emphasizing constant
terms all the way > down. P(x) = 14706125 x^3 - 900375 x^2 +
17640 x + 1078Q(x) = 61255 x^3 + 6125 x^2 - 6370 x + 1078 wch
has a constant term that is 1078. Well P(x) can also be
written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1
+ 49 x )(5)(7^2) + 7^3(it has been noted by somebody else, ts
does *not* match your originalpolynomial above. I tnk in the
original is should be - 17640 x.)Q(x) = 7^2 ((2 x^2 - 3
x)(5^3) - 3(-1 + m)(5)(7^2) + 7^3 so I can factor to get P(x)
= (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)Q(x) = (5 c_1 + 7)(5 c_2 +
7)(5 c_3 + 7) where the a's are roots of a^3 + 3(-1 + 
49x)a^2
- 49(2401 x^3 - 147 x^2 + 3x).c^3 + 3(-1 + x)c^2 - 49(2 x^2 -
3 x). Notice it *appears* that the constant terms for the
three factors are > all 7, wch can't be right, as the 
constant
term of P(x) is 1078, so > setting x=0, reveals P(0) = (5(0) +
7)(5(0) + 7)(5(3) + 7) = 7(7)(22)Q(0) = (5(0) + 7)(5(0) +
7)(5(3) + 7) = 7(7)(22) as the cubic defining the 
a's with x=0
is a^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked 
a_1
and a_2c^3 - 3c^2 > for 0, so that leaves a_3 with a value of
3 when x=0. So let a_3 = b_3 + 3, where I keep indices
matched. Then I haved_3 = c_3 + 3 P(x) = (5 a_1 + 7)(5 a_2 +
7)(5 b_3 + 5(3) + 7)Q(x) = (5 c_1 + 7)(5 c_2 + 7)(5 d_3 + 5(3)
+ 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)Q(x) = (5 c_1 +
7)(5 c_2 + 7)(5 d_3 + 22) and now my constant terms work out
correctly. But P(x) has 49 as a factor as every term in P(x) =
14706125 x^3 - 900375 x^2 + 17640 x + 1078Q(x) = 6125 x^3 +
6125 x^2 - 6370 x + 1078 has 49 as a factor, so I can divide
by 49, and dividing 1078 by 49 > gives me 22, as the new
constant term. Well that means that P(x)/49 = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 b_3 + 22)Q(x)/49 = (5 c_1/7 + 1)(5 c_2/7 + 1)(5
d_3 + 22) is the only way that the constant terms keep
matcng.Apart from terminology, you are entirely correct upto
ts point. It isjust your tnking that somehow a1/7 and a2/7
should be algebraic integersthat is wrong. Your factorisation
of P(x)/49 is perfectly valid (as ismy factorisation of
Q(x)/49) *in the algebraic numbers*, not in thealgebraic
integers.It has been shown that for each x, 7 can be factored
as u(x).v(x).w(x) with(again for each x) u(x), v(x) and w(x)
pairwise coprime such that thefactorisation: P(x)/49 = (5
a_1/u(x).v(x) + w(x))(5 a_2/u(x).w(x) + v(x)) (5 a_3/v(x).w(x)
+ u(x))is a factorisation in the algebraic integers. When x =
0, u(x) = 7 andv(x) = w(x) = 1. Sometng similar occurs with my
Q(x). For Q, whenx = 0, u(x) = 7, v(x) = w(x) = 1, when x = 1,
u(x) = v(x) = w(x) = 7^{1/3},and when x = 2, u(x) = 1, v(x) =
(-sqrt(5) + 3i)/sqrt(2) andw(x) = (-sqrt(5) - 3i)/sqrt(2).Your
basic problem is that wle b_3(x) is coprime to 7 in general,(5
b_3(x) + 22) is in general *not* coprime to 7 in the algebraic
===
Math dependency logic REVISED> Here's the math once again.>
Notice how I'll be strongly emphasizing constant terms all 
the
way> down.> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078>
wch has a constant term that is 1078.> Well P(x) can also be
written out as> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) -
3(-1 + 49 x )(5)(7^2) + 7^3No, it can't. No matter how many
times you repost ts error, it is still false. The correct
factorization isP(x)= 7^2(2401 x^3 - 147 x^2 - 3x) (5^3) + 3(1
+ 49 x )(5)(7^2) + 7^3Your factorization expands to a
polynomial with the wrong sign for the linear term in 
Ôx'. You
can keepreposting it, if you wish. Maybe it will become true
after enough repeats.--There are two tngs you must never
attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
===
Puzzle> so what happened to Zeno Mr Gerl?> HercSo is Paul the
answer? The average?Should't the problem have the same 
answer
as starting at 12:45 withPaul starting with the bar as 12:30
===
homeomorpsms in
halfspacerwDs-4ZvgeaacRmJ2YpNSjU807NnG8rDC+
Kl0ITRqt409bvM5TxOr2Let H^n = { x in R^n | x_1 >= 0 } be the
halfspace where the firstcoordinate is non-negative. Suppose
U_1 and U_2 are open in H^n and thereis a homeomorpsm f : U_1
--> U_2.Prove or give a counterexample: for y in U_1 that lies
on the boundary ofH^n, also f(y) lies on the boundary.Here the
boundary of H^n is { x in R^n | x_1 = 0 }I wanted to show that
when y is not on the boundary, then also f(y) is not.The result
would follow when applying ts proposition to f^(-1).So my idea
was to lay a sphere so that, by looking at the image of
thesphere, we conclude that f(y) cannot be on the boundary.
But I do not knowhow to make ts accurate. Can anyone help?--
===
R^n | x_1 >= 0 } be the halfspace where the first> coordinate
is non-negative. Suppose U_1 and U_2 are open in H^n and
there> is a homeomorpsm f : U_1 --> U_2.> Prove or give a
counterexample: for y in U_1 that lies on the boundary of>
H^n, also f(y) lies on the boundary.> Here the boundary of H^n
is { x in R^n | x_1 = 0 }Ts can be done be considering local
===
theorem> Where can I find the best (the simplest, elegant)
proof of the prime number> theorem> (a) with complex variables
allowed> (b) without complex variables (the so called
elementary proof)?proof of the prime number theorem (American
===
number theorem>Where can I find the best (the simplest, 
elegant)
proof of the prime number>theorem>(a) with complex variables
allowed>(b) without complex variables (the so called
elementary proof)?> proof of the prime number theorem
(American Mathematical Monthly 76 pp.> 225-245 (1969)).> Best
elementary proofs withits good survey is Andrew Granville's
paper On elementary proofs ofthe prime number theorem for
arithmetic progressions, withoutcharacters. I am pretty sure
===
proof of prime number theorem>Where can I find the best (the
simplest, elegant) proof of the primenumber>theorem>(a) with
complex variables allowed>(b) without complex variables (the
so called elementary proof)?>> a wle back. He gives a very
elegant and short proof (due to D.J.> Newman), using barely
more complex analysis than Cauchy's theorem:> Zagier, D.>
Newman's short proof of the prime number theorem.> Amer. 
Math.
Monthly 104 (1997), no. 8, 705--708.> Another excellent account
of ts proof can be found in> Hlawka, Edmund; Schoissengeier,
Johannes; Taschner, Rudolf> Geometric and analytic number
theory.> Translated from the 1986 German edition by Charles
Thomas.> Universitext.> Springer-Verlag, Berlin, 1991. x+238
pp.> ISBN 3-540-52016-3... for ts info, Paul. Google found ts
as well:http://www.math.uga.edu/~mbaker/pdf/pnt2.pdfIt's a
17-page pdf, covering the background about zeta and the
===
Laplacetransform, as well as Newman's proof of 
PNT.LHSubject:
(hook@nas.nasa.gov) commentsRe: remainder of a connected
subspace > Theorem: If Y is a connected subspace of the
connected space X > and A,B are disjoint >>open sets in X<<
such that XY = A cup B, >> OP did not say and there is no need
to assume Y is closed. >> It is enough that A and B are open in
Acup B. > True -- > then A cup Y and B cup Y are connected. >
Proof: It suffices to prove that A cup Y is connected (since
the > same argument will work for B cup Y). Argue by
contradiction - > suppose A cup Y is not connected. Then there
is a continuous > surjection f: A cup Y --> 2. Since Y _is_
connected, f is > constant on Y -- WLOG, assume Y subset
f^{-1}(0). Then > f^{-1}(1) subset A and is relatively open in
A cup Y -- > since A is open in X, it follows that >> Yes,
f^(-1){A} is relatively open in A cup Y ( >> even if Y is not
closed) > f^{-1}(1) is an open subset of X. Also, there is
some open > O subset X such that f^{-1}(0) = O cap (A cup Y).
> Extend f to F: X --> 2 by defining F(x) = 0 if x in B. > 
Then
F is continuous (since we have F^{-1}(1) = f^{-1}(1) > and
F^{-1}(0) = O cup B, both of wch are open) >> Ts will also
work even if Y is not closed (B is not open) >> because B is
contained in an open set that misses all of A.Indeed, B subset
int B/Y. But how do I use that to show Fis continuous? The
problem I'm having is showing F^-1(0)is open in X because 
all
else I have is A open in A/B andf^-1(0) open in A/Y wch seems
inadequate./ / = cap cup > and F is a surjection (since f is).
Ts contradicts the > fact that X is conected and establishes
the result. [] >> Two comments: >> (1) There is no need to
argue by contradiction. >> Nowhere do you use the fact that
both f(-1){0} and f(-1){1} >> are both non-empty until the
very last line. >> Simply declare f is continuous and extend
to F. >> Then argue that F is constant, hence f must be
constant. >> Then X must be connected since f was arbitrary. >
OK -- I'll buy that ... >> (2) Much to my dismay, involving 
the
function is just a >> distraction. The crux of your proof is to
cover X by >> disjoint open sets. Ts is painfully obvious when
you >> note both of wch are open >> in order to argue that F
is continuous.I find using the function to show cl A connected
when A connectedand a union of connected subspaces with common
point is connectedmakes for quick, easy and readily apparent
proofs. > but ... I don't buy that -- I guess 
that's why
Baskin-Robbins > has to have 31 ßavors -- I really *do*
understand what's > going on better when the proof is
presented in ts particular > way. YMMV, of course ...Is that
also called the pasting or gluing lemma ?A function continuous
on two closed sets is continuous on bothA function continuous
on a bunch of open sets is continuous on union the open sets
of the bunch.Are their other versions?A corollary to the
theorem is for closed(or open) A,Bconnected A/B and A/B
implies A,B connectedConversely it can be use to show A / cl Y
===
14706125 x^3 - 900375 x^2 - 17640 x + 1078Well I had a sign
error with an expression I've been using for 
awle.I've
givenP(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078but as a
poster that goes as C. Bond pointed out in a post that
isincorrect.The correct polynomial isP(x) = 14706125 x^3 -
900375 x^2 - 17640 x + 1078wch you can verify for yourself
usingP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3wch is, and has been correct.Note: Change, of
course, also applies to previous posts where I usedm instead
of x.Sorry for the error, but I guess that sometimes the
calculator, simpleaddition, multiplication, and I are not
===
900375 x^2 - 17640 x + 1078> Sorry for the error, but I guess
that sometimes the calculator, simple> addition,
multiplication, and I are not friends.> There is a great deal
of mathematics, besides the above mentioned, that JSH is not
===
CORRECTION: P(x) = 14706125 x^3 - 900375 x^2 - 17640 x +
1078In sci.physics,
<jstevh@msn.com><3c65f87.0310290350.6e977b28@
expression I've been using for a> wle.> I've 
given> P(x) =
14706125 x^3 - 900375 x^2 + 17640 x + 1078> but as a poster
that goes as C. Bond pointed out in a post that is>
incorrect.> The correct polynomial is> P(x) = 14706125 x^3 -
900375 x^2 - 17640 x + 1078> wch you can verify for yourself
using> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3> wch is, and has been correct.> Note: Change,
of course, also applies to previous posts where I used> m
instead of x.> Sorry for the error, but I guess that sometimes
the calculator, simple> addition, multiplication, and I are not
friends.> Well, that's one error down, but what about the 
cubic
wch thea's satisfy after decomposition of P(x)? :-)--
===
17640 x + 1078> Well I had a sign error with an expression 
I've
been using for a> wle.Your conclusion did not alter with the
reversed data. You are anineducable psychotic troll. A noted
single incident doesn't begin todo you
justice.http://w0rli.home.att.net/youare.swfhttp://
www.mazepath.com/uncleal/sunsne.jpghttp://www.you-moron.com/
http://www.apa.org/journals/psp/psp7761121.htmlhttp://
insti.physics.sunysb.edu/~siegel/quack.html<http://
www.firehead.org/~jessh/film/kubrick/Kubrick-Psych
o.html><http:/
===
JSH: Sign corrected short core error argumentNotice how I'll 
be
strongly emphasizing constant terms all the waydown.P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078wch has a constant
term that is 1078.Well P(x) can also be written out asP(x)=
7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 +
7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).Notice it *appears* that the constant terms
for the three factors areall 7, wch can't be right, as the
constant term of P(x) is 1078, sosetting x=0, revealsP(0) =
(5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining
the a's with x=0 isa^3 - 3a^2, wch has roots, 0, 0 and 3, 
and
I've picked a_1 and a_2for 0, so that leaves a_3 with a 
value
of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices
matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3)
+ 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my
constant terms work out correctly.But P(x) has 49 as a factor
as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x +
1078has 49 as a factor, so I can divide by 49, and dividing
1078 by 49gives me 22, as the new constant term.Well that
means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is
===
Re: JSH: Sign corrected short core error argument> Notice how
I'll be strongly emphasizing constant terms all the way>
down.> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078> wch
has a constant term that is 1078.> Well P(x) can also be
written out as> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) -
3(-1 + 49 x )(5)(7^2) + 7^3> so I can factor to get> P(x) = (5
a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)> where the a's are roots of> 
a^3
+ 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice it
*appears* that the constant terms for the three factors are>
all 7, wch can't be right, as the constant term of P(x) is
1078, so> setting x=0, reveals> P(0) = (5(0) + 7)(5(0) +
7)(5(3) + 7) = 7(7)(22)> as the cubic defining the 
a's with x=0
is> a^3 - 3a^2, wch has roots, 0, 0 and 3, and I've picked 
a_1
and a_2> for 0, so that leaves a_3 with a value of 3 when
x=0.> So let a_3 = b_3 + 3, where I keep indices matched. Then
I have> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7)> P(x) =
(5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)> and now my constant terms
work out correctly.> But P(x) has 49 as a factor as every term
in > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078> has 49
as a factor, so I can divide by 49, and dividing 1078 by 49>
gives me 22, as the new constant term.> Well that means that>
P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)> is the only
way that the constant terms keep matcng. Really? You tnk so?
Let a1 = 1, a2 = 1562, and b3 = 25. Then (5*a1 + 7)*(5*a2 +
7)*(5*b3 + 22) = 12 * 7817 * 147 = 13789188 Now it so happens
that P(1) = 14706125 - 900375 - 17640 + 1078 = 13789188. Note
that a1 = 1 is not divisible by 7, and a2 = 1562 is also not
===
core error argumentnow, we're getting some where. the 
trouble
is,mister Harris apparently cannot read -- just type and
quote.[NB: I didn't check or read it, myself;I just 
wan't to
see what Math Hero says, before I retirefrom the 10-year
programme to prove ... what ever.] > Let a1 = 1, a2 = 1562,
and b3 = 25. Then (5*a1 + 7)*(5*a2 + 7)*(5*b3 + 22) = 12 *
7817 * 147 = 13789188 Now it so happens that P(1) = 14706125 -
900375 - 17640 + 1078 = 13789188.> Note that a1 = 1 is not
divisible by 7, and a2 = 1562 is > also not divisible by
===
corrected short core error argument>Notice how I'll be
strongly emphasizing constant terms all the way>down.P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078wch has a constant
term that is 1078.Well P(x) can also be written out asP(x)=
7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 +
7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).Notice it *appears* that the constant terms
for the three factors are>all 7, wch can't be right, as the
constant term of P(x) is 1078, so>setting x=0, revealsP(0) =
(5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining
the a's with x=0 isa^3 - 3a^2, wch has roots, 0, 0 and 3, 
and
I've picked a_1 and a_2>for 0, so that leaves a_3 with a 
value
of 3 when x=0.So let a_3 = b_3 + 3, where I keep indices
matched. Then I haveP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3)
+ 7)P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22)and now my
constant terms work out correctly.But P(x) has 49 as a factor
as every term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x +
1078has 49 as a factor, so I can divide by 49, and dividing
1078 by 49>gives me 22, as the new constant term.Well that
means thatP(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)is
the only way that the constant terms keep matcng.>> Really?
You tnk so? Let a1 = 1, a2 = 1562, and b3 = 25. ThenYou 
can't
just stick in any value you want Nora Baron as the a's 
arethe
roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)as
told in the argument given.So are you claiming that you've
given roots for some particular x?If so, give the x, wch will
give the cubic that has roots 1, 1562and 28, as a_3 = b_3 +
===
argumentIn sci.math,
<jstevh@msn.com><3c65f87.0310290406.45d409df@
constant terms all the way> down.> P(x) = 14706125 x^3 -
900375 x^2 - 17640 x + 1078> wch has a constant term that is
1078.> Well P(x) can also be written out as> P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> so I
can factor to get> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)>
where the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).[1] Why are the x's in the above, wch is
presumably a cubic with unknown Ôa'?[2] A 
polynomial doesn't
have roots, an equation has roots, if the equation has an
unknown variable (in ts case Ôa'). One can of 
course set a^3 +
3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0 and work on
that basis.[3] You're not exactly clear on how 
you're deriving
ts cubic. If one defines Q(x) = P(7/(5x)) and then generates
Q(a), one might get somewhere. As it is, how one got ts cubic
is clear as mud.> Notice it *appears* that the constant terms
for the three factors are> all 7, wch can't be right, as the
constant term of P(x) is 1078, so> setting x=0, reveals> P(0)
= (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic
defining the a's with x=0 is> a^3 - 3a^2, wch 
has roots, 0, 0
and 3, and I've picked a_1 and a_2> for 0, so that leaves 
a_3
with a value of 3 when x=0.> So let a_3 = b_3 + 3, where I
keep indices matched. Then I have> P(x) = (5 a_1 + 7)(5 a_2 +
7)(5 b_3 + 5(3) + 7)> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 +
22)> and now my constant terms work out correctly.One would
hope so. :-)> But P(x) has 49 as a factor as every term in >
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078> has 49 as a
factor, so I can divide by 49, and dividing 1078 by 49> gives
me 22, as the new constant term.> Well that means that>
P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)> is the only
way that the constant terms keep matcng.If one is restricted
to Q, perhaps. Since the originaldiscussion was in the ring of
algebraic integers a numberof tngs get a little strange.For
example, if we takeK(x) = x^3 - 49wch can be factorizedK(x) =
(x - r_1) (x - r_2) (x - r_3)it's clear that r_1, r_2, and 
r_3
have two factors of 7 between them,so we could do sometng like
ts:K(x) = (x - 7 s_1)(x - 7 s_2)(x - s_3)The trouble is that
ts wouldn't guarantee anytng.At best, one would get one
algebraic integer (probably7^(2/3), but it doesn't matter) 
and
two algebraic numbersthat are provably not integers (as they
are two roots ofthe equation K2(y) = 7y^3 - 1 = 0).> --
===
sha1:i2Vj0Hu2ZZmEx8NAV/8oR0fB8=> [2] A polynomial doesn't 
have
roots, an equation has roots,Surely, a polynomial has roots,
and an equation has solutions, and theroots of a polynomial
P(x) are the solutions of the equation P(x) = 0?-- Jesse
HughesSurround sound is going to be increasingly important in
futureoffices. -- Microsoft marketing manager displays s keen
===
argumentIn sci.math, Jesse F.
Hughes<jesse@pwumbda.org><87smlc7wpr.fsf@pwumbda.org>:>> [2] A
polynomial doesn't have roots, an equation has roots,> 
Surely,
a polynomial has roots, and an equation has solutions, and
the> roots of a polynomial P(x) are the solutions of the
equation P(x) = 0?An interesting point, but it bothers me. The
roots of the equationx^2 = 1are different from the roots of the
equationx^2 = 4although the left side of both equations is the
===
core error argument>> [2] A polynomial doesn't have roots, 
an
equation has roots,> Surely, a polynomial has roots, and an
equation has solutions, and the> roots of a polynomial P(x)
are the solutions of the equation P(x) = 0?He's right. :-) A
polynomial has zeroes and an equation has roots, and thezeroes
of a polynomial P are the roots of the equation P(x) = 0.--
===
argumenttaek a rest. do tngs other than it's just math.
although that might be the True Viewof an aristotelean
programme of mathematics,such as at Vanderbilt U., just math
is silly.>[2] A polynomial doesn't have roots, an equation 
has
roots,> Surely, a polynomial has roots, and an equation has
solutions, and the> roots of a polynomial P(x) are the
===
corrected short core error argument Adjunct Assistant
Professor at the University of Montana.>Notice how I'll be
strongly emphasizing constant terms all the way>down.>P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078>wch has a constant
term that is 1078.>Well P(x) can also be written out as>P(x)=
7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) +
7^3>so I can factor to get>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3
+ 7)>where the a's are roots of>a^3 + 3(-1 + 49x)a^2 - 
49(2401
x^3 - 147 x^2 + 3x).>Notice it *appears* that the constant
terms for the three factors are>all 7, wch can't be right, 
as
the constant term of P(x) is 1078, so>setting x=0,
reveals>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>as
the cubic defining the a's with x=0 is>a^3 - 
3a^2, wch has
roots, 0, 0 and 3, and I've picked a_1 and a_2>for 0, so 
that
leaves a_3 with a value of 3 when x=0.>So let a_3 = b_3 + 3,
where I keep indices matched. Then I have>P(x) = (5 a_1 + 7)(5
a_2 + 7)(5 b_3 + 5(3) + 7)>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3
+ 22)>and now my constant terms work out correctly.>But P(x)
has 49 as a factor as every term in >P(x) = 14706125 x^3 -
900375 x^2 - 17640 x + 1078>has 49 as a factor, so I can
divide by 49, and dividing 1078 by 49>gives me 22, as the new
constant term.>Well that means that>P(x)/49 = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 b_3 + 22)>is the only way that the constant terms
keep matcng.Here's a different way in wch they keep matcng:
nonconstant term constant termP(x)/49 = [ (5a_1(x)/w_1(x) +
(7/w_1(x)) -1) + 1] [ (5a_2(x)/w_2(x) + (7/w_2(x)) -1) + 1] [
(5b_3(x)+22)/w_3(x) ) -22 ) + 22]where w_1(x) is any algebraic
integer common factor of a_1(x) and 7;w_2(x) is any algebraic
integer common factor of a_2(x) and 7; andw_3(x) is any
algebraic integer common factor of a_3(x) and 7 (or
of5b_3(x)+22 and 7), subject to the condition
thatw_1(x)*w_2(x)*w_3(x) = 49.We have produced explicit
situations in wch ts holds, and wherew_1(x) is not 7, w_2(x)
is not 7, and w_3(x) is not 1; in fact, wherenone of them are
and a long purse, wch does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great many people are staggered to ts extent, that they
imagine there must be the indefinite sometng in the mysterious
all ts. They are brought to the point of suspicion that the
mathematicians ought not to treat all ts with such undisguised
contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
===
guys,I've got a question to discuss with you.Suppose xxx...x
is a string of fixed length, where x takes
0,1,...,9randomly.Consider the
sum0.xxx...x0.0xxx...x0.00xxx...x...0.0..0xxx...x
+...------a.bcdefh....It's quite easy to show that the sum 
IS
a convergent using Cauchy'scriterion. But how to interpret 
it
===
intuitively? The carriesembarrassed me.Anyone read me?Subject:
discuss with you.> Suppose xxx...x is a string of fixed 
length,
where x takes 0,1,...,9> randomly.> Consider the sum>
0.xxx...x> 0.0xxx...x> 0.00xxx...x> ...> 0.0..0xxx...x +> ...>
------> a.bcdefh....> It's quite easy to show that the sum IS 
a
convergent using Cauchy's> criterion. But how to interpret 
it
intuitively? The carries> embarrassed me.> Anyone read
me?Suppose your string is x1 x2 x3 ... xn.Then your sum:0.x1
x2 x3 ... xn0.0 x1 x2 x3 ... xn0.0 0 x1 x2 x3 ... xn...0.0 0 0
... 0 x1 x2 x3 ... xn+...-----------can be rewritten as:0.x1 x2
x3 ... xn x1 x2 x3 ... xn x1 ...0.0 x1 x2 x3 ... xn x1 x2 x3
... xn x1 ......0.0 0 0 ... x1 x2 x3 ... xn x1 ...-----Ts is a
sum of finitely many rational numbers, wch is clearly a 
specific
===
got a question to discuss with you.> Suppose xxx...x is a
string of fixed length, where x takes 0,1,...,9> randomly.>
Consider the sum> 0.xxx...x> 0.0xxx...x> 0.00xxx...x> ...>
0.0..0xxx...x +> ...> ------> a.bcdefh....> It's quite easy 
to
show that the sum IS a convergent using Cauchy's> criterion.
But how to interpret it intuitively? The carries> embarrassed
me.> Anyone read me?If I understand your question, the sum
appears to be:0.xxx...x * 1.1111... (recurring), wch equals
x.xx...x / 9-- Using M2, Opera's revolutionary e-mail 
client:
===
explanation?> guys,> I've got a question to discuss with 
you.>
Suppose xxx...x is a string of fixed length, where x takes
0,1,...,9> randomly.> Consider the sum> 0.xxx...x> 0.0xxx...x>
0.00xxx...x> ...> 0.0..0xxx...x +> ...> ------> a.bcdefh....>
It's quite easy to show that the sum IS a convergent using
Cauchy's> criterion. But how to interpret it intuitively? 
The
carries> embarrassed me.> Anyone read me?I'm not sure I
exactly understand your problem but assuming you are tryingto
add0.xxxx...x+ 1/10 0.xxxx...x+ 1/10^2 0.xxxx...x+...where
each x in 0.xxxx...x is a some digit. Then if carries worry
you it canbe rewritten
as0.xxxx...x(1+1/10+1/10^2+1/10^3+...)=0.xxxx...x
(1-1/10)^-1=0.xxxx...x/0.9 (not quite as many to carries to
===
sure I exactly understand your problem but assuming you are
trying> to add> 0.xxxx...x+ 1/10 0.xxxx...x+ 1/10^2
0.xxxx...x+...> where each x in 0.xxxx...x is a some digit.
Then if carries worry you it can> be rewritten as>
0.xxxx...x(1+1/10+1/10^2+1/10^3+...)> =0.xxxx...x (1-1/10)^-1>
=0.xxxx...x/0.9 (not quite as many to carries to worry about),
I did not made it clear. Let me be more exact.The string
xxx...x stands for a random string, say, each time itappears,
it takes a random value. So, the sum actually
reads0.S(1)+0.0S(2)+0.00S(3)+...,where S(i) is a random string
like xxx...x.Anyway, even if all S(i)'s are identical, 
carries
still worried me:how do we explain that the far later terms
will eventually unable tochange the leading digits of the sum?
===
explanation?>> I'm not sure I exactly understand your 
problem
but assuming you are trying>> to add>> 0.xxxx...x+ 1/10
0.xxxx...x+ 1/10^2 0.xxxx...x+...>> where each x in 0.xxxx...x
is a some digit. Then if carries worry you it can>> be
rewritten as>> 0.xxxx...x(1+1/10+1/10^2+1/10^3+...)>>
=0.xxxx...x (1-1/10)^-1>> =0.xxxx...x/0.9 (not quite as many
to carries to worry about)>, I did not made it clear. Let me
be more exact.>The string xxx...x stands for a random string,
say, each time it>appears, it takes a random value. So, the
sum actually reads>0.S(1)+0.0S(2)+0.00S(3)+...,>where S(i) is
a random string like xxx...x.>Anyway, even if all S(i)'s are
identical, carries still worried me:>how do we explain that
the far later terms will eventually unable to>change the
leading digits of the sum? Observing from the
analytic>result.It may be that one cannot determine a
particulardigit except by knowing the numbers (not just
aslimits of finite decimals). For example, .9 + .09 + .009 +
... = 1.0,but none of the digits is determined by
merelyknowing a finite number of the digits of eachof the
summands. However, the real number,considered as the limit of
===
Re: Intuitive explanation?>> I'm not sure I exactly 
understand
your problem but assuming you are trying>> to add0.xxxx...x+
1/10 0.xxxx...x+ 1/10^2 0.xxxx...x+...where each x in
0.xxxx...x is a some digit. Then if carries worry you it can>>
be rewritten as0.xxxx...x(1+1/10+1/10^2+1/10^3+...)>>
=0.xxxx...x (1-1/10)^-1>> =0.xxxx...x/0.9 (not quite as many
to carries to worry about)>, I did not made it clear. Let me
be more exact.>The string xxx...x stands for a random string,
say, each time it>appears, it takes a random value. So, the
sum actually reads>0.S(1)+0.0S(2)+0.00S(3)+...,>where S(i) is
a random string like xxx...x.>Anyway, even if all S(i)'s are
identical, carries still worried me:>how do we explain that
the far later terms will eventually unable to>change the
leading digits of the sum?Because the finite string xxx..x is
bounded, by a string of all 9's.Thus, you can bound the 
value
===
explanation?
sha1:RnjZYYscxKP/QAIq7sMBtb5/mBE=> , I did not made it clear.
Let me be more exact.> The string xxx...x stands for a random
string, say, each time it> appears, it takes a random value.
So, the sum actually reads> 0.S(1)+0.0S(2)+0.00S(3)+...,>
where S(i) is a random string like xxx...x.Note that, whatever
xxx...x is for a particular i, it's less
than1000...0.Therefore, your sum is strictly less than: 1.0 +
0.1 + 0.01 + ... = 1.111...Whatever the carries do, they 
can't
force the partial sums to t orexceed ts amount.> Anyway, even
if all S(i)'s are identical, carries still worried me:> how 
do
we explain that the far later terms will eventually unable to>
change the leading digits of the sum?Well, you can't, in the
sense I tnk you mean. Suppose the randomnumber generator
malfunctions and start generating the string 9 overand over.
The partial sum then looks like 0.99999...9. At any
point,someone could smack the side of the random number
generator and get itto generate sometng more interesting, like
94 followed by 6,say. At ts point, all the leading digits
change (since the newpartial sum is 1). That is, there's no
fixed step that you canspecify beforehand after wch the 
leading
digits won't change.On the other hand, if you ever observe a
digit smaller than 9 two ormore steps to the left of where the
leading digit of your next stringwill be added, all digits to
the left of the non-9 are safe fromcarries. That's because, 
as
above, the sum of the remaining termsmust be strictly less
than: 0.0...0111111...where the leading 1 is to the right of
the non-9 digit. No matterwhat the remaining terms are, at
most all their carries can increasethe non-9 digit by 1---and
ts carry can happen only once. Since anon-9 digit increased by
1 won't carry, the digits to its left aresafe.-- Kevin
===
TheoryDavenport can be had for less than Amazon's price.See
http://www.bookfinder.com and several used and new prices
ofvarious dealers are presented. Ames> The first half of
Ireland-Rosen was written for Brown's undergraduate> math
major number theory course, and is often used for that
purpose.> It presupposes a basic undergrad algebra course
(groups, rings, a> little about fields, but not Galois 
theory).
The latter half was added> later, and is suitable for a
beginning graduate course.> Another very enjoyable, short,
elementary introduction to number> theory is The gher
Arithmetic by Davenport. Also has the advantage> that as a
paperback, it's relatively inexpensive. Well, looks like 
$30>
on Amazon, so maybe not all that cheap. Anyway, when I took
number> theory (from Rosen, using Ireland-Rosen), he also had
us buy> Davenport's book to get another look at the 
subject.>
JHS> > <snip> >An introduction to Algebraic Number Theory,
though a bit more advanced>than Apostol's intro to analytic
number theory, is Kenneth Ireland and>Michael Rosen's _A
Classical Introduction to Modern Number Theory_,>2nd Edition
(Springer Verlag Graduate Texts in Mathematics); I've 
seen>it
used for an undergraduate course once (at
===
about a problem wch is very hard to me. The problem is
asbelow: Assume I have n different primes now. How many
results can I get if Iexecute m multiplication?Who can tell me
how to solve ts problem or where to find some usefulmethods? 
in
===
tnking about a problem wch is very hard to me. The problem is
as> below:> Assume I have n different primes now. How many
results can I get if I> execute m multiplication?> Who can
tell me how to solve ts problem or where to find some useful>
methods?Very difficult. Even the case that n = 1 is quite
difficult; it is basically the question: How many different
powers of a number x can be calculated using m
multiplications? You might check Donald Knuth, The Art of
Computer Programming, Vol. 2 (Seminumerical Algorithms)for the
===
results can I get?> I'm tnking about a problem wch is very
hard to me. The problem is as> below:> Assume I have n
different primes now. How many results can I get if I> execute
m multiplication?> Who can tell me how to solve ts problem or
where to find some useful> methods?> in advance!> Z GanThe
resulting set of value depends on how you carry out your
mmultiplications. Do you mean the set of values acevable witn
msteps or exactly m steps?I'll assume the later in wch 
number
of results can be calculatedroughly as
n^2*(n^2+1)^2*((n^2+1)^2+1)^2*... (m terms). n is number
ofdistinct prime you initially have. Is there any repreated
===
about a problem wch is very hard to me. The problem is as>
below:> Assume I have n different primes now. How many results
can I get if I> execute m multiplication?> Who can tell me how
to solve ts problem or where to find some useful> methods?> in
advance!> Z GanHow about an example of what you mean?If I have
two primes {2,3} and am allowedtwo multiplications, what would
be the results?Do you mean only: 2*2*2 2*2*3 2*3*3 3*3*3or do
you mean to include sometng like ts: Using one multiplication
I get 2*3, and that result with another multiplication
(2*3)*(2*3), so 2*2*3*3 is still only two
===
mean the second problem: (2*3)*(2*3) is still 2
multiplications. for pointing out my inaccuracy!>>I'm tnking
about a problem wch is very hard to me. The problem is
as>>below:>> Assume I have n different primes now. How many
results can I get if I>>execute m multiplication?>>Who can
tell me how to solve ts problem or where to find some
useful>>methods?>> in advance!>>Z Gan> How about an example of
what you mean?> If I have two primes {2,3} and am allowed> two
multiplications, what would be the results?> Do you mean
only:> 2*2*2> 2*2*3> 2*3*3> 3*3*3> or do you mean to include
sometng like ts:> Using one multiplication I get 2*3, and that
result with> another multiplication (2*3)*(2*3), so 2*2*3*3> is
===
simple question:What's the inverse of (2n over n), that is,
===
over n)> Just a simple question:Simple? If it were simple,
then the answer would be (n over 2n). ;-)> What's the 
inverse
of (2n over n), that is, (2n)!/(n!*n!) ?Would you settle for
an approximation wch is good for large n? (An exactexpression
for the inverse is given later if n is a positive integer.)If
so, then I suggest that the inverse you desire is
approximately W(-1, -4*Log(2)/(Pi*x^2)) / (-4*Log(2))Note that
the above uses the nonprincipal real branch of the Lambert
Wfunction. More information on the Lambert W function can be
found
at<http://mathworld.wolfram.com/LambertW-Function.html>.How
good is ts approximation? Here are two examples:If x = (20
over 10), then the approximation gives 9.9906...If x = (200
over 100), then the approximation gives 99.99909...Since my
derivation of the approximation was based on
Stirling'sapproximation for n!, the approximation is, of
course, not good whenx is small. BTW, I've never seen ts
approximation before. Has anyoneelse? Or does anyone have a
better approximation?OTOH though, if you're only interested 
in
integers n >= 1,then I can give you the _precise_ inverse:
Ceiling( W(-1, -4*Log(2)/(Pi*x^2)) / (-4*Log(2)) )In other
words, Ceiling( W(-1, -4*Log(2)/(Pi*(2n over n)^2)) /
===
Re: The inverse of (2n over n)Below I give an improvement,
again giving the precise inverse, but nowvalid for integers n
>= 0 and not requiring the Lambert W function.>What's the
inverse of (2n over n), that is, (2n)!/(n!*n!) ?> Would you
settle for an approximation wch is good for large n? (An>
exact expression for the inverse is given later if n is a
positive> integer.) If so, then I suggest that the inverse you
desire is> approximately(*) W(-1, -4*Log(2)/(Pi*x^2)) /
(-4*Log(2))> Note that the above uses the nonprincipal real
branch of the Lambert W> function. More information on the
Lambert W function can be found at>
<http://mathworld.wolfram.com/LambertW-Function.html>.> How
good is ts approximation? Here are two examples:> If x = (20
over 10), then the approximation gives 9.9906...> If x = (200
over 100), then the approximation gives 99.99909...> Since my
derivation of the approximation was based on Stirling's>
approximation for n!, the approximation is, of course, not
good when> x is small. BTW, I've never seen ts approximation
before. Has anyone> else? Or does anyone have a better
approximation?> OTOH though, if you're only interested in
integers n >= 1,> then I can give you the _precise_
inverse:(**) Ceiling( W(-1, -4*Log(2)/(Pi*x^2)) / (-4*Log(2))
)> In other words,> Ceiling( W(-1, -4*Log(2)/(Pi*(2n over
n)^2)) / (-4*Log(2)) ) = n> for all integer n >= 1.The Lambert
W function can be approximated in various ways in terms
ofelementary functions. For our purposes, we will use only the
crudeapproximation of the nonprincipal real branch of the
Lambert W functiongiven by W(-1, y) ~= Log(-y) -
Log(-Log(-y)).Denoting ts approximation on the right side as
F(y), we couldapproximate the requested inverse by
F(-4*Log(2)/(Pi*x^2)) / (-4*Log(2))but ts yields an
approximation wch is substantially worse than (*)above.
_However_, if we are only interested in x = (2n over n)
forpositive integer n, ts is still good enough!
Specifically,(***) Ceiling( F(-4*Log(2)/(Pi*x^2)) / 
(-4*Log(2))
)is the precise inverse, and it avoids requiring the Lambert W
function._Furthermore_, as opposed to (**) above, wch fails
when n = 0, we findserendiptously that (***) also happens to
work when n = 0. In other words, Ceiling( F(-4*Log(2)/(Pi*(2n
over n)^2)) / (-4*Log(2)) ) = nfor all integer n >= 0.
===
wch F(y) was defined as Log(-y) - Log(-Log(-y)) ]>
_Furthermore_, ... we find> serendip tously that (***) also
happens to work when n = 0. In other words, ^ i> Ceiling(
F(-4*Log(2)/(Pi*(2n over n)^2)) / (-4*Log(2)) ) = n> for all
integer n >= 0.BTW, Henrik, would you please tell us why you
wanted the inverse, at leastif it's an interesting
===
sha1:nTCPDIQnrI/8CFxvdxZB5WL9Ejk=> [ snip, in wch F(y) was
defined as Log(-y) - Log(-Log(-y)) ]>> _Furthermore_, ... we
find>> serendip tously that (***) also happens to work when n 
=
0. In other words,> ^> i>> Ceiling( F(-4*Log(2)/(Pi*(2n over
n)^2)) / (-4*Log(2)) ) = n>> for all integer n >= 0.> BTW,
Henrik, would you please tell us why you wanted the inverse,
at least> if it's an interesting 
application?Sure,I'm looking
into Covering Arrays of strength 2, that is, arrays withthe
property that for any pair of columns, any combination of
valuesin the two columns appear at least once. As an example,
ts is aCovering Array of strength 2 with the values 0,
1:
000000000000001111110111000111101101100111011010101110110100One
 of the main challenges regarding Covering Arrays, is to
constructCAs with as few rows as possible. If the array only
contains twovalues (i.e. 0 and 1), an optimal construction is
known. The abovearray is optimal.As you can probably see, the
array is constructed by setting the firstrow to 0s, and then
use all the possible ways to select two out of theremaining
five rows to 0 (and set 1 on the other rows).I'm 
investigating
the array you get if no longer set the first row to0s, but 
just
construct columns by selecting half the rows as 0 and theother
half as 1. You no longer get a Covering Array, but it's not
faraway either. Ts construction gives you (2n over n) columns
if youhave 2n rows. What I would like to know is how many rows
you get fora given number of columns, instead of how many
columns you can makefor a given number of lines.Your answer
seems quite simple, probably I can get sometng out ofthat. --
===
explaining your application.> Your answer seems quite simple,
probably I can get sometng out of> that.As a final comment,
here's a form of the answer wch looks slightlynicer:Letting 
q
denote -Log(Log(16)/(m^2*Pi)),the desired inverse function can
be expressed precisely as Ceiling( (q + Log(q)) / Log(16) )In
other words, if m = the central binomial coefficient C(2n, n)
forsome integer n >= 0, then the above expression yields
===
n)http://www.giganews.com/info/dmca.html>!>Just a simple
question:>What's the inverse of (2n over n), that is,
(2n)!/(n!*n!) ?(n!*n!)/(2n)!You're right, that _was_
===
sha1:Fe6yhrjf1zpCb/1r6FUOXvEjEGY=>>!>>Just a simple
question:>>What's the inverse of (2n over n), that is,
(2n)!/(n!*n!) ?> (n!*n!)/(2n)!> You're right, that _was_
simple.Er, no, that was not what I meant...I'm looking for 
the
inverse function of f(n) = (2n over n), the sameway as x=ln y
===
inverse of (2n over n)!Just a simple question:What's the
inverse of (2n over n), that is, (2n)!/(n!*n!)
?>(n!*n!)/(2n)!>You're right, that _was_ simple.> Er, no, 
that
was not what I meant...> I'm looking for the inverse 
function
of f(n) = (2n over n), the same> way as x=ln y is the inverse
of y=e^x etc.> --> HenrikIf you are just looking for small
values:http://www.research.att.com/projects/OEIS?Anum=A000984(
Central binomial coefficients: C(2n,n) =
(2n)!/(n!)^2.)Otherwise use Cantrell's approximation.Hugo
===
simple question:What's the inverse of (2n over n), that is,
(2n)!/(n!*n!) ?>(n!*n!)/(2n)!>You're right, that _was_
simple.> Er, no, that was not what I meant...> I'm looking 
for
the inverse function of f(n) = (2n over n), the same> way as
x=ln y is the inverse of y=e^x etc.Small values of n:
Calculate a table, use a lookup procedure.Moderate to large
values of n: Use the Stirling approximationfor n!. log(n!) ~
n*log(n) - n + 0.5*log(2*pi*n)So log[(2n)!/(n!*n!)] = log(2n!)
- 2*log(n!) = 2n*log(2n) - 2n + 0.5*log(2*pi*2n) - 2n*log(n) +
2n - log(2*pi*n) = 2n*log(n) + 2n*log(2) - 2n +
0.5*log(2*pi*n) + 0.5*log(2) - 2n*log(n) + 2n - log(2*pi*n) =
(2n+0.5)*log(2) - 0.5*log(2*pi*n) (2n)!/(n!*n!) ~
sqrt(2)*2^(2n)/sqrt(2*pi*n)I don't see any good way to 
invert
that analytically, buta computer procedure could probably zero
in on the rightvalue of n pretty quickly using the log form.
===
sqrt(2)*2^(2n)/sqrt(2*pi*n)> I don't see any good way to
invert that analytically,See my previous response in ts thread
===
Ts is one of the methods that can be used to draw a circle
passing>> through 3 non colinear points, it even works in 3d,
with the added>> requirement that the 3 points must lie in the
same plane of course.>Aren't any three points always in one
plane?Yes you are right. What I really meant to say is that
the 2 middlenormals of the segments AB and BC have to be in
the plane defined bythe 3 points if they are to intersect at
the center of the circle. In2d there is no problem because
there is only one possible plane, butin 3d each of these lines
can be in an infinity of planes, so you haveto specify ts 
added
===
Simple geometry question> Given four points p1, p2, q1, q2 in
a plane and suppose >that there is a line segment from p1 to
p2 and from q1 to q2.>How will I know if the two line segments
intersect? Is there a >simple formula for ts given only the
coordinates of the fourpoints?By standard methods of
Coordinate Geometry,compute intersection pointof line (p1,p2)
and line (q1,q2), and let it be (x,y). [ x willinvolve x,y
coordinates of p1,p2,q1,q2]. Only one projection need
beconsidered. The intersection will be inside if
(xp1-x)*(x-xp2) > 0 .Else no intersection, as the intersection
point is outside theinterval. If zero, one line sits on the
===
Luckily for me the mathematical argument that proves that 
I've
been correct all along can be further simplified by the use of
*numbers* instead of variables, as wle algebra as an idea is
well-founded, so letter symbols should do, when mathematicians
are lying about the mathematics, you need to use what you can,
and pray. 1. First the problematic definition: Algebraic
integers are defined to be roots of monic polynomials with
integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17,
where monic refers to the leading coefficient.If the 
definition
is so problematic, what should the definition be?>>The problem
has to do with *assuming* that the ring doesn't 
have>>certain
interesting properties, so it's not like the 
definition
needs>>to be changed.Huh??? You've been going on for months
about how there's an>error with ts definition. 
Now in spite of
that it doesn't need to>be changed?> The 
definition doesn't
need to be changed.> I've already given the fix, 
wch is the
object ring.> Algebraic integers are included in the object
ring.What are the differences between your object ring and
algebraicintegers? Are there some elements in your object ring
wch is not inthe algebraic integers? If so, name one of those
elements.> Now then, here's the math argument emphasizing
*CONSTANT TERMS* wch> is key in showing there is a problem.>
Notice how I'll be strongly emphasizing constant terms all 
the
way> down.> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078I
still want to see P(x)=x^3-x+6.> wch has a constant term that
is 1078.> Well P(x) can also be written out as> P(x)= 7^2(2401
x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3> so I
can factor to get> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)>
where the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 
x^3
- 147 x^2 + 3x).> Notice it *appears* that the constant terms
for the three factors are> all 7, wch can't be right, as the
constant term of P(x) is 1078, so> setting x=0, reveals> P(0)
= (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)> as the cubic
defining the a's with x=0 is> a^3 - 3a^2, wch 
has roots, 0, 0
and 3, and I've picked a_1 and a_2> for 0, so that leaves 
a_3
with a value of 3 when x=0.> So let a_3 = b_3 + 3, where I
keep indices matched. Then I have> P(x) = (5 a_1 + 7)(5 a_2 +
7)(5 b_3 + 5(3) + 7)> P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 +
22)> and now my constant terms work out correctly.> But P(x)
has 49 as a factor as every term in > P(x) = 14706125 x^3 -
900375 x^2 + 17640 x + 1078> has 49 as a factor, so I can
divide by 49, and dividing 1078 by 49> gives me 22, as the new
constant term.> Well that means that> P(x)/49 = (5 a_1/7 + 1)(5
a_2/7 + 1)(5 b_3 + 22)> is the only way that the constant terms
===
provenwell, maybe finding such a non-algebraic integerin the
Object Ring is undecidable (in the Godel sense,not in the
Harris sense, wch I forget .-) > What are the differences
between your object ring and algebraic> integers? Are there
some elements in your object ring wch is not in> the algebraic
integers? If so, name one of those elements. > I still want to
===
see P(x)=x^3-x+6.Subject: Re: Simple principle, core error
that I've been>correct all along can be further 
simplified by
the use of *numbers*>instead of variables, as wle algebra as
an idea is well-founded, so>letter symbols should do, when
mathematicians are lying about the>mathematics, you need to
use what you can, and pray.1. First the problematic
definition:Algebraic integers are defined to be 
roots of monic
polynomials with>integer coefficient e.g. x^3 + 3x + 1 or 
x^234
- 34x^12 + 17, where>monic refers to the leading
coefficient.>>If the definition is so problematic, 
what should
the definition be?> The problem has to do with *assuming* that
the ring doesn't have> certain interesting properties, so 
it's
not like the definition needs> to be changed.Would you name 
one
of these mysterious properties please?> It's just that its
limitations must be noted.Then note them.> Here the problem is
that the emphasis on monic polynomials clips some> numbers.Such
as?> Not a big deal, if you know that's happening, but
mathematicians> didn't realize it for over a hundred years, 
so
now it's a big deal.Except you still haven't 
shown any of ts.--
===
did an algorithm in Applied maths known asNewton's method. 
Ts
is an iterative method for finding roots ofespecially
non-linear equations. Is there a similar iterative methodfor
finding the polynomial, given the roots of the polynomial?
(Irecall that we did some algorithm where we computed the
function_xwch had a_0,a_1,..,a_x as roots and then in the next
iteration wecomputed the function_{x+1} such that a_{x+1} would
also be a root aswell as a_0,..a_x, however I can't remember
===
Inverse of Newton's method?>Is there a similar iterative
method for finding the polynomial, given the >roots of the
polynomial? I'm not sure why you'd need an 
iterative method
for ts.If x1,x2,x3,...,xn are the roots of the polynomial
f(x), then f(x) maybe written asf(x) =
===
method?>f(x) = (x-x1)(x-x2)(x-x3)...(x-xn)I should, of course,
have included an arbitrary multiple C of ts:f(x) =
C(x-x1)(x-x2)(x-x3)...(x-xn)since there are a continuum of
===
of Newton's method? ,Ts is actually much easier. Just 
multiply
out (x-root1)(x-root2)(x-root3)....(x-rootn)Since the roots by
themeselves are an incomplete specification, youget to also
multiply the resulting poly by any non zero number.Now there
is Newton's method for evaluating a poly given its roots.Ts 
is
akin to LaGrange's interpolation. And you may be recalling
theNewton form.IHTH,Clay> In our second year, we did an
algorithm in Applied maths known as> Newton's method. Ts is 
an
iterative method for finding roots of> especially non-linear
equations. Is there a similar iterative method> for finding 
the
polynomial, given the roots of the polynomial? (I> recall that
we did some algorithm where we computed the function_x> wch
had a_0,a_1,..,a_x as roots and then in the next iteration we>
computed the function_{x+1} such that a_{x+1} would also be a
root as> well as a_0,..a_x, however I can't remember how it
===
Inverse of Newton's method?> In our second year, we did an
algorithm in Applied maths known as> Newton's method. Ts is 
an
iterative method for finding roots of> especially non-linear
equations. Is there a similar iterative method> for finding 
the
polynomial, given the roots of the polynomial? (I> recall that
we did some algorithm where we computed the function_x> wch
had a_0,a_1,..,a_x as roots and then in the next iteration we>
computed the function_{x+1} such that a_{x+1} would also be a
root as> well as a_0,..a_x, however I can't remember how it
was done or the> name of the algorithm).The function f_n(x) =
(x - a_1)...(x-a_n) is a polynomialwith a_1, ..., a_n as
roots. If you want it in the formf_n(x) = b_0 + b_1*x + ... +
b_n*x^n, it's almost trivialto write an algorithm to 
multiply
any given polynomialby a term (x - a).In fact, here's the
algorithm:Define a polynomial as a variable-length list of
coefficientsin the order [b_0, b_1, ..., b_n] wch are the
coefficientsof x^0, x^1, ..., x^n.Suppose you want the
polynomial that results from multiplying[b_0, b_1, ..., b_n]
(in algebraic notation, the expressionnotation, a_0 +
a_1*x).The result is a new polynomial with n+2 coefficients
[c_0, c_1, ..., c_(n+1)] given by c_0 = a_0*b_0 c_1 = a_1*b_0
+ a_0*b_1 c_2 = a_1*b_1 + a_0*b_2 ... c_n = a_1*b_(n-1) +
a_0*b_n c_(n+1) = a_1*b_nNow the answer to your question:
f_n(x) = (x-a_1)(x-a_2)...(x-a_n) is a polynomial withroots
a_1, a_2, ..., a_n. f_n(x)*(x-a_(n+1)) has all those roots,
===
Newton's method?> In our second year, we did an algorithm in
Applied maths known as> Newton's method. Ts is an iterative
method for finding roots of> especially non-linear equations.
Is there a similar iterative method> for finding the
polynomial, given the roots of the polynomial? (I> recall that
we did some algorithm where we computed the function_x> wch had
a_0,a_1,..,a_x as roots and then in the next iteration we>
computed the function_{x+1} such that a_{x+1} would also be a
root as> well as a_0,..a_x, however I can't remember how it
was done or the> name of the algorithm).Why iterate? If (R) is
a root of a polynomial in x, then (x - R) is afactor of that
polynomial. Just multiply together all the factors soobtained.
One factor -- the one that sets the scale -- isn't given 
bythe
set of roots. It needs to be determined some other way.--
===
year, we did an algorithm in Applied maths known as> 
Newton's
method. Ts is an iterative method for finding roots of>
especially non-linear equations. Is there a similar iterative
method> for finding the polynomial, given the roots of the
polynomial? (I> recall that we did some algorithm where we
computed the function_x> wch had a_0,a_1,..,a_x as roots and
then in the next iteration we> computed the function_{x+1}
such that a_{x+1} would also be a root as> well as a_0,..a_x,
however I can't remember how it was done or the> name of the
algorithm).If you know the roots of a polynomial, the
polynomial is explicitlygiven by:P(x) = k (x - root1) (x -
root2) (x - root3) ... (x - rootn)To determine the constant k,
you need to know the value of thepolynomial somewhere other
than at a root. Otherwise the constantpolynomial 0 trivially
solves all such systems.Ts generalizes to the notion of an
interpolating polynomial.Given a function f(x) and n points,
x1, x2 ... xn, there is aunique polynomial of degree n-1 that
that matches the valueof f(x) at those points.Finding for the
coefficients of the polynomial should be a simplematter of
solving a system of n linear equations in n unknowns.As long
as the x values are all unique, ts system of equationsshould
always have a unique solution.If the x values are not all
unique, you can further generalizeby requiring that the
polynomial's first derivitive match 
f'(x)at all doubled x's,
that the second derivitive match f''(x)at all 
tripled x's and
so on. At the extreme of n duplicatedpoints you get the n term
Taylor expansion.Again, you should obtain a solvable system of
n linear equationsin n unknowns.My single published
contribution to mathematical lore is the affirmativeanswer to
the question of whether, given a continuous (andcontinuously
differentiable to all degrees as I recall) functionthat is
strictly positive over [most of] an open interval, there isan
interpolating polynomial of any chosen degree that is
strictlypositive over that same interval.The proof technique I
employed was iterative (or inductive if youprefer). Given a
suitable interpolating polynomial of degree n,you adjust that
polynomial until at least one more intersection pointis
obtained, thus producing an interpolating polynomial of degree
n+1.Interpolating polynomials tend not to be good numerical
approximations.They are usually very sensitive to small
measurement errors and oscillatewildly above and below the
function that they are supposed to approximate.If you're
trying to fit a curve to a function, you are typically
betterserved by selecting from a family of functions with a
much smallerparameter set and choosing those parameters using
===
Newton's method?> My single published contribution to
mathematical lore is the affirmative> answer to the question 
of
whether, given a continuous (and> continuously differentiable
to all degrees as I recall) function> that is strictly
positive over [most of] an open interval, there is> an
interpolating polynomial of any chosen degree that is
strictly> positive over that same interval.> The proof
technique I employed was iterative (or inductive if you>
prefer). Given a suitable interpolating polynomial of degree
n,> you adjust that polynomial until at least one more
intersection point> is obtained, thus producing an
interpolating polynomial of degree n+1.MR0613872 (82e:41005),
;Interpolation by nonnegative polynomials.J. Approx. Theory 30
(1980), no. 3, 160--168.41A05C. K. Chui posed the following
problem: Let $f$ be a continuous and real-valued function on
$[0,1]$; if $f$ is nonnegative, is it possible to choose $n+1$
points on the graph of $f$ so that the interpolating polynomial
$p_n$ is also nonnegative? The main result of the paper under
review reads as follows: Let $f$ be a continuous and
nonnegative real function on the closed interval $[0,1]$. For
each $n=0,1,cdots$, there exist $n+1$ points $0leq
x_0<x_1<cdots<x_nleq 1$ and a polynomial $p_n$ of degree $leq
n$ that is also nonnegative on $[0,1]$ such that
$p_n(x_k)=f(x_k)$ for $k=0,1,cdots,n$.Reviewed by E. Neuman--
===
roots of a polynomial, the polynomial is explicitly> given
by:> P(x) = k (x - root1) (x - root2) (x - root3) ... (x -
rootn)An interesting case is polynomials of infinite degree.
For example,sin(pi*x) has roots at all the integers, so:
sin(pi*x) = pi * x * (1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3)
...(Ts statement is true, although I haven't proven
it.)Anyway, you can pair the positive roots wch their
corresponding negativeroots: sin(pi*x) = pi * x *
(1-x^2)(1-x^2/4)(1-x^2/9) ...Now compare ts with the Taylor
series for sin(pi*x): sin(pi*x) = pi*x - (pi*x)^3/3! +
(pi*x)^5/5! - ...The coefficients of the two series have to be
equal. The x coefficientof each series is pi. Tngs get
interesting with the gher coefficients.The 
coefficient of x^3 in
the first series is the sum -( 1 + 1/4 + 1/9 + ... + 1/k^2 + 
...
)The coefficient of x^3 in the second series is -pi^3/6. So
theequivalence of the two series implies that the sum of the
reciprocalsof the squares of the integers is pi^2/6. It gets
messier, but youcan find the sums of the reciprocals of all 
the
even powers by evaluatingsuccessively gher terms in these
===
know the roots of a polynomial, the polynomial is explicitly>>
given by: >> P(x) = k (x - root1) (x - root2) (x - root3) ...
(x - rootn)>An interesting case is polynomials of infinite
degree.There is no such tng. By definition a polynomial has
finite degree.There are infinite series, and 
infinite products,
but they're notthe same tng.> For example,>sin(pi*x) has 
roots
at all the integers, so:> sin(pi*x) = pi * x *
(1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3) ...>(Ts statement is
true, although I haven't proven it.)Ts formula is true, but
the so is not. How do you know that ts product isn't, e.g.,
sin(pi x) exp(x^2)? In general, for any analytic functions f
and g, f(x) and f(x) exp(g(x)) have the same roots.So the
===
Newton's method? If you know the roots of a polynomial, the
polynomial is explicitly given by:> P(x) = k (x - root1) (x -
root2) (x - root3) ... (x - rootn)An interesting case is
polynomials of infinite degree.> There is no such tng. By
definition a polynomial has finite degree.> There 
are infinite
series, and infinite products, but they're not> 
the same tng.>
For example,sin(pi*x) has roots at all the integers, so:>
sin(pi*x) = pi * x * (1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3)
...(Ts statement is true, although I haven't proven it.)> Ts
formula is true, but the so is not. How do you know that ts >
product isn't, e.g., sin(pi x) exp(x^2)? In general, for any
analytic > functions f and g, f(x) and f(x) exp(g(x)) have the
same roots.> So the roots don't determine the function.You 
are
correct. Please forgive my inexact language, leading to
falsestatements. I just find it interesting that certain
functions can beso easily characterized by the product of
===
Inverse of Newton's method?> The coefficients of 
the two series
have to be equal. The x coefficient> of each series is pi. 
Tngs
get interesting with the gher coefficients.> The 
coefficient of
x^3 in the first series is the sum> -( 1 + 1/4 + 1/9 + ... +
===
Inverse of Newton's method?> In our second year, we did an
algorithm in Applied maths known as> Newton's method. Ts is 
an
iterative method for finding roots of> especially non-linear
equations. Is there a similar iterative method> for finding 
the
polynomial, given the roots of the polynomial? It's not
similar, but if you know that a_0, a_1, ... , a_n are
theroots, then one polynomial that has those roots isp(x) = (x
===
Generators of cyclic groupI tnk you can use a counting
argument. You have alreadyproved that quadratic residues mod p
cannot be generators.You also proved that -1 cannot be a
generator. How manyelements are left? And how many generators
are their mod pwhere p = 2q+ 1 with q prime (Euler p function
===
groupGzQNbkZOYe8w1hsVCM2zo62QiEgJ1Lfw4+ddKuvSWLiCoW1a0qjY4Z> I
tnk you can use a counting argument. You have already> proved
that quadratic residues mod p cannot be generators.> You also
proved that -1 cannot be a generator. How many> elements are
left? And how many generators are their mod p> where p = 2q+ 1
with q prime (Euler p function will help> out here...).See the
===
group>I tnk you can use a counting argument. You have
already>proved that quadratic residues mod p cannot be
generators.>You also proved that -1 cannot be a generator. How
many>elements are left? And how many generators are their mod
p>where p = 2q+ 1 with q prime (Euler p function will help>out
here...).> See the post of Gerry Myerson.I missed that post.
===
Generators of cyclic group> ,> Given that p = 2q + 1 where p
and q are odd prime. Prove that the> generators of the group
Z*p are QNRp - { -1 }.> Yes, ts is an assignment question.> 1)
I've been able to prove that QRp cannot be generators.> 2) 
I've
also been able to prove that -1 (i.e. 2q) cannot be a>
generator either.> 3) However, I'm unable to prove that the
rest of the group members are> all generators.As you know that
Z*p is cyclic of order 2q, you can surely listthe orders of all
the possible non-trivial subgroups. For cyclicgroups you should
have further information about the number ofsubgroups of each
possible size. Finally, an element is a generator,iff it is
===
Generators of cyclic group> Given that p = 2q + 1 where p and
q are odd prime. Prove that the> generators of the group Z*p
are QNRp - { -1 }.> Yes, ts is an assignment question.> 1)
I've been able to prove that QRp cannot be generators.> 2)
I've also been able to prove that -1 (i.e. 2q) cannot be a>
generator either.> 3) However, I'm unable to prove that the
rest of the group members are> all generators.> For (1):> If g
is a generator, then for all b, there exist an i such that b =>
g^i.> However, we know that g is a quadratic residue, so g=a^2.
Hence, b => a^2^i = a^i^2. Hence, b must be a quadratic residue
as well! So, g> only generates other quadratic residue (not the
whole group).> For (2): Ts is rather trivial.> (3) I'm not
getting anywhere for that one. Any idea? (I would prefer a>
nt, not a complete solution). The order of an element (defined
as the lowest nontrivial number of copies of the element that
multiplied together produce the identity) of a group has to
divide the order of the group (defined as the number of
===
Generators of cyclic
groupbcIMzkZXQekAoXMylkA67CLJFU1OdDpdBo4393yua4B7jMe2jV4oQc>
The order of an element (defined as the lowest nontrivial
number of> copies of the element that multiplied together
produce the identity) of> a group has to divide the order of
the group (defined as the number of> elements in the group). 
Ts
is only a mild nt.I suppose one has to follow the way indicated
byGerry Myerson. If there is another solution of theproblem I
===
Re: Generators of cyclic groupIf p = 2q+1, and q is a QNR,
then what do Euler's Criterion & Lagrangetheorem tell
you?GREG> ,> Given that p = 2q + 1 where p and q are odd
prime. Prove that the> generators of the group Z*p are QNRp -
{ -1 }.> Yes, ts is an assignment question.> 1) I've been 
able
to prove that QRp cannot be generators.> 2) I've also been 
able
to prove that -1 (i.e. 2q) cannot be a> generator either.> 3)
However, I'm unable to prove that the rest of the group
members are> all generators.> For (1):> If g is a generator,
then for all b, there exist an i such that b => g^i.> However,
we know that g is a quadratic residue, so g=a^2. Hence, b =>
a^2^i = a^i^2. Hence, b must be a quadratic residue as well!
So, g> only generates other quadratic residue (not the whole
group).> For (2): Ts is rather trivial.> (3) I'm not getting
anywhere for that one. Any idea? (I would prefer a> nt, not a
===
groupGreg Doyle a .8ecrit :> If p = 2q+1, and q is a QNR, then
what do Euler's Criterion & Lagrange ^ q?> theorem tell 
you?mm