mm-62
===
I am having a terrible time understand a mathematical
derivation in this computer graphics
paper:<http://www.vis.uni-stuttgart.de/~roettger/data/Papers/
TERRAIN.PDF>I know this is asking a lot, but if someone could
go to page 4, and see ifthey understand how the constant K is
inferred, I would very much appreciate it. I have been trying
to gure this out with no success and I have postedon several
graphics forums, but no one seems to be able to gure it
out.Here are some specic questions regarding the
paper:confused for the number 1):Thus the following
condition must hold for the decision variable f1 of
anadjacent block in order to limit the level differences:(4)
f1 < f2 <=> ln1 / (d*d21) < ln2 / [(d/2) *
d22]-Ok, I think he is saying that if we are
going to subdivide f2, which is alevel lower than f1, then we
must also subdivide f1. But, couldnt therelationship also be
f1 <= f2, that will still ensure they are bothsubdivided,
right?Also, the original equations of f is(3) f = ln /
[d*C*max(c*d2, 1)How is it that he infers (4) from (3). I can
see that we can cancel C,but I do not follow how he
simplied the max. function.He continues:For a point of view
falling inside the rectangular region Equation (3) [f <1] is
always satised, since ln1 / d is always less than the
minimumresolution C.-Now when he says for a
point of view falling inside the region, does he meanthe
view point (camera position) lies inside the region, or does
he mean theregion is inside the viewing frustum? I also do
not follow how he says ln1/ d < C.He continues:Outside this
region the value of the fraction ln1 / (2*ln2) is bounded
by1/2 and the constant K:1/2 < ln1 / (2*ln2) < K (C > 2)K =
L1 / (2*L2) = C / [2*(C - 1)]Im totally
lost here. First of all, from the gure in the paper, I
cannotmake out what ln1 / (2*ln2) means geometrically.
Perhaps if I knew where heobtained that relationship I would
understand the rest; specically how itis bounded. Moreover, I
do not understand how he obtains the relationship K= L1 /
(2*L2) = C / [2*(C - 1)] from that.
===
A brief scan of Dave
Rusins archive and the group didnt seem to turnup any
solutions to these two problems that looked much like
mine:A2: Proceed by induction. If n=1, then the inequality is
clear (infact, its an equality...)To get from n-1 to n, apply
Holders inequality with p=n to the
twovectors((a_1a_2...a_{n-1})^{1/n},(b_1b_2...b_{n-1})^{1/n})
) and(a_n^{1/n},b_n^{1/n}).A6: Let b(n) be the number of 1s
in the base-2 expansion of n; letf(n)=(-1)^b(n).Let
A=f^{-1}(1), B=f^{-1}(-1). Then clearly A and B partition the
setof nonnegative integers. Also, f(2n)+f(2n+1)=0 for all n.
Thussum_{i=0}^n f(i)=0 if n is odd, and f(n) if n is
even.Suppose p+q=n, with p neq q. If p and q are both in A,
thenf(p)+f(q)=2; if p and q are both in B, then f(p)+f(q)=-2;
if one is ineach set, then f(p)+f(q)=0. It therefore sufces
to show thatsum_{p+q=n,p neq q} f(p)+f(q)=0 for all n. If
n is odd this sum isjust2sum_{i=0}^n f(i)=0; if n is even,
it is 2(sum_{i=0}^nf(i)-f(n/2))=2(f(n)-f(n/2)). By
construction, f(n)=f(n/2) for alleven n; thus this sum is
zero. So we are done.-Micah Smukler
===
Let R denote the real
line and let S denote the cartesian product ofR with the
three element set {-1,0,+1} and give S the order topology
forthe lexicographical order.I think it is very likely that
this space has a name. If you know thename of this space,
please let me know. I think it has probably beengood for
something, such as studying functions f:R->R with left and
righthand limits at every point. If you know where that might
have been done,and whether in fact that is the reason the
space was cooked up, Id liketo know that
also.Ignorantly,Allan
Adlerara@zurich.ai.mit.edu**
** Intelligence
Lab. My actions and comments do not reect ** in any way on
MIT. Moreover, I am nowhere near the Boston ** metropolitan
area. **
**
===
> Let R denote the real line and let S
denote the cartesian product of> R with the three element set
{-1,0,+1} and give S the order topology for> the
lexicographical order.I think it is very likely that this
space has a name. If you know the> name of this space, please
let me know. I think it has probably been> good for something,
such as studying functions f:R->R with left and right> hand
limits at every point. If you know where that might have been
done,> and whether in fact that is the reason the space was
cooked up, Id like> to know that also.> I have seen R x
{0,1} lexicographic. It is sometimes known as the twoarrows
space. Or the top and bottom of the lexicographic square.
Ithink this is the one you want if you require left and right
limits atevery point...although it doesnt provide an isolated
point in betweenwhere you can provide a value different from
those two limits.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
> Let R denote the
real line and let S denote the cartesian product of> R with
the three element set {-1,0,+1} and give S the order topology
for> the lexicographical order.> S = lex Rx{-1,0,1}which has
isolated pts (r,0) or S = lex {-1,0,1}xRwhich is disconnected
as {-1}xR, {0}xR, {1}xR> I think it is very likely that this
space has a name. If you know the> name of this space, please
let me know. I think it has probably been> good for something,
such as studying functions f:R->R with left and right> hand
limits at every point. If you know where that might have been
done,> and whether in fact that is the reason the space was
cooked up, Id like> to know that also.>The closest I come to
it is the dictonary order unit square.
===
Suppose you have two
dictionaries, D1 and D2. Suppose that D1 has a muchnarrower
selection of words than D2 but is generally regarded as being
morereliable. Suppose further that D1 is sufciently long that
it would beimpractical to go through D1 to see whether D2
handles at least the wordsin D1 correctly for the most part.
It would, however, be practical tosample the words in D1 and
compare the treatments those words receivein D1 and D2.How
can one design an experiment in which one samples the words
in D1,compares the results with the entries of D2, and
arrives at a measure ofhow accurately D2 denes the words
that occur in D1?It is very likely that some problem
equivalent to this is standard instatistics. Id be glad to
have a reference to where it is treated
indetail.Ignorantly,Allan
Adlerara@zurich.ai.mit.edu**
** Intelligence
Lab. My actions and comments do not reect ** in any way on
MIT. Moreover, I am nowhere near the Boston ** metropolitan
area. **
**
===
> Yes, . means multiplication. I
thought that that was a kind of universal> notation.The only
universal is that there are no universals.Jon Miller
===
In
the context of Innocents proof number obviously means
integer, Doug.Please see my reply to Thomas Bushnell for an
improved version of her proof.>prove if x is an irrational
number then sqroot of x is irrational??>Any rational number
has a nite number of digits> You mean, like 1/3 =
0.3333333333333333...?Doug
===
Arent you unnecessarily
confusing this proof rather than improving it?First, why
insist on using base 10?Second, fractions seem irrelevant to
the original question.Third, pi is already irrational.Here
are my improvements on Innocents proof: Any rational integer
plus one is also rational, so that may be taken as a base x
such that (x-1) squared is obviously also a rational integer
with twice as many digits (in base x) just as she claimed.
Thus, since Innocent has shown that any rational integer has
a rational square, with the obvious corollary for any
rational fraction, it follows that there are no rational
candidates for sqroots of irrationals.Q.E.D. Its interesting
that for all functions f where root is an integer greater than
one, p is prime and i is an irrational, then: f(root p)=i Of
course for any integer n not an integral power of an integer
> 1 (all primes and composites): f(root n)=iThis simple
truth has philosophical import in that if reality is
fundamentally digital then innity is rational whereas if
its continuous then innity is irrational - another way of
phrasing that old either/or duality vs. bothand.>prove if x
is an irrational number then sqroot of x is
irrational??>Any rational number has a nite number of
digits, so when>multiplied by itself it would have at most
twice a many while any>irrational number requires innite
digits.> What do you mean any rational number has a nite
number of digits?> For example, in base 10, the number 1/3
has innitely many digits.Or if you mean nitely many
different digits (so that 1/3 has one> digit), then pi has
only digits.Thomas
===
> Arent you unnecessarily confusing
this proof rather than improving it?My goal was to correct an
incorrect proof. You give a perfectly goodproof, but solving
peoples homework for them is not my game.Thomas
===
I
apologize, Thomas. Unnecessarily confusing was a poor
choice for describing your clarication of loosely dened
terms in Innocents proof. Its better to appreciate that
your correction game leads me to look deeper.Emit
===
>As a
matter of fact, the entire universe could be a Swiss cheese
of>giant expanding VDk voids, and we would be...oblivious.
For we could>not observe any of them until one hit us,
because of the fact they are>expanding at light speed.>
[snip - more crap]> This isnt a relevant subject for
either sci.math or rec.puzzles...>Whether or not the
hypothesis is preposterous, the question asked is>perfectly
serious. Given the distributions of radii, velocities
and>rates of expansion, he asks a perfectly valid
mathematical question.>So get off your high horse. Youre a
SELF-APPOINTED censor<indignation>How is it that I am a
censor exactly? Did I censor something or someone?Do you know
what the word censor means? I snipped some of LeRoys
crapout of my own post, but I didnt remove his post from the
newsgroup...Unfortunately, I dont have that power...And if
LeRoys question is perfectly valid then why didnt you
answer it, smart guy.</indignation>who has contributed
nothing to the newsgroup, so far as I can see,<sarcasm>Ouch!
It really hurts me when you say I have contributed nothing to
thisnewsgroup... man, you sure have me pegged right; Im a
high-horse ridingcensor who has contributed nothing to this
newsgroups... </sarcasm>as opposed to LeRoy, who is a
regular and frequently asks valid (if >usually less quixotic
questions.>--Ron Bruck><middle nger>have a nice
day,</middle nger>adam
===
> Depends on which
multiplication(s) you use. If o is the convolution>
operator (sort of multiplication between functions) and *
is the> normal multiplication operator, thenL(f(x) o g(x)) =
L(f(x)) * L(g(x))> F(f(x) o g(x)) = F(f(x)) * F(g(x))where
L(f(x)) is the Laplace transform and F(f(x)) is the Fourier>
transform of f(x).Im using bad notation here I guess, what
if the operator itself isconvolution?ie.f(x) o ( S1(x) +
S2(x) ) = ( f(x) o S1(x) ) + ( f(x) o S2(x) )says that
convolution is linear given proper boundary conditions, but
are there conditions under which this is true:f(x) o ( S1(x)
* S2(x) ) = ( f(x) o S1(x) ) * ( f(x) o S2(x) )Im coming
from a signal processing angle, so the functions S1 & S2are
signals, and f() is a lter kernel. Which i think means
thatnumbers....? ( Have I got the right end [any end] of the
stick this? )SX-Received: (from approve@localhost) by
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===
Perhaps theres
another name for this besides what Im calling it here.Def.
Let X be a set. Call *: X times X -> X an associative if
(a*b)*c = a*(b*c) for all a,b,c in X.Let Z be the whole
numbers. Then normal multiplicationand ab = a+b-ab (ab normal
multiplication) are associatives.Earlier, I had asked if the
set of all associatives ofa given set had a particular name
(still not sure): Let As(Z) be such a set on Z.Every x in N
(naturals) has a unique representation asx
=(x_1)(x_2)...(x_n) wherex_1, x_2, ...,x_n are prime and x_1
<= x_2 <= ... <= x_n (normal multiplication)If x in Z and x <
0, then x also has a unique primerepresentation asx
=-(x_1)(x_2)...(x_n) wherex_1, x_2, ...,x_n are prime and x_1
<= x_2 <= ... <= x_n I would like to show that * in As(Z)
where * is dened as:For x,y > 0 x*y = [(-1)^j]xy =
[(-1)^j](x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) =
[(-1)^j](z_1)(z_2)...(z_{n+m}) = [(-1)^j]zwhen
(x_1)(x_2)...(x_n) (y_1)(y_2)...(y_m) (z_1)(z_2)...(z_{n+m})
are the (unique) prime representations of x, y and z (=xy)
respectively and j is the minimum number of shifts
necessary to order (x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m)into
(z_1)(z_2)...(z_{n+m}).If x or y = 0, then dene x*y = 0.If x
< 0 or y < 0, then do the multiplication exactly as ifboth
were positive, but substitute j with (j+k) werek = 1 <-> (x <
0 and y > 0) or (x > 0 and y < 0)k = 2 <-> x,y <
0.Examples:(3*10)*14 = (3*((2)(5))*14 = (-(2)(3)(5))*14 =
(-(2)(3)(5))*((2)(7)) = -(2)(2)(3)(5)(7) = -420Reason: in the
2nd equation, 1 shift was necessary; (2) was shifted once to
the left. in the 4th equation, 2 shifts were necessary; (2)
was shifted twice to the left. Note: Whether the shift is to
the left or the right does not play a role, i.e. j is the
same number in either case. 3*(10*14) = 3*((2)(5)*(2)(7)) =
3*(-(2)(2)(5)(7) = -(2)(2)(3)(5)(7) = -420Reason: in the 2nd
equation, (2) was shifted once to the left. alternatively,
(5) was shifted once to the right. in the 3rd equation, (3)
was shifted twice to the right.C. Dement
===
> Def.> Let X be
a set. Call *: X times X -> X an> associative if (a*b)*c =
a*(b*c) for all a,b,c in X.>(X,*) is a semi-group. 2x2
matrices is an example ofa noncommunative monoid, ie
semi-group with identity.> Let Z be the whole numbers. Then
normal multiplication> and ab = a+b-ab (ab normal
multiplication) are associatives.>a#b = a+b - ab is know as
symmetric difference in set theory.Symmetric differences of a
boolean ring produce a boolean algebra.<snip>, Im allergic to
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===
>Today I landed
a pretty idea. Whether it ys or not is unknown as>yet. It
came to me whilst thinking of those cylindrical farm bins
used>to store grains. Suppose one were to build one not out
of steel but>instead from concrete block or brick. Expensive
I know but consider>it. And suppose the builder was given a
3/8 joint for mortar.>Pretty Idea: then given a standard
size brick would dictate the>circumference and diameter of
this building. So that given the brick>size would determine
the size of the building where no sawing of brick>is allowed.
Granted of course the building is not really a circle but>we
can consider the midpoint inside each brick once the building
is>completed sribes a circle.>What is pretty about this idea
is the fact that a rectangle determines>a circle. So that the
circle is quantized from a rectangle.>I have not thought of
the reverse where given a circle determines or>quantizes>a
rectangle or a square. It maybe even possible that a circle
just>does not quantize a square or rectangle and if that is
the case gives>deep mystery and deep implications as to why
this Cosmos is built>where rectangles can quantize circles
but not the reverse. And if that>is the case would go back to
the idea that the cosmos is one gigantic>plutonium atom which
is cosmically spherical and cylindrical and that>life sees
Euclidean geometry not as a reection of reality but as
a>result of the mind trying to make sense of the
world.>Euclidean Geometry in this sense does not exist in
Nature but only as>a gment of the imagination of a mind.
Well ,This is true .Do lines exist,dots and surfaces ?All
this lot is timeless and immaterial.Even third dimention
Eucledian/Platonic Forms bear the same qualities.After all
this is Pure Geometry.This is the absolute truth that
cannotbe violated.When matter embodies this backbone
structure of Geometry ,naturally deformations occur for its
accomodation to form the real world.I, UNDERSTAND THAT
EXCHANGE OF FORMSBETWEEN ALL KINDS OF LINEAR ONES
(SQUARES,RECTANGLESTRIANGLES)AND CIRCLE ,IF ACCEPTED IN ONE
DIRECTIONIT IS TRUE FOR THE
REVERSE.http://www.stefanides.gr/quadbig.htmPanagiotis
Stefanideshttp://www.stefanides.grEuclidean Geometry would
be>like those many Optical Illusions we see such as the
artist Escher.>I doubt that Euclidean Geometry is a illusion
for zero is a true>existing number and that Euclidean Geometry
is the zero curvature>geometry.>But, if rectangles and
squares can generate circles but the reverse is>not true
would have deep implications upon Euclidean
Geometry.>Archimedes Plutonium>whole entire Universe is just
one big atom where dots>of the electron-dot-cloud are
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===
Dear All,I am
trying to solve a system of nonlinear equations which is very
complicated. The resulting function is a stochastic variable,
which I have to approximate using the Monte Carlo methods.
Therefore evaluation of the function is very time consuming
(one evaluation is about one minute). My problem is to nd
an appropriate algorithm which solves the nonlinear system
of equations without approximation of the Jacobian/Hessian
matrix. Due to the complexity of the stochastic function it
does not make sense to use methods like Newton,
Newton-Raphson, etc. I was thinking about something like
bisections or regula falsi, eventually secant method. They
work ne in one dimensional space. But I do not know about
higher dimensions.PRzemyslawX-Received: (from
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===
>prove if x is an irrational number then
sqroot of x is irrational???Let y = exp(1) and x = exp(2).x
is irrational (For any rational a>0, exp(a) is irrational),
and sqrt(x) = exp(1) is irrational :)X-Received: (from
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===
>Perhaps theres another name for this
besides what >Im calling it here.>Def. >Let X be a set.
Call *: X times X -> X an >associative if (a*b)*c = a*(b*c)
for all a,b,c in X.>Let Z be the whole numbers. Then normal
multiplication>and ab = a+b-ab (ab normal multiplication) are
associatives.>Earlier, I had asked if the set of all
associatives of>a given set had a particular name (still not
sure): >Let As(Z) be such a set on Z.>Every x in N
(naturals) has a unique representation as>x
=(x_1)(x_2)...(x_n) where>x_1, x_2, ...,x_n are prime and
x_1 <= x_2 <= ... <= x_n >(normal multiplication)>If x in Z
and x < 0, then x also has a unique prime>representation as>x
=-(x_1)(x_2)...(x_n) where>x_1, x_2, ...,x_n are prime and
x_1 <= x_2 <= ... <= x_n >I would like to show that * in
As(Z) where * is dened as:>For x,y > 0>x*y = [(-1)^j]xy =
[(-1)^j](x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m)> =
[(-1)^j](z_1)(z_2)...(z_{n+m})> = [(-1)^j]z>when
(x_1)(x_2)...(x_n)> (y_1)(y_2)...(y_m)>
(z_1)(z_2)...(z_{n+m}) are the (unique) prime
representations>of x, y and z (=xy) respectively and j is
the >minimum number of shifts necessary to order
>(x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m)>into
(z_1)(z_2)...(z_{n+m}).>If x or y = 0, then dene x*y =
0.>If x < 0 or y < 0, then do the multiplication exactly as
if>both were positive, but substitute j with (j+k) were>k = 1
<-> (x < 0 and y > 0) or (x > 0 and y < 0)>k = 2 <-> x,y <
0.>Examples:>(3*10)*14 = (3*((2)(5))*14 = (-(2)(3)(5))*14 =
(-(2)(3)(5))*((2)(7))> = -(2)(2)(3)(5)(7) = -420>Reason: in
the 2nd equation, 1 shift was necessary;> (2) was shifted
once to the left.> in the 4th equation, 2 shifts were
necessary;> (2) was shifted twice to the left.>Note:
Whether the shift is to the left or the right does not> play
a role, i.e. j is the same number in either case.>3*(10*14)
= 3*((2)(5)*(2)(7)) = 3*(-(2)(2)(5)(7)> = -(2)(2)(3)(5)(7) =
-420>Reason: in the 2nd equation, (2) was shifted once to
the left.> alternatively, (5) was shifted once to the right.>
in the 3rd equation, (3) was shifted twice to the right.>We
can now assign each x in N a parity according to:x in P^{+}
<-> (x*x)/(xx) = 1x in P^{-} <-> (x*x)/(xx) = -1Ex. 10 in
P^{-}, 70 in P^{-}, but 560 in P^{+}.Let R be the reverse
operator, i.e.if x = (x_1)(x_2)...(x_n)=
(x_1)*(x_2)*...*(x_n) is the prime representation of x, then
Rx := (x_n)*...*(x_2)*(x_1)A few properties and conjectures
(in the order of decreasing obviousness, as I see it now):i)
RRx = xii) x*(Rx) in P^{+}iii) x in P^{+}, y in P^{-} -> x*y
in P^{-} x in P^{-}, y in P^{+} -> x*y in P^{-} x in P^{-}, y
in P^{-} -> x*y in P^{+} x in P^{+}, y in P^{+} -> x*y in
P^{+}iv) x*y = y*x <-> x and y of same parityv) x*(y+z) = x*y
+ x*z <-> x*y and x*z of same parity>C. Dement>X-Received:
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===
>prove if x is an irrational number then
sqroot of x is irrational???What was I thinking !? My
previous message was purely NOTHING.Assume that x =
irratoinal, and x^(1/2) = (x)^(1/2) = p/qfor some positve
integers p, q. Then,ln(x) = 2ln(p/q)ln(x)/(ln(p/q)) = 2.If
p/q = 2/3, thenln(x) = 2ln(2/3) x = exp(2*ln(2/3)).Sorry if
my message was out of present discussion (I would have todo
some homework and so there is no time to read the bunch of
messages; this is really a bad attitude for a person whowants
to be a mathematician).I think the real line is very dense
where denotes that the term is used
informally.H.S.X-Received: (from approve@localhost) by
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===
>prove if x is
an irrational number then sqroot of x is
irrational???PROOF:Assume that(a)x is irrational;(b)x^(1/2)
is p/q for some natural numbers p, q.Putx^(1/2) = p/qln(x) =
2ln(p/q))x = exp(ln((p/q)^2)) = (p/q)^2.But this contradicts
(a). This completes the proof.
===
>prove if x is an
irrational number then sqroot of x is irrational???>
PROOF:Assume that> (a)x is irrational;> (b)x^(1/2) is p/q for
some natural numbers p, q.Putx^(1/2) = p/qln(x) = 2ln(p/q))> x
= exp(ln((p/q)^2))> = (p/q)^2.But this contradicts (a). This
completes the proof.> Is there some reason youre using
logarithms and exponentialsin this problem? I mean, its not
incorrect, but its trulyirrelevant to what you want to
show.Why not prove the contrapositive: If sqrt(x) is
rational, then x is rational.or more simply put: If y is
rational, then y^2 is rational.???A word to the wise: If you
use things (e.g., concepts, functions, constructions,
calculations) that are not required in, or relevant to, an
argument, it shows that you dont understand what the
argument is about.I suppose that principle may apply more
generally thanwithin mathematics.DaleX-Received: (from
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===
As part of a software engineering group
project to design a math tutor/testing program for middle
school pre-algebra students, I have written a C program to
generate random angles (to be sent to the GUI) for 3 to 8
sided polygons. Is there a (relatively) simple way of
calculating the length of all the sides, assuming that one
side is set to an arbitrary length of one Unit?Assume that
the rst angle would be on the bottom left, that the angles
progress clockwise, the rst angle splits side 0 and side
1, that side 0 is a horizontal line, that line 1 goes up,
and that the polygon is regular?, e.g. describes a more or
less circular pattern.X-Received: (from approve@localhost) by
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===
Please, send me
an order form of ChiWriter word-processor by Horstmann
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I am trying to
do some research on alcuins sequence, but all the references
that I can nd quote the terms of the sequence as the
coefcients in the Maclaurin expansion of
[(1-x^2)(1-x^3)(1-x^4)]^-1. I nd this surprising, as I
believe the sequence to have been named after Alcuin of York
(735-804) who lived many years before Maclaurin expansions
were discovered. This sequence also gives the number of
different triangles that have integral sides and perimeter n.
I would think that it is much more likely that this is what
Alcuin discovered, and years later someone found that the
above expansion acted as a generating function. That is, of
course, if the Alcuin who the sequence is named after is
Alcuin of York and not someone else with the same name.I
would be most grateful if you could either conrm or refute
this, or point me in the direction of any references.Many
thanksTony
===
> This sequence also gives the number of
different triangles that have > integral sides and perimeter
n. I would think that it is much more > likely that this is
what Alcuin discovered, and years later someone > found that
the above expansion acted as a generating function. This is
the most plausible explanation. Certainly Maclaurin
serieswere unknown to Alcuin!Alcuin did write texts on
arithmetic, geometry and astronomy, for thepurpose of
teaching students.Alcuin is never reported as being a
wonderful intellectual innovator,but rather, an exceptional
teacher, and of course, calligrapher.(Dont discount that
last--his calligraphy was as important in its dayas computers
are today, and for much the same reasons.) He was alsoan
important liturgist and a strong (but not very
creative)theologian.juvenes, with twenty-one simple problems
for sharpening the minds ofyouth.The text of Alcuins book
is available
athttp://www.thelatinlibrary.com/alcuin.propos.html. There is
aand Technology Bulletin, and also an annotated translation by
DavidSingmaster and John Hadley.This is one of his most famous
puzzles from the book, number 18 (withmy translation): Homo
quidem debebat ultra avium transferre lupum, capram, et
fasciculum cauli. Et not potuit aliam navem invenire, nisi
quae duos tantum ex ipsis ferre valebat. Praeceptum itaque ei
fuerat ut omnia haec ultra illaesa omnino transferret. Dicat,
qui potest, quomodo eis illaesis transire potuit. A certain
man must bring across a river a wolf, a goat, and a bundle of
plants. But he also cannot nd any boat, except one that is
only able to carry two of them. He was told that he must
bring all these across entirely unharmed. Let him say, if he
can, how he can go across with them unharmed.And Alcuins
answer is: Solutio. Simili namque tenore ducerem prius capram
et dimitterem foris lupem et caulum. Tum deinde venirem,
lupumque transferrem: lupoque foris misso capram navi
receptam ultra reducerem; capramque foris missam caulum
transveherem ultra; atque iterim remigassem, capramque
assumptam ultra duxissem. Sicque faciendo facta erit
remigatio salubris, absque voragine lacerationis. Solution.
Similarly [to the answer to the previous problem], for with a
noose I would lead the goat and leave behind the wolf and the
plants. Then when I returned, I would carry the wolf over,
release the wolf, and return across again leading the goat;
and releasing the goat, transfer the plants across; and then
again row, and taking the goat, lead it across. And by doing
thus, the rowing will be done safely, free from the disaster
of tearing teeth.None of the problems are in general form,
but all could obviously beadapted to be in such. For example,
number XXI: Est campus qui habet in longitudine pedes CC, et
in latitudine pedes C. Volo ibidem mittere oves; sic tamen ut
unaquaeque ovis habet in longo pedes V, et in lato pedes IV.
Dicat, rogo, qui valet, quot oves ibidem locari possint?
Solutio. Ispe campus habet in longitudine pedes CC. Et in
latitudine pedes C. Duc bis quinquenos de CC, unt XL. At
deinde C divide per IIII. Quarta pars centenarii XXV. Sive
ergo XL vicies quinquies; sive XXV quadragies ducti,
millenarium implent numerum. Tot ergo ibidem oves collocari
possunt. There is a eld which is two hundred feet long and
one hundred feet wide. I want to place sheep there; but so
that each sheep has ve feet long and four feet wide. Let him
say, I ask, if he can, how many sheep can be placed there?
Solution. The eld is two hundred feet long. And in width,
one hundred feet. Compute twice: in fths from two hundred,
that makes forty. And then divide one hundred by four. A
fourth part of a hundred is twenty-ve. Therefore is computed
either forty, twenty-ve times; or twenty-ve, forty times,
the full number of a thousand. Therefore so many sheep can be
placed there at once.That gives you the avor. I dont know
enough about the sequence tobe sure if any of the problems
can be easily generalized to somethingsimilar. I would guess
that the annotated edition by Singmaster andHadley includes
such interesting tidbits.ThomasX-Received: (from
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> Michigan State University grad student
Michael Shafer has succeeded in> identifying the largest
known prime number to date, using a distributed> computer
network of more than 200,000 computers located around the
world.> The new number is 6,320,430 digits long and is only
the 40th Mersenne>prime> to have ever been discovered
(Mersenne primes are an especially rare breed> that take
the form of 2-to-the-power-of-P, where P is also a
prime>number).>Sounds like a waste of processor power (and
electricity) to me. There is no>reason>in discovering these
large numbers (even if you want to use them for>encryption.>I
believe the encryption process would need veeeery much time
;-}} ).>KarlAnd everyone now knows what the primes
are,,,duh!X-Received: (from approve@localhost) by
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>Dan
schrieb:>Strange how (e^2) has no order in its cf but the (n)
in my equation> has.>cf(e^2) => [ 7, 2, 1,> 1, 3, 18, 5,
1,> 1, 6, 30, 8, 1,> 1, 9, 42, 11, 1,> 1, 12, 54, 14, 1,> 1,
15, 66, 17, 1,> 1, 18, 78, 20, 1,> 1, 21, 90, 23, 1,> 1, 24,
102, 26, 1,> 1, 27, 114, 29, 1,> 1, 30, 126, 32, 1,> ...
]>computed by the gnu-program maxima>Gottfried HelmsI was
wrong on that one. I didnt check deep enough.Interesting how
the center column progress by 12 and the two columnson either
side progress by 3. I guess in some way there is a spinoff
after all with my constant(n)s cf being the most orderly of
e and e^2 cf.Dan
===
>Interesting how the center column
progress by 12 and the two columns>on either side progress by
3. Im unable really to read Perrons book on continued
fractionsbecause I dont know any German, but Perron appears
to makea study of the numbers whose continued fraction is
sort of closeto periodic in this way: [a1,...,an, b1,...,bn,
b1+c1, b2+c2,b3+c3,..., bn+cn, b1+2c1, b1+2c2,..., bn+2cn,
...], with eachelement of the cycle increasing in an
arithmetic progression.Gosper has explained how to compute
with continued fractionnumbers, and you can calculate the
expansion of multiplesof e and so on easily enough, and
youll nd a bunch ofrelated numbers which have the same type
of continued fraction.See for
examplehttp://www.inwap.com/pdp10/hbaker/hakmem/cf.html#
item101bKeith RamsayX-Received: (from approve@localhost) by
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> As noted in
the marginalia of the AMS Notices, the cover sheet is
intended> to help departments process job applications. As
such, it serves a purely> bureaucratic purpose and could
perhaps be forgiven on that basis if it were> then not also
used to determine how one is to be treated.You seem to be
using the word bureaucratic like a curse word. There needs
to be *some* way to classify the hundreds of applications that
ood into every mathematics department with jobs available.
The purpose of the coversheet and the cover letter is to
match applicants with faculty members who can evaluate
them.> The one category that sounds sufciently ambiguous for
me to feel Im> not misrepresenting myself is 00, i.e.
General. Perhaps by its nature,> after looking at
publications with that classication in Math-Sci, I cant>
gure out what it means, but that doesnt matter. What does
matter is what> assumptions a hiring committee will make
about someone who describes his/her> primary interest as 00.
And what would they assume?I think what would happen is that
the secretaries would put your application in the were not
sure pile. At X University, where I am, they employ someone
to go through that pile to check who would be best suited to
evaluate the unknowns. Maybe at some other places they
dont bother. > If someone can look over my publications on
Math-Sci and gure out what> the appropriate codes are for my
primary and secondary interests for the> AMS Standard Cover
Sheet, I would appreciate the advice. I dont seem to> have
been programmed with this capability.Accroding to MathSciNet,
at least some of your publications are 14G35 (algebraic
geometry of modular varieties). Many of your other
publications are in the general area of number theory (the
11-XX hierarchy). So you could put 14G35 and 11-XX as your
primary and secondary interests. Maybe you could ask your
co-authors for their advice too.The big question is: would
you feel happier if a number theorist or (say) a Bayesian
statistician read your application? Which would be able to
better appreciate your work?X-Received: (from
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>prove if x is an irrational number then
sqroot of x is irrational???>PROOF:>Assume that>(a)x is
irrational;>(b)x^(1/2) is p/q for some natural numbers p,
q.>Put>x^(1/2) = p/q>ln(x) = 2ln(p/q))>x =
exp(ln((p/q)^2))> = (p/q)^2.>But this contradicts (a). This
completes the proof.Actually, what is wrong with the proof?
Certainly, I did not have touse the natural logarithm
:)Putx^(1/2) = p/q.x = (p/q)^2and so x must have been
rational, a contradiction.If x is rational, then of course
there exists some x whose squareis yet a rational:x :=
rational -> x^(1/2) is (1) rational, (2) irrational.If x is
irrational, then by the proof,x := irrational -> x^(1/2) is
(3) irrational.This discussion seems to have been already
over, but for making sure.X-Received: (from
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At 10,000 digitsthe following
{0,1,2,3,4,5,6,7,8,9} occur in the following
amounts{987,1015,973,993,1016,1055,996,1022,943,1000}At
100,000 digitsthey occur
at{9983,10178,9947,9926,10035,10186,9929,9845,9938,10033}
Seems fairly regular to me.
===
The mathematician comes home
early from a conferenceand nds his wife...eh, you know the
routine.Off all the...With a traveling salesman?Well, at
least he was P-space hard.-- Hauke Reddmann <:-EX8 For our
chemistry workgroup,remove math from the addressFor
spamming, remove anything else
===
Why is it required that
Hom(A,B) = Hom(C,D) <=> (A,C) = (C,D) in thedenition of a
category?Most of the time were dealing with sets and
mappings, so that axiom istrivially
veried.Thks,--J.S
===
|Why is it required that Hom(A,B) =
Hom(C,D) <=> (A,C) = (C,D) in the|denition of a category?As
someone else pointed out, this implies that each arrow has
aunique domain and codomain. Often categories are dened so
thateach arrow has a designated domain and codomain, and
thenHom(A,B) is dened as the set of arrows with A as domain
and Bas codomain, which makes the above property
automatic.|Most of the time were dealing with sets and
mappings,Whether were dealing with sets and mappings most of
the time dependson what kind of category theory were doing.
Lot of categories aresubcategories of the category of sets,
but this is not always the mostappropriate context to put
them in.|so that axiom is|trivially veried.Its a very light
assumption, basically. If we started out witharrows f dened
so that they could be both from A to B andC to D, without A=C
and B=D, then we could switch insteadto considering our
arrows to be triples (f,A,B) where f is oneof our original
arrows from A to B. Then the axiom wouldbecome
automatically true.I suppose its not so obvious why we want
arrows to haveunique domains and codomains. But I think it
amounts toa way to smooth out notation. If category is a
subcategoryof the category of sets, then its tempting to
think of anf:A->B as equal to f:A->C where B is a subset
of C, andf(a)=f(a) for all a. Yet a preference exists for
distinguishingthem, perhaps partly because in category theory
we dontwant to have to say things like B is a subset of C
whichrely upon equality between elements of one set and
elementsof another set. Instead we want to say only that
there existsan injection g:B->C, so that f = g o f.If the
category is just an arbitrary abstract category,
however,permitting f:A->B to be equal to f:C->D is not of
much use.Keith Ramsay
===
>Why is it required that Hom(A,B) =
Hom(C,D) <=> (A,C) = (C,D) in the>denition of a
category?>Most of the time were dealing with sets and
mappings, so that axiom is>trivially veried.So that the
domain and codomain are functions?--Dan Grubb
===
A space S is
monotoniclly normal when foreach x in S and each open U nhood
x, theres assigned mx(x,U) another open nhood of x,for which
for all x,y, open U,V x in U, y in V, mu(x,U) / mu(y,V) not
empty ==> x in V or y in U.Theorem: monotonically normal ==>
completely normal ==> normalProblem: to show a linear order
space is monotonically orderedProof with weakness marked **
below. Whats to be done about **?Let S be linear order
spaceLet < be order of S for the order-topologyLet <_w
well-order Sdene mu(x,U) as follows: let x in (c,d) be
interval inside UIf (c,x) nonnul, let x_l = <_w-rst in
(c,x), otherwise let x_l = cIf (x,d) nonnul, let x_r =
<_w-rst in (x,d), otherwise let x_r = dlet mu(x,U) = (x_l,
x_r) which is an open nhood of xIf mu(x,U) / mu(y,V) nonnul
thenx in (x_l, x_r); y in (y_l, y_r)some z in I = (x_l,x_r)
/ (y_l,y_r) = (max x_l,y_l, min x_r,y_r)Consider when x_l
<= y_l. [(y_l <= x_l) similar] if y_r <= x_r: I = (y_l, y_r);
y in I subset U if x_r < y_r: I = (y_l, x_r) assume x,y not in
Ithen we have x <= y_l < z < x_r <= yif x < y_l < z < x_r < y,
then comes desired contradiction x_r <=_w y_l; y_l <=_w x_r;
y_l = x_r; z in I = nulset** Otherwise, what to do when x =
y_l or x_r = y ?For indeed when x = y_l, then y_l not in
nonnul (x,d) so one cannot claim x_r <=_w y_l
===
in
message <bra2br$cg6q$1@netnews.upenn.edu>:[...]> Dear
all,> Trying to gure out the 1-dimensional
representations of D_n, the> dihedral group of order 2n.>
Should be easy...[...]> I guess the last thing I would have
to do is analyze the conjugacy classes> of D_n and that is
all. (I can do this on my own)In the end youll nd this is a
specic instance of a generaltheorem in character theory which
you might also enjoy trying toprove:The number of 1-d complex
linear reps of a nite group G is|G/[G,G]|.-- Jim
Heckman
===
in message
<83f64ab8.0312102356.424eb9c9@posting.google.com>:> im trying
to nd all groups of order eight. i already know what these>
groups are, but i am trying to derive it, mainly from the
sylow> theorems. will this approach work? for example, I was
able to work> with groups of order 21=3*7, because I was able
to nd a semidirect> product representation of the group, but
8=2^3 doesnt really lead me> anywhere. is there another
method?The approach Ive always liked is based on the fact
that everynon-cyclic group P of order p^n for p prime is
generated by 2subgroups of order p^{n-1}, each necessarily
normal in P,intersecting in a subgroup of order p^{n-2}. So
look for allpossible ways in which 2 groups of order 4 can
intersect in agroup of order 2 such that the 4-groups
normalize each other.(In practice, you need only consider one
of the 4-groupsnormalizing the other.)-- Jim Heckman
===
hi,
and thank you for the suggestion. i worked by case analysis
onmaximal order of an element and it came out quite well,
although i raninto quite a few complications.in any case, i
feel that trying to describe all the groups of order 8turned
out to be more work than what was actually required for what
iam doing. in the end i am trying to show that Aut(Z16) is
isomorphicto Z2xZ8, and I dont think I want to use the fact
that I know allgroups of order 8.I have the curious fact that
there is only one sylow subgroup, namelyH2 the 2-Sylow
subgroup, which of course H2=G. does it seem like anyof
sylows theorems could help me in this restricted case?
otherwise isuppose i will continue with the rigmarole of
describing groups oforder 8.thanks againim trying to nd all
groups of order eight. i already know what these> groups are,
but i am trying to derive it, mainly from the sylow> theorems.
will this approach work? Trouble is, as you say later, is that
2 is a prime power> so Sylow tells you nowt.for example, I was
able to work> with groups of order 21=3*7, because I was able
to nd a semidirect> product representation of the group, but
8=2^3 doesnt really lead me> anywhere. is there another
method?Two possible attacks> (i) the exponent of the group:
this is the smallest n such that a^n = 1> for all a in the
group. For a group of order 8 it is 2, 4 or 8.> Divide into
these three cases.(ii) Use the centre. A standard theorem
states that the centre Z> of a p-group G is nontrivial. Split
into cases according> to the structure of the centre.
===
Given
a set S = { (1,2),(1,3),(1,4)...(k-1,k)}, is it possible to>
partition the above set into k subsets such that the
elements of> each subset are pairwise-disjoint?Yes. You can
nd this in most any introductory textbook on graph> theory.
Keywords: edge-coloring, edge-chromatic number, chromatic>
index, complete graph. Is this homework?larger problem I was
attempting.One of the questions asked by Erd.9as is - Let
K_1, K_2,..., K_k be complete graphs, all of size at most k,
andassume that for any two of them the intersection is either
empty orexactly one vertex. Let G be the union of K_1,
K_2,..., K_k. Is Gk-colorable?Proof by reverse induction - If
P = [0,m] and P(n) is a statement andthefollowing two
conditions hold - (1) P(m) is true. (2) for all m <= n <= 0,
P(n+1) is true implies P(n) is truethen P(n) is true
for all 0 <= n <= m.Let us assume that the complete graphs in
G are numbered from 1 to kand theintersection point of two or
more graphs is represented as a tupleconsisting of the number
of the intersecting graphs. For instance, ifG contains a
intersection point between K_1, K_2 and K_3, theintersecting
point is represented as (1,2,3).We note that G can be colored
in k colors if the intersecting pointscan be colored in k
colors such that each of the complete graph, K_i,is properly
(vertex) colored.Basic step [to show P(m) is true] :Let G
consist of intersecting points between exactly two
completegraphs (i,j) for all i,j <= k. G contains maximum
number ontersecting points.G can be colored in k colors by
coloring each intersection point with[(i+j) mod k].We note
that for two tuples (i,j) and (m,n) - (i+j) mod k = (m,n) mod
k only if i,j,m,n are all different and (i+m) mod k != (i,n)
mod k for m != n; which implies, the colorassigned to (i,m)
is unique for all m <= k; which implies that thecoloring of i
th complete graph is proper. Hence, the coloring of Gis
proper.Induction hypothesis - If the k graphs intersect at
exactly (n+1)vertices, there is a k-coloring of G.Induction
step - To prove that if the k graphs intersect at nvertices,
there is a k-coloring of G.I have not been able to complete
the proof but I suppose I would haveto show that I can create
a graph G from G with (n+1) intersectingpoints and that, a
k-coloring of G is also a k-coloring of G.I would appreciate
any thoughts on this or the whole argument so far.Many thanks
for your suggestions.--Pradip
===
Induction hypothesis - If
the k graphs intersect at exactly (n+1)> vertices, there is a
k-coloring of G.We note that : m = k(k+1)/2Induction step - To
prove that if the k graphs intersect at n vertices(n < m),
there is a k-coloring of G.Consider an intersection point
(i,j,...,l) [There is atleast oneintersecting point
(i,j,...,l) because n < m]. We construct G from Gsuch that
the intersection point (i,j,...,l) in G is split as(i,j,...)
and (i,l) in G. By induction hypothesis, G has ak-coloring.
A k-coloring of G is also a k-coloring of G by assigningthe
color of (i,j,...) in G to (i,j,...,l) in G.--
Pradip
===
Here is my answer:We can say that at least the
sequence ...11111111 is not in the list,for
example:...00000000 <--> 1...00000001 <--> 2...00000010 <-->
3...00000011 <--> 4 ...00000100 <--> 5...00000101 <-->
6...00000110 <--> 7...00000111 <--> 8...Let us examine the
innite from another point of view.When we have ...11111 AND
...00000 in an ordered combinations list, itmeans that the
list is complete.But this is the whole point, innitely many
objects cannot becompleted, otherwise they are
nite.Therefore ...11111 AND ...00000 are not in the list of
innity manyobjects.In other words [...000, ...111) XOR
(...000, ...111] .There are 2 possible structural types of
innitely many 01 notations:(?...0](?...1]We know how some
innitely long combination starts, but its oppositeside
isunknown (can be 0 XOR 1) and this missing information is
essential tothe existence of the induction.Therefore we can
nd a meaningful missing result by CantorsDiagonalization
method, only in a nite combinations list.For more details
please look
at:http://www.geocities.com/complement...
iemannsBall.pdfBecause I don&#8217;t know how to write my
idea in the common formal> way, I am going to do it in a
non-formal way, but I will do my best to> write it in the
clearest way (PLEASE READ ALL OF IT, OTHERWISE IT> CANT BE
UNDERSTOOD).So here it is: Let us check these lists. P(2) =
{{},{0},{1},{0,1}} = 2^2 = 4 and also can be represented as:
00 > 01 > 10 > 11 > P(3) =
{{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8 and also
can be represented as: 000 > 001 > 010 > 011 > 100 > 101 > 110
> 111 Let us call any full 01 list, combinations list. Now,
let us use Cantors Diagonalization method on some nitely
long> combinations list, for example, the combinations list
of number 3:000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 We
can change the order of the rows, and then use Cantors>
Diagonalization method, for example:001 > 011 > 010 > 000 >
101 > 100 > 111 > 110 The input for Cantors Diagonalization
method in the rst example is> 000 and the output is 111.of
course that choice was completely arbitrary. you could have
started> with any string. and any ordering of the strings. >
The input for Cantors Diagonalization method in the second
example is> 010 and the output is 101.ne, so youre ignoring
the bottom 2^n - n rows then?> In both examples we nd that
the result is already in the> combinations list, and this
combination, which is already in the list,> is one of the
combinations that Cantors Diagonal does not cover.The number
of the combinations, which are out of the range of Cantors>
diagonal is:2^n - n > so thats where that number came from.
now i understand what it means.> incidentally, youd (still
be wrong but) be better off saying domain. > Every column,
which belongs to some combinations list is a sequence of> 01
notations, based on some periodic frequency changes, for
example:the right column of number 3 combinations list, is
based on 2^0(=1). Therefore the periodic frequency changes
are 1, and the result in this> case is:> 01010101. > but a
completely arbitrary ordering of the rows wouldnt look so
nice> would it? i mean 01110100 is possible under some
ordering, so whats> your point?The result of the middle
column is based on 2^1(=2), therefore the> sequence is:>
00110011. The result of the left column is based on 2^2(=4),
therefore the> sequence is:> 00001111. and we get the full
combinations list of number 3: 000 > 001 > 010 > 011 > 100 >
101 > 110 > 111 We can get a combinations list of innitely
many places, by using the> ZF Axiom of innity induction, on
the left side of our combinations> list, by using the
induction on the power_value of each column, for>
example:2^0, 2^1, 2^2, 2^3, ... > still passing to innite
constructions without demonstrating its valid.In this stage
we have proven, by induction, that Cantors diagonal> cannot
cover any full 01 combinations list, nite or innite.> what
do you mean cover? at best youd need transnite induction
wouldnt you to pass from a> nite cardinal to a countably
innite cardinal?otherwise:the sum of 1/r from 1 to n is
nite. therefore, using your assumption,> the innte sum is
nite. prove thats a problem if you dont already know
Therefore its result is not a new combination (that has to be
added to> the list).Because Cantors diagonal cannot cover the
full 01 combinations list,> (of aleph0 places for each
combination) we can conclude that 2^aleph0> aleph0.But each
innitely long sequence of 01 notations can be mapped with>
some natural number, for example:...000 <--> 1 > ...001 <-->
2 > ...010 <--> 3 > ...011 <--> 4 > ...100 <--> 5 > ...101
<--> 6 > ...110 <--> 7 > ...111 <--> 8 > ... what does the
string entirely made of ones get sent to? i can see how
that> works for strings with a nite number of ones in
them:the string (x_i) i in N gets sent tosum x_i.2^ibut
where do the ones with innitely many 1s get sent to? >
Therefore we can conclude that 2^aleph0 = aleph0, and we come
to> contradiction.(2^aleph0 >= aleph0) = {}, and we have a
proof saying that Boolean> Logic cannot deal with innitely
many objects.> Doron> I think that at best youve proven the
set of nite subsets of a> countable set is countable. But
thats not surprising is it?> Youve not got the innite
subsets in your enumeration.if you insist on doing this. note
that it is not required to talk about> all those sets and
strings. just think about all strings of 0s and 1s,> forget
that set stuff, its just confusing and unnecessary.
===
As
its a friday lunch time and the ofce party is happening
soon, idont feel too guilty about taking time off to analyse
your pdf.Fortunately, it is clearer than the posting to this
newsgroup, andtherefore it is simple to point out where you
are still going wrong.You make some conclusion about nite
sets. In fact you are showing thereare no bijections from a
nite set to its power set. Good. You pass,through an
induction, to a set of countably innitely many elements.
Theresult fails because simple induction cannot do that:I
repeat: The product of a nite number of sets with 2 elements
in eachis nite. The innite product, if it exists, is not
nite. It is noteven countable. As CAntors argument tells
us.There is a bijection between the collection of all nite
subsets of acountable set and itself. You are using that
fact: thats where yoursending things to binary expansions
comes in.You conclude that cantors diagonal argument must
produce something not inyour list because it doesnt in the
nite case. That may or may not betrue. And isnt important.
You argue that as 2^n - n is not zero, this must pass through
in thelimit. Im not sure what you want to mean by that. The
number of sequencesin the limit is not 2^aleph_0,
absolutely not. It has cardinality aleph_0as does the number
of entries in each row. There are no missing rows notin the
range of the diagonalization argument. If there were there
would bea rst such, where is it? Either the diaginalization
produces something on the list, and all isne, or it doesnt,
and so what? Im almost certain it would be aninnite set, but
why does that matter?
===
Thought of an even better way of
saying it.Your entire argument depends upon claiming that
that list of 0s and 1scompletely lists all the power sets
of a countable set. How is it produced? It is by taking a
(ltered direct) limit over thepower sets of an nested
increasing collection of nite sets.There are no innite sets
in this limit. Any element of the limit lies inone of the
nite power sets over which you are taking your limit
(andthen in all the succeeding ones), and is thus nite. So
you dont have alist of all the elements in the power
set.Amazingly you have correctly asserted that 2^aleph_0 >
aleph_0 but theproof is not valid. And youve demonstrated
that the nite subsets of acountably innite set are
countable. But youve put those things togetherwrongly and
got some odd conclusion.
===
Here is my answer:We can say that
at least the sequence ...11111111 is not in the list,> for
example:...00000000 <--> 1> ...00000001 <--> 2> ...00000010
<--> 3> ...00000011 <--> 4 > ...00000100 <--> 5> ...00000101
<--> 6> ...00000110 <--> 7> ...00000111 <--> 8> ...> So you
are saying that there are no innite sets on the list? And
thismeans what? The nite subsets of a countable set are
countable. You can list them.You could do that thing you
cantors diagonalization method. It doesntfollow that what
you have is necessarily some innite set from whatyouve
written. And it doesnt matter.> Let us examine the innite
from another point of view.When we have ...11111 AND ...00000
in an ordered combinations list, it> means that the list is
complete.> No, ...00000 is on the list, it is the empty
set.And we can simply add ....1111 to the list by placing it
in the 0thposition. > But this is the whole point, innitely
many objects cannot be> completed, otherwise they are
nite.Therefore ...11111 AND ...00000 are not in the list of
innity many> objects.> No, ..00000 is on the list, and
..11111 isnt on the list,but as ...1111 represents some
innite set, and youve only got thenite ones there that is
not a contradiction with anything.> In other words [...000,
...111) XOR (...000, ...111] .There are 2 possible structural
types of innitely many 01 notations:(?...0]> (?...1]We know
how some innitely long combination starts, but its opposite>
side is> unknown (can be 0 XOR 1) and this missing information
is essential to> the existence of the induction.Therefore we
can nd a meaningful missing result by Cantors>
Diagonalization method, only in a nite combinations list.>
For more details please look
at:http://www.geocities.com/complement...iemannsBall.pdf>
>So, what are you trying to show? That cantor is wrong? that
there isaleph-0 and aleph_1 are the same?By the way, you
still cannot pass from a nite cardinal to an innitecardinal
and say results hold by induction on the nite
cardinalitiesalone. You seem want to assume that there are
strictly more objects in thenatural numbers union a nite
set, than in the natural numbers, from allthe 2^n - n stuff
which I still dont see the relevance of.
===
Give a
point-by-point exposition to support your case without
insults,> innuendo, or any other attempts at sly or
back-handed insults, but> with *mathematics* and logic.> When
are you going to get round to answering my questions, James?
All ingiven in an insult free manner> The challenge is to you
now, show that you can reason and discuss> mathematics
reasonably, if you can.> James Harrisstill waiting.
===
Hash:
SHA1[Follow-up reset]> Some of you may have noticed these
free-wheeling mathematical> discussions Ive been having with
all these people, and wondered.Well, lots of posters have
tried to educate you about math, and youve beenfreewheeling
for everyone.[..]> Now then, use your common sense.Is it
simpler that Im really some crank, Yes, you are a crank. If
not, then youre the best fake Ive seen outsidethe circles
of professional crank-imitators, but thats a different
story.[..]> And dont worry, this post is very unlikely to>
change anything.> You mean that the responses are unlkikely
to make you change your behaviour?All too true :-(- -- Key
ID 0x09723C12,
j.tingleff@ieee.org/jens_tingleff@yahoo.comhttp://
www.imaginet.fr/~jensting/ +44 1223 211 585I told you to
write a fake letter, not fake writing a letter
CerebusiD8DBQE/2W28imJs3AlyPBIRAjq0AKCOQ/
Z3MPZcwWTDdhnShx0Xiow/bwCgoNvMsfCVf60yBeVZCmUnVXqaAfU==snCZ==
==I just nished reading an by Manfred Schroeders excellent
1992 (93) math book Fractals, Chaos and Power Laws, and
he makes the argument that any function is chaotic if its
spectral transform is a power law. That means that if P(w) ~
w^x, where w is the frequency omega, P(w) is the power
spectrum and x is any small number, typically either rational
or convenient round decimals, like 0.5, 1, 1.33, 1.4, 1.45,
1.5, 1.6, 1.75, 1.8, 2, etc. I dont think that x ever
exceeded 3 in the many examples he gave.Of course, you can
always t a polynomial of arbitrary order n to any multimodal
function if n is the number of peaks in the function, but it
makes no sense if n is too large. Besides, power laws and
polynomials are not the same thing.If you make the basis
function expansion and the result is still extremely noisy,
youre on the wrong track. If you use a low-pass lter,
youve probably also ruined the math, since all of
Schroeders functions had positive exponents, making them
rise monotonically with frequency. In other words, the
highest order modes are the most populated states, making
them the most important modes.And if you do low pass
ltering, you arent doing wavelets, youre doing sums of
polynomials or trigonometrics. Im still very loyal to the
FFT exclusively, since trig functions are the only functions
I know that arent polynomials, and you can get free source
codes on-line. Im just an analyst, I dont do math very
well.In fact, Ill bet you can generalize Schroeders result
to the conclusion that a function is chaotic if P(v) ~ v^x,
where v is the working parameter of any basis function you
choose (Phi(v)), not necessarily frequency. Phi could be a
polynomial or your wavelet, and wavelets Ive never seen the
math for, so I dont know how they x get past the low-pass
problem. And that is out of my reach.> power law ... would
you mind to give a little bit> more than you already said?> I
think that the idea would then be to simply examine the
spectral> function and see if it obeys a power law. Has that
ever been> observed in the data? Power law dependence would
mean that it is aperiodic, that basis> function expansions
are a waste of time (not counting wavelets) and> that active
control wouldnt work.Passive control might still work, (ie.
setting interest rates to> neutral and forgetting about
them) but I wouldnt bet on that either.> Im not up on the
math, but a few years ago there was a breakthrough> in
aerodynamics, where engineers realized when a ghter jet
went> into a tumble (chaotic trajectory) the pilot was
causing it by> ghting the stick. If he let go of the stick,
the plane would> spontaneously come out of the tumble. A
sloppy person might be inclined to simply try the same
solution for> stabilizing the business cycle, but that would
be a bad idea. It> might be that there is no stable limit
cycle in the available range> of parameters except the one
where all activity stops dead.This would be the same as if
the time constant for recovery were so> long that everybody
starved before the economy came back. In the> absence of
actual state equations for the process model, I think> youd
have to assume that this is the way things are and that the>
business cycle is the best solution allowed by the system.>
Yes, a long time ago.> The business cycles has a varying
period and amplitude. Also cycles> are not always that clear
when you get double dips and the such> like.>
===
>How
about zero? As in There are no prime numbers which
are>powers of Mersenne primes. None at all. Aint gonna
happen.>End of story. Next.The poster is wrong of course, but
I have discovered powers ofMersenne primes that are neither
primes nor composites. I will get mymoney once the
mathematical community acknowledges my discovery or theyear
3017, whichever comes rst.
===
> Michigan State University
grad student Michael Shafer has succeeded in> identifying the
largest known prime number to date, using a distributed>
computer network of more than 200,000 computers located
around the world.> The new number is 6,320,430 digits long
and is only the 40th Mersenne prime> to have ever been
discovered (Mersenne primes are an especially rare breed>
that take the form of 2-to-the-power-of-P, where P is also a
prime number).AP is reporting that this genius had 2
gigahertz of memery.I dont get how such a simple story can be
mangled by Arts majors so badly.Oh. I just gured it
out.Shaun
===
Michigan State University grad student Michael
Shafer has succeeded in> identifying the largest known prime
number to date, using a distributed> computer network of more
than 200,000 computers located around the world.> The new
number is 6,320,430 digits long and is only the 40th Mersenne
prime> to have ever been discovered (Mersenne primes are an
especially rare breed> that take the form of
2-to-the-power-of-P, where P is also a prime number).AP is
reporting that this genius had 2 gigahertz of memery.2
billion cycles per second of memory? Thats a rate, not a
quantity.I dont get how such a simple story can be mangled by
Arts majors so badly.> Oh. I just gured it out.> Shaun
===
>
AP is reporting that this genius had 2 gigahertz of memery.>
2 billion cycles per second of memory? Thats a rate, not a
quantity.Welcome to the genius club.-Mike
===
>Michigan State
University grad student Michael Shafer has succeeded
in>identifying the largest known prime number to date, using
a distributed>computer network of more than 200,000 computers
located around the world.>The new number is 6,320,430 digits
long and is only the 40th Mersenne prime>to have ever been
discovered (Mersenne primes are an especially rare breed>that
take the form of 2-to-the-power-of-P, where P is also a
primenumber).> AP is reporting that this genius had 2
gigahertz of memery....and a 5 1/4 processor running at 31.2
kps of USB keyboard cache memory.
===
>Michigan State
University grad student Michael Shafer has succeeded
in>identifying the largest known prime number to date, using
a distributed>computer network of more than 200,000 computers
located around the world.>The new number is 6,320,430 digits
long and is only the 40th Mersenne prime>to have ever been
discovered (Mersenne primes are an especially rare
breed>that take the form of 2-to-the-power-of-P, where P is
also a primenumber).>AP is reporting that this genius had 2
gigahertz of memery.> ...and a 5 1/4 processor running at
31.2 kps of USB keyboard cache memory.Ill have a prime rib,
please. Rare. Baked potato. Salad with Italian dressing. And
bacon for dessert.--Bill-- The World Wide Web is the hugest
vanity press in the history of the human
race!http://billwilkinson.home.mindspring.com/index.html
===
>
Michigan State University grad student Michael Shafer has
succeeded in> identifying the largest known prime number to
date, using a distributed> computer network of more than
200,000 computers located around the world.> The new number
is 6,320,430 digits long and is only the 40th Mersenne
prime> to have ever been discovered (Mersenne primes are an
especially rare breed> that take the form of
2-to-the-power-of-P, where P is also a prime
number).
===
Every place I look at leaves it as an
exercise.I know that the proof involves thinking of A (mxn
matrix) as a collection ofcolumn vectors, so that[A1 .. An]
X B is a subpace of the column space of A1..An.But how do we
know that the product is a subspace? How do we know
itdoesnt increase the dimension?
===
> Every place I look at
leaves it as an exercise.I know that the proof involves
thinking of A (mxn matrix) as a collection of> column
vectors, so that[A1 .. An] X B is a subpace of the column
space of A1..An.But how do we know that the product is a
subspace? How do we know it> doesnt increase the
dimension?of a matrix equals the dimension of the range of
the associated linear transformation.
===
> By O(1), I was
dening the _computational_ complexity with regard to> the
variable x and not dening the size of an error term.Even
that sense of O(1) is difcult -- Your forumula contains
anumber of computations using numbers that are approximately
as largeas x, so each operation takes at least O(log x) time.
I also see 2logarithms, 2 exponentiations, and 1 cosh. How
long does it take toexecute those, in the general case? (What
degree of accuracy are youdemanding?)I think what you really
mean is Here is an approximation to pi(x)that is in closed
form and gets remarkably close.Dale
===
> The error for
10^23 appears to be 57867510952120449. Why do> you call this
an O(1) approximation? You have some reason to> think the
error is never larger than 57867510952120450 or what?>My
apologies.>By training I am a computer scientist, and so had
forgotten that the >O(x) notation is more commonly interpreted
as an indication of the size >of difference from some required
value.>By O(1), I was dening the _computational_ complexity
with regard to >the variable x and not dening the size of
an error term.>Simple formulae are not normally used for
calculating, say, the Riemann >Prime Counting function or the
Logarithmic integral, as both are usually >calculated by
summing an innite series.>Summing an innite series is
computationally O(p) where p is some >chosen limit of
precision.>Again, sorry for any confusion.No problem.
Although of course you must be aware that thereare well-known
approximations to pi(x) which are O(1) in thesame sense as
what you gave above, for example
x/log(x).>CarlDavid C. Ullrich
===
>
No problem. Although of course you must be aware that there>
are well-known approximations to pi(x) which are O(1) in
the> same sense as what you gave above, for example
x/log(x).It was that very approximation that led to the one
described.Suppose there exists a solution (as a function of
x) for N to: pi(x) = x/log_N(x) == log(N)*x/log(x)or even an
approximation to such a function, then pi(x) will be
calculable to the same degree.The actual value of N tends
toward e as x -> oo, but for the values of pi(x) available it
appears that N is not less than e for x >= ~100. It was
assumed that this would remain true upto innity.Graphing N
over the values of pi(10^n) from the Mathworld page resulted
in something resembling that for the function f(x)=x^(1/x),
yet converging towards e rather than 1. This idea eventually
became: a = L^[2/(3L)]A modied hyperbolic secant was
introduced to attempt to eliminate an aberration in the
difference between the graphs for N and e*a, peaking at
around L = e^2. Why this should be when L is a logarithm to
base ten is unclear, but this led to the formula: b = 1 +
K1*sech(L-e^2)^K2 ; 0 < K1 < 1 ; 0 < K2 << 1The 1 +
intending to make the curve converge to 1 (as L->oo) so that
a/b would converge to 1.After some experimentation, the sech
formula became the cosh formula seen in the original post.k
is then set to the approximated value of log(N).The 1 + in
this case adjusts log(a/b) so that it again converges to 1,
allowing N = e^k to converge to e as required.And thus we
have a marginal improvement on the standard x/log(x)
approximation.Carl
===
Hallo all,Does anybody have a clue how
to solve the following denite integralInt( cos x / (1 + 2C^2
- 4C cosh x + 2 C^2 cosh 2x ), x,0..pi )Karjanto
===
> It is
kind of curious to note the observed frequency of the digit
5> is *always* less than expected. Is there an explanation
for this> wierdness?Id start by asking if the deviations
from the expectation values arelarger than would be expected
from a random variable with thatdistribution. It may not be
so weird.Dale
===
> Suppose x is a real number and 0<x<1.
Set x[0]=x and for i>=0 compute:1: a[i]:=the largest integer
k s.t. oor(k*x[i])=1> 2: x[i+1]:=a[i]*x[i]-1Its not too
difcult to see: (1) 2<=a[i]<=5 if i>0, and (2)>
x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the
ith digit of>x.In a very nice post from several years ago,
Rob Johnson showed the expected> distribution of the digits
of a random x is [206,81,30,20]/337> (tinyurl.com/ys8t).
Thus p(a[i]=5) is 20/337 or about .059347. The exact>
distribution of the rst n digits of log(2) is as follows
(according to> Pari/GP): n freq1000:[587,276,87,50]>
2000:[1179,537,178,106]> 3000:[1783,793,267,157]>
4000:[2404,1035,360,201]> 5000:[3008,1274,448,270]>
6000:[3618,1529,528,325]> 7000:[4240,1753,625,382]>
8000:[4886,1943,727,444]> 9000:[5490,2194,912,562]>
10000:[6113,2413,912,562]> 20000:[12236,4817,1780,1167]>
30000:[18337,7266,2678,1719]> 40000:[24449,9716,3545,2290]>
50000:[30564,12105,4443,2888]>
60000:[36692,14483,5368,3457]>
70000:[42847,16819,6291,4043]It is kind of curious to note
the observed frequency of the digit 5 is> *always*> less
than expected. Is there an explanation for this wierdness?>
>Always? Or just when n ends in 000?>Good question. Pari/GP
sez for 51<=n<=70000 the observed frequency is lessthan
20/337. rich
===
> Suppose x is a real number and 0<x<1.
Set x[0]=x and for i>=0 compute:> 1: a[i]:=the largest
integer k s.t. oor(k*x[i])=1> 2: x[i+1]:=a[i]*x[i]-1> Its
not too difcult to see: (1) 2<=a[i]<=5 if i>0, and (2)>
x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the
ith digit>of>x.> In a very nice post from several years
ago, Rob Johnson showed the>expected> distribution of the
digits of a random x is [206,81,30,20]/337>
(tinyurl.com/ys8t). Thus p(a[i]=5) is 20/337 or about
.059347. The exact> distribution of the rst n digits of
log(2) is as follows (according to> Pari/GP):> n freq>
1000:[587,276,87,50]> 2000:[1179,537,178,106]>
3000:[1783,793,267,157]> 4000:[2404,1035,360,201]>
5000:[3008,1274,448,270]> 6000:[3618,1529,528,325]>
7000:[4240,1753,625,382]> 8000:[4886,1943,727,444]>
9000:[5490,2194,912,562]> 10000:[6113,2413,912,562]>
20000:[12236,4817,1780,1167]> 30000:[18337,7266,2678,1719]>
40000:[24449,9716,3545,2290]> 50000:[30564,12105,4443,2888]>
60000:[36692,14483,5368,3457]> 70000:[42847,16819,6291,4043]>
> It is kind of curious to note the observed frequency of the
digit 5 is> *always*> less than expected. Is there an
explanation for this wierdness?>Always? Or just when n
ends in 000?>Good question. Pari/GP sez for 51<=n<=70000
the observed frequency is less>than 20/337. >However, if
n=179569 the observed frequency is greater than 20/337.
rich
===
> May I suggest SymbMath for $20:> It is an online
symbolic math and computer algebra system.> It can perform
exact, numeric, symbolic and graphic computation,> e.g.
arbitrary-precision calculation, solve equation, plot data
and> user-dened functions, linear regression, symbolic
differentation> and integration, pattern-match. It is a
programming language, in> which you can dene conditional,
case, piecewise, recursive,> multi-value functions and
procedures, derivatives, integrals> and rules.>
www.SymbMath.com> www.mathHandbook.com> For various obscure
reasons Ive needed an inexpensive shareware> software which
would calculate basic arithmetic functions, ln(n), e^n,>
roots, and the basic trig. functions,...all out to pretty
high decimal> points,> say 50 or so (or more depending on how
playful I get).> For a while Ive tested the Haxial
Calculator which is OK but laborious,> esp. since it has, as
yet, no trig. functions. For such trig stuff Ive> used>
basic formulae to let it calculate actual trig. values,
though it kicksup> a> fuss when I stuff certain numbers into
it. The Hax. Calc. costs maybe> $15 or 20.> Does anyone know
of a good shareware, perhaps even a bit more expen-> sive,
which will do ALL the functions I described, to fairly large>
arbitrary> accuracy?? (There is no way I can afford anything
like Mathematica> unless I don a mask and take to the
highroad ;- )> Gene>
===
> I was wondering whether there
be a name for a faithful functor F: C> --> D such that given
any isomorphism f: F(a) --> b in D, there is> unique object
b of C such that for some isomorphism f: a --> b, F(f) => f?
In other words, when F is into the category of sets, given a>
structure on a and a bijection f from a into b, there is a
unique> structure on b such that f is an isomorphism from a
into b. This> seems to be the categorical notion with which
you would want to> replace Bourbakis notion of structure and
morphism (as dened in> Theories des Ensembles, Ch. IV) if you
want all his results to work> out yet (understandably) want to
use a categorical approach. It is> what I prefer to think of
when he uses the concepts of morphism and> structure.Such a
functor is called uniquely transportable in the terminology
ofAdamek,Herrlich,Strecker: Abstract and concrete
categories(see Def 5.28 for details and a funny
picture).More precisely, in the above situation F: C --> D
they callC a concrete category over D, when F is
faithful.In this setting they call F: C --> D uniquely
transportablewhen the above lifting property holds.Unlike a
op-bration, only _isomorphisms_ are asked to be liftable.So
it is easy to produce UT-Functors where you cannot
liftarbitrary morphisms, e.g. restrict C by a
size-conditionon the F(c) (c in C). However these examples
look a bit articial.If you are interested in more of this
stuff, the Adamek/Herrlich/Streckerbook might be a good
starting point. There is also an earlier bookManes:
Algebraic Theories, where some of this is
discussed.Marc
===
|Such a functor is called uniquely
transportable in the terminology
of|Adamek,Herrlich,Strecker: Abstract and concrete
categories|(see Def 5.28 for details and a funny
picture).Subtitled, The Joy of Cats. The illustration at
the beginning ofeach chapter features a cat doing various
things. :-) The illustrationof unique transportability is of
the machine that you should imaginedoing the unique
transportation, I think.I looked at that book once out of
curiosity about the denitions ofalgebraic functor and
topological functor. The forgetful functorfrom the category
of groups to the category of sets is algebraic.The forgetful
functor from the category of topological spaces tothe
category of sets is topological. The forgetful functors from
thecategory of topological groups to the categories of groups
andtopological spaces are, if I remember correctly,
topological andalgebraic, respectively. These are concepts of
a similar avor.Keith Ramsay
===
> |Such a functor is called
uniquely transportable in the terminology of>
|Adamek,Herrlich,Strecker: Abstract and concrete
categories> |(see Def 5.28 for details and a funny
picture).> Subtitled, The Joy of Cats. The illustration at
the beginning of> each chapter features a cat doing various
things. :-) The illustration> of unique transportability is
of the machine that you should imagine> doing the unique
transportation, I think.>...> Keith Ramsay>Cat-egory
theory--I get it. Or maybe a main inspiration behind the
bookwas a girl named after a cat, which can happen sometimes.
So the cat isreally altogether quite various in the things it
is doing? Meow.
===
post of Tom
Leinster(http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&
oe=UTF-8&edition=us&selm=73362839.0308250747.33206a21%
40posting.google.com):This is a special case of the notion
of discrete bration, or moreaccurately, discrete opbration.
A functor Q: J ---> K is called adiscrete opbration if: for
any object j of J and any morphism g: Q(j) ---> k in K, there
is a unique morphism f: j ---> j in J such that Q(f) = g.I
guess what you are saying is that my diagram condition is
equivalentto the discrete opbration condition holding for
the faithful functorobtained by restricting both the domain
and range of the faithfulfunctor F I mentioned to the
respective subcategories containing asmorphisms precisely the
isomorphisms. I get what you are saying now.
===
|I see the
following denition of discrete opbration in
a|(http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-
8&edition=us&selm=73362839.0308250747.33206a21%
40posting.google.com):||This is a special case of the notion
of discrete bration, or more|accurately, discrete opbration.
A functor Q: J ---> K is called a|discrete opbration if:||
for any object j of J and any morphism g: Q(j) ---> k in K,
there | is a unique morphism f: j ---> j in J such that Q(f)
= g.||I guess what you are saying is that my diagram
condition is|equivalent to the discrete opbration condition
holding for the|faithful functor obtained by restricting both
the domain and range of|the faithful functor F I mentioned to
the respective subcategories|containing as morphisms
precisely the isomorphisms.yes. its quite possible that
theres also some special terminologyfor a functor between
(possibly non-groupoid) categories that gives adiscrete
bration on the groupoid level, but i dont know what itmight
be offhand.also, although i dont think that ive read the
section of bourbakithat you mentioned, i had gotten the
impression from second-handsources that in that section
bourbaki was actually doing somethingcloser to groupoid
theory than to category theory. it might have beena false
impression though.--
===
if im not misunderstanding what you
said too badly, then it sounds> like you really want to
consider only the case where c and d are> groupoids rather
than arbitrary categories. (in particular when you> take the
objects of d to be the sets it seems that you want the>
morphisms to be the bijections rather than the arbitrary
functions.)> in that case i think the standard name for a
functor of the type> youre describing would be something
like discrete bration, and> yes theyre used for exactly
the purpose that youre describing.No. For instance, F might
be the forgetful functor from the categoryof topological
spaces (with morphisms the continuous maps) to thecategory of
sets. Given any bijection f from the underlying set of
atopological space U into a set S, there is a unique topology
on Smaking f a homeomorphism. Thus, F would be a functor
having theproperties I enumerated, notwithstanding neither
the category of setsnor the category of topological spaces
are groupoids. It is true thatBourbaki (in E.IV) only
considers F into the category of sets--in facthe always takes
F to be a forgetful functor. But he demands more thanthat F be
a forgetful functor (assuming forgetful is dened asMacLane
vaguely denes it)--e.g., to be a legitimate concept
ofmorphism in the Bourbaki sense, the identity map is an
isomorphismbetween two objects having the same underlying set
if and only if thestructures on these objects are
identical.
===
|> if im not misunderstanding what you said
too badly, then it sounds|> like you really want to consider
only the case where c and d are|> groupoids rather than
arbitrary categories. (in particular when|> you take the
objects of d to be the sets it seems that you want the|>
morphisms to be the bijections rather than the arbitrary|>
functions.) in that case i think the standard name for a
functor|> of the type youre describing would be something
like discrete|> bration, and yes theyre used for exactly
the purpose that|> youre describing.|||No. For instance, F
might be the forgetful functor from the category|of
topological spaces (with morphisms the continuous maps) to
the|category of sets. Given any bijection f from the
underlying set of a|topological space U into a set S, there
is a unique topology on S|making f a homeomorphism. Thus, F
would be a functor having the|properties I enumerated,
notwithstanding neither the category of sets|nor the category
of topological spaces are groupoids.you might as well just
consider u(F) where u is the forgetful functorfrom the
category of categories to the category of groupoids, as faras
i can tell from the description youve given.--
===
>I have
the following application: >I have a force sensor probe that
touches 3 points on a plate. At each>of the points, I can
calculate an XYZ point representing the point>that the probe
detectes the plate.>I can easily calculate the equation of
the plane Ax+By+Cz+D = 0. What>I really need is the rotation
matrix/Euler angles as represented by>the plane so I can
correctly translate positions along the slope of>the
plane.What do you mean by rotation as represented by the
plane? There isno preferred orientation in 3d, so you have
to specify which is theinitial orientation relative to which
the rotation of the plane is tobe calculated.So for example,
if the initial plane is the standard xy plane, withnormal
vector n1=(0,0,1), and the destination plane equation
isAx+By+Cz+D=0 with normal vector n2=(A,B,C) then you can
rst createthe necessary rotation in axis/angle format, then
convert to matrixformat to extract euler angles.
Seehttp://www.geocities.com/scroussette/rotplana.html#rotn>
Internally, I am storring all my locations and frames of
reference as>XYZ points and Euler angles in terms of DZ, DX,
DZ rotations.>Any thoughts would be appreciated
===
>Youve
been smoking that stuff for far too many years!!!!!!>...> --
Two sets A and B are the same size, |A| = |B|, if and only
if> there exists a bijection between the two sets.>If size
is a natural number, then you cant take the size of>an
innite set (without additional axioms and denitions).>At
the moment you say that innite sets are of the same
size>when there exists a bijection, then you already
introduce some>of the choices Cantor made. But those are
choices, not proofs.> -- If B is a subset of A, then A is
at least the same size> as B, that is, |B| =< |A|.> (There
is more that needs to be shown. For example, for> all sets A
and B, either |A| =< |B| or |B| =< |A|, and> that |A| =< |B|
and |A| =< |B| together imply |A| = |B|. I think> those two
points are all we have to assume, though.)> We have shown
_there does not exist_ a bijection between> the natural
numbers and the reals. Remember we have looked> at _all_ the
potential bijections and _all_ of them failed> at at least
one point.>This is indeed proven.> As we understand the
size of innite sets, that means N> and R are different
sizes. Its easy to show |N| =< |R|,> since N is a subset of
R. Thus |N| < |R|.>This is Cantors choice, not a proof.>You
may also conclude that the logical system you are
currently>using is not capable of denining all irrational
numbers and>leave it there. So, there are at least two
choices you can>make:>a. You may refer to R even when there
is no logical system> that can list all its elements (this is
Cantors choice).>b. You must be aware that R is never be
complete in your> logical system (alternative to Cantor).>I
dont think choice b is very attractive, but to my opinion>it
is a way you can try to go.>However, I always questioned if
saying that |N| < |R|>has any more meaning than saying that
irrational numbers>are green.>Lucas>
===
Greens functions
often have logarithmic terms in them, so they areoften
irrational functions.So irrational( function)s can be
Green(s fuctions).-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
> ...> -- Two sets
A and B are the same size, |A| = |B|, if and only if> there
exists a bijection between the two sets.> If size is a
natural number, then you cant take the size of> an innite
set (without additional axioms and denitions).At the moment
you say that innite sets are of the same size> when there
exists a bijection, then you already introduce some> of the
choices Cantor made. But those are choices, not proofs.-- If
B is a subset of A, then A is at least the same size> as B,
that is, |B| =< |A|.> (There is more that needs to be shown.
For example, for> all sets A and B, either |A| =< |B| or |B|
=< |A|, and> that |A| =< |B| and |A| =< |B| together imply
|A| = |B|. I think> those two points are all we have to
assume, though.)> We have shown _there does not exist_ a
bijection between> the natural numbers and the reals.
Remember we have looked> at _all_ the potential bijections
and _all_ of them failed> at at least one point.> This is
indeed proven.As we understand the size of innite sets, that
means N> and R are different sizes. Its easy to show |N| =<
|R|,> since N is a subset of R. Thus |N| < |R|.Although the
cardinality of the naturals is smaller than thecardinality of
the reals, it is not because the naturals are a subsetof the
reals. The natural numbers are also a subset of the
rationalnumbers yet these two sets have the same cardinality.
In fact,innite sets are sometimes dened as sets of elements
such thatthere is a one-to-one correspondense between the
elements of the setand the elements of a proper subset of
that set.Darren> This is Cantors choice, not a proof.You may
also conclude that the logical system you are currently> using
is not capable of denining all irrational numbers and> leave
it there. So, there are at least two choices you can> make:>
a. You may refer to R even when there is no logical system>
that can list all its elements (this is Cantors choice).> b.
You must be aware that R is never be complete in your> logical
system (alternative to Cantor).I dont think choice b is very
attractive, but to my opinion> it is a way you can try to
go.However, I always questioned if saying that |N| < |R|> has
any more meaning than saying that irrational numbers> are
green.> Lucas
===
...> -- Two sets A and B are the same size,
|A| = |B|, if and only if> there exists a bijection between
the two sets.> If size is a natural number, then you cant
take the size of> an innite set (without additional axioms
and denitions).Then size is not a natural number for innite
sets. Is thisa problem?> At the moment you say that innite
sets are of the same size> when there exists a bijection,
then you already introduce some> of the choices Cantor made.
But those are choices, not proofs.OK, those are choices (or
denitions, as I would say).They are, however, very good
denitions. They allow usto extend the idea of sets size to
innite sets and preserve many of the prperties of size that
we are familiar with from nite sets.> -- If B is a subset of
A, then A is at least the same size> as B, that is, |B| =<
|A|.> (There is more that needs to be shown. For example,
for> all sets A and B, either |A| =< |B| or |B| =< |A|, and>
that |A| =< |B| and |A| =< |B| together imply |A| = |B|. I
think> those two points are all we have to assume, though.)>
We have shown _there does not exist_ a bijection between> the
natural numbers and the reals. Remember we have looked> at
_all_ the potential bijections and _all_ of them failed> at
at least one point.> This is indeed proven.As we understand
the size of innite sets, that means N> and R are different
sizes. Its easy to show |N| =< |R|,> since N is a subset of
R. Thus |N| < |R|.> This is Cantors choice, not a proof.You
may also conclude that the logical system you are currently>
using is not capable of denining all irrational numbers and>
leave it there. So, there are at least two choices you can>
make:> a. You may refer to R even when there is no logical
system> that can list all its elements (this is Cantors
choice).> b. You must be aware that R is never be complete in
your> logical system (alternative to Cantor).If I redene R
using other axioms, it may be that this newR could be
listed, but so what? There are many sets that are countable.
(Many is a bit of an understatement.) The factthat R is
countable does not remove the fact that R, describedby the
original axioms, is not countable.> I dont think choice b is
very attractive, but to my opinion> it is a way you can try to
go.I nd choice b very unattractive, too. Its as though
someonedecides the consequences of some axioms are wrong or
uglyor something. Say they dont like Eulers formula exp(i
theta) = cos(theta) + i sin(theta)so they mess around with
the axioms for complex analysisuntil it goes away. It doesnt
seem like a very good way todo things.> However, I always
questioned if saying that |N| < |R|> has any more meaning
than saying that irrational numbers> are green.I know what I
mean when I say |N| < |R|. I mean something aboutthe
existence of functions between N and R. I also know Ican
often get away with thinking about |N| and |R| like thesizes
of nite sets and when I need to be morecareful. We often
deal with things in math by pretending wereimagining things
we cant really imagine, like innite-dimensional vector
spaces, or the empty set.What can you say about irrationals
being green?Jim Burns
===
> However, I always questioned if
saying that |N| < |R|> has any more meaning than saying that
irrational numbers> are green.thats obviously because they
are purple!
===
> However, I always questioned if saying
that |N| < |R|> has any more meaning than saying that
irrational numbers> are green.> thats obviously because they
are purple!Dude...take off those colored glasses... Isee them
clearly as Chartreuse.... with pink spots...>
===
>
However, I always questioned if saying that |N| < |R|> has
any more meaning than saying that irrational numbers> are
green.The irrationals are green, and justiably so IMO, out
of envy of therationals.> thats obviously because they are
purple!> Dude...take off those colored glasses... I> see
them clearly as Chartreuse.... with pink spots...Now somewhat
seriously:Besides being a mathematician, I am also a musician.
As such, I am awarethat some people experience synesthesia.
The composer Scriabin was famousfor this, experiencing
certain color sensations when he heard certainpitches. I
would not be greatly surprised if there were some people
whoexperience certain color sensations when contemplating
certain numbers.Does anyone know of any instances of
such?David
===
Now somewhat seriously:> Besides being a
mathematician, I am also a musician. As such, I am aware>
that some people experience synesthesia. The composer
Scriabin was famous> for this, experiencing certain color
sensations when he heard certain> pitches. I would not be
greatly surprised if there were some people who> experience
certain color sensations when contemplating certain numbers.>
Does anyone know of any instances of such?DavidThere is a
chapter on this sort of thing in Underwood Dudleys
bookMathematical Cranks. That material (number forms I
think they arecalled?) is a bit different than the rest of
the book.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
> However, I
always questioned if saying that |N| < |R|> has any more
meaning than saying that irrational numbers> are green.> The
irrationals are green, and justiably so IMO, out of envy of
the> rationals.> thats obviously because they are purple!>
Dude...take off those colored glasses... I> see them clearly
as Chartreuse.... with pink spots...> Now somewhat
seriously:> Besides being a mathematician, I am also a
musician. As such, I am aware> that some people experience
synesthesia. The composer Scriabin was famous> for this,
experiencing certain color sensations when he heard certain>
pitches. I would not be greatly surprised if there were some
people who> experience certain color sensations when
contemplating certain numbers.> Does anyone know of any
instances of such?> DavidI was only refferancing the subject
line of irrationaland why I see Chartreuse.... with pink
spots...
===
|-|erc says...>You said that it misses non
computable numbers, like r = Halt(1) + 1/10 Halt(2)>+ 1/100
Halt(3)...>since Halt(n) is unknown for all n.>i.e. r is not
computable and will not be listed in UTM(Z).>I tried to argue
r is just not known, you tried to show r is not computable
at>all (standard view),>now I am now suggesting r it not
existent! so Computables = Reals.Yes, it is perfectly
consistent to believe that the only real numbersthat exist
are computable reals. However, Cantors theorem is
*still*valid: Under this interpretation, it says that there
is no computablelisting of all computable reals.--Daryl
McCulloughIthaca, NY
===
> There is only 1 innity type, the
answer is NO, heres a similar exposition of
diagonalisation.Not true. There are an innite number of
transnite cardinals.Bob Kolker
===
> What is an irrational
number? Can you count to it?> Can you pin point it? There is
no such thing.Not true. The reals can be built from the
rationals in several equavalent ways. 1. Dedikind Cuts 2.
Equivalence Classes of rational Cauchy SequencesAll numbers
are the result of computable functions, sqrt(2) is the>
result of turing machine(x), where x is some integer,
probably> under a million with any crude mapping technique.A
Cauchy Sequence converging to sqrt(2) can be generated by a
non-stopping turing machine.>Bob Kolker
<3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>
<1Ff9oJOLyx1$Ew6T@baesystems.com>
<2a0cceff.0312101407.2225db91@posting.google.com>
===
In
message <2a0cceff.0312101407.2225db91@posting.google.com>,
Edward > In message
<3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern><...>
sqrt(2) is the> result of turing machine(x), where x is some
integer, probably> under a million with any crude mapping
technique.>Thats incomprehensible. Could you rephrase
this, please?> He means that any Turing machine can be
represented by a tape fed to a> universal Turing machine,
and the contents of that tape can be> represented by some
number x.>Well, lets forget about the Turing machine for a
second. Lets just>note that sqrt(2) is computable in the
sense that, armed with some>nite algorithm and an indenite
amount of storage, we can compute>an arbitrary number of
digits of it.>Ok ... that does sound a lot like a Turing
machine after all. ;-)>This stirs some vague recollection:
there are some privileged>irrational numbers which can be
specied with a nite amount of>information. Obviously _all_
irrational numbers cannot be so>specied, or they in fact
would be countable. So most irrational>numbers are poor lost
souls which not only have non-repeating
decimal>representations, but cant even be named in any
meaningful way -- they>are unknowable. (This was the subject
of some (IBM?) news release>within the past decade, possibly
one of those over-hyped news releases>which appear regularly,
repeating essentially known results as if were>fresh
revelation).>Can somebody remind me what this concept is
called?I think youre thinking of the distinction between
algebraic (roots of a polynomial) and transcendental (the
rest) ?<thinks> I wonder if there are any other possibilities
- read on...>Do these>unknowable irrational numbers correspond
to non-computable functions?Or to unknown unknowns? ;-)> That
only leaves two types of numbers left that qualify for
irrational,> non computable and random numbers.>Well, I
would say that irrational numbers are not computable; why
do>you think otherwise?> Because hes confused about the
halting problem.>So am I, evidently -- would you refresh my
memory of its signicance>here?In essence, you cant devise a
universal algorithm which, given _any_ algorithm as input,
will tell you whether or not the other halts. That doesnt
mean that for any particular algorithm you cant determine
what it does.>And, repeating the argument above, surely
_some_ irrational numbers>are in fact computable?Only the
ones you can write down ;-)Dont forget, the complete answer
hasnt been computed until the algorithm halts.> Non
computable numbers is not a proof that irrationals exist,
IMO,> that no halting function exists does not clearly dene
what the halting> number is, it states it is impossible.
There is no gap on the number line> from non computable
numbers.>Thats incomprehensible again.> I think hes
trying to argue that because there exist TMs for which we>
cant determine to which of the sets halts or does not
halt they> belong, the does not halt set must be
empty.>Ah yes ... irrational numbers correspond to the
output of a machines>which neither halts nor loops.;-) I have
one here that does that - sometimes it just freezes. Perhaps
its running in imaginary time?> As to what he is trying to
say, OTOH,>I would feel comfortable leaving it in the
incomprehensible bin.Aha! Thats the missing category:
algebraic, transcendental, or incomprehensible.>Ah,
knowledge, ah, intellect, ah, thought, I miss you ... you
jade.-- Richard Herring
===
Richard Herring <junk@[127.0.0.1]>
most amusingly scrivened in message> In message
<2a0cceff.0312101407.2225db91@posting.google.com>, Edward >
In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>,
Bjoern> ... there are some privileged>irrational numbers
which can be specied with a nite amount of>information.
Obviously _all_ irrational numbers cannot be so>specied, or
they in fact would be countable. So most irrational>numbers
are poor lost souls which not only have non-repeating
decimal>representations, but cant even be named in any
meaningful way -- they>are unknowable. ...>Can somebody
remind me what this concept is called?I think youre thinking
of the distinction between algebraic (roots of a > polynomial)
and transcendental (the rest) ?> <thinks> I wonder if there
are any other possibilities - read on...Maybe -- Feuerbacher
says the same. But I think Im thinking ofessentially what I
am thinking of, if you follow my thought.To put us on the
same page, I understand computable to meanproduceable as
the output of a Turing machine. The term is normallyapplied
to integer functions of the integers, but obviously a
decimalrepresentations of numbers can be considered a
function on theintegers.I presume all algebraic numbers are
computable, but at least some ofthe rest of em are too:
anything we could write down a niteinstruction set for --
given sufcient time and paper -- calculatingany needed
number of digits of, is computable: a class which includespi
and e.The argument for the existence of the unknowable
numbers is that allnite algorithms or specications form a
countable set: so almostall of the irrationals can be neither
unambiguously specied norcomputed....> That only leaves two
types of numbers left that qualify for irrational,> non
computable and random numbers.I was silent the rst time, but
this claim is obviously completenonsense: the denition of
irrational is ... well, you know what itis ... not non
computable, and a random number is a meaninglessidea in
this context. The number line contains numbers, not
randomnumbers. But the OP is still to be thanked for giving
an occasion todisplay foils of learnedness.>Well, I would say
that irrational numbers are not computable; why do>you think
otherwise?> Because hes confused about the halting
problem.>So am I, evidently -- would you refresh my memory
of its signicance>here?In essence, you cant devise a
universal algorithm which, given _any_ > algorithm as input,
will tell you whether or not the other halts. That > doesnt
mean that for any particular algorithm you cant determine
what > it does.Right ... but -- given the admited silliness
of asking for details ofa hypothetical nonsensical argument
-- what do you _think_ he thinksthis has to do with the
existence of irrational numbers?>And, repeating the argument
above, surely _some_ irrational numbers>are in fact
computable?Only the ones you can write down ;-)Exactly -- if
write down means give a nite algoritm to compute. >
Dont forget, the complete answer hasnt been computed until
the > algorithm halts.Hmm... maybe thats right. The Turing
machine in general is free togo back and erase parts of the
tape, so that you can never be sure anygiven bit will stand
until the machine stops? Is that it?Yet I think there is a
concept of computable for innite strings ofdigits; Im
just not sure how this ts together. We can imagine amachine
with two tapes, one non-erasable, such that the n-th digit
isnot written on that tape until the algorithm is sure thats
the realn-th digit. Thats not a Turing machine, but anything
any bastardizedcomputing machine can do, a general Turing
machine can do better ...right? ... so we ought to be able to
emulate this machine with aTuring machine.Maybe this is a
red-herring: as you say, the fact that no generalhalting test
exists doesnt prevent us from being able to analyze_some_
algorithms with certainty to know that they halt.
Similarly,although a general Turing machine might back up and
write over anyspace in the tape before the algoritm stops,
some algorithms might beanalyzable to prove that, for
example, only the last 5 writtenpositions on the tape are
labile, so at any given stage all earilerpositions are
immutable.Clearly a Turing machine which prints a complete
function on N (whichis not trivially zero above some range)
is never going to stop, yet Idont think computable
functions are limited to those with onlynite information:
this is some semantic problem here.>Ah yes ... irrational
numbers correspond to the output of a machines>which neither
halts nor loops.;-) I have one here that does that -
sometimes it just freezes. Perhaps > its running in
imaginary time?Pshaw. You know perfectly well that your
frozen machine is runningin a loop down on the CPU level. >
As to what he is trying to say, OTOH,>I would feel comfortable
leaving it in the incomprehensible bin.Aha! Thats the
missing category: algebraic, transcendental, or >
incomprehensible.:-)))Similar to a serious classication of
statements though: right,wrong, and nonsense.
===
> In
message <2a0cceff.0312101407.2225db91@posting.google.com>,
Edward ... >This stirs some vague recollection: there are
some privileged >irrational numbers which can be specied
with a nite amount of >information. Obviously _all_
irrational numbers cannot be so >specied, or they in fact
would be countable. So most irrational >numbers are poor lost
souls which not only have non-repeating decimal
>representations, but cant even be named in any meaningful
way -- they >are unknowable. (This was the subject of some
(IBM?) news release >within the past decade, possibly one of
those over-hyped news releases >which appear regularly,
repeating essentially known results as if were >fresh
revelation). >Can somebody remind me what this concept is
called? > I think youre thinking of the distinction
between algebraic (roots of a > polynomial) and
transcendental (the rest) ? > <thinks> I wonder if there are
any other possibilities - read on...Wrong. Some
transcendentals can be calculated. Pi and e to name
two.(Where I mean calculated to mean that you can give a
nite algorithmto calculate an arbitrary amount of digits.)
The concept is computable.And indeed, the set of
computable numbers is countable. There is, however,a
beautiful problem when you try to apply a Cantor like
argument to a(complete) list of computable numbers. This
would tell that you get anew computable number (you just
computed it) that is not on the list.The problem here is that
a complete list of computable numbers is itselfnot
computable.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland; http://www.cwi.nl/~dik/
<2a0cceff.0312101407.2225db91@posting.google.com>
<DO2A6HOhHa2$Ew1x@baesystems.com> <HpsE39.9p0@cwi.nl>
===
> In
message <2a0cceff.0312101407.2225db91@posting.google.com>,
Edward>...>This stirs some vague recollection: there are some
privileged>irrational numbers which can be specied with a
nite amount of>information. Obviously _all_ irrational
numbers cannot be so>specied, or they in fact would be
countable. So most irrational>numbers are poor lost souls
which not only have non-repeating decimal>representations,
but cant even be named in any meaningful way -- they>are
unknowable. (This was the subject of some (IBM?) news
release>within the past decade, possibly one of those
over-hyped news releases>which appear regularly, repeating
essentially known results as if were>fresh revelation).>Can
somebody remind me what this concept is called?> I think
youre thinking of the distinction between algebraic (roots
of a> polynomial) and transcendental (the rest) ?> <thinks>
I wonder if there are any other possibilities - read
on...>Wrong.And seasons greetings to you too. _Whats_
wrong? Last I heard, those were (near enough) the denitions
of algebraic and transcendental. Or are you saying that what
I think Ed is thinking is wrong? Unless you are reading his
mind, you cant possibly tell.Actually I think you missed the
joke. Never mind, its not worth labouring it.> Some
transcendentals can be calculated. Pi and e to name
two.>(Where I mean calculated to mean that you can give a
nite algorithm>to calculate an arbitrary amount of
digits.)To use your own word, Wrong. All you have
calculated is a rational approximation to the number.
Thats not the number itself, though it can approximate it as
closely as you would like.>The concept is computable.>And
indeed, the set of computable numbers is countable. There is,
however,>a beautiful problem when you try to apply a Cantor
like argument to a>(complete) list of computable numbers.
This would tell that you get a>new computable number (you
just computed it) that is not on the list.>The problem here
is that a complete list of computable numbers is itself>not
computable.So you agree that the computable numbers are
incomprehensible.-- Richard Herring
===
> In message
<2a0cceff.0312101407.2225db91@posting.google.com>,
Edward>...>This stirs some vague recollection: there are
some privileged>irrational numbers which can be specied
with a nite amount of>information. Obviously _all_
irrational numbers cannot be so>specied, or they in fact
would be countable. So most irrational>numbers are poor lost
souls which not only have non-repeating
decimal>representations, but cant even be named in any
meaningful way -- they>are unknowable. (This was the
subject of some (IBM?) news release>within the past decade,
possibly one of those over-hyped news releases>which appear
regularly, repeating essentially known results as if
were>fresh revelation).>Can somebody remind me what this
concept is called?> I think youre thinking of the
distinction between algebraic (roots of a> polynomial) and
transcendental (the rest) ?> <thinks> I wonder if there are
any other possibilities - read on...>Wrong.And seasons
greetings to you too. _Whats_ wrong? Last I heard, those
were (near enough) the denitions > of algebraic and
transcendental. Or are you saying that what I think Ed > is
thinking is wrong? Unless you are reading his mind, you cant
> possibly tell.Actually I think you missed the joke. Never
mind, its not worth > labouring it.> Some transcendentals
can be calculated. Pi and e to name two.>(Where I mean
calculated to mean that you can give a nite algorithm>to
calculate an arbitrary amount of digits.)To use your own
word, Wrong. All you have calculated is a rational >
approximation to the number. Thats not the number itself,
though it can > approximate it as closely as you would
like.He is talking about a perfectly reasonable notion of
calculated.Pick a number.It is computable if and only if
there is an algorithm A taking inputn such that A(n) returns
the rst n digits of your number.Your point is that for all
n, A(n) is merely a rational approximation.And thats
certainly correct.in this sense then A is an algorithm that
calculates the exactreal number. Not just an approximation.
The real thing.If you prefer, we could have A() return
rational numbers such thatthe sequence A(i), i = 1 to oo is
Cauchy. Or we could have A()take rational numbers as input
and return MEMBER or NON-MEMBERso that the set {x: x in Q
and A(x) = MEMBER} is an upperDedekind cut.There are many
formalisms that encompass the notion of computable.Deciding
whether a particular algorithm calculates a particularreal
number ought not depend on whether the printer has aninnite
supply of ink. John Briggs
===
> In message
<2a0cceff.0312101407.2225db91@posting.google.com>, Edward...
> So most irrational >numbers are poor lost souls which not
only have non-repeating decimal >representations, but cant
even be named in any meaningful way -- they >are unknowable.
(This was the subject of some (IBM?) news release >within the
past decade, possibly one of those over-hyped news releases
>which appear regularly, repeating essentially known results
as if were >fresh revelation). >Can somebody remind me
what this concept is called? > I think youre thinking of
the distinction between algebraic (roots of a > polynomial)
and transcendental (the rest) ? > <thinks> I wonder if there
are any other possibilities - read on... >Wrong. > And
seasons greetings to you too. > _Whats_ wrong? Last I
heard, those were (near enough) the denitions > of
algebraic and transcendental. Or are you saying that what I
think Ed > is thinking is wrong? Unless you are reading his
mind, you cant > possibly tell.Well, this is incorrect. See
the word named in a meaningful way? > Actually I think
you missed the joke. Never mind, its not worth > labouring
it. > Some transcendentals can be calculated. Pi and e to
name two. >(Where I mean calculated to mean that you can give
a nite algorithm >to calculate an arbitrary amount of
digits.) > To use your own word, Wrong. All you have
calculated is a rational > approximation to the number.
Thats not the number itself, though it can > approximate
it as closely as you would like.The same holds for quite a
few algebraic numbers. Or do you have someway to calculate
(or name) the roots of, say, x^5 + 5x + 3, and todistinguish
the roots?nite space you can calculate an arbitrary amount
of digits.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland; http://www.cwi.nl/~dik/
<xTRcN6n8qd2$EwjF@baesystems.com> <HpsH1x.5FM@cwi.nl>
===
> In
message <2a0cceff.0312101407.2225db91@posting.google.com>,
Edward>...> So most irrational>numbers are poor lost
souls which not only have non-repeating decimal>
>representations, but cant even be named in any meaningful
way -- they>are unknowable. (This was the subject of some
(IBM?) news release>within the past decade, possibly one of
those over-hyped news releases>which appear regularly,
repeating essentially known results as if were>fresh
revelation).>Can somebody remind me what this concept is
called?> I think youre thinking of the distinction between
algebraic (roots of a> polynomial) and transcendental (the
rest) ?> <thinks> I wonder if there are any other
possibilities - read on...>Wrong.> And seasons greetings
to you too.> _Whats_ wrong? Last I heard, those were (near
enough) the denitions> of algebraic and transcendental. Or
are you saying that what I think Ed> is thinking is wrong?
Unless you are reading his mind, you cant> possibly
tell.>Well, this is incorrect._What_ is incorrect?>See the
word named in a meaningful way?Yes, thats in Ed Greens
posting, not mine. Take it up with him, dont yell wrong at
me.> Actually I think you missed the joke. Never mind, its
not worth> labouring it.> Some transcendentals can be
calculated. Pi and e to name two.>(Where I mean calculated to
mean that you can give a nite algorithm>to calculate an
arbitrary amount of digits.)> To use your own word, Wrong.
All you have calculated is a rational> approximation to the
number. Thats not the number itself, though it can>
approximate it as closely as you would like.>The same holds
for quite a few algebraic numbers.Of course.>Or do you have
some>way to calculate (or name) the roots of, say, x^5 + 5x +
3, and to>distinguish the roots?No.>nite space you can
calculate an arbitrary amount of digits.Thats not very
precise. I think you mean that it is said to be computable if
your algorithm can calculate a rational which approximates to
it with an arbitrarily small error bound. Note that the thing
you _calculate_ is rational.Isnt that what I said above?--
Richard Herring
===
>nite space you can calculate an
arbitrary amount of digits.Thats not very precise. I think
you mean that it is said to be> computable if your algorithm
can calculate a rational which> approximates to it with an
arbitrarily small error bound. Note that> the thing you
_calculate_ is rational.Isnt that what I said above?I think
you two are violently agreeing on this point. Still, Diks
isthe usual denition, though I think yours is certainly
equivalent.Thomas
===
Richard Herring <junk@[127.0.0.1]>
says...>So most irrational numbers are poor lost souls
which not only>have non-repeating decimal representations,
but cant even be>named in any meaningful way -- they are
unknowable...>Can somebody remind me what this concept is
called?Richard Herring responds:> I think youre thinking of
the distinction between algebraic> (roots of a polynomial)
and transcendental (the rest) ?> <thinks> I wonder if there
are any other possibilities - read on...Dik Winter said
Wrong because Ed Green is clearly talking aboutnumbers that
are unknowable that cant even be named in anymeaningful
way. Those descriptions are not true of the numberspi and e,
for example. pi and e are just as knowable assquare-root(2).
So the algebraic/transcendental distinctiondoesnt capture
what Ed was talking about.> Some transcendentals can be
calculated. Pi and e to name two.>(Where I mean calculated
to mean that you can give a nite algorithm>to calculate an
arbitrary amount of digits.)> To use your own word, Wrong.
All you have calculated is a rational> approximation to the
number. Thats not the number itself, though it can>
approximate it as closely as you would like.Thats what it
*means* to say that a real number is computable.A number like
pi or e is computable if and only if we canapproximate them as
closely as you like. So it really isnt correctto call them
unknowable, we can know quite a bit about them.In contrast,
there are some numbers that are *not* computable. Theycannot
be approximated as closely as you like. If a number r
isnoncomputable, there does not exist an algorithm for
computing itsdigits. So there is a real sense that a
noncomputable number ismuch less knowable than e or pi.>The
same holds for quite a few algebraic numbers.>Of course.Then
it follows that the algebraic/transcendental distinctionis not
what Ed was getting at. Algebraic numbers are noless
unknowable than computable transcendental numbers.--Daryl
McCulloughIthaca, NY <W1o1UUxame2$EwUJ@baesystems.com>
<brcrae01srt@drn.newsguy.com>
===
In message
<brcrae01srt@drn.newsguy.com>, Daryl McCullough >Richard
Herring <junk@[127.0.0.1]> says...>So most irrational
numbers are poor lost souls which not only>have
non-repeating decimal representations, but cant even be>
>named in any meaningful way -- they are unknowable...>Can
somebody remind me what this concept is called?>Richard
Herring responds:> I think youre thinking of the distinction
between algebraic> (roots of a polynomial) and transcendental
(the rest) ?> <thinks> I wonder if there are any other
possibilities - read on...>Dik Winter said Wrong because
Ed Green is clearly talking about>numbers that are
unknowable that cant even be named in any>meaningful
way. Those descriptions are not true of the numbers>pi and
e, for example. pi and e are just as knowable
as>square-root(2). So the algebraic/transcendental
distinction>doesnt capture what Ed was talking
about.clarication._I_ was referring to Eds earlier words in
the same paragraph:>This stirs some vague recollection: there
are some privileged>irrational numbers which can be specied
with a nite amount of>information.where I took the nite
amount of information to mean coefcients of a [nite]
polynomial. If you take the name of a number to be the set
of polynomial coefcients which denes it, then clearly the
transcendentals, even pi and e, dont have a name in that
sense.> Some transcendentals can be calculated. Pi and e to
name two.>(Where I mean calculated to mean that you can give
a nite algorithm>to calculate an arbitrary amount of
digits.)> To use your own word, Wrong. All you have
calculated is a rational> approximation to the number.
Thats not the number itself, though it can> approximate it
as closely as you would like.>Thats what it *means* to say
that a real number is computable.>A number like pi or e is
computable if and only if we can>approximate them as closely
as you like. So it really isnt correct>to call them
unknowable, we can know quite a bit about them.>In
contrast, there are some numbers that are *not* computable.
They>cannot be approximated as closely as you like. If a
number r is>noncomputable, there does not exist an algorithm
for computing its>digits. So there is a real sense that a
noncomputable number is>much less knowable than e or pi.>The
same holds for quite a few algebraic numbers.>Of
course.>Then it follows that the algebraic/transcendental
distinction>is not what Ed was getting at. Algebraic numbers
are no>less unknowable than computable transcendental
numbers.>But they are less unnameable.-- Richard
Herring
===
So you agree that the computable numbers are
incomprehensible.Nonsense. Computability can be clearly
dened hence the concept is comprehensible, as is the concept
of a computable number. The square root of two is computable.
I know an algorithm that will give you the n-th digit of the
decimal expansion for any n. The sqare root of two is totally
comprehnsible and easily constructed with a compass and
straightedge.Bob Kolker
<2a0cceff.0312101407.2225db91@posting.google.com>
<DO2A6HOhHa2$Ew1x@baesystems.com> <HpsE39.9p0@cwi.nl>
<xTRcN6n8qd2$EwjF@baesystems.com>
<brcnmd$2023c$1@ID-76471.news.uni-berlin.de>
===
In message
<brcnmd$2023c$1@ID-76471.news.uni-berlin.de>, Robert J.
>And indeed, the set of computable numbers is countable.
There is, however,>a beautiful problem when you try to apply
a Cantor like argument to a>(complete) list of computable
numbers. This would tell that you get a>new computable number
(you just computed it) that is not on the list.>The problem
here is that a complete list of computable numbers is
itself>not computable.> So you agree that the computable
numbers are incomprehensible.>Nonsense. Computability can be
clearly dened hence the concept is >comprehensible, as is the
concept of a computable number.<fx:
whooooooooooooooooooosssssssssssssssssshhhhhhhhhhh!>
Comprehensible has more than one
meaning:http://www.ccel.org/creeds/athanasian.creed.htmlThe
computable numbers cannot be _comprehended_ within a
computable list.>The square root of two is computable. I know
an algorithm that will >give you the n-th digit of the decimal
expansion for any n.Thats some puny rational approximation.>
The sqare root of two is totally comprehnsible and easily
constructed >with a compass and straightedge.>Ah. Thats more
like it.-- Richard Herring
===
> In message
<brcnmd$2023c$1@ID-76471.news.uni-berlin.de>, Robert J. >The
square root of two is computable. I know an algorithm that
will >give you the n-th digit of the decimal expansion for
any n.Thats some puny rational approximation.No. Thats a
set of puny rational approximations, one for each n.Take a
course in real analysis someday and see what a real numberis
dened to be in the Cauchy model. John Briggs
<3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>
<1Ff9oJOLyx1$Ew6T@baesystems.com>
<3FD83D0A.1BF67E01@ix.urz.uni-heidelberg.de>
===
In message
<3FD83D0A.1BF67E01@ix.urz.uni-heidelberg.de>, Bjoern > In
message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>,
Bjoern>[snip]> sqrt(2) is the> result of turing
machine(x), where x is some integer, probably> under a
million with any crude mapping technique.>Thats
incomprehensible. Could you rephrase this, please?> He
means that any Turing machine can be represented by a tape
fed to a> universal Turing machine, and the contents of that
tape can be> represented by some number x.>How can the
content of the whole tape be represented with a single>number
x?John Briggs answered this.>And where does he get the
probably under a million from?Heaven knows.>[snip]> Non
computable numbers is not a proof that irrationals exist,
IMO,> that no halting function exists does not clearly dene
what the halting> number is, it states it is impossible.
There is no gap on the number> line from non computable
numbers.>Thats incomprehensible again.> I think hes
trying to argue that because there exist TMs for which we>
cant determine to which of the sets halts or does not
halt they> belong, the does not halt set must be
empty.>*scratches head* Does anyone understand this
logic?>Thats my point. Logic it is not.-- Richard
Herring
===
<SNIP> It seems odd Information Theory is wiped
out of existence from a singleproof> that is self
referential, can anyone give me an actual number that
isntcomputable?> Herc>You would think that it was
impossible to specify a number that wasntcomputable - the
act of specifying the number would seem to requirespecifying
the algorithmic process which generates it.Well, you would be
wrong. Here is exactly what you asked
for:http://mathworld.wolfram.com/ChaitinsConstant.html
===
>
<SNIP> It seems odd Information Theory is wiped out of
existence from a single> proof> that is self referential,
can anyone give me an actual number that isnt>
computable?> Herc> You would think that it was
impossible to specify a number that wasnt> computable - the
act of specifying the number would seem to require>
specifying the algorithmic process which generates it.>
Well, you would be wrong. Here is exactly what you asked
for:> http://mathworld.wolfram.com/ChaitinsConstant.htmlIs
this completely fair? If there is a Turing Halting problem
thatis undecidable within our own system of reasoning? Is it
then stilla proper dened number?Lucas
===
> You would think
that it was impossible to specify a number that wasnt>
computable - the act of specifying the number would seem to
require> specifying the algorithmic process which generates
it.Well, you would be wrong. Here is exactly what you asked
for:http://mathworld.wolfram.com/
ChaitinsConstant.htmlBeautiful!
===
In sci.physics, Uncle
Al<UncleAl0@hate.spam.net><3FD74A11.E8852E20@hate.spam.net>:
What is an irrational number? Can you count to it?> Can you
pin point it? There is no such thing.> sqrt(2)>
http://mathforum.org/library/drmath/view/57117.html>
http://medialab.dyndns.org/bignum/> All numbers are the
result of computable functions, sqrt(2) is the> result of
turing machine(x), where x is some integer, probably> under
a million with any crude mapping technique.> The concepts
irrational and transcendental are rigorousy dened> and
rigorously provable. Do some reading. If you think sqrt(2)
is> not irrational, supply integers for the ratio. Note that
sqrt(2) has> been taken to ridiculous lengths - though not to
the 1.241 trillion> places of pi by Yasumasa Kanada - to do
statistical correlations.Of course, any such pair of integers
has a minor problem.Assume sqrt(2) = p/q, then 2 = p^2/q^2.
Assume gcd(p,q) = 1,p must be even. Writing p^2/q^2 = 4
r^2/q^2 and ipping,we nd 2 = q^2/r^2, and therefore q must
be even as well.Whoops.Or one could pull a James Harris and
let p and q be algebraicintegers. Since sqrt(2) is an
algebraic integer (x^2 - 2 = 0is irreducible and monic),
ta-daah! Except that that doesntdo much for sqrt(2)s
(ir)rationality.>
ftp://metalab.unc.edu/pub/docs/books/gutenberg/etext94/
2sqrt10a.zip> sqrt(2) to 5 million decimal places>
http://antwrp.gsfc.nasa.gov//htmltest/gifcity/sqrt2.10mil>
sqrt(2) to 10 million decimal places>
http://www.sciencenews.org/20021214/mathtrek.asp> sqrt(A):
> 1) Choose a rough approximation G of sqrt(A). > 2) Divide
A by G and then average the quotient with G, > G* =
((A/G)+G)/2> 3) If G* is sufciently accurate, stop.
Otherwise, let G = G* and> return to step 2.> The number of
correct decimal places roughly doubles with each> repetition
of step 2.Thats a cute one. Ill have to remember that
technique. :-)If G = sqrt(A) + e, then G* = ((A/(sqrt(A) +
e)+sqrt(A)+e))/2= sqrt(A)/2 - e/2 + e^2/(sqrt(A)) + ... +
sqrt(A)/2 + e/2= sqrt(A) + e^2/(sqrt(A)) + ...and Als right
again, damnit. :-)> That only leaves two types of numbers
left that qualify for irrational,> non computable and random
numbers.> Jesus you do whine. The digit occurances of sqrt(2)
are demonstrated> to be random (with one exception) with a
huge battery of statistical> qualiers including string poker
hands (no ushes), but fractionally> less so than those of pi
(also with the obvious excpetion).> [snip]>-- #191,
ewill3@earthlink.netIts still legal to go .sigless.
===
>
Thats a cute one. Ill have to remember that technique.
:-)Its a special case of something more general: Newtons
Method.If G is an approximate solution of f(x)=0, you can
often get a betterapproximation with G* = G - f(G)/f(G)In
this case, f(x) = x^2 - A and f(x) = 2x, so it reduces to
the givenformula.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
> For example, I might not
know whether the least upper bound> of some complicated set
is computable, but I do know that,> if the set is bounded,
the LUB exists. Under your scheme,> I might not know whether
the LUB exists, but I would know> that, if it existed, it
would be computable. You might> prefer to know the second and
not the rst, but generally> speaking, knowing the rst and
not the second is more> useful in real analysis.>Its not more
useful, its a superimposed mutually sustaining> impression
that mathematics lies outside the realMs of >
computability.You say its not more useful; I say it is more
useful. How should we decide this point? I could give you
many exampleswhere the existence of the LUB is important to
the conclusiondrawn _in real analysis_. Can you do the same
with the computability property?The second sentence here is
difcult to understand. I thinkyoure saying the LUB property
is part of the whole complexthat gives the (false) impression
that mathematics extendsbeyond the computable.Why is this
impression false? You havent gone into your reasonsfor
asserting this yet, unless youre counting That wouldmean
the reals are countable as a reason. Im going toassume you
agree with me that this would be a bad reason.> Your proof
jumps to the fact of a contradiction, what is this >
contradiction?At worst, my proof _derives_ a contradiction.
It does not pullit out of thin air. Its true that I glossed
over a somethe details, but they were details youve been
over many times.Im willing to ll them in, if you really
want me to.However, several people have pointed out that you
do not need acontradiction for this result. In fact, Cantor
did not usecontradiction. In the latest version, the one you
most recently snipped, there is no contradiction, just a
demonstration thatevery list misses at least one real number.
This is a contradictiononly to the assumption that |N| = |R|.>
Is it self referencing? YES> Does the formulation of extra
numbers occur purely depending on > the original formulation
of all numbers? YESI dont know what you mean by
self-referencing, but I suspect thatif I did, I would
disagree. There is no spooky Goedel-stylemixing of levels
going on here.OK, the not-found number Cantor(L) is a
function of the list, L,it is not found in. I dont see how
it could be otherwise.Every real is in some list or other;
there is no real whichis missing from every list. > Does this
contradiction rely on specifying the new number is >
*different* from the original forumlation at nite set
points? > YES> Does the extra number reinsert into the
original list to > demontrate the imcompatibility? YESNO. The
extra number is not re-inserted into the original list.If you
think I did this, show me where.> The extra number is simply
this :> Given a number on a list, change it and put it back
in the list > without changing the list.The most generous
interpretation of this I can make is thatyou meant to say
something else. (1) The extra number is not one ofthe numbers
on the list (hence extra). (2) Changing a numberon the list
and putting it back in the list must change thelist. (3) I
havent done anything like this.I suggest you try again. This
time say what you mean.> Does the diagonal transform form a
well dened number?> NO -> computables = reals> YES ->
computables C realsWell, it would be handy for me to agree
with your implicationhere. The Cantor diagonal number is very
well dened, for a well-dened list, so it would follow that
the computables area proper subset of the reals.However, you
have not made any argument for the pairof implications. I
think I saw you _assert_ in a separate postthat uncomputable
nnumbers are not reals, but you dont givea reason for that
beyond allowing you to conclude what you want toconclude,
that |N| = |R|.> There is only 1 innity type, the answer is
NO, heres a similar > exposition of diagonalisation.Apart
from anything else in this thread, there is more thanone
innity type. You would have to re-work set theory to getrid
of that conclusion, if even that would do it. Questionsof
computability just do not enter into it.For any set A, |A| <
|P(A)|, where P(A) is the powerset of A.Original Format>
alt.paranormal, alt.sci.physics> I dont see anything new
here.Jim
Burns
===
--> For example, I might not
know whether the least upper bound> of some complicated set
is computable, but I do know that,> if the set is bounded,
the LUB exists. Under your scheme,> I might not know whether
the LUB exists, but I would know> that, if it existed, it
would be computable. You might> prefer to know the second and
not the rst, but generally> speaking, knowing the rst and
not the second is more> useful in real analysis.> Its not
more useful, its a superimposed mutually sustaining>
impression that mathematics lies outside the realMs of>
computability.> You say its not more useful; I say it is
more useful. How> should we decide this point? I could give
you many examples> where the existence of the LUB is
important to the conclusion> drawn _in real analysis_. Can
you do the same with the> computability property?do any
examples apply outside of number theory? what isthe
contribution to technology of uncountability? anything?> The
second sentence here is difcult to understand. I think>
youre saying the LUB property is part of the whole complex>
that gives the (false) impression that mathematics extends>
beyond the computable.> Why is this impression false? You
havent gone into your reasons> for asserting this yet,
unless youre counting That would> mean the reals are
countable as a reason. Im going to> assume you agree with
me that this would be a bad reason.> Your proof jumps to the
fact of a contradiction, what is this> contradiction?> At
worst, my proof _derives_ a contradiction. It does not pull>
it out of thin air. Its true that I glossed over a some> the
details, but they were details youve been over many times.>
Im willing to ll them in, if you really want me to.>
However, several people have pointed out that you do not need
a> contradiction for this result. In fact, Cantor did not use>
contradiction. In the latest version, the one you most
recently> snipped, there is no contradiction, just a
demonstration that> every list misses at least one real
number. This is a contradiction> only to the assumption that
|N| = |R|.misses an innitely long number, at a digit that it
species itself mustbe contradictory. the number is dened in
terms of itself.you treat an innitely long sequence like a
simple object you can manipulate.where is the contradictory
bit? can you point it out? why cannot the numberbe computed
to any precision specied? its a simple trick of
enumeration,the number is part of the list and poorly dened
at its own referential digit,it doesnt lie outside the
list.why does logic conne itself to non self reference yet
mathematics yields it.there are simple limitations to
computer programs when they self reference,its a practical
base of limitations. natural language and set theory both
getaround self referential paradoxes by specifying before
hand their use in thedomains. limitations to computers are
very limited in scope, all other elds ofscience focus on
them, they are trivial.> Is it self referencing? YES> Does
the formulation of extra numbers occur purely depending on>
the original formulation of all numbers? YES> I dont know
what you mean by self-referencing, but I suspect that> if I
did, I would disagree. There is no spooky Goedel-style>
mixing of levels going on here.> OK, the not-found number
Cantor(L) is a function of the list, L,> it is not found in.
I dont see how it could be otherwise.> Every real is in some
list or other; there is no real which> is missing from every
list.> Does this contradiction rely on specifying the new
number is> *different* from the original forumlation at nite
set points?> YES> Does the extra number reinsert into the
original list to> demontrate the imcompatibility? YES> NO.
The extra number is not re-inserted into the original list.>
If you think I did this, show me where.right!depends on how
you conclude the proof as an assertion or contradiction.>
The extra number is simply this :> Given a number on a list,
change it and put it back in the list> without changing the
list.> The most generous interpretation of this I can make
is that> you meant to say something else. (1) The extra
number is not one of> the numbers on the list (hence
extra). (2) Changing a number> on the list and putting it
back in the list must change the> list. (3) I havent done
anything like this.> I suggest you try again. This time say
what you mean.Would it help to tally the list of reals as a
single number? DIGIT 1 2 3 4 5
6__________________________UTM(1) 4 3 6 4 2 4UTM(2) 7 4 3 4 3
2UTM(3) 0 1 0 1 1 1UTM(4) 1 2 2 2 2 2UTM(5) 7 7 7 7 7 7Working
along the diagonal in this fashion :1 2 4 73 56that would give
the single real number :437640....> Does the diagonal
transform form a well dened number?> NO -> computables =
reals> YES -> computables C reals> Well, it would be handy
for me to agree with your implication> here. The Cantor
diagonal number is very well dened, for a> well-dened list,
so it would follow that the computables are> a proper subset
of the reals.> However, you have not made any argument for
the pair> of implications. I think I saw you _assert_ in a
separate post> that uncomputable nnumbers are not reals, but
you dont give> a reason for that beyond allowing you to
conclude what you want to> conclude, that |N| = |R|.> There
is only 1 innity type, the answer is NO, heres a similar>
exposition of diagonalisation.> Apart from anything else in
this thread, there is more than> one innity type. You would
have to re-work set theory to get> rid of that conclusion, if
even that would do it. Questions> of computability just do not
enter into it.> For any set A, |A| < |P(A)|, where P(A) is
the powerset of A.> Original Format> alt.paranormal,
alt.sci.physics> I dont see anything new here.>Its a
diagonalisation proof of Godels theorom by Roger Penrose,
Steven Hawkingsassistant. Probably accepted as a correct
proof by mainstream mathematiciansto this day, yet the
denition of the diagonal function is obviously awed andnot
well dened.Read the capitalisation part of the post and tell
me if its a valid proof.When a proposed theorem passes
certain credulity test it is automaticallygiven the label of
well formed. A second examination will bring down thewhole
encapsulation to naive eyes, rather than expose the single
theorem.Herccredulity ( P ) Pronunciation Key (kr-dl-t,
-dy-)n.A disposition to believe too readily.
===
|-|erc
says...>misses an innitely long number, at a digit that it
species itself must>be contradictory. the number is dened
in terms of itself.No, thats not true. Cantors method works
like this:if you give me a list of real numbers, then I can
giveyou a real number that does not appear anywhere on the
list.It always works, which implies that there cannot be a
listthat contains every real number.For example. Consider the
following list ofreal numbers (in decimal notation): 0.1 0.01
0.001 0.0001 0.00001 ...Cantors method produces a real
number that is not on the list,namely 0.222... = 2/9This is a
perfectly ordinary real number. It isnt denedin terms of
itself---it is dened in terms of the abovelist of real
numbers. Given *any* list of numbers, Cantorsmethod will
allow you to construct a real number that doesntappear
anywhere on the list. The meaning of construct a realnumber
is to give a method for computing its decimal places.Which
number Cantors method returns depends on which listyou
provide. Give me a different listing of reals, and Iwill give
you a different real that is not on that list.So *every* list
misses some real number, althoughdifferent lists miss
different reals.So what Cantors method does is to constructa
function Diagonal from lists of reals to reals with
theproperty: forall f: list of reals, Diagonal(f) is not on
the list fwhich logically implies forall f: list of reals,
exists r: real, r is not on list fwhich logically implies
there does not exist a list containing all realsNotice that
if we replace real by rational, theresult is not true.
There *does* exist a list of allpossible rational numbers:
0/1, -1/1, 0/2, 1/1, -2/1, -1/2, 0/3, 1/2, 2/1, -3/1, -2/2,
-1/3, 0/4, 1/3, 2/2, 3/1, -4/1, -3/2, -2/3, -1/4, 0/5, 1/4,
2/3, 3/2, 4/1, ...This list contains every possible rational
number.--Daryl McCulloughIthaca,
NY
===
--www.StealthHostiing.com You rule Truman.
http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE
the Truman I heard him. Very spooky! >Is the truman living in
Townsville? Ive been hearing stuff, yeah.Webmasters help the
TRUEman by joining www.theBanner.net Current:1
Goal:1000---
---> |-|erc says...>misses an
innitely long number, at a digit that it species itself
must>be contradictory. the number is dened in terms of
itself.> No, thats not true. Cantors method works like
this:> if you give me a innite list of real numbers,
then I can givenote *> you a real number that does not
appear anywhere on the list.> It always works, which implies
that there cannot be a list> that contains every real
number.It doesnt work on every list. It wont work on a list
that is dened tobe complete because then it self
references.> For example. Consider the following list of>
real numbers (in decimal notation):> 0.1> 0.01> 0.001>
0.0001> 0.00001> ...> Cantors method produces a real number
that is not on the list,> namely> 0.222... = 2/9> This is a
perfectly ordinary real number. It isnt dened> in terms of
itself---it is dened in terms of the above> list of real
numbers. Given *any* list of numbers, Cantors> method will
allow you to construct a real number that doesnt> appear
anywhere on the list. The meaning of construct a real>
number is to give a method for computing its decimal
places.> Which number Cantors method returns depends on
which list> you provide. Give me a different listing of
reals, and I> will give you a different real that is not on
that list.> So *every* list misses some real number,
although> different lists miss different reals.> So what
Cantors method does is to construct> a function Diagonal
from lists of reals to reals with the> property:> forall f:
list of reals, Diagonal(f) is not on the list f> which
logically implies> forall f: list of reals,> exists r:
real,> r is not on list f> which logically implies> there
does not exist a list containing all reals> Notice that if
we replace real by rational, the> result is not true.
There *does* exist a list of all> possible rational
numbers:> 0/1,> -1/1, 0/2, 1/1,> -2/1, -1/2, 0/3, 1/2, 2/1,>
-3/1, -2/2, -1/3, 0/4, 1/3, 2/2, 3/1,> -4/1, -3/2, -2/3, -1/4,
0/5, 1/4, 2/3, 3/2, 4/1,> ...> This list contains every
possible rational number.> DIGIT 1 2 3 4 5
6__________________________UTM(1) 4 3 6 4 2 4UTM(2) 7 4 3 4 3
2UTM(3) 0 1 0 1 1 1UTM(4) 1 2 2 2 2 2UTM(5) 7 7 7 7 7 7Here is
the procedure :1/ Construct a UTM and an array of processors
to evaluate each applied to increasing integers.2/
Eventually you will construct a TM ~ UTM(7777) say ~ that
performs the diagonal transformcalculation.The output of
UTM(7777) starts by emulating UTM(1), get the 1st digit 4,
adds 1 = 5.The next output of UTM(7777) is UTM(2), gets the
2nd digit 4, adds 1 = 5,Then the next output of UTM(7777) is
UTM(3), gets the 3rd digit 0, adds 1 = 1.So far UTM(7777) =
5 5 1UTM(7777) seems clearly dened, but when it reaches row
7777 it has to emulate that functionjust like it emulated
UTM(1) and UTM(2). UTM(7777) emulates UTM(7777), whichagain
emulates UTM(7777). The digit at (7777,7777) never gets
calculated, it isnot a properly dened TM or a properly
dened number.When you dene a number you are dening the
algorithm to exploit it.You *know* that that algorithm will
appear on the list of algorithms, all permutationsof
computer programs are there, so you *know* you are dening
the number toreference itself.There are 2 approaches to
diagonalisation :1 The number does not appear on the list.->
The diagonal forms a new number-> The new number does not
appear on the list(nothing new here)2 The number appears on
the list-> The diagonal forms a self referencing number-> The
contradiciton refutes A The diagonal was on the list -> the
list is not complete -> sci.math thinks Cantors proofworks B
The diagonal is well formed -> the list is complete for well
formed numbers -> single ooHerc
===
|-|erc says...> |-|erc
says...>misses an innitely long number, at a digit that
it species itself must>be contradictory. the number is
dened in terms of itself.> No, thats not true. Cantors
method works like this:> if you give me a innite list
of real numbers, then I can give>note *> you a real
number that does not appear anywhere on the list.> It always
works, which implies that there cannot be a list> that
contains every real number.>It doesnt work on every
list.Yes, it does.>It wont work on a list that is dened
to>be complete because then it self references.How do you
create a list that is dened to be complete?A list of decimal
numbers is by denition a function which,given a natural
number n, returns a decimal expansion for a real.In turn, a
decimal expansion is a function which given a naturalnumber
i, returns an integer between 0 and 9.So, how can there be a
list of decimal numbers that is denedto be complete?> DIGIT
1 2 3 4 5 6>__________________________>UTM(1) 4 3 6 4 2
4>UTM(2) 7 4 3 4 3 2>UTM(3) 0 1 0 1 1 1>UTM(4) 1 2 2 2 2
2>UTM(5) 7 7 7 7 7 7>Here is the procedure :>1/ Construct
a UTM and an array of processors to evaluate each applied
to>increasing integers.>2/ Eventually you will construct a TM
~ UTM(7777) say ~ that performs>the diagonal transform
calculation.No, there is no such TM. If your list includes
only *total* TMs(that is, they halt on all possible inputs),
then there cannotbe a UTM that performs the diagonal
calculation.>The output of UTM(7777) starts by emulating
UTM(1), get the 1st>digit 4, adds 1 = 5.>The next output of
UTM(7777) is UTM(2), gets the 2nd digit 4, adds 1 = 5,>Then
the next output of UTM(7777) is UTM(3), gets the 3rd digit 0,
adds 1 = 1.>So far UTM(7777) = 5 5 1>UTM(7777) seems
clearly dened, but when it reaches row 7777 it>has to
emulate that function just like it emulated UTM(1) and
UTM(2).>UTM(7777) emulates UTM(7777), which>again emulates
UTM(7777). The digit at (7777,7777) never gets calculated,>it
is not a properly dened TM or a properly dened number.Thats
right. If you broaden your collection of Turing Machines to
includeboth total functions and partial functions (functions
that may fail to halton some input), then you *can* enumerate
all possible such Turing Machines.Then, as you say, there will
be a number (say 7777) such that digit 7777if UTM(7777) is
undened.>When you dene a number you are dening the
algorithm to exploit it.>You *know* that that algorithm will
appear on the list of algorithms,>all permutations of
computer programs are there, so you *know* you>are dening
the number to reference itself.If you want to include
nonhalting Turing Machines in your list, thenyou have to
modify the diagonalization procedure slightly. Given
aninnite list UTM(i) of turing machines, you dene a decimal
expansionr according to the rule: If digit number n produced
by UTM(n) is undened, then make digit n of r be 0. If digit
number n produced by UTM(n) is 9, then make digit n of r be
0. Otherwise, make digit n of r be equal to 1 + digit number
n of UTM(n).It is still the case that r cannot be equal to
the output of any Turingmachine on the list.>There are 2
approaches to diagonalisation :>1 The number does not appear
on the list.>-> The diagonal forms a new number>-> The new
number does not appear on the list>(nothing new here)>2 The
number appears on the list>-> The diagonal forms a self
referencing numberThat isnt possible.--Daryl
McCulloughIthaca, NY
===
> where is the contradictory bit? can
you point it out? why cannot the number> be computed to any
precision specied? its a simple trick of enumeration,> the
number is part of the list and poorly dened at its own
referential digit,> it doesnt lie outside the list.Were
talking about a list of computable numbers. For every number
onthe list [at least] _one_ Turing Machine that can correctly
compute_every_ digit in the number. Thats what it means for a
number to becomputable.There are, of course, other equivalent
denitions.If you choose to relax your denition of
computability to say thata number is computable if no matter
how many digits you want thereis at least one Turing Machine
that computes those digits correctlythen you get a different
set. Its obvious that all real numbers arecomputable under
that denition.Heck, under that denition, all real numbers
are computable by nitestate machines. You want 100 digits,
pick a machine that spits out100 digits and halts. You want
200 digits, pick a machine that spitsout 200 digits and
halts. No matter how many digits you want, theresalways a
machine that will give them to you.But under that denition
of computability the fact that there are onlycountably many
nite state machines no longer implies that there areonly
countably many real numbers. Because now the correspondenceis
between real numbers and equivalence classes of
[Cauchy-like]sequences of nite state machines and not
between real numbers andTuring machines.The cardinality of
the set of equivalence classes of sequences ofnite state
machines is greater than the cardinality of the set of
allnite state machines. John Briggs
===
> What is an
irrational number? Can you count to it?> Can you pin point
it? There is no such thing.Really? Hmmm...lets try creating
an isosceles right triangle witheach side 1. What do you do
with the length of the hypotenuse irrationals dont exist.In
fact, for *ANY* right triangle, if any two sides are of
integerlength, the remaining length must be either another
integer or else(usually) an irrational number!Good Luck with
your rewriting of Geometry.Jonathan HoyleOlder and Wiser than
Age 11
===
--www.StealthHostiing.com You rule Truman.
http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE
the Truman I heard him. Very spooky! >Is the truman living in
Townsville? Ive been hearing stuff, yeah.Webmasters help the
TRUEman by joining www.theBanner.net Current:1
Goal:1000---
---> What is an irrational
number? Can you count to it?> Can you pin point it? There is
no such thing.> Really? Hmmm...lets try creating an
isosceles right triangle with> each side 1. What do you do
with the length of the hypotenuse if> irrationals dont
exist.> In fact, for *ANY* right triangle, if any two sides
are of integer> length, the remaining length must be either
another integer or else> (usually) an irrational number!>
Good Luck with your rewriting of Geometry.The answer would be
sqrt(1^2 + 1^2) = sqrt(2).That is a terminating, consise and
usable representation of the solution.Hercwell you can keep
rationals separate to functions if you like, a divisionfor
the division_is_special_computation crowd.
===
> The answer
would be sqrt(1^2 + 1^2) = sqrt(2).That is a terminating,
consise and usable representation of the solution.Huh??? How
does that change the fact that sqrt(2) is irrational?
Youclaimed that irrationals did not exist...now you seem to
be agreeingthat sqrt(2) exists.So, are you saying that
sqrt(2) is rational? Or are you admittingthat your original
statement was wrong?Jonathan HoyleEducating adults with 11
year old child-like minds
===
Jonathan Hoyle> Educating adults
with 11 year old child-like mindsYou time would be better
spent in vertically urinating up an elongated brerous
fastening device.Bob Kolker
===
> You time would be better
spent in vertically urinating up an elongated > brerous
fastening device.No doubt. It reminds me of the adage to
never argue with a fool, aspeople might not recognize which
one the fool is. Ah well, my futileattempts to bring
intelligence to our adolescent impersonator.
===
You are
confused because you believe the fact that a TM can
enumerateall the algebraic numbers contradicts the
uncountability of the reals.Algebraic numbers are countable,
hence there is no
contradiction.James
===
--
>
-->To extend,>basically Im reopening
a 75 post thread from several months ago>with Daryl McCollough
(I recognise Jim Burns and BruceS my skeptic>counterparts
too).> Sorry, I dont recall what it was about.> I think
the Cantor diagonalisation across UTM(Z) is more a concern
anyway,> but the thread is here : http://tinyurl.com/yo9w>
I said reals are countable by UTM(Z)> You said that it
misses non computable numbers, like r = Halt(1) + 1/10
Halt(2) + 1/100Halt(3)...> since Halt(n) is unknown for all
n.> i.e. r is not computable and will not be listed in
UTM(Z).> I tried to argue r is just not known, you tried to
show r is not computable at all (standardview),> now I am
now suggesting r it not existent! so Computables = Reals.>If
r is the sequence of Halt values along all indexed functions,
then parts of rare unknown. We can nd out more values of r
as we go along but it willnever be complete, some computer
programs are impossible to test!This make r a loci of numbers
not a number itself.Herc
===
> I found this on a web search,
the result was stated without the> details of
derivation:gravitational eld around a central mass m, the
eccentricity (e) is> given by e=sqrt(1 + 2Eh^2 / m^2)where E
= v^2 / 2 - m/r is the total> energy (kinetic plus
potential), h = rv_t is the angular momentum, v> is the total
speed, v_t is the tangential component of the speed, and> r is
the radial distance from the center of the mass. (Note v_t is
v> subscript t).> Id suggest you reference Beckers book
Introduction to Theoretical> Mechanics, Section 10-5 in
which he addresses Equation of the Orbit> by the Energy
Method. While the equations of the derivation are a> bit too
much to post on Usenet, essentially Becker begins with>
expressions for the total energy and angular-momentum : T + V
= W> mr^2*Theta(dot) = JAfter a bit of integration and routine
substitutions, Becker imports a> previously derived
relationship derived earlier in chapter 10: r = 1/((1/ep) -
(1/p)Cos(Theta))and obtains: e = sqrt((2WJ^2)/mk^2) + 1)This
is a remarkably similar to form as the equation that you
posted,> and by expanding to equivalent terms as that you
posted, I suspect> that they will or should turn out to be
equivalent. I wasnt> sufciently ambitious to perform this
nal step. Harry C.p.s. My copy of Beckers text is dated
1954 (Lord, how time ies!),> so you may have a difcult time
locating a copy. If so, Id be happy
===
> I found this on a
web search, the result was stated without the> details of
derivation:> gravitational eld around a central mass m, the
eccentricity (e) is> given by e=sqrt(1 + 2Eh^2 / m^2)where E =
v^2 / 2 - m/r is the total> energy (kinetic plus potential), h
= rv_t is the angular momentum, v> is the total speed, v_t is
the tangential component of the speed, and> r is the radial
distance from the center of the mass. (Note v_t is v>
subscript t).>Solve Newtons Equations of motion for such
orbits and you will see that itcomes out of the wash.This is
a standard rst year classical mechanics problem, dealt with
in anyworth while text book on the subjectFranz
Heymann
===
>I found this on a web search, the result was
stated without the>details of derivation:>gravitational eld
around a central mass m, the eccentricity (e) is>given by
e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the
total>energy (kinetic plus potential), h = rv_t is the
angular momentum, v>is the total speed, v_t is the tangential
component of the speed, and>r is the radial distance from the
center of the mass. (Note v_t is v>subscript t).>formulas,
which with a small bit of work, verify this formula. In
thatthe initial position, p, and velocity, v, of the
satellite along withthe gravitational constant, G, and the
mass of the primary, M. 1 k = - |p x v| [3a] 1 2 1 2 GM k = -
|v| - --- [6a] 2 2 |p|k_1 is the rate of area swept out per
unit time; it is also your h/2.k_2 is the energy per unit
mass of the satellite; it is also your E. derive the formula
of the orbit as 1 r = [10] b + c
cos(@-k_3)where b = GM/(4 k_1^2) and c^2 = b^2 + k_2/(2
k_1^2). I match thatformula with the formula in terms of the
semimajor axis, a, andeccentricity, e. a(1-e^2) r =
[10a] 1 + e cos(@)Looking at these formulas, c e
= - b k_2 = sqrt( 1 + --- ) 2 k_1^2 b^2 8 k_2 k_1^2 =
sqrt( 1 + --- ) (GM)^2 2 E h^2 = sqrt( 1 + --- )
(GM)^2which is your equation using m = GM. This implies that
if E < 0, theorbit is elliptical and if E > 0, the orbit is
hyperbolic.Rob Johnson <rob@trash.whim.org>take out the trash
before replying
===
> I found this on a web search, the result
was stated without the> details of derivation:gravitational
eld around a central mass m, the eccentricity (e) is> given
by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the
total> energy (kinetic plus potential), h = rv_t is the
angular momentum, v> is the total speed, v_t is the
tangential component of the speed, and> r is the radial
distance from the center of the mass. (Note v_t is v>
subscript t).> This is quite adequately derived in most
theoretical mechanicstextbooks (at either the undergrad or
grad level). The one Irecommend is Herbert Goldstein. Its
the most thorough treatment oforbital motion Im aware
of.
===
> Epilog> The Interesting Question has already been
satisfactorily answered now.> But I thought that some people
might enjoy a nice, pertinent passage> (which I only just now
discovered) in the excellent _Concrete Mathematics_> by
Graham, Knuth, and Patashnik.> On page 481 (2nd ed.), where
m is the number of terms of Stirlings> asymptotic expansion
of ln(n!) to be used, they say> ... if n is xed and m
increases,> the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1))
decreases to a certain> point and then begins to increase.
Therefore the approximation reaches a> point beyond which a
sort of uncertainty principle limits the amount by> which n!
can be approximated.> DavidYou can nd an asymptotic series
for n!rather than that for ln(n!) inA new derivation of
Stirlings Approximation to n!, G. and J. Marsaglia, Am.
Math. Monthly, 97,No. 9, 1990.Mixing TeX and * for
multiplication:n! = n^n e{-n}sqrt{2pi n}* [1 + 3a_3/n +
3*5a_3/n^2 + 3*5*7a_5/n^3 + 3*5*7*9a_9/n^4+...]where the as
satisfy the recursion, (n+1)a_n = a_{n-1} - 2a_2a_{n-1} -
3a_3a_n-2- ... - (n-1)a_{n-1}a_2initialized with a_1=1,
a_2=1/3.The rst few as are
1,1/3,1/36,-1/270,1/4320,1/17010,-139/5443200Thusn! = n^n e^{
1+ c_1/n + c_2/n^2 + c_3/n^3 + ...with the rst nine cs:1/12,
1/288, -139/51840, -571/2488320,163879/209018880,
5246819/75246796800,-534703531/902961561600,
-4483131259/86684309913600,432261921612371/514904800886784000
,(sign pattern +,+,-,-),If S_k = n^n exp{-n}sqrt(2pi
n}*[1+c_1/n+...+c_k/n^k]sums the series through k terms, then
these examplesshow the ultimately chaotic behaviour of the
asymptoticseries, but for which early terms may be quite
useful:With n=4, 4!= 24 S_1 = 23.995888... S_2 =
24.000988281... S_3 = 24.000003472778576 S_4 =
23.999982402422244949 ... S_24=
23.99999999997607404482055067828621313524 S_25=
24.00000000002184951216601742396435065363 S_26=
24.00000000002280566142889597320458957977 S_27=
23.99999999997569894109331075218544444326 ... S_65=
24.07902081946226925298335074817255083162 ... S_70=
28.02611844348535334766765116652722021875 S_71=
-6.281830892605180810481922429588177923873 ... S_75=
-2098.109030254598135895004889287527652778 S_76=
-2147.591820869521698236087744269317078347 S_77=
19280.21714886293068959493792076383534356 ... With n=10, 10!
= 3628800 S_1 = 3628684.748897... S_2 = 3628809.703606... S_3
= 3628800.054325... S_4 = 3628799.971745... ... S_10 =
3628800.0000000672569... ... S_23 =
3628799.99999999999999936190019... ... S_40 =
3628799.999999999999999999987543840408047... ... S_70 =
3628800.000000000000000000000130930739823... ... S_130=
3628800.000000000038873676252536002626173... ... S_150=
3628800.00029462746487456259981407627620... ... S_160=
3628797.730095008848234664534843871313008... S_161=
3628814.537100312802960926738475176189011... ... S_166=
3629460.850717572983759390565123042386732... ... S_170=
3661581.5010228672... ... S_174= 5416236.0003953219... S_175=
-9909389.3801847535217... S_176= -10037108.898098820886747...
S_177= 109529312.221996667266697...You may nd this
interesting, or, if you have Mapleor Mathematika or other
such package for rational arithmetic,do some examples
yourself.George Marsaglia
===
> Suppose I want to studdy a
function f : G -> H> where H is the quaternion division
algebra, hypercomplex> numbers of dimension 4 and G is a
given (commutative) group,> for example a nite one e.g.
Z/nZ.Is there a kind of representation theory with values
on H ?> What can be said for the caracter of Z/nZ with
walue in> H ?I think they are of the form> phi_k : l -> e^{
j*2*pi*k*l/n }for j a unit pure quaternion.> Once you have
xed j, you get a complete set> of caracter, and you can
compute Fourier-like decomposition,> with an inversion
formula.Is it that simple ?You can certainly dene
representations of Gin a vector space V over H.Such a
representation can also be consideredas a representation over
C of twice the dimension/degree.There are fairly
straightforward criteria for determiningif a representation
over C arises in this way.(To start with, the dimension must
be even -- if nite --and the character must be real.)I dont
think one can usefully dene the characterof a representation
over H as a function G -> H,but I could be mistaken.Normally
one considers the characterof the corresponding
representation over C, I think.(Im not clear what k,l are in
your formula above.)-- Timothy Murphy tel: +353-86-2336090,
+353-1-2842366
===
I think that2^n = sum (nCk) (k ranges from
0 to n)(by nCk I mean n!/(k!(n-k)!))but cant prove it. does
anyone have any hints? also, is there any standard way to
represent sigma notation in ascii?the above expression is a
little vague as Ive written it.cheers,Gareth
===
> I think
that2^n = sum (nCk) (k ranges from 0 to n)(by nCk I mean
n!/(k!(n-k)!))but cant prove it. does anyone have any
hints?binomial theorem? > also, is there any standard way to
represent sigma notation in ascii?> the above expression is a
little vague as Ive written it.I would write that in
pseudoTeX as2^n = sum_{k=0}^n {n choose k}.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
Oh
please, you NPD a?ted loser! You are such a LIAR and CHEAT
and you> are a super-troll and a super-crank!Just look at
some of the quotes from you below! This has been going on
for> 8 years now and you are about as close to a proof as
you were before you> started, you ignoramus!> Whereas James
makes valuable contributions to sci.math and math worldthe
detractors and their negativity add nothing, you might try
notposting for a month and instead really read and try to
understandJamess posts.
===
<fake post>Pauvre andouille===>
If you do a Google Search on my name James Harris and math,
like you> put in James Harris math, youll currently see a
link to> www.crank.net coming up third, and I dont know how
many of you know> that site is run by a guy named Erik Max
Francis who used to be a> regular sci.math poster.> I want
you to do that search, look at what he has on me, and please>
explain to me exactly why this person is calling me names,
like, what> makes me a crackpot.> Can *any* of you go there
to what he has and tell me whats supposed> to be so bad
about what Im doing?The evidence lies in the history of your
posting.That makes it clear that you are a minor, self
important, cranklet.Franz Heymann
===
> That makes it clear
that you are a minor, self important, cranklet.I am not an
expert on cranks, but I think that JSH deserves betterthan
minor cranklet. I think he is a primary gure of
modernUsenet crankery.Hes no founder of Lawsonomy, but his
work is not without a certainkind of merit. I think that his
name ranks up there as a pioneer ofsorts (but not one without
peers, even on Usenet -- still, Ive alwaysfound him more
engaging than the competition).(The follow .sig was chosen
merely coincidentally.)-- Jesse F. HughesMy proofs are out
there. -- James S. Harris
===
> If you do a Google Search on
my name James Harris and math, like you> put in James
Harris math, youll currently see a link to> www.crank.net
coming up third, and I dont know how many of you know> that
site is run by a guy named Erik Max Francis who used to be a>
regular sci.math poster.> I want you to do that search, look
at what he has on me, and please> explain to me exactly why
this person is calling me names, like, what> makes me a
crackpot.> Can *any* of you go there to what he has and tell
me whats supposed> to be so bad about what Im doing?The
evidence lies in the history of your posting.> That makes it
clear that you are a minor, self important, cranklet.Franz
HeymannThe question has to do with the website
http://www.crank.net/harris.htmland what Erik Max Francis
presents there to defend his insult.Im curious about whether
or not anyone can nd justication in what he presents.Why
dont you try again given that further explanation of what is
requested?James Harris
<3c65f87.0312120708.bb9dace@posting.google.com>
===
The
evidence lies in the history of your posting.> That makes it
clear that you are a minor, self important, cranklet.Franz
Heymann> The question has to do with the website >
http://www.crank.net/harris.html> and what Erik Max Francis
presents there to defend his insult.Thats curious.You are
*not* alleging that the total evidence available
isinsufcient to prove youre a crank? You are only claiming
thatFrancis failed to provide enough evidence that youre a
crank?I guess you could check Google, then. Maybe that would
shed light onwhether Franciss claim is justied regardless
of the evidence hissite includes.-- Jesse HughesSurround
sound is going to be increasingly important in futureofces.
-- Microsoft marketing manager displays his keen insight
===
>
As can be seen by the number of posts in this thread,> and the
references to his web site in thousands of other posts,> a
computer programmer, who took some data processing classes>
at a third rate California college, has become a highly
regarded expert> in math, physics, and other science
disciplines, and> many people, who pretend to be rational,
intelligent, open-minded> scientists (Or at least, pretend to
have a scientic mind.),> frequently use this programmer as a
major reference.> So what-if it is true? What is your
point?My point is,that only an fool,would use a web site like
that as a reference.And only a sociopath would put up such a
web site.There are many web sitesthat reference
non-conventional ideas,in an informative, entertaining
way,and the web site owner does not try toget an ego trip by
demeaning other folks.Better questions would be:Whats the
point of posters in science newsgroupsusing that web site as
a reference?Whats the point in that web site?(Ego
tripping, hate, neurosis, sociopathy, revenge, etc?)--Tom
Potter http://tompotter.us
===
=WHO instigates
conict and war for power and wealth?WHO instigated the class
wars of the 1900s?WHO is instigating the religious wars of
the 2000s?WHO has a well organized propaganda machine?WHO
gang attacks all who expose their agenda and methods?Visit my
web site, and download the worlds best physics
tutorial!
===
> As can be seen by the number of
posts in this thread,> and the references to his web site in
thousands of other posts,> a computer programmer, who took
some data processing classes> at a third rate California
college, has become a highly regarded expert> in math,
physics, and other science disciplines, and> many people, who
pretend to be rational, intelligent, open-minded> scientists
(Or at least, pretend to have a scientic mind.),> frequently
use this programmer as a major reference.So what-if it is
true? What is your point?It is true. The point is that Erik
Max Francis tells a lot of peoplewhat they want to hear, and
few seem to care that he lacks theexpertise to do a real
evaluation of the work he pans.So they ignore his lack of
credentials and support him.Its fascinating behavior from a
sociological viewpoint.James Harris
===
> Its fascinating
behavior from a sociological viewpoint.> James HarrisIf any
behavior in this NG is fascinating from that viewpoint, it is
JSHs sociopathy.
===
> Its fascinating behavior from a
sociological viewpoint.> James Harris> If any behavior in
this NG is fascinating from that viewpoint, it is> JSHs
sociopathy.Everyone has there own diagnosis for him. :)
===
>
As can be seen by the number of posts in this thread,> and
the references to his web site in thousands of other posts,>
a computer programmer, who took some data processing classes>
at a third rate California college, has become a highly
regarded expert> in math, physics, and other science
disciplines, and> many people, who pretend to be rational,
intelligent, open-minded> scientists (Or at least, pretend to
have a scientic mind.),> frequently use this programmer as a
major reference.> So what-if it is true? What is your
point?> It is true. The point is that Erik Max Francis tells
a lot of people> what they want to hear, and few seem to care
that he lacks the> expertise to do a real evaluation of the
work he pans.One does not have to be a chicken to identify a
bad egg -- Thomas A. EdisonHow would you know? Did a
scientist or mathematician tell you that? Or, more likely,
did you just make that up to make*yourself* feel better?> So
they ignore his lack of credentials and support
him.Predictable explanation, except for your cryptic
reference to support. But heres a better one. Namely,
that you are apompous, arrogant, unsufferable idiot and this
observable fact is obvious to everyone but you.> Its
fascinating behavior from a sociological viewpoint.Theres
nothing fascinating about it -- unless you nd it fascinating
that people run from a deluded paranoiac who ingsmanure.>
James HarrisWacky, isnt it? But, hey, its just basic math.
Yup, yup, yup.--There are two things you must never attempt
to prove: the unprovable -- and the obvious.--Democracy: The
triumph of popularity over
principle.--http://www.crbond.comX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oft
===
at 01:38 PM,
Bruce Harvey <newsforbruce@bearsoft..co.uk> said:>I like
the denition of a tensor as the set of all
possible>representations of a vector in all possible
co-ordinate systems.Thats not a denition. Further, it seems
to be an alternate way oooking at vectors and to say nothing
about tensors of higher rank.>Lets face it, we are dealing
with maths developed by guys who could>not dene the length
of a vector in the complex plane becaue they>missed the
lectures on complex numbers.WTF is that supposed to refer
to?>If you just assume that space is Euclidian and the
universe knows>only the moment of now, then its rulers that
shrink and clocks that>slow and the Schwarzchild metric pops
out in a few lines.Please show us thos few lines.>All this
tensor stuff is just a way of hidding the fudges.ROTF,LMAO!
This tensor stuff is just as important to
ClassicalMechanics as it is to Relativity. -- Shmuel
(Seymour J.) Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.orgX-Cise:
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===
at 06:58 PM,
Boris Schaefer <boris@uncommon-sense.net> said:> A
multilinear function T : V^k --> R is called a k-tensor on V
and> the set of all k-tensors, denoted J^k(V), becomes a
vector space> (over R), if for S,T in J^k(V) and a in R we
dene:I assume that he gives a more general denition latter,
for a mixedtensor T: V^kxV*^l ->R.>Now my problem is that I
cannot reconcile this with my previous>notion of tensors.
Thats because you have not encountered tensors before, but
rathertensor elds on a manifold, and your texts dened them
in terms ofcomponents in particular coordinate systems. That
doesnt really makemuch sense once you start working with
local coordinate systems.>Also, these quantities have to
transform in a certain way.What does that mean when you dont
have global coordinate systems?>I thought a second-rank tensor
was a matrix No. If you pick a particular basis for a vector
space then you canexpress the components of a second rank
tensor with respect to thatbasis. As a Physicist, you should
understand that neither 0 nor 32 isa temperature, but that
you can express the temperature as eithernumber by picking an
appropriate scale, e.g., C versus F.>and an (m x n)-matrix is
a linear mapping from V^n --> V^m.Likewise, the matrix just
groups component in terms of a particularbasis. Change the
basis and you need a different matrix.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
>Yes, very much. This got me
interested. Do you know any good books>and/or webpages that
explain these two different viewpoints on>tensors and their
relationship.Unfortunately, there are not many good sites for
such a comparison.Walds book on General Relativity is better
than most.Another thing to point out is that most physics
texts do tensorson manifolds and the change of basis for the
vector space comes froma change of coordinate system at the
point of the manifold. Thats whyall those partial
derivatives come up. However, it is quite possible(and
useful) to consider bases of the tangent space that are not
derivedfrom some coordinate system at the point. If this is
done, those partialderivatives are not relevant, even for
tensors ona manifold.--Dan GrubbX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
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===
at 11:09 PM,
Matthew Senn <msenn77@yahoo.com> said:>Ive always been
strong in math,ave you had any exposure to Mathematics, or
only what passes for it inthe public stool system?>but
matrices and vectors spaces have got me entirely
stumped.Matrices? That sounds like youre taking a
computation course. Try astrictly Mathematical text like
Halmoss Finite Dimensional VectorSpace. If you can get
through that, it might help to put thecomputational material
in context. OTOH, if your problem is graspingabstractions,
then you might have trouble with it.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.orgX-Cise:
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===
>I need help
on the following problem. >such that>Kerv={h in H|
v(h)=0}={0} and v(H)is dense in H but does not coincide>with
it. Is it true that v(H)=Hspace the answer is no, because you
specied but does not coincidewith it. Drop one or the other
and the answer is that such an-- Shmuel (Seymour J.) Metz,
SysProg and JOATnot reply to spamtrap@library.lspace.org
<3ik3tvosm69hgkat84jr0dpuur49s8gatf@4ax.com>
<Xns94495AB5CC5BBgoddardb@129.250.170.82>
<br2lg8$30lk@odds.stat.purdue.edu>
<ac6dnVcJe6ZOtEuiRVn-jg@comcast.com>X-Cise:
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admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
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===
at 02:01 PM,
Jonathan Miller <jonmillere1@comcast.net> said:> All
degrees, credentials, etc., should be by comprehensive>
examinations ALONE. Anything else is
anti-educational.>Written by someone with a degree which is a
result of a long>paper.But none the less valid.
Multiple-choice questions are almost uselessin testing
comprehension. And *this* is written by someone with adegree
that is *not* the result of a long paper ;-)-- Shmuel
(Seymour J.) Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
> All degrees, credentials,
etc., should be by comprehensive> examinations ALONE.
Anything else is anti-educational.Perhaps the appropriate
method depends on the degree. For example, onecould imagine
that a degree in music composition could have a nalhurdle
where the candidate composes a musical composition... A Ph.D.
inmathematics could have a nal hurdle where the candidate
producesoriginal theorems. And so on.X-Cise:
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===
>Youll need
to convince the graduate faculty that you really do >have the
stuff, not such an easy task, unfortunately. Why is that
unfortunate? Certainly it should not be oppressivelydifcult,
but neither should it be so easy as to allow
unqualiedapplicants to get in.-- Shmuel (Seymour J.) Metz,
SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
|>Youll need to convince the
graduate faculty that you really do |>have the stuff, not
such an easy task, unfortunately. ||Why is that unfortunate?
Certainly it should not be oppressively|difcult, but neither
should it be so easy as to allow unqualied|applicants to get
in.Indeed.Some people here may not be aware of how much of a
problemit can be when someone enters a graduate program
withoutreally being ready to benet from it the way they want
to. Iveheard of students who scraped their way through
several yearsof unpromising work and lost a lot of income
without gettingwhat they were after. It really isnt just
bureaucratic or just forthe convenience of the department
that the people decidingadmissions are cautious.A BA is not
an absolute requirement. I know one very wellrespected
mathematician who never got his (although he didattend
college). Its also not enough-- some people with oneare not
ready enough.Keith Ramsay
===
Been too busy this week to read
sci.logic, but heres a belated response.>the category of
all sets IS AMBIGUOUS.Absolute unambiguity is probably an
unattainable ideal in any context.What one needs is to be
unambiguous enough for the purposes at hand.It is perfectly
reasonable to assume that a professional mathematicianwill
understand the category of all sets accurately enough in
thiscontext---i.e., he will not mistakenly latch on to some
other thing,or inquire with puzzlement as to what you are
talking about.>You may NOT just gloss over these>in the name
of professional mathematical context IF the reason WHY
you>are alleging that all categories are imitative of the
category of sets>is to HELP people achieve a BASIC
understanding!Depends on what basic understanding
means...more on this below.> : No. I use the word naturally
here in the sense that it is found,> : implicitly at
least,>Nice dodge. If we had to deal with explicit mentions
of the word>natural, especially as it relates to category
theory, youd>just be lying here.Ah, the word it in my
sentence has an ambiguous antecedent! I meantit to refer to
category of sets, not the word `naturally. Thatis, I
meant to say: When I said that the category of sets is found
naturally, I meant that the category of sets is found,
implicitly at least, in the mathematical literature.I did not
mean to say: When I used the word naturally, I was using the
word in the same way that the word naturally is used in the
mathematical literature.>Once you say all sets, one>becomes
tempted to think of a function as a set of ordered pairs!No, I
dont think so. Its true that if one says, everything is a
set,then we are tempted to think of a function as a set. But
if one does notalready think of a function as a set, then the
phrase all sets is notgoing to start making you think that
way.>GETTING him familiar with it would REQUIRE>him to notice
the difference between a function-set as an object>and a
non-function-set being isomorphic to the
identity-function-arrow>on the same set. It INVOLVES DETAILS
that hed RIGHTLY refuse to be>BOTHERED with!Familiarity is
again a relative thing. I dont have to understand everylast
detail and every theorem in number theory to have some
familiaritywith the natural numbers. Similarly, these
details about the category ofsets arent really needed to
grasp james dolans point---a supercialfamiliarity is
enough.> : Open up Jacobsons Basic Algebra II and you will
see that the category> : of sets is one of the rst examples
he gives.>There are a LOT of misguided textbooks in the
world.In my opinion, this isnt one of them. In any case,
whether or not BasicAlgebra II is misguided doesnt matter so
much, since Im citing it not asan example of stellar
pedagogy, but to illustrate how working mathematicianstalk
with each other. They may talk with each other in misguided
ways, buteven if thats the case, you still need to listen if
you want to understandhow they think.>But in general, if one
wants to understand category theory,>understanding why it is
a good alternative foundation is>simply indispensable.And
here again, we hit the fundamental issue. For a working
mathematician,a basic understanding of category theory
means an understanding that issufcient for him to follow the
way it is used in topology or algebraicgeometry or other
subjects that he is more interested in. He can easilyspend
his entire career using the language of basic category theory
andits simplest theorems, without ever even realizing that it
can be used asan alternative foundation. I daresay many
mathematicians do.I repeat, working mathematicians study
category theory *not* because it canbe used as a foundation
for mathematics, but because its language and basicresults
are useful for proving/formulating theorems and solving
problemsthat they are interested in. Thats the same reason
they study calculusor group theory or any other core topic in
mathematics.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range
of our projectiles---even ... the artillery---however great,
willnever exceed four of those miles of which as many
thousand separate us fromthe center of the earth. ---Galileo,
Dialogues Concerning Two New Sciences
===
tchow@lsa.umich.edu
says...>Been too busy this week to read sci.logic, but
heres a belated response.>the category of all sets IS
AMBIGUOUS.>Absolute unambiguity is probably an unattainable
ideal in any context.>What one needs is to be unambiguous
enough for the purposes at hand.Actually, when people talk
about some sort of collection of *all*whatevers, it is
usually (always?) good enough that it satisfycertain closure
properties. We just want to make sure that anywhatever that
is going to come up will be included. Saying all setsis
just a way to avoid saying precisely *which* closure
properties.Its closed under whatever operations youre going
to be interestedin.--Daryl McCulloughIthaca, NY
===
> A
category has two sorts, objects and arrows. (In a pinch, you
can do> without the objects.) It has operations of domain
and codomain and a> partial operation of composition of
arrows, whose domain is denable> in terms of the domain and
codomain operations. > That takes about 6 axioms.Actually,
only one structure is required: the product, dened as a>
partial operation.Most categories DO NOT HAVE products.The
product you MEANT to be referring to hereis composition. In
the context of category theory,product ALREADY means
something ELSE. : Maybe around 4 axioms are
required.Classical FOL doesnt do partial function
application.And if you are going to count axioms then youll
be needingto outlaw conjunction except in special cases.I
will be happy to post some actual axiom-sets ifyou persist in
tilting at this windmill.> Among the categories> we single out
some very special ones called toposes. > That takes at least 3
more axioms, because [...]You have to be careful about
reasoning this way. Axioms> dont add.For example,
non-Abelian groups are characterized by 1> operation (/)
and 3 axioms:> (a/c)/(b/c) = a/b> a/(b/b) = a> (a/a)/(b/b) =
b/b> or something like that. I dont recall the exact forms,
but> it was posted it to sci.math.research several years ago
and> you can probably nd it there.Abelian groups, however,
are characterized by 1 operation> (-) and only 2 axioms:>
a-(a-b) = b> a-(b-c) = c-(b-a).More stuff, here, means fewer
axioms.You can characterize groups by a single axiom if you
arewilling to make it complicated enough. More
importantly,though, for purposes of THIS discussion, you ALSO
CAN *make*the axioms add IF you make them SIMPLE enough.If
youll excuse me for overloading (0-ary identity and 1-ary
inverse) i,you can dene the group axiomsbyi*a = a i(a)*a = i
(a*b)*c = a*(b*c)Then,you CAN add, to THAT.In particular, for
Abelian groups,you can adda*b = b*a.
<br8sum$bg57$1@netnews.upenn.edu>
<Pine.LNX.4.56.0312111129110.29797@cauchy.math.duke.edu>
<bra7nk$col3$1@netnews.upenn.edu>
===
> Consider the events
A_n = {X_n > a n log n} where a is a constant.>This is where
my problem is....how do I calculate P(A_n)? I understandShow
for your distribution that P(X_n > t ) geq (const)/tfor all
t.
===
> For the rst part, see Fellers An Introduction
to Probability> Theory and Its Applications: Volume 1. In
my 3rd edition (1968),> its in section X.4 starting on p.
251.> Where can I see your 3rd edition? What is your
textbook called?> (more text below)> Sorry, I meant my 3rd
edition copy of Fellers An Introduction to> Probability
Theory and Its Applications: Volume 1.> Now, is the
following proposition true?> For any sequence of independent
random variables {Z_k}, if> Summation (from k = 1 to innity)
of P( |Z_k| > epsilon) = innityfor> all epsilon > 0,> then
Z_k converges almost surely to innity. (Maybe I can prove
thiswith> borel-cantelli lemma of independent events?)> No,
its not true. Constant random variables (i.e., random>
variables that are always equal to a specic constant) are>
independent of anything, so any deterministic sequence is a
sequence> of independent random variables, too, but your
proposition fails for> Z_k=-k (where the limit is -innity
instead of innity) or Z_k=k for> k even and =0 for k odd
(where there is no limit).> It *is* true that a sequence of
nonnegative, independent random> variables {Z_k} satisfying>
sum_{k=1}^innity P(Z_k > epsilon) = innity> for all epsilon
> 0 will have limsup Z_k = innity, and this can be> proved by
Borel-Cantelli. (Note that we have added the hypothesis of>
nonnegativity and changed the conclusion to specify the
limsup rather> than the limit.)> If true, then Id like to
use Markovs inequality, by saying that> P((X_1 + X_2 + ....
+ X_n)/nlog(n) > epsilon) <= (1/epsilon) *E((X_1 +> X_2 +
.... + X_n)/nlog(n))) or some variation> of markovs
inequality, and then sum up these values to possibly get>
innity and our proof would be done.> The problem is that
the inequality goes the wrong way. To apply the> above
result, you need to show that the sum of the probabilities
is> innite, but the inequality will only let you conclude
that this sum> is less than or equal to something, and that
wont help.> If this doesnt work, how can I view the events
{X_n > nlog(n)}? Iagree> this would be helpful.> Well, if
n=6, say, then nlog(n) is about 10.75. So, how would you go>
about calculating P(X_n > nlog(n)) = P(X_6 > 10.75)? You
should be> able to generalize this to an arbitrary n.>To make
sure were on the same page,$P(X_1 > 1 $log(1))$ = 1$$P(X_2 >
2 $log(2))$ = 1$$P(X_3 > 3 $log(3))$ = 1/2$But the results do
not follow a simple pattern...to write out some
moreterms....$P(X_4 > 4 $log(4))$ = 1/4$$P(X_5 > 5 $log(5))$
= 1/8$$P(X_6 > 6 $log(6))$ = 1/8$$P(X_7 > 7 $log(7))$ =
1/8$$P(X_8 > 8 $log(8))$ = 1/16$So I would like to say the
sum of these probabilities is innite as we takethe innite
series...but it seemsto be a bit messy.In particular, P(X_n >
nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ...,wherea_1 is
the rst integer that is greater than [log(nlog(n))]/log(2)
sincewe want 2^a_1 > nlog(n) where a_1 is the rst integer
that satisies thatinequality.So this gets a bit messy it
seems...> I dont see how I can tell if they occur innitely
often.> Furthermore, if this were true, how would this imply
that the limsup> is innite?> Well, okay, I spoke too
glibly. It still wouldnt be *quite* enough,> but it would
tell you that, innitely often (i.e., whenever X_n >
nlog(n)), youd have to have:> (X_1 + ... + X_n)/nlog(n) >
X_1/nlog(n) > 1> Now, if you could do the same thing with
{X_n > a n log(n)} for every> a > 0 (instead of just for
a=1), *then* that would be enough. Do you> see why?>I see why
this would be enough. The trouble is proving this now. But if
Ihavent proved it for a = 1, theres no chanceI can get it
for arbitrary a. So you must be thinking of an easier
methodto show thatgiven the events A_n = {X_n > n log(n)},
that the sum_{n=1}^{infty}P(A_n) = infty.> And which
Borel-Cantelli lemma would you use?> Which do you have? The
two I know about are the one that applies when> the sum of
the probabilities is innite and the other that applies> when
the sum of the probabilities is nite. So, once youve>
determined whether the sum of the probabilities is innite or
nite,> youll know.>(I had gured this part out before you
posted)> -- > Kevin <buhr@telus.net>Cancel-Lock:
sha1:P2d0jr69ANVq7W3pvQWQpG5jfcY
===
=> In particular, P(X_n
> nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ...,> where>
a_1 is the rst integer that is greater than
[log(nlog(n))]/log(2) since> we want 2^a_1 > nlog(n) where
a_1 is the rst integer that satisies that> inequality.> So
this gets a bit messy it seems...assume you meant a_2=a_1+1,
and so on.Lets introduce some standard notation here: for a
real number c, thequantity [c] is the greatest integer thats
less than or equal to c.So, [3]=3 and [3.6]=3. So, for any
real constant c, you have: P(X_n > c) = P(X_n =
2^{[log(c)/log(2)]+1}) + P(X_n = 2^{[log(c)/log(2)]+2}) +
...c=nlog(n).If you work out that right-hand side, it should
be a geometric seriesthat sums to a pretty nice function of
[log(c)/log(2)]. Now, sinceyou know that log(c)/log(2) <=
[log(c)/log(2)] < log(c)/log(2)+1, youcan bound that nice
function of [log(c)/log(2)] above and below byeven nicer
(after simplication) functions of c that dont involvethis
new [...] notation.Finally, if you substitute in c=n log n,
you get a value for P(X_n > nlog n) thats a pretty nice
function of [log(nlog(n))/log(2)] and isbounded above and
below by very nice functions of n. If you look atthe innite
sum of those probabilities over all n, youll be able tobound
it above and below by innite sums that involve only n
(withoutthe [...] notation), and you can use the
convergence or divergenceof those sums to show whether or not
your probabilities sum toinnity.-- Kevin
<buhr@telus.net>
===
> In particular, P(X_n > nlog(n)) =
P(X_n = 2^a_1) + P(X_n = 2^a_2) + ...,> where> a_1 is the
rst integer that is greater than [log(nlog(n))]/log(2)since>
we want 2^a_1 > nlog(n) where a_1 is the rst integer that
satisiesthat> inequality.> So this gets a bit messy it
seems...> assume you meant a_2=a_1+1, and so on.> Lets
introduce some standard notation here: for a real number c,
the> quantity [c] is the greatest integer thats less than or
equal to c.> So, [3]=3 and [3.6]=3. So, for any real constant
c, you have:> P(X_n > c) = P(X_n = 2^{[log(c)/log(2)]+1})> +
P(X_n = 2^{[log(c)/log(2)]+2}) + ...> c=nlog(n).> If you
work out that right-hand side, it should be a geometric
series> that sums to a pretty nice function of
[log(c)/log(2)]. Now, since> you know that log(c)/log(2) <=
[log(c)/log(2)] < log(c)/log(2)+1, you> can bound that nice
function of [log(c)/log(2)] above and below by> even nicer
(after simplication) functions of c that dont involve> this
new [...] notation.> Finally, if you substitute in c=n log
n, you get a value for P(X_n > n> log n) thats a pretty nice
function of [log(nlog(n))/log(2)] and is> bounded above and
below by very nice functions of n. If you look at> the
innite sum of those probabilities over all n, youll be able
to> bound it above and below by innite sums that involve only
n (without> the [...] notation), and you can use the
convergence or divergence> of those sums to show whether or
not your probabilities sum to> innity.>Dear Kevin,Ok, I am
getting closer and closer...By the way in Feller where you
mention X.4 for the rst original problemthat we havent
discussed, it is proven thatP{ |S_n/nlog(n) - 1| > epsilon }
---> 0which is equivalent to sayingS_n/nlog(n) converges in
probability to 1.However, in Feller, the log is base 2.Is it
henceforth equivalent to saythatS_n/nlog(n) converges in
probability to log(2)where log is base e? My problem
originally was toShow that (S_n)/(n*log(n)) converges in
probability tolog(2), which is different from what Feller
says in X.4 edition 3.(I have edition 2, but I believe
edition 3 is just a reprint).Robert> -- > Kevin
<buhr@telus.net>
===
> In particular, P(X_n > nlog(n)) =
P(X_n = 2^a_1) + P(X_n = 2^a_2) +...,> where> a_1 is the rst
integer that is greater than [log(nlog(n))]/log(2)> since> we
want 2^a_1 > nlog(n) where a_1 is the rst integer that
satisies> that> inequality.> So this gets a bit messy it
seems...> assume you meant a_2=a_1+1, and so on.> Lets
introduce some standard notation here: for a real number c,
the> quantity [c] is the greatest integer thats less than or
equal to c.> So, [3]=3 and [3.6]=3. So, for any real constant
c, you have:> P(X_n > c) = P(X_n = 2^{[log(c)/log(2)]+1})> +
P(X_n = 2^{[log(c)/log(2)]+2}) + ...> c=nlog(n).> If you
work out that right-hand side, it should be a geometric
series> that sums to a pretty nice function of
[log(c)/log(2)]. Now, since> you know that log(c)/log(2) <=
[log(c)/log(2)] < log(c)/log(2)+1, you> can bound that nice
function of [log(c)/log(2)] above and below by> even nicer
(after simplication) functions of c that dont involve> this
new [...] notation.> Finally, if you substitute in c=n log
n, you get a value for P(X_n > n> log n) thats a pretty nice
function of [log(nlog(n))/log(2)] and is> bounded above and
below by very nice functions of n. If you look at> the
innite sum of those probabilities over all n, youll be able
to> bound it above and below by innite sums that involve only
n (without> the [...] notation), and you can use the
convergence or divergence> of those sums to show whether or
not your probabilities sum to> innity.> Dear Kevin,> Ok,
I am getting closer and closer...> By the way in Feller
where you mention X.4 for the rst original problem> that we
havent discussed, it is proven that> P{ |S_n/nlog(n) - 1| >
epsilon } ---> 0> which is equivalent to saying> S_n/nlog(n)
converges in probability to 1.> However, in Feller, the log
is base 2.> Is it henceforth equivalent to say> that>
S_n/nlog(n) converges in probability to log(2)> where log is
base e? My problem originally was to> Show that
(S_n)/(n*log(n)) converges in probability to> log(2), which
is different from what Feller says in X.4 edition 3.> (I have
edition 2, but I believe edition 3 is just a reprint).>
Robert>I believe what I have just said above is true, but
please afrm this if youwould.Robert> -- > Kevin
<buhr@telus.net>
===
n distinct parallel lines are given in
the euclidean plane (n>2).construct a regular n-gon having
each of its vertices on a differentparallel line.any
idea?m
===
>n distinct parallel lines are given in the
euclidean plane (n>2).>construct a regular n-gon having each
of its vertices on a different>parallel line.>any
idea?>m>Its clear that there must be constraints on the set
of parallel linesfor this to be possible. For example, for n =
4, draw a unit square(the problem is invariant under
similarity transformations), thenconsider the space of
quadruples of parallel lines, one line passingthrough each
corner of the square. The rst line determines theremaining
lines, so the set of such quadruples is smoothlyparametrized
by the angle of the rst line. The set of all quadruplesof
parallel lines, modulo similarities, is a two-dimensional
space(dened by ratios of distances between the lines). The
set ofquadruples that arise from a square form a curve in
this space, sothere is no solution for most such
quadruples.The case n = 3 is special. The space of triplets
of parallel lines,modulo similarities, is one-dimensional,
and its easy to see that allsuch triplets arise from some
equilateral triangle.John Mitchell
===
John Mitchell
<john.mitchell.deletethispart@autodesk.com> ha scritto
nel>n distinct parallel lines are given in the euclidean
plane (n>2).>construct a regular n-gon having each of its
vertices on a different>parallel line.>any idea?>m> Its
clear that there must be constraints on the set of parallel
lines> for this to be possible. For example, for n = 4, draw
a unit square> (the problem is invariant under similarity
transformations), then> consider the space of quadruples of
parallel lines, one line passing> through each corner of the
square. The rst line determines the> remaining lines, so the
set of such quadruples is smoothly> parametrized by the angle
of the rst line. The set of all quadruples> of parallel
lines, modulo similarities, is a two-dimensional space>
(dened by ratios of distances between the lines). The set
of> quadruples that arise from a square form a curve in this
space, so> there is no solution for most such quadruples.>
The case n = 3 is special. The space of triplets of parallel
lines,> modulo similarities, is one-dimensional, and its
easy to see that all> such triplets arise from some
equilateral triangle.very good. actually i found a very
simple way to construct the triangle(case n=3), but couldnt
nd a way for the square; now you convinced me itwas somehow
impossible...
===
> n distinct parallel lines are given in the
euclidean plane (n>2).construct a regular n-gon having each of
its vertices on a different> parallel line.any idea?Yeah, I
have the idea that this is impossible for the general case.
Only a few special cases can be solved.A triangle will always
t; but for instance imagine four parallel lines with
successive distances d1, d2 and d3 with d1 <> d3, then I
think that no square will t.Unless I read this quastion in
the wrong way...(?)-- M.vr.gr. Dave
(d-dot-langers-at-wxs-dot-nl)
===
Dave Langers
<spam.messages@annoy.us> ha scritto nel messaggio> n distinct
parallel lines are given in the euclidean plane (n>2).>
construct a regular n-gon having each of its vertices on a
different> parallel line.> any idea?> Yeah, I have the idea
that this is impossible for the general case. Only> a few
special cases can be solved.> A triangle will always t; but
for instance imagine four parallel lines> with successive
distances d1, d2 and d3 with d1 <> d3, then I think that> no
square will t.> Unless I read this quastion in the wrong
way...(?)no you read it right! thanks for your hint!
===
I
havent tried to prove it, but it occurred to me after
posting that if onedoes the same construction with the
closed interval [a,b], but leaves out(a,-1) and (b,+1), then
one ought to get the maximal ideal space of thealgebra of
functions on [a,b] with right and left hand limits at every
point,which ought to be a Banach algebra in the sup norm.Of
course, even if that is correct, it isnt a
name...Ignorantly,Allan
Adlerara@zurich.ai.mit.edu**
** Intelligence
Lab. My actions and comments do not reect ** in any way on
MIT. Moreover, I am nowhere near the Boston ** metropolitan
area. **
**
===
F
===
I am working with Calculus of Variations
by Gelfand and Fomin. I am stuckon canonical transformations
and Hamilton-Jacobi equations. Do you guys haveany pointers
(web site, book)I would like to have sort of the intuition
and the proof.Tony
===
>I am working with Calculus of
Variations by Gelfand and Fomin. I am stuck>on canonical
transformations and Hamilton-Jacobi equations. Do you guys
have>any pointers (web site, book)>I would like to have
sort of the intuition and the proof.>Tony>You can try
Mathematical Methods of Classical Mechanics by V. I.Arnold,
Graduate Texts in Mathematics, Springer-Verlag, 1978.I dont
know if it will be any clearer than Gelfan and Fomin, but
itcant hurt to take a look.John Mitchell
===
Id like to get
some opinions about this problem, please.The dependable
capacity C available in an electric system in a monthof the
future, say Dec 2008, is a random variable with
densityfunction f dened on [0, Cmax]. We can admit f is
continuous on thisinterval. On the cited month you have to
meet a known anddeterministic demand r<=Cmax. Dene a random
variable decit, D, byD= r-C if C<r and D=0 if C>=r.
Supposing the density function f isindependent of r, the
expected value of D corresponding to a demand ris given by
E(r) = Integral [0,r] (r-c) f(c) dc = r*Integral
[0,r]f(c) dc - Integral [0,r] c f(c) dc. Since f is
continuous, E(r)exists for r<=Cmax and E is differentiable
wrt r. Applying the F.T. OfInt. Calculus, we get E(r) = r*
f(r) + Integral [0,r] f(c) dc -r*f(r) = Integral [0,r]
f(c) dc = Probability (C<=r) = Probability(D>=0). Though I
dont have f in a closed form, I can estimateProbability
(D>=0) by means of simulation models, using a processsimilar
to Monte Carlos. Then, for variations on r of about 5%, I
canmake the estimate Delta E(r) = (Delta r) * Probability
(D>=0). All Ineed to know is that f exists and is continuous
on [0, Cmax]. I dontneed to know how exactly f sends c into
f(c).In some situations this is reasonable, but there are
cases when itsnot admittable at all to suppose f is
independent of r. In such cases,for each r theres a
particular density function f_r. You can stillsuppose r is
known, but it affects the distribuition of C on [0,
Cmax](Cmax is always known and independent of r]. Then I
think I havesomething like E(r) = Integral [0,r] (r-c)
g(r,c) dc where g isdened on R^2 with values on R and, for a
given r, g(r,c) = f_r(c).Supposing g is continuous, is that OK
if I use Leibiniz formula tocompute E(r)? Anyway, if f_r
depends on r than that beautifulconclusion I came to before
is no longer true, right? Im a bitconfused here.Artur
===
The
surface area of the base of a non-regular tetrahedron is
unknown,but the surface areas of the three other faces are
known (A1, A2, A3respectively).Why is the surface area of the
base equal to:Ai/{cosine(angle between the normal of the base
and the normal to facei)}where i=1,2 or 3?
===
The surface
area of the base of a non-regular tetrahedron is unknown,>
but the surface areas of the three other faces are known (A1,
A2, A3> respectively).Why is the surface area of the base
equal to:Ai/{cosine(angle between the normal of the base and
the normal to face> i)}where i=1,2 or 3?> Let vectors a,b,c
from one vertex dene three edges.Then the oriented surface
vector of the adjoining facesis one half ofC == a X b , A ==
b X c, B == c X aand the oriented surface vector of the
fourth face is( c - a ) X ( b - a ) = c X b - a X b - c X a =
- b X c - a X b - c X a == D( a form of Stokes theorem )Now
the magnitude of this vector is twice the area of the
face,and is given by the sqrt ( D dot D ), but|D|^2 = D dot D
= - D dot A - D dot C - D dot Band dividing both sides by |D|
gives you your answer.... Well, almost, you wanted the sum of
the terms you described.Note that if you dene the normal
vectors as all pointing outwardsthe normal to the base points
down.Lew Mammel, Jr.
===
>The surface area of the base of a
non-regular tetrahedron is unknown,>but the surface areas of
the three other faces are known (A1, A2,
A3>respectively).>Why is the surface area of the base equal
to:>Ai/{cosine(angle between the normal of the base and the
normal to face>i)}>where i=1,2 or 3?>Any one of the three
faces F1, F2, F3 projects orthogonally onto thebase face.
Think about how orthogonal projection from one plane
toanother affects area. If you imagine viewing the planes in
thedirection of the planes intersection line (if the planes
areparallel, the answer is trivial), this reduces to thinking
about howprojection from one line to another affects lengths,
which should beeasy to gure out.John Mitchell
===
Can anyone
solve this?Use the implicit function theorem to show that as
a subspace ofR^{n+1} an n-surface M is locally homeomorphic
to an open subset of R^n.That is, for each p in M, there
exists a neighborhood of p(call it O, which is a subset of M)
and an open set U (subset of R^n)and a homeomorphismf : U -->
ONote : An n-surface M (which is a subset of R^{n+1}) is a
non-empty subsetsuchthat M = f^{-1}(c) wheref : U > R (U
is a subset of R^{n+1}) , f is C-innity, and suchthat for
all p in M, pis a regular point of f (i.e. its not a
critical point of f). Thats all Iam given.Steven
Rossi
===
>Can anyone solve this?>Use the implicit function
theorem to show that as a subspace of>R^{n+1} an n-surface M
is locally homeomorphic to an open subset of R^n.>That is,
for each p in M, there exists a neighborhood of p>(call it O,
which is a subset of M) and an open set U (subset of R^n)>and
a homeomorphism>f : U --> O>Note : An n-surface M (which is
a subset of R^{n+1}) is a non-empty subset>such>that M =
f^{-1}(c) where>f : U > R (U is a subset of R^{n+1}) , f
is C-innity, and such>that for all p in M, p>is a regular
point of f (i.e. its not a critical point of f). Thats all
I>am given.>Steven Rossi>You got an answer when you posted
this question yesterday. Why start anew post on the same
question?John Mitchell
===
>Can anyone solve this?>Use the
implicit function theorem to show that as a subspace
of>R^{n+1} an n-surface M is locally homeomorphic to an open
subset of R^n.>That is, for each p in M, there exists a
neighborhood of p>(call it O, which is a subset of M) and an
open set U (subset of R^n)>and a homeomorphism>f : U -->
O>Note : An n-surface M (which is a subset of R^{n+1}) is a
non-emptysubset>such>that M = f^{-1}(c) where>f : U > R
(U is a subset of R^{n+1}) , f is C-innity, and such>that
for all p in M, p>is a regular point of f (i.e. its not a
critical point of f). Thatsall I>am given.>Steven Rossi>
You got an answer when you posted this question yesterday. Why
start a> new post on the same question?>My apologies. I
wouldnt have posted again if I had seen your last
responsein the previous thread...I was still confused.> John
Mitchell
===
> Can anyone solve this?Use the implicit function
theorem to show that as a subspace of> R^{n+1} an n-surface M
is locally homeomorphic to an open subset of R^n.> That is,
for each p in M, there exists a neighborhood of p> (call it
O, which is a subset of M) and an open set U (subset of R^n)>
and a homeomorphismf : U --> ONote : An n-surface M (which is
a subset of R^{n+1}) is a non-empty subset> such> that M =
f^{-1}(c) where> f : U > R (U is a subset of R^{n+1}) , f
is C-innity, and such> that for all p in M, p> is a regular
point of f (i.e. its not a critical point of f). Thats all
I> am given.We have grad(f)(p) nonzero. Write p = (p1, p2,
..., p(n+1)). WLOG, the partial derivative of f with respect
to the last variable is nonzero. The IFT then says that near
p, M is the graph of a C^oo function dened near (p1, p2,
..., pn) in R^n. The graph function will give you your
homeorphism.
===
> Can anyone solve this?> Use the implicit
function theorem to show that as a subspace of> R^{n+1} an
n-surface M is locally homeomorphic to an open subset of
R^n.> That is, for each p in M, there exists a neighborhood
of p> (call it O, which is a subset of M) and an open set U
(subset of R^n)> and a homeomorphism> f : U --> O> Note :
An n-surface M (which is a subset of R^{n+1}) is a
non-emptysubset> such> that M = f^{-1}(c) where> f : U >
R (U is a subset of R^{n+1}) , f is C-innity, and such>
that for all p in M, p> is a regular point of f (i.e. its
not a critical point of f). Thatsall I> am given.> We have
grad(f)(p) nonzero. Write p = (p1, p2, ..., p(n+1)). WLOG,
the> partial derivative of f with respect to the last
variable is nonzero. The> IFT then says that near p, M is the
graph of a C^oo function dened near> (p1, p2, ..., pn) in
R^n. The graph function will give you yourhomeorphism.When
you say M is the graph, what do you mean by graph? And
later whenyou saygraph function, what do you mean by graph
function?Steven Rossi
===
I have a problem that I cant seem
to get...can anyone help please?Consider : X_1, X_2, ... is a
sequence of independent random variables withnite variances
and a common distribution F such that F(0) = 0.What is the
relationship between2*E[(X_n)^2]andE[1 / ( (X_{n-1})^2 +
(X_{n-2})^2 ) ]?Is one always at least as large as the other?
(E denotes expectation) If Ireplaced2*E[(X_n)^2] by
C*E[(X_n)^2], what value of C wouldstill make this true? (C
is a real number)Henrique
===
> Consider : X_1, X_2, ... is a
sequence of independent random variables with> nite
variances and a common distribution F such that F(0) = 0.What
is the relationship between> 2*E[(X_n)^2]> and> E[1 / (
(X_{n-1})^2 + (X_{n-2})^2 ) ]?Is one always at least as large
as the other?If the random variables take values in [a,b],
where, 0 < a < b < oo, then your rst expression lies between
2a^2 and 2b^2, and your second expression lies between
1/(2b^2) and 1/(2a^2). If the rst expression is <= the
second, then we have 2a^2 <= 1/(2a^2) for all a > 0,
nonsense. Same for the other way around.
===
> Consider :
X_1, X_2, ... is a sequence of independent random
variableswith> nite variances and a common distribution F
such that F(0) = 0.> What is the relationship between>
2*E[(X_n)^2]> and> E[1 / ( (X_{n-1})^2 + (X_{n-2})^2 ) ]?>
Is one always at least as large as the other?> If the random
variables take values in [a,b], where, 0 < a < b < oo, then>
your rst expression lies between 2a^2 and 2b^2, and your
second> expression lies between 1/(2b^2) and 1/(2a^2). If the
rst expression is> <= the second, then we have 2a^2 <=
1/(2a^2) for all a > 0, nonsense. Same> for the other way
around.Dear World Wide Wade,So you are saying that I cant
say anything about the relationship betweenthese
expectations? What if I replace the 2 in front of
2*E[(X_n)^2] by anarbitrary constant C?What I am trying to
get at eventually is to look at :Z_n = product (from k = 1 to
n) of { (C(X_k)^2) / ( (X_{k-1})^2 +(X_{k-2})^2 ) }and to
determine which values of C would guarantee that Z_n
converges withprobability one to a limit W which may be a
constant or a random variable.What do you think?Henrique
===
I want to ask how i can proof this theorm:If R is a
semi-prime ring with maximum condition on annihilators
thenevrey annihilator ideal K contain a nit generated ideal
N such thatannihilator K equal annihilator N.. or send to me
where can i found the useful thing suspened withgoldie ring.
thanks
===
>Problem :>Use the implicit function theorem
to show that as a subspace of>R^{n+1} an n-surface M is
locally homeomorphic to an open subset of R^n.>That is, for
each p in M, there exists a neighborhood of p>(call it O,
which is a subset of M) and an open set U (subset of
R^n)>and a homeomorphism>f : U --> O>Steven
Rossi> The details depend on how your surface is dened.
If its dened (at> least locally) by an equation of the form
F(x) = 0, with x in R^{n+1}> and F: R^{n+1} -> R being
non-singular on the zero-set M, the implicit> function
theorem says that F locally looks like a linear function
L:> R^{n+1} -> R. Clearly, the zero-set of a linear function
is globally> (and therefore locally) diffeomorphic to an open
subset of R^{n+1}, so> the same is true for F. The phrase
looks like means that there are> locally diffeomorphisms
of the domain and range space that transform F> to L.>My
surface is dened as follows :>An n-surface M (which is a
subset of R^{n+1}) is a non-empty subset such>that M =
f^{-1}(c) where>f : U > R (U is a subset of R^{n+1}) , f
is C-innity, and such>that for all p in M, p>is a regular
point of f (i.e. its not a critical point of f). Thats all
I>am given.> John Mitchell>John
Mitchell
===
Prinzip:Ich habe einen Spielautomaten entwickelt,
das einGeschicklichkeitsspiel anbietet, so wie z.B. ein
J.8ager geschickt seinmuss sein Wild zu fangen - das Wild
wiederum wiederum manchmal einfachund manchmal schwieriger zu
kriegen ist, was nicht eindeutigvoraussehbar ist f.9fr den
J.8ager, einmal wird die Geschicklichkeit mehrgefordert,
manchmal weniger, im Wesentlichen abh.8angig von
denErfolgen/Misserfolgen der anderen J.8ager (Konkurrenten),
die durchDezimieren des Wildes den Wildbestand mitbestimmen,
d.h. dieSchwierigkeit zu jagen.Zum Ger.8at selber:F.9fr die
Spieler ist es m.9aglich mit dem Einsatz von einer Einheit
Geldmax 20x der Einheiten (pro Spiel) zu gewinnen -
mitGeschicklichkeit/Strategie/etc.Die Spieler k.9annen
jederzeit und Intensit.8at teilnehmen wie sie wollen.Das
heisst es ist eine offene Konkurrenz. Das Ger.8at
funktioniert wieein Messger.8at, das die Geschicklichkeit
der Spieler in angepassterSchwierigkeit wiedergibt. Wir
haben folgende Kriterien die mathematisch relevant w.8aren,
meinerAnsicht nach:Einsatz: 1 Einheit Auszahlung: mal 20
Einheiten Geschicklichkeit der Spieler (GamblerSkill):
Anfangs unbekannt, dasGer.8at startet bei 50% der technisch
m.9aglichenSchwierigkeit/Aufgabenstellung, lernt aber
hinzu und berechnet einenDurchschnitt bei jeder
Aktion.Schwierigkeit des Ger.8ats (MachineSkill): 120%
des Durchschnittes(kommerziellen Umsetzung 20%) des
GamblerSkills.Erkl.8arung: Die Bandbreite der m.9aglichen
Schwierigkeit ist bekannt undrein technisch vorgegeben. Dies
wird in Prozenschritten am Ger.8atangezeigt. Von 0%
(leichtestet Stufe) bis 100% (schwierigste Stufe) -intern
aber im Promille-Bereich berechnet.Bandbreite der
Kaskadierungsgeschwindigkeit in welcher sich dieSchwierigkeit
dem Durchschnitt anpasst ist gleich 50fache derm.9aglichen
Hochstauszahlung (hier 20 Einheiten) will heissen 1000.Das
w.9frde sich dann folgendermassen auswirken: ein Ger.8at
hat eineSchwierigkeitsstufe von 52%, ein Anderes (Lokal) 56%.
Sind die
Spieler(Spielergemeinschaft/(Versicherungsgemeinschaft)) an
einem Ortkonstant geschickter, pendelt sich das Ger.8at auf
56% ein. Die(durchschnittliche) Geschicklichkeit l.8asst
sich direkt am erziehltenErfolg der gewonnen oder verlorenen
Spiele ablesen. Trainingsspiele (mit weniger Einsatz --.20)
basieren auf demaktuellen Stand des Ger.8ates ohne diesen
zu ver.8andern - weil auchkeine Gewinne m.9aglich sind. Die
Herausforderung ist - wen wunderts - ein Spiel zu entwickeln,
dasf.9fr m.9aglichst viele Spieler attraktiv ist, eine
interessante Safari zuveranstalten - um auf die J.8ager
zur.9fckzukommen. In der Praxis funktioniert es - mit
empirischen Versuchen ermittelt.Mein Wunsch w.8are es dies
mittels einer mathematischen Formel oder/undModell-Theorie
darzustellen. Im Grunde empnde ich dies alsMarktwirtschaft
in spielerischer Weise umgesetzt - auf dem Markt
derGeschicklichkeit/Strategie - Der Spielautomat als Spiegel
desGeschicklichkeits-Marktes. Wobei ich mich selber wiederum
in diesemMarkt selber bew.8ahren muss (kommerzielle
Umsetzung).Es geht mir um die Darstellung des Systems
(Spieler/Automat)(Geschicklichkeit/Schwierigkeit) mittels
einer math.-formel undentsprechender Umschreibung damit auch
ein Jurist sie verstehen kann,inkl. mich selber.Juristen
behaupten, dass es keine math. Formel gibt die
funktioniert,denen w.9frde ich gerne das Gegenteil beweisen,
mangels math. Talent ichkeine Chance habe dies zu tun. Wer
kann mir dabei weiterhelfen?Karl
FayCH-Z.9frichinfo@fay.ch
===
I tried to solve another task
on this issue today and failed. {/-(First of all, I wanted to
know if there is a general approach to solvetasks of the form:
You have given a fraction of the form P(x)/Q(x).There P(x)
and Q(x) are polynoms. You have to express this fractionin
the form sum(a_k*(m*x+n)^k, k, 0, innity). ?[m, n are
constants and a_k represents k constants.]If somebody knows a
general approach or a website (understandablefor a beginner)
there this is explained then please tell me. (I somehowmust
understand this matter.)Anyway, this was my task
today:###task:^^^^^^Express 14 / ( 8*(x^2-1) ) as an innite
sequence of the form:sum(a_k*(2-x)^k, k, 0,
innity).____________|my attempt(s):^^^^^^^^^^^^^^^^^^14 /
( 8*(x^2-1) ) = (7/4) * ( 1 / ( (x-1)(x+1) ) )partial
fraction decomposition (Do I name it
right?):^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^1 = A*(x+1) + B(x-1)if x = 1 => A = 1/2if x = -1 => B
= -1/2==> (7/4) * ( 1 / ( (x-1)(x+1) ) )= (7/4) * ( 1/(
2*(x-1) ) - 1/( 2*(x+1) ) )= (7/8) * ( 1/(x-1) - 1/(x+1) )
Attempt N 1:^^^^^^^^^^^^^^^^ = (7/8) * ( -1/(1-x) -
1 / ( 1-(-x) ) )= (7/8) * ( -sum( x^k, k, 0, innity) - sum(
(-x)^k, k, 0,innity) )= (7/8) * ( -sum( x^k - (-1)^k * x^k,
k, 0, innity) )= (7/8) * ( -sum( ( 1 - (-1)^k ) * x^k, k,
0, innity) ) = ?I didnt know how to transform this to the
demanded form.Attempt N 2:^^^^^^^^^^^^^^^^ = (7/8) * (
1/(1-2+x) + 1/(1-2-x) )= (7/8) * ( 1 / ( 1 - (2-x) ) + 1 / (
1-(2+x) ) )= (7/8) * ( sum( (2-x)^k, k, 0, innity ) + sum(
(2+x)^k, k, 0, innity ) ) = ?I didnt know how to simplify
(2-x)^k + (2+x)^k, sothat I get (..)*(2-x)^k {the demanded
form}.###Thats it. It would be very nice if you told me a
generalapproach for these tasks.Karl
===
[snip]> Express 14 /
( 8*(x^2-1) ) as an innite sequence of the form:>
sum(a_k*(2-x)^k, k, 0, innity).> ____________> |my
attempt(s):> ^^^^^^^^^^^^^^^^^^> 14 / ( 8*(x^2-1) ) = (7/4) *
( 1 / ( (x-1)(x+1) ) )partial fraction decomposition (Do I
name it right?):>
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>
1 = A*(x+1) + B(x-1)if x = 1> => A = 1/2if x = -1> => B =
-1/2==> (7/4) * ( 1 / ( (x-1)(x+1) ) )> = (7/4) * ( 1/(
2*(x-1) ) - 1/( 2*(x+1) ) )> = (7/8) * ( 1/(x-1) - 1/(x+1) )
Partial fractions is the right name. 1/(x+1) =
1/(3-(2-x)) = (1/3) / (1 - (2-x)/3 )Jim Buddenhagen-- To
reply copy jbuddenh@REMOVEtexas.net to address bar and edit
out REMOVECall an innite subset X of the real numbers
recognizable if and only if thereis a bijection f:R->R such
that for all x in X there is an integer k (possiblydependent
on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.]Say a real number x
is recognizable iff x in X for some recognizable X.What
numbers are recognizable?rich
===
Call an innite subset X of
the real numbers recognizable if and only> if there is a
bijection f:R->R such that for all x in X there is an>
integer k (possibly dependent on x) with f^k(x)=0.
[f^2(x)=f(f(x)),> ect.] Say a real number x is recognizable
iff x in X for some recognizable> X. What numbers are
recognizable?If b is a positive real number, let X be the set
of all positiveinteger multiples of b, and let f:R->R be f(x)
= x-b. Then X isrecognizable, and so is b. The negative reals
are handled similarly.0 is of course recognizable. So the
answer is all of them.
===
>Call an innite subset X of the
real numbers recognizable if and only> if there is a
bijection f:R->R such that for all x in X there is an>
integer k (possibly dependent on x) with f^k(x)=0.
[f^2(x)=f(f(x)),> ect.] Say a real number x is recognizable
iff x in X for some recognizable> X. What numbers are
recognizable?>If b is a positive real number, let X be the
set of all positive>integer multiples of b, and let f:R->R be
f(x) = x-b. Then X is>recognizable, and so is b. The negative
reals are handled similarly.>0 is of course recognizable. So
the answer is all of them.>Yes. Kind of a dumb denition, I
suppose. Wait a minute, what if X isrequired to be a *dense*
subset of R? The rationals are recognizable
withf:x->1/x-oor(1/x). Is sqrt(2) recognizable? How about
that awful numberlog(2)?thanks,rich
===
> Call an
innite subset X of the real numbers recognizable if and
only> if there is a bijection f:R->R such that for all x in X
there is an> integer k (possibly dependent on x) with
f^k(x)=0. [f^2(x)=f(f(x)),> ect.] > Say a real number x is
recognizable iff x in X for some recognizable> X. > What
numbers are recognizable?>If b is a positive real number,
let X be the set of all positive>integer multiples of b, and
let f:R->R be f(x) = x-b. Then X is>recognizable, and so is
b. The negative reals are handled similarly.>0 is of course
recognizable. So the answer is all of them.>Yes. Kind of
a dumb denition, I suppose. Wait a minute, what if X
is>required to be a *dense* subset of R? The rationals are
recognizable with>f:x->1/x-oor(1/x). No, they are not, at
least, not with this function: your f is notinjective: both 3
and 3/4 map to 1/3: 3|-> (1/3) - oor(1/3) =(1/3)-0 =
1/3.(3/4) |-> (4/3) - oor(4/3) = 4/3 - 1 = 1/3.--
===
===
===
Its not denial. Im just very selective about what
I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
> Call an
innite subset X of the real numbers recognizable if and
only> if there is a bijection f:R->R such that for all x in
X there is an> integer k (possibly dependent on x) with
f^k(x)=0. [f^2(x)=f(f(x)),> ect.] > Say a real number x
is recognizable iff x in X for some recognizable> X. >
What numbers are recognizable?>If b is a positive real
number, let X be the set of all positive>integer multiples of
b, and let f:R->R be f(x) = x-b. Then X is>recognizable, and
so is b. The negative reals are handled similarly.>0 is of
course recognizable. So the answer is all of them.>Yes.
Kind of a dumb denition, I suppose. Wait a minute, what if X
is>required to be a *dense* subset of R? The rationals are
recognizable with>f:x->1/x-oor(1/x). >No, they are not,
at least, not with this function: your f is not>injective:
both 3 and 3/4 map to 1/3: 3|-> (1/3) - oor(1/3) =>(1/3)-0 =
1/3.>(3/4) |-> (4/3) - oor(4/3) = 4/3 - 1 = 1/3.>You are,
of course, correct! Ok, forget about all of R and just
consider theinterval [0,1). Then f:x->1/x-oor(1/x), f(0)=0
recognizes the rationals in[0,1). Are there any more
recognizable numbers? [X a dense subset of [0,1); positivek;
bijective f:[0,1)->[0,1)]thanks,rich
===
>Call an innite
subset X of the real numbers recognizable if and only if
there>is a bijection f:R->R such that for all x in X there
is an integer k (possibly>dependent on x) with f^k(x)=0.
[f^2(x)=f(f(x)), ect.]And I assume f^{-2}(x) means
f^{-1}(f^{-1}(x)), etc...>Say a real number x is recognizable
iff x in X for some recognizable X.>What numbers are
recognizable?THEOREM. All real numbers are
recognizable.Proof. Given an arbitrary real number r, dene
g_r:R->R to be thefunction g_r(x) = x+r, and let X_r be:X_r =
{kr: k an integer}.g_r is a bijection; and (g_r)^k(x) = x+kr
for every integer k. Inparticular, for every x in X_r, x=kr
for some k, we have (g_r)^{-k}(x)= x -kr = 0.Since r is in
X_r, it follows that any real number is in arecognizable set,
and therefore, that every real number isrecognizable. QED--
===
===
===
Its not denial. Im just very selective about what
I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
>Call an innite
subset X of the real numbers recognizable if and only
if>there>is a bijection f:R->R such that for all x in X
there is an integer k>(possibly>dependent on x) with
f^k(x)=0. [f^2(x)=f(f(x)), ect.]>And I assume f^{-2}(x)
means f^{-1}(f^{-1}(x)), etc...Fair assumption. I was
thinking k>0, but didnt include that restriction.>Say a
real number x is recognizable iff x in X for some
recognizable X.>What numbers are recognizable?>THEOREM.
All real numbers are recognizable.>Proof. Given an arbitrary
real number r, dene g_r:R->R to be the>function g_r(x) = x+r,
and let X_r be:>X_r = {kr: k an integer}.>g_r is a
bijection; and (g_r)^k(x) = x+kr for every integer k.
In>particular, for every x in X_r, x=kr for some k, we have
>(g_r)^{-k}(x)= x -kr = 0.>Since r is in X_r, it follows
that any real number is in a>recognizable set, and therefore,
that every real number is>recognizable. QEDYes, thanks.Are
there any simple restrictions on k and/or f that will make
the recognizablenumbers a proper subset of R that includes
the algebraic numbers?rich
===
>Call an innite subset X
of the real numbers recognizable if and only if>there>is a
bijection f:R->R such that for all x in X there is an integer
k>(possibly>dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)),
ect.]>And I assume f^{-2}(x) means f^{-1}(f^{-1}(x)),
etc...>Fair assumption. I was thinking k>0, but didnt
include that restriction.Even so all numbers are
recognizable: given an arbitrary real numberr, the set X_r
would become {-kr : k a positive integer}, and then X_ris the
set recognized by g_r(x) = x+r. Then -r is in X_r, so
everyreal number is in some X_r, hence every real number is
recognizable.>Are there any simple restrictions on k and/or f
that will make the>recognizable numbers a proper subset of R
that includes the algebraic>numbers?I doubt it. The algebraic
numbers are zeros of polynomials withrational coefcients, but
polynomials are not usually one-to-oneanyway...--
===
===
===
Its not denial. Im just very selective about what
I accept as reality. --- Calvin (Calvin and
Hobbes)
===
=
===
==Arturo
Magidinmagidin@math.berkeley.edu
===
The question was
whether there are> necessarily arbitrary choices involved in
constructing the> Riemann surface for a function...>Then as
a technique analytical continuation>is only practical with the
aid of a computer program. >If cuts are>allowed I can treat
some algebraic functions (I do not claim all)>without using
it.Huh? Nobodys said anything about cuts not being allowed.>
The question was [see above]. So I am allowed to use cuts
after all. Perhaps now it can be agreedthat there are two
divergent theories, one in analysis, one intogeometry. Then
the question which relates to functions was answeredin the
rst paragraph of my rst post.
===
>The question was
whether there are> necessarily arbitrary choices involved in
constructing the> Riemann surface for a function...>Then
as a technique analytical continuation>is only practical
with the aid of a computer program. >If cuts are>allowed
I can treat some algebraic functions (I do not claim
all)>without using it.Huh? Nobodys said anything about cuts
not being allowed.> The question was [see above].> So I am
allowed to use cuts after all. Perhaps now it can be
agreed>that there are two divergent theories, one in
analysis, one into>geometry. Then the question which
relates to functions was answered>in the rst paragraph of my
rst post.Youre not making any
sense.David C. Ullrich
===
> James
Takayama posts under the names of Thomas, Nospam and Maui
Cop. He> posted libel against me using the name of Nospam in
the WaxyOrg web site and<<<snipped a lot of psycho crap from
Shawn pedophile-boy Kabatoff>The following (courtesy of
Waxy.org) is sort of an unofcial FAQexplaining the
psychotic nonsense posted to Usenet by Shawn DarylKabatoff
AKA Dar, AKA Probababbilities. And now AKA marcia
andme.WARNING: Read below before even thinking about
responding to
thistwit.http://www.waxy.org/archive/2002/05/21/dar_
kaba.shtml#000643Usenet has the tendency to provide a public
forum for those who wouldnormally be scribbling in a closet.
For example, take Daryl ShawnKabatoff. For the last few
years, hes methodically gatheredstatistics from various
sources, ranging from local newspaperobituary pages to the
food court of the Saskatoon Midtown Plaza mall.With all the
raw data hes collected, hes attempting to prove dailythat
our full names are in mathematical harmony with our
birthdays.about, starting with calculations related to their
birthdate and fullnames, blending in whatever other personal
information about theirfamily members, spouses, birthplace,
and career hes been able tozealotry, and personal torment.
Ive never seen anything like it.With all the prime numbers,
Fibonacci sequences and biblicalreferences, its like reading
the notebooks of Maximillian Cohen andJohn Nash combined.
Unsurprisingly, several posts unfold to reveal ahistory of
painful mental illness. If you have some time, take a
look.Ive detailed his posting history and a several sample
posts below. January 27, 1999 to July 5, 2000 as
Catsco@home.comDecember 9, 2000 to May 4, 2001 as
s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as
kabatoff@the.link.caposts have been removed from Google
Groups archive)Selected Posts:Tessa Lynne SmithDastageer
Sakhizai and Helen SmithBrett David MakiAndrew Meredith
CottonAmanda Dawn NewtonMona Marie EtcheverryTony Peter
NusplLisa Charlene McMillanGrant Allyn Wood Comments scarier
still is that saskatoon is my hometown, though not my
currentresidence. and every single place hes mentioned in
his posts (mostnotably nervous harolds and the roastary)
were either places ivebeen (as its a small city of 200K) or
hangouts, ie. the two placesout if they know of him, they (my
friends that is) being of thebroadway-centred slacker ilk.
myself, too, until i got out of there.eh, anyways. thought it
odd to see all this. midtown mall. i ate mymeals there, whilst
waiting several days in line for star wars episodeone, at the
theatre across the street.posted by andy raad on May 22, 2002
06:20 PMFascinating. Its like hes trying to take chaos and
bind it intowhatever rules he can nd, religious, logical and
otherwise. Numbersand math have a reliable pattern, something
that can always be provento true or false. People and
religion do not. It reminds me of DarrenAronofskys movie Pi.
Its the story of an paraniod genius who istrying to nd a
pattern in Pi. A group that takes interest in hiswork is
convinced that the existence of Pi, a number whose
existencecan be proven but no quantied, is proof of the
existence of God.Kabatoffs hunt for patterns in something as
random as name selectionis a way to reconcile his deeply
logical thought process with hisconicting religious
views.posted by matt on May 23, 2002 11:19 AMasking him if
hed be willing to create a numerological analysis forme. I
also asked him if he had seen either Pi or A Beautiful Mind,
andwhat he thought of them. If he replies, Ill be sure to
post it.posted by Andy Baio on May 23, 2002 11:24 AMI baked
many pumpkin pies for Shawn (he likes pumpkin pies). I
rubbedpumpkin pie all over my breasts for him, and my breasts
turned orange.I am a pumpkin for Shawn.posted by Trisha
Blondie on July 24, 2002 10:41 PMUm, thats swell. So, youre
in love with him?posted by Andy Baio on July 25, 2002 07:10
AMShawn once went to a funeral for a Jehovah Witness that
shot himselfand the lemon tarts were very bad, they were not
only sour but wererubbery as well. Shawn said that the guy
was some kind of JehovahWitness prophet, he saw in advance
that the lemon tarts at his funeralwere to be very very bad,
and so he shot himself. Shawn said that henever ate pumpkin
pie at a funeral but would like to some day. Shawnlikes
pumpkin pie and so I have been practicing to make very
goodpumpkin pies.posted by Trisha Blondie on July 25, 2002
02:49 PMShawn said that the lemon tarts were sour, bitter and
rubbery.posted by Trisha Blondie on July 30, 2002 12:32 AMI
dont think this guy takes notes. I think he has Total
Recall, andit has driven him insane...posted by Todd Smith on
December 26, 2002 11:00 AMOh... I almost forgot... I didnt
spend thousands of dollars a daytormenting Daryl... We got a
deal on tormenting that scal year, itonly came to about
37cents a day....posted by Dr Claw on December 30, 2002 01:56
AMMr. Kabatoff attempts to portray himself as a victim, but in
fact heis a violent predatory pedophile who is well known to
his local lawenforcement. In his post to multiple newsgroups
with the subjectCollecting Mail For The Coming Anti-Christ,
he encourages mothers tosend him photos of their naked
daughters. Mr Kabatoff explains, IAnt-Christ) that were of
underage children unless the parent wassigning consent. He
is banned from virtually all the shopping mallsin his
community because he stalks young people and sexually
harassesthem. He has an extensive arrest record which
includes sexualmolestation charges. Hes been hospitalized in
mental institutionsabout his contact with young girls in many
posts. Search newsgrouparchives for posts by him containing
the word nubile. As part of hisharrassment, he provides
personal details in a public forum, such asthe real names of
real children, in these and other posts. About onewanted her
and her sister
dead.http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+
dead+or+in+my+bed&hl=en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%
241%40ID-136124.news.dfncis.de&rnuHe not only curses children
and prays for their death in his posts, healso enjoys
attending the funerals of young people: And so, sincenubile
sweeties are found in greatest abundance at the funerals
ofhigh school students, then it is the funerals of high
school studentsthat make the very very best funerals,
especially if there is food...I stuff my face (and my
pockets) with all the good food and look atall the pretty
nubile sweeties and have the time of my
life...http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff
+nubile+sex&hl=en&lr=&ie=UTF-8&scoring=d&selm=LfXN8.63042%
24R53.25142039%40twister.socal.rr.com&rnum=1 Many of his
posts are sent to alt.teens.advice. However, he
liberallyspams, oods and crossposts his off-topic
threatening and offensivemissives to countless newsgroups.
Some people HAVE problems and somefolks ARE problems. Dont
dismiss Mr. Kabatoff as a harmless nut. Whenhe sends these
posts to any newgroup, please help by reporting him toI knew
of him when I was attending the University of
Saskatchewan.Hed hang out in the Arts computer lab and all
youd see is screens ofnumbers racing by on his laptop. I
have an original copy of hisCollecting Mail for the Coming
Anti-Christ pamphlet, and have seenhim be hauled away by
campus security on more than one occasion. Myfriends and I
refer to him as Crazy Number Man.Ive been posting to (and
about) Shawn for over two years with biggaps in between. He
has seen Pi and didnt like it and didnt think itresembled
him at all. (Wrong, it ts him to a tee) He doesnt havetotal
recall and has stated that he travels with a lap top to
notateitems. Also, he uses cut n paste a lot if you read
all the waythrough his ramblings. He is anti-social as shown
by his angrystatements towards those who, by his own
admission, have been kind(but not kind enough) to him. Still,
hes intelligent and seems to beable to take a joke on
occassion. Thats where I came in. ALOHA if it comes from
anyone not already in my addressbook. Ill never even see
it)
===
It was a dark and stormy night, and Marcia and Me
managed to scribble:> James Takayama posts under the names of
Thomas, Nospam and Maui Cop. He> posted libel against me using
the name of Nospam in the WaxyOrg web site> and then quotes
the material using the name Thomas, quoting your libel and>
attributing it to someone else is just more libel. He has
repeatedly cited> my interest in Callie to support his
unfounded allegations that I am a> pedophile but Callie is
not a child but is instead a woman. I have never> been
hospitalized in mental institutions for accosting women (nor
do I> accost women as he purports), but only for daring to
criticize Protestant> and Catholic churches and daring to try
to show people that God provides> them with their names (and
for failing to abide by my parents good> advice). -Daryl S.
Kabatoff HALOHA!!!Well I only quote Daryl Shawn Kabatoffbut
year after year after year I tell people of my plight and of
their lthy penises,The homosexuals will say that I am
embarrased about my penisAt some appeal panel hearings I am
drugged so heavily that I do not know that I am even
therehttp://groups.google.com/groups?selm=fPID6.299%24pp
.2818127%40tomcat.sk.sympatico.cagtoomey
===
> What with
the war on terror, and Canadas puny military and corrupt>
police force, the only way to insure peace and safety for
that> country, is for it to be annexed by America.> It is
the only way foward for Canada aka Canuckistan.>You seem
to forget that the Canadians kicked our ass in two
conicts....> Really? Which
two?>1775>http://www.civilization.ca/cwm/chrono/
1000invasion_e.htmlThere was no country then called Canada.
Ergo there were no Canadians.The people that lived there were
simply British subjects just liketheir neighbors to the south
pre-revolution. BTW the so-calledcanadians kicked so much
ass the their neighbors to the south becametruly independent
while the Canadians had to wait another 80+ years.Well even
longer considering Britain ran their foreign affairs wellinto
the twentieth
century.>1812>http://www.civilization.ca/cwm/chrono/
1774invasion_repelled_e.html>Your fantasizing again (see
above explanation). Even if it was true itjust shows how
pathetic you are that you have to go back almost twohundred
years to nd something.dkp
===
> What with the war on
terror, and Canadas puny military and corrupt> police force,
the only way to insure peace and safety for that> country, is
for it to be annexed by America.> It is the only way foward
for Canada aka Canuckistan.>You seem to forget that the
Canadians kicked our ass in two conicts....> Really?
Which
two?>1775>http://www.civilization.ca/cwm/chrono/
1000invasion_e.htmlyou apparently didnt read it very well.
PLUS Canada didnt existthen, it was just another British
puppet and the forces that workedtheir way north went after
the British, not the so called Canadians.The area then was
commonly called New France and the name Kanatais an
indian word for cluster of villages or houses.Canada was
formed in 1867 with the Constitution Act of 1867, and
inCanadas original constitution the name of the new dominion
would be .. . Canada.
>1812>http://www.civilization.ca/cwm/chrono/1774invasion_
repelled_e.htmlthat was a war with the British. Canada did
not exist then.THOM>
===
> What with the war on terror,
and Canadas puny military and corrupt> police force, the only
way to insure peace and safety for that> country, is for it to
be annexed by America.> It is the only way foward for Canada
aka Canuckistan.>You seem to forget that the Canadians
kicked our ass in twoconicts....> Really? Which
two?>1775>http://www.civilization.ca/cwm/chrono/
1000invasion_e.html> you apparently didnt read it very
well. PLUS Canada didnt exist> then, it was just another
British puppet and the forces that worked> their way north
went after the British, not the so called Canadians.By that
logic, those that were repelled werent american, either.>
The area then was commonly called New France and the name
Kanata> is an indian word for cluster of villages or
houses.And...?> Canada was formed in 1867 with the
Constitution Act of 1867, and in> Canadas original
constitution the name of the new dominion would be .> . .
Canada.And...?>1812>http://www.civilization.ca/cwm/chrono/
1774invasion_repelled_e.html> that was a war with the
British. Canada did not exist then.> THOMWow, you really are
a nation in denial...
===
Thom <thomandpam@yahoo.com.au> a
.8ecrit dans le message de> you apparently didnt read it
very well. PLUS Canada didnt exist> then, it was just
another British puppet and the forces that worked> their way
north went after the British, not the so called Canadians.>
The area then was commonly called New France and the name
Kanata> is an indian word for cluster of villages or
houses.> Canada was formed in 1867 with the Constitution Act
of 1867, and in> Canadas original constitution the name of
the new dominion would be .> . .
Canada.>1812>http://www.civilization.ca/cwm/chrono/
1774invasion_repelled_e.html> that was a war with the
British. Canada did not exist then.>
THOM>---
--Jacques Cartier was the rst
Frenchman to write the word Canada, in 1535.--
http://www.civilisations.ca/vmnf/reper/glossair/r-ge-02.htm
Anemic Combatant
===
How do I use the sinh, cosh functions on
my calculator? Im not sure whatbuttons to use. I have a
Casio fx-991MS.
===
> How do I use the sinh, cosh functions on
my calculator? Im not sure what> buttons to use. I have a
Casio fx-991MS. Should I really encourage to hyp(e) before
sin? :-) I never used a casio but... Raymond
===
> How do I
use the sinh, cosh functions on my calculator? Im not sure
what> buttons to use. I have a Casio fx-991MS.Is there an
instruction booklet that came with it?
===
> How do I use the
sinh, cosh functions on my calculator? Im not surewhat>
buttons to use. I have a Casio fx-991MS.> Is there an
instruction booklet that came with
it?http://ftp.casio.co.jp/pub/world_manual/edu/en/fx570MS_
991MS_E.pdf
===
>How do I use the sinh, cosh functions
on my calculator? Im not surewhat>buttons to use. I have a
Casio fx-991MS.>Is there an instruction booklet that came
with it?>
http://ftp.casio.co.jp/pub/world_manual/edu/en/fx570MS_991MS_
E.pdf For simpler operations try rather this one :
http://ftp.casio.co.jp/pub/world_manual/edu/en/fx115MS_991MS_
E.pdf For sinh do hyp sin x = For sinh^(-1) do hyp shift
sin-1 x = and so on... Raymond
===
> I bet you already knew
that cos pi = -1 and such. But I was shocked to nd,> just
now, that there appears to be a _vegetable_ in the domain of
cos x.> (Yes, you read that correctly. Obviously this is a
silly pseudo-mathematical> post, but then you could tell
that from the threads title.)What is that vegetable? What is
the value of cos x there?(I guess I should wait for other
people to try to answer rst. But this is> far too silly to
be drawn out that way, and so Ive given the answers
below.)DavidThe vegetable is lettuce.According to my
dictionary:cos lettuce = romaine>Youre wrong, lettuce is NOT
in the domain of cos. It would be if it>was cos(lettuce) =
romaine, which is not the case.I really like puns, which is
why im feeling annoyed now - im not-- -johnFebruary 28
1997: Last day libraries could order catalogue cardsfrom the
Library of Congress.
===
> I bet you already knew that cos pi
= -1 and such. But I was shocked to> nd, just now, that
there appears to be a _vegetable_ in the domain> of cos x.
(Yes, you read that correctly. Obviously this is a silly>
pseudo-mathematical post, but then you could tell that from
the> threads title.)> What is that vegetable? What is
the value of cos x there?> (I guess I should wait for
other people to try to answer rst. But> this is far too
silly to be drawn out that way, and so Ive given the>
answers below.)> The vegetable is lettuce.> According
to my dictionary:> cos lettuce = romaine> I really like
puns, which is why im feeling annoyed now - im notSorry. I
didnt want anyone to be annoyed. Theres nothing deep
here._The American Heritage Dictionary of the English
Language_ says thatcos lettuce is the same as romaine.
[The reason for the name coslettuce is that that type of
lettuce was originally exported from theisland of Cos (or
Kos).] And so we have the well known cos pi = -1,
etc.,together with the silly (and perhaps not well known) cos
lettuce = romaine,which appears to place lettuce in the domain
of the cosine function.David
===
Does anyone know why for
example my calculate can produce the argument andmodulus for
1 - 3i no problems at all, but if I enter (1-3i)^4 I get a
MathError?