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Richard Tobin
science forum Guru Wannabe
Joined: 02 May 2005
Posts: 165
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 Posted: Sat Apr 22, 2006 10:12 pm Post subject:
Re: JSH: Extreme mathematics reminder
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In article <1145709606.219841.293550@i40g2000cwc.googlegroups.com>,
Larry Lard <larrylard@hotmail.com> wrote:
| Quote: | Andrezj Unpronounable
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That would be someone who can't be referred to by a pronoun?
-- Richard |
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Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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| Posted: Thu Apr 27, 2006 1:17 am Post subject:
Re: Hamiltonian cycle
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On 26 Apr 2006 04:31:26 -0700, <ugoren@gmail.com> wrote in
<news:1146051086.159006.105800@v46g2000cwv.googlegroups.com>
in alt.math.undergrad:
| Quote: | Hello,
I need to prove that every simple graph with 10 edges and 6 vertices
has a Hamiltonian cycle.
I think it's related to Ore's theorem, according to which:
Every simple graph G with n vertices that satisfies d(v)+d(u)<=n for
every non-adjant 2 vertices (u and v) has a Hamiltonian cycle.
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It's false: there's at least one graph with 10 edges and 6
vertices that has a vertex of degree 1, and hence no
Hamilton circuit. (It does have a Hamilton path, however.)
Brian |
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Tim Peters
science forum Guru
Joined: 30 Apr 2005
Posts: 426
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| Posted: Thu Apr 27, 2006 2:24 am Post subject:
Re: JSH: Now you can see it happening
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[quasi]
| Quote: | Ok, it is easily proved, although a minor correction needs to be made
to your statement, namely:
Change "quadratic residue mod p1" to "square mod p1".
Proposition:
...
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[jshsucks@yahoo.com]
| Quote: | The above is a quote from a post made by quasi in another thread.
He basically changes the whole conjecture that James has been making,
he totally took out the part about quadratice residues and replaced it
by square mod p1.
James then responds to his post by saying that he agreed, and then
basically took credit for what quasi did, and said how he has made many
discoveries like it, yet what quasi posted wasn't what James came up
with.
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As quasi said, it was a minor correction. The only difference between
"quadratic residue" and "square" here is that most definitions of the former
explicitly exclude 0, while square does not. Since 0 is in fact a
possibility here, that makes "square" at least clearer.
Which James already knew: 0 came up in one of his examples, and someone
already pointed out to him that therefore he shouldn't be saying "quadratic
residue". Rather than take that as an opportunity to improve his
presentation, James made another of those goofy "math doesn't care about
human quirks" arguments. Even with a correct result, the guy seems
determined to slit his own throat -- one of his more endearing qualities ;-)
| Quote: | James, are you trying to steal quasi's work.
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I don't think that's fair here.
Indeed. |
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Mike Amling
science forum Guru
Joined: 05 May 2005
Posts: 525
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| Posted: Thu Apr 27, 2006 3:35 am Post subject:
Re: Now you can see it happening
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<jstevh@msn.com> wrote in message
news:1146098606.500164.48070@v46g2000cwv.googlegroups.com...
| Quote: | I'm reminding now with this quadratic residue result of when I came up
with my prime counting function.
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Why don't you post any Mathematics ?
This is a Mathematics group. |
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David Moran
science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252
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| Posted: Thu Apr 27, 2006 5:15 pm Post subject:
Re: JSH: Now you can see it happening
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"Justin" <no@spam.com> wrote in message
news:e2qncs$fbk$1@grapevine.wam.umd.edu...
| Quote: | jstevh@msn.com wrote:
: There was this stunned pause on the newsgroups for a while, as some
: posters realized I was right, and a few even admitted it was this neat
: thing.
You should understand here that what you consider a "stunned pause" might
just be the fact that people here have lives and only get around to
addressing your posts in their spare time, for a bit of a laugh.
Justin
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James doesn't understand what a life is. He's wasted all of his on useless
crap that won't get him any of the things he thinks he'll get. I can tell
you that Oprah, for one, certainly won't give a crap about his "research"
whether or not it's correct. And secondly, women won't be impressed with
being famous for being a crank.
Dave |
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Mike Amling
science forum Guru
Joined: 05 May 2005
Posts: 525
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| Posted: Thu Apr 27, 2006 7:17 pm Post subject:
Re: JSH: Now you can see it happening
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"David Moran" <dmoran21@cox.net> wrote in message
news:d%64g.14697$fG3.10282@dukeread09...
| Quote: |
"Justin" <no@spam.com> wrote in message
news:e2qncs$fbk$1@grapevine.wam.umd.edu...
jstevh@msn.com wrote:
: There was this stunned pause on the newsgroups for a while, as some
: posters realized I was right, and a few even admitted it was this neat
: thing.
You should understand here that what you consider a "stunned pause" might
just be the fact that people here have lives and only get around to
addressing your posts in their spare time, for a bit of a laugh.
Justin
James doesn't understand what a life is. He's wasted all of his on useless
crap that won't get him any of the things he thinks he'll get. I can tell
you that Oprah, for one, certainly won't give a crap about his "research"
whether or not it's correct. And secondly, women won't be impressed with
being famous for being a crank.
Dave
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He has figured out the right statements to post to always get a response.
And that is something |
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Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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| Posted: Thu Apr 27, 2006 11:06 pm Post subject:
Re: Block Graph
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On 27 Apr 2006 15:01:09 -0700, <ugoren@gmail.com> wrote in
<news:1146175268.985920.128580@j33g2000cwa.googlegroups.com>
in alt.math.undergrad:
| Quote: | Hello,
I need to find a graph, that it's Block graph is Cn (a cycle with n
vertices)
Or on the other hand, prove that such graph doesn't exist.
For n=3, the graph is K1,n the block graph is K3 (aka C3)
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What's your definition of the block graph of a graph G? I
understand it to be the graph whose vertices are the cut
vertices and blocks of G, where a block of G is a maximal
connected subgraph with no cut vertices, and edges vb, where
v is a cut vertex of G, b is a block of G, and v is in b.
By that definition K(1,n) is its own block graph: it's a
star with n rays, the central vertex is a cut vertex, and
each ray is a block containing the central vertex.
Are you using a different definition, or am I missing
something obvious? By the definition I'm using, the block
graph of G is always acyclic (and therefore a forest).
[...]
Brian |
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ugoren@gmail.com
science forum beginner
Joined: 26 Apr 2006
Posts: 14
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| Posted: Fri Apr 28, 2006 12:19 am Post subject:
Re: Block Graph
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thank you for your quick reply,
we both agree on the definition of a Block,
however, I'm not sure we both agree on the definition of 'Block Graph'.
The Block Graph of G (BL(G)) is a graph that has a vertex for every
block of G.
two vertices are connected, if and only if the two blocks in G share a
vertex.
in K(1,n) for example, every edge is a block, and all the blocks share
the central vertex.
therfore, every two vertices in the Block Graph of K(1,n) should have
an edge between them.
in fewer words, BL(K(1,n))=Kn (the complete graph with n vertices.) |
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Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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| Posted: Fri Apr 28, 2006 12:39 am Post subject:
Re: Block Graph
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On 27 Apr 2006 17:19:32 -0700, <ugoren@gmail.com> wrote in
<news:1146183572.799555.76630@i39g2000cwa.googlegroups.com>
in alt.math.undergrad:
| Quote: | thank you for your quick reply,
we both agree on the definition of a Block,
however, I'm not sure we both agree on the definition of 'Block Graph'.
The Block Graph of G (BL(G)) is a graph that has a vertex for every
block of G.
two vertices are connected, if and only if the two blocks in G share a
vertex.
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Okay, that's definitely a different definition from the one
I've seen before.
| Quote: | in K(1,n) for example, every edge is a block, and all the blocks share
the central vertex.
therfore, every two vertices in the Block Graph of K(1,n) should have
an edge between them.
in fewer words, BL(K(1,n))=Kn (the complete graph with n vertices.)
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Agreed.
Let's see; suppose that your block graph has four blocks,
b(1), b(2), b(3), and b(4), forming a 4-cycle in that order.
Doesn't that imply that in G the blocks b(1) and b(2)
intersect in a cut point v(1), b(2) and b(3) intersect in a
cut point v(2), b(3) and b(4) intersect in a cut point v(3),
and b(4) and b(1) intersect in a cut point v(4)? Suppose
that v(1) = v(2); then the block graph ought to have an edge
between b(1) and b(3), and it doesn't. Similar problems
arise if any two of these cut vertices are equal, so they
must all be distinct. But then there is a cycle in G
containing all four of the v(i), which can't be cut points
after all.
If I didn't make a mistake working with the unfamiliar
definition, this idea should generalize fairly easily.
Brian |
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ugoren@gmail.com
science forum beginner
Joined: 26 Apr 2006
Posts: 14
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| Posted: Fri Apr 28, 2006 1:34 am Post subject:
Re: Block Graph
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Proposition: Blocks in G cannot form a cycle,
- Proof:
Let's assume that B(1),....B(n) are blocks of Graph G , such that
B(1),....B(n) form a cycle.
( B(i) and B(i+1) have a vertex in common for every 1<i<n-1, and B(1)
and B(n) also share a vertex. )
But, the subgraph of G, C=U B(i) (the union of all the blocks)
is also a block, since no vertex of C in a cut vertex of C.
By definition, B(1),....B(n) are no longer blocks, since they are not
maximal.
-
One might think that the proposition above is enough to prove that
BL(G) cannot contain a cycle.
that's not true, Since BL(K(1,3))=K(3)=C(3)
I have a sneaking suspicion that the block graph of G cannot be
C(4)....
.... |
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Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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| Posted: Fri Apr 28, 2006 1:20 pm Post subject:
Re: Block Graph
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On 27 Apr 2006 18:34:43 -0700, Clocker <ugoren@gmail.com>
wrote in
<news:1146188083.766482.279850@e56g2000cwe.googlegroups.com>
in alt.math.undergrad:
| Quote: | Proposition: Blocks in G cannot form a cycle,
- Proof:
Let's assume that B(1),....B(n) are blocks of Graph G , such that
B(1),....B(n) form a cycle.
( B(i) and B(i+1) have a vertex in common for every 1<i<n-1, and B(1)
and B(n) also share a vertex. )
But, the subgraph of G, C=U B(i) (the union of all the blocks)
is also a block, since no vertex of C in a cut vertex of C.
By definition, B(1),....B(n) are no longer blocks, since they are not
maximal.
-
One might think that the proposition above is enough to prove that
BL(G) cannot contain a cycle.
that's not true, Since BL(K(1,3))=K(3)=C(3)
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And that example shows that the proposition fails for n = 3.
The problem is that for n = 3 the blocks can all have the
*same* vertex in common. This can't happen for n > 3.
| Quote: | I have a sneaking suspicion that the block graph of G cannot be
C(4)....
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It can't, and I think you'll find that the argument that I
outlined last time proves this.
Brian |
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Titus Piezas III
science forum Guru Wannabe
Joined: 10 Mar 2005
Posts: 102
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| Posted: Sat Apr 29, 2006 6:20 am Post subject:
Re: Useful identity for ax^2+bxy+cy^2=dz^2
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Jyrki Lahtonen wrote:
| Quote: | titus_piezas@yahoo.com wrote:
[snipped description of an identity I cannot comment on]
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You can use www.quickmath.com to verify the identity. :)
| Quote: | Example. The equation x^2+xy+y^2 = 7z^2 has small solution (x1,y1,z1) =
(m,n,p) = (2,1,1) thus giving the parametrization,
x = 3u^2+2uv-2v^2
y = -u^2+4uv+3v^2
z = u^2+uv+v^2
Neat, huh? Seen this identity before? (Might be in Dickson, though not
sure.)
So essentially you are looking at a projective variant of a quadratic
curve, IOW a conic section. There the following well-known trick is
available (explaining why these curves give rise to the rational
function field):
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(snipped nice explanation)
I was doing some work that led to several equations of form
ax^2+bxy+cy^2=dz^2 (eq.1) and being familiar with the result that if
one non-trivial solution exists, then more can be found, I figured
there might be a "template identity" such that by simply plugging in
those initial values, it will give a parametric solution to eq.1.
(Lazy, yes, but efficient.)
P.S. The identity I found in fact has 4 free parameters, the one in the
original post is a simplified version. Note that it always has a sum
that is a multiple of the initial solution z_1 = p, though the more
general identity does not necessarily lead to that result.
-Titus |
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G.E. Ivey
science forum Guru
Joined: 29 Apr 2005
Posts: 308
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| Posted: Sat Apr 29, 2006 9:22 pm Post subject:
Re: JSH: How can they care?
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| Quote: | I'm sure you people want to keep the faith, but come
on. How can any
of these people you think are actually competent and
excellent
mathematicians keep so quiet if they really were?
I like to try and contact some of those big names
every once in a
while, as I try to find a way to break the impasse,
which is why Mazur
and Granville got early drafts of a key paper of
mine.
Recently, I sent them some stuff about n^2 - r while
I was on my way to
figuring out my latest result, but got no reply THIS
time.
They've learned.
If they reply, I talk about them on Usenet, when they
sit quiet
afterwards.
I fear that they do not give a damn about
mathematics, and why should
they?
It's screwed them over.
These people grew up being told mathematical ideas
that I can shoot
down with simple quadratic equations were gold.
They built their careers on research that my research
shows is invalid.
What do they have left?
They just have the faith of the world, which keeps
them in their
positions, and keeps them getting paychecks.
What does Wiles have if the full story comes out?
Not only does he lose credit for proving FLT, but
it's likely that ALL
of his research over his entire career goes out, as
not being valid
mathematics.
These people get shot back to zero.
More and more I can understand why they would choose
to sit quiet as in
their position, would I do any different?
Maybe luckily for me I've been disillusioned so many
times in life that
it's hard for me to believe in anything any more,
except what I can
personally and simply prove down to basic axioms so
that there is
absolutely no room for error.
Then what happened to Wiles, Ribet, Taylor,
Granville, Mazur and so
many of you cannot happen.
If you all had gone through your lessons, proving
everything back to
basic axioms, you might possibly have found a flaw in
ideal theory, and
saved yourselves a lot of grief.
James Harris
James, I have sworn to myself that I would never respond to a JSH thread (it's a blood pressure thing) but I just can't control myself! |
Over and over, you have given us things that you said were obvious, then berated people for saying you were wrong, then, eventually, admitted they were right. Then you repeat the whole pattern! Yes, they care! They care about mathematics and they care about the truth one hell of a lot more than you do!
Do you remember when you told us that you were intentionally writing things that you knew were wrong, that is you were "tricking" people into corecting you so you could then print their corrections as your own? |
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David Moran
science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252
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| Posted: Sat Apr 29, 2006 10:30 pm Post subject:
Re: How can they care?
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<jstevh@msn.com> wrote in message
news:1146335284.645276.217450@g10g2000cwb.googlegroups.com...
| Quote: | I'm sure you people want to keep the faith, but come on. How can any
of these people you think are actually competent and excellent
mathematicians keep so quiet if they really were?
I like to try and contact some of those big names every once in a
while, as I try to find a way to break the impasse, which is why Mazur
and Granville got early drafts of a key paper of mine.
Recently, I sent them some stuff about n^2 - r while I was on my way to
figuring out my latest result, but got no reply THIS time.
They've learned.
If they reply, I talk about them on Usenet, when they sit quiet
afterwards.
I fear that they do not give a damn about mathematics, and why should
they?
It's screwed them over.
These people grew up being told mathematical ideas that I can shoot
down with simple quadratic equations were gold.
They built their careers on research that my research shows is invalid.
What do they have left?
They just have the faith of the world, which keeps them in their
positions, and keeps them getting paychecks.
What does Wiles have if the full story comes out?
Not only does he lose credit for proving FLT, but it's likely that ALL
of his research over his entire career goes out, as not being valid
mathematics.
These people get shot back to zero.
More and more I can understand why they would choose to sit quiet as in
their position, would I do any different?
Maybe luckily for me I've been disillusioned so many times in life that
it's hard for me to believe in anything any more, except what I can
personally and simply prove down to basic axioms so that there is
absolutely no room for error.
Then what happened to Wiles, Ribet, Taylor, Granville, Mazur and so
many of you cannot happen.
If you all had gone through your lessons, proving everything back to
basic axioms, you might possibly have found a flaw in ideal theory, and
saved yourselves a lot of grief.
James Harris
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If you really cared about mathematics, you'd be willing to actually learn
something. How do you plan on learning if you don't even try to pick up a
book? If you took an interest in learning, maybe you'd be taken seriously.
Dave |
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porky_pig_jr@my-deja.com1
science forum Guru Wannabe
Joined: 08 May 2005
Posts: 102
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| Posted: Sun Apr 30, 2006 12:44 am Post subject:
Re: JSH: How can they care?
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| Quote: | Then what happened to Wiles, Ribet, Taylor, Granville, Mazur and so
many of you cannot happen.
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So what would the opposite of "If I have seen farther than others, it
is because I was standing on the shoulders of giants."? |
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