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8:00 AM
no, I am assuming nothing, i am assuming the leading coefficients to be $a_0,a_1,a_2$ like that
@robjohn are you there? I need your help here...
 
@Swadhin what's up?
 
I need help with this problem, please help
$P(x)$ be a polynomial of degree $3n$ such that $P(0)=P(3)=…=P(3n)=2,P(1)=P(4)=…=P(3n−2)=1,P(2)=P(5)=…=P(3n−1)=0$ and $P(3n+1)=730$. Find $n$
@robjohn
 
@Swadhin Have you looked at Lagrange Interpolation?
 
@robjohn No, I haven't, let me see...
 
i think n=3
wait
no n=2
 
8:09 AM
@robjohn Please don't mind, its a little above my level, the book I am doing is of standard 9 olympiad level, would you relook it please?
$P(7)=1\ne 730$
 
OH
I see what to do
$P(x)=(x−2)(x−5)⋅...⋅(x−(3n−1))Q(x)$
huh
if you plug in $3n+1$ you get $0$
 
ok, please go on
 
that makes no sense
 
@ForeverMozart if you plug in $3n-1$ you get $0$
 
oh lol
ok what happens if you plug in 3n+1
 
8:16 AM
@robjohn I tried an example of lagrange interpolation and tried to use it here, I get $\dfrac{\cdot}{0}$ form, which is absurd, so please help
 
$730=P(3n+1)=(3n-1)(3n-4)...(2)Q(x)$
 
yes, we get that, but that doesn't really help, does it?
 
$365=P(3n+1)=(3n−1)(3n−4)...Q(x)$
arrgh
$730=P(3n+1)=(3n−1)(3n−4)...(2)Q(3n+1)$
oh
$P(3n)=(3n−2)(3n−5)...(1)Q(3n)$
$P(3n+1)P(3n)P(3n-1)=(3n!)Q(3n+1)Q(3n)Q(3n-1)$
$730*2*1=(3n!)Q(3n+1)Q(3n)Q(3n−1)$
So $3n!$ divides 1460
 
doesn't the $q(3n)$ and $q(3n-1)$ cancel out?
 
how?
i am still stuck
 
8:27 AM
no, mistake, but I do not understand your statement too
 
but I think this is better
you need to use the roots first
then use the other information later
 
yes, but i am thinking in a loop, i would come back to this problem later...
but see this one please, $(2n)!<(n(n+1))^n$ for any natural number $n>1$
I couldn't proceed a bit here
I tried induction, but that was a futile approach for me
 
yes I think induction
 
I think it is done, but do you think it is correct?
$2n+1<n^2+n$
and we have $(2n)!<(n(n+1))^n$
 
what happened to the $4n^2$?
 
8:36 AM
where is $4n^2$?
 
multiplying we get our desired result, just need a little tweaking
 
yeah I think you got it
Suppose true for $n-1$. Then $(2n)!=(2n)(2n-1)(2(n-1))!<(2n)(2n-1)((n-1)(n))^{n-1}$
are you sure you wrote the problem correctly?
 
yes, very sure
I believe the problem is incorrect
 
8:50 AM
isn't it?
 
i dont know, but induction does not work
I think it is probably true though
 
Yes, I checked it, it is quite true
Ok, I have another one
$f(x)=ax^2+bx+c$ where $a$ and $4a+3b+2c$ have the same sign. Prove that the equation cannot have both its roots lying in the interval $(1,2)$. (you may assume $f(x)$ has all real roots).
@ForeverMozart
 
9:06 AM
Hey!
@Swadhin
I solved that problem
 
which one?
 
the last one
the one before last
 
P(3n+1)?
 
$(2n)!<(n(n+1))^n$
want to see how?
 
ok...tell me how you do it
 
9:07 AM
@Swadhin forward differences take a while, but show that $n=4$
 
Look at the series $\sum _{n=1} ^\infty \frac{(2n)!}{(n(n+1))^n}$
Apply the Ratio Test
the series converges
 
@robjohn using lagrange interpolation?
 
so the sequence $\frac{(2n)!}{(n(n+1))^n}$ goes to $0$
 
@ForeverMozart converges to 1 definitely?
 
it does matter, it the sequence converges to 0
so eventually the sequence is $<1$
and then you have the desired inequality
that leaves only a finite number of cases to check by computer
;)
I am good
oh you know what
I think I'm wrong though
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
dude
I think that is false
 
9:20 AM
But case study cannot find a loop, atleast i couldn't find yet
 
that is a frustrating problem
 
9:32 AM
@Swadhin
 
yeah...
could you solve it?
 
how about rearranging $(2n)!$ like this...
$(2n)!=(1\cdot 2n)(2(2n-1))(3(2n-2))...((n-1)n)(n(n+1))$
you have $n$ factors
all are $<n(n+1)$ except for the last
 
yes...i thought like this, does it help?
 
I think that does it
 
how?
 
9:36 AM
the factors are all $<n(n+1)$
 
$6n-6<n^2+n$?
 
@Swadhin No... the $n^{\text{th}}$ forward difference of a degree $n$ polynomial is a constant...
$$
\begin{array}{c}
2\\
&-1\\
1&&0\\
&-1&&3\\
0&&3&&-9\\
&2&&-6&&18\\
2&&-3&&9&&-27\\
&-1&&3&&-9&&27\\
1&&0&&0&&0&&0\\
&-1&&3&&-9&&27&&-81\\
0&&3&&-9&&27&&-81&&243\\
&2&&-6&&18&&-54&&162&&-486\\
2&&-3&&9&&-27&&81&&-243&&729\\
&-1&&3&&-9&&27&&-81&&243\\
1&&0&&0&&0&&0&&0&&\color{#C00}{0}\\
&-1&&3&&-9&&27&&-81&&\color{#C00}{243}\\
0&&3&&-9&&27&&-81&&\color{#C00}{243}\\
&2&&-6&&18&&-54&&\color{#C00}{162}\\
2&&-3&&9&&-27&&\color{#C00}{81}\\
&-1&&3&&-9&&\color{#C00}{27}\\
1&&0&&0&&\color{#C00}{0}\\
&-1&&3&&\color{#C00}{-9}\\
 
look, every factor is of the form $(n-j)(n-k)$ which is less than $n^2+n$
that is easy to see
 
$n^2-5n+6=(n-2)(n-3)$
which is less than 0 when $2<n<3$
 
9:40 AM
For a degree 12 polynomial, the $729$ will be repeated instead of the $\color{#C00}{0}$
 
we get a contradiction! @ForeverMozart
 
So every red element will be $729$ bigger.
This leaves the value of the polynomial at $730$
 
@robjohn sir, this seems very complicated, can't you give me something easier, understandable for my age...please don't take it otherwise, its a humble request
 
@Swadhin forward differences are about the simplest way to look at this. It is the way that one often extrapolates polynomials.
 
@robjohn okay then, i will try harder to understand.
 
9:46 AM
@Swadhin the consecutive values of the polynomial are in the left column. The differences of those values are in the second column. The differences of the second column are in the third column, etc.
 
@robjohn thank you, i will try to understand
 
Hi. Is $F(z)= (z-1)/(z-2)$ an LFT sending infinity to 1? If not, how can I construct an LFT on the extended complex plane that sends infinity to 1?
Doh, nevermind. It is 1 by definition of an LFT at infinity.
 
10:06 AM
Bounty expired with no answer :c
 
10:29 AM
1
Q: Can we use the Nullstellensatz?

Mary StarIn $\mathbb{C}[x, y, z]$ we have that $V=\{y-x^2, z-x^3)=\{(t, t^2, t^3) | t \in \mathbb{C}\}$. To show that $$I(V(y-x^2, z-x^3))=\langle y-x^2, z-x^3\rangle $$ can we use the Nullstellensatz?? EDIT: To show that $\langle y-x^2, z-x^3\rangle$ is prime do we have to do the following?? We co...

Could you tell me if the following step is correct??

If $p(x, y, z) \in ker \phi$, $$\phi(p(x, y, z))=0 \Rightarrow \phi(g(x, y, z)(y-x^2)+(z-x^3)a(x, z)+b(x))=0 \Rightarrow \phi(g(x, y, z)) \phi((y-x^2))+\phi((z-x^3))\phi(a(x, z))+\phi(b(x))=0\Rightarrow \phi(b(x))=0 \Rightarrow b(\phi(x))=0 \Rightarrow b(x)=0$$
 
10:58 AM
@TedShifrin Why d'you need me?
 
@KajHansen Could you take a look at my post? I added that's what I have tried.. math.stackexchange.com/questions/1031013/…
 
11:27 AM
Greetings
@robjohn @DanielFischer @Hippalectryon do you see a nice way of computing this one? $$\lim_{\large \epsilon \to 0}\int_0^{\infty}\int_0^{\large \epsilon}\arctan\left(\frac{1}{y}\right) e^{-y x}\ dx \ dy$$
 
Greetings my friend.
 
@infinitesimal Hello :-)
 
12:12 PM
@Chris'ssis hello =) how are you?
 
@user153330 Hi. How am I? In a great shape I think.
 
@Chris'ssis lol you're not sure? : P
 
@user153330 hehe, I'm sure I'm in a great shape! ;)
 
@Chris'ssis your new integral looks good, kinda familiar yet no idea.
 
@user153330 What I post here are not my daily creations since I don't do that for a long time. :-) Just tiny "details".
Yeah, this is an interesting limit "detail". :-)
@user153330 I think expressing $\arctan(1/y)$ as an integral it might be useful.
 
12:30 PM
Does the alternating series test apply to series of the form $\sum_{k=1}^{\infty}(-1)^{n}a_n$?
(my book only mentions $\sum_{k=1}^{\infty}(-1)^{n-1}a_n$)
 
@Chris'ssis No $1/\epsilon$ or anything?
 
@robjohn No.
I'm preparing to visit some relatives.
 
What will happen if we apply Diekstra's method in graph theory on directed graph?
 
Somebody please help me with this problem, if $2^n+1$ is a prime then $n$ is a power of $2$.
 
@Karlo You will become a shortest path tree.
 
12:41 PM
@evinda and if it is not connected the method won't work?
 
@Karlo I think that it works for both cases..
 
@DanielFischer if $2^n+1$ is a prime then $n$ is a power of $2$. Please help.
@robjohn if $2^n+1$ is a prime then $n$ is a power of $2$.
Please help
 
@evinda suppose the graph is a tree and the root has 2 leaf's one has value 1 and the other is a graph if we go to the 1 we haven't calculated the other values
@evinda or we can go back to r and go on?
 
@Ted @ABeau I have set up study groups of Facebook for all of my classes next semester and I have a few people in each that are willing to study and hang out outside of uni, thank you both for encouragement!
@Everyone

Is Dummit and Foote the most recommended from your experiences for abstract algebra
 
12:58 PM
@Swadhin Start with a proof by contradiction. Note that if a number is not a power of 2 ($1 = 2^0$ included), then it is a power of 2 times some odd number.
@Answer It's probably the most intensive and broad book, but I don't know if it gives the best or most down-to-earth explanations of certain concepts.
 
1:22 PM
Hi
 
Hi
 
@Swadhin another method would be to note that for a polynomial of degree $13$, we have $$\sum_{k=0}^{13}(-1)^k\binom{13}{k}P(k)=0$$ That is, the $13^{\text{th}}$ forward difference is $0$. We can compute $P(13)$ from the previous $13$ values
 
@robjohn A much more elegant method than mine, certainly, haha.
 
@Swadhin Do you know how to factor $2^n+1$ if $n$ has an odd factor?
@Swadhin $(2^b+1)(2^{(a-1)b)}-2^{(a-2)b}+2^{(a-3)b}-\dots+1)=2^{ab}+1$ when $a$ is odd.
 
1:35 PM
@Karlo So you mean that the tree has a height of 1?
0
Q: $V$ is irreducible exactly then when $I(V )$ is a prime ideal

evindaIf $V$ is an algebraic set of $K^n$, show that $V$ is irreducible exactly then when $I(V )$ is a prime ideal of $K[X_1, X_2, \dots, X_n]$ . Let $V$ be irreducible. We suppose that $I(V)$ is not a prime ideal of $K[X_1, X_2, \dots, X_n]$. That means that there are polynomials $f,g \in K[X_1, ...

Could someone take a look at my question?
 
@evinda no suppose it has root r and left branch only 1 edge with length 10 for example and the right branch is a graph for example with height 10
 
@Karlo I will think about it and I'll tell you...
 
@evinda ok
 
What's a good book to start out proof-based probability with?
 
@bd1251252: I figured it out. You were watching the second semester lecture on integration :P
 
1:44 PM
0
Q: Intersection points of curves

user159870In my lecture notes there is the following example for intersection points of curves: $$F(x, y, z)=xz^3-y^4 \\ G(x, y, z)=xz^2-y^3$$ in $\mathbb{P}^2(\mathbb{C})$, where $\mathbb{P}^2(\mathbb{C})=U_2 \cup H$ where $U_2=\{[x, y, 1] | x, y \in \mathbb{C}\}, H=V(z)=\{[x, y, z] | x, y \in \mathbb...

Can someone of you help me?
 
@TedShifrin Hi!!! Do you maybe have an idea about the following?

http://math.stackexchange.com/questions/1128902/v-is-irreducible-exactly-then-when-iv-is-a-prime-ideal
 
Hello @Chris'ssis. What are you doing now?
 
@ABeautifulMind Hey. How are you doing? I'm trying to visit some relatives, but still being caught with some research. Anyway, I'll leave in 15 min.
 
@Chris'ssis have fun :D
 
@infinitesimal Thanks! :-)
 
1:59 PM
@infinitesimal Yo. Yesterday you called me a drama queen. =(
 
@ABeautifulMind I did not
 
@infinitesimal OK, forget what I said then.
 
@ABeautifulMind I said "and then you become..."
 
@infinitesimal OK.
 
@ABeautifulMind it was a prediction
in English Language & Usage, 24 hours ago, by infinitesimal
Then you become drama queen :D
@ABeautifulMind note: the ":D"
 
2:03 PM
@infinitesimal I hate it when they call me drama queen in that room. It's not funny when you are suffering.
 
@ABeautifulMind there's nothing wrong with a little drama as long as you don't over do it
 
The things I say are not drama, they are real. They just have no idea what I am experiencing.
 
we did some drama on the metas...remember?
 
They think it is drama because they only experience ten per cent of my pain maybe, and they think they know what my life is like.
 
they are the drama queens
 
2:05 PM
@infinitesimal Sure, but when I talk about my problems, it is certainly not in that category.
 
I know pal.
 
@infinitesimal That is why I will only talk to say Kit and Matt about my problems now.
 
@ABeautifulMind Focus is important.
 
@infinitesimal I think what has happened is that a lot of damage has been done to my mind, and it takes a long time to fix.
 
Once a mind has been stretched by a new idea...
@ABeautifulMind do you know the rest^
 
2:10 PM
@infinitesimal I think so, but what has it got to do?
 
You will never be the same as you once were.
 
One talks about a new idea, the other is mental problems, quite different.
 
@ABeautifulMind But they both influence the brain.
'Once the mind has been stretched by a new idea, it will never again return to its original size.'
'Once the mind has been stretched by a new idea damaged by mental illness, it will never again return to its original size.'
 
I want to find the taylor expansion of $f(x, y)=x^2 (3y-2x^2)-y^2 (1-y)^2$ at the point $(0, 1)$ and I got the following:

$$f(x, y)=3x^2+(y-1)^2+3x^2 (y-1)+\frac{2}{3} (y-1)^3-2x^4$$

but a friend of mine got the following result:

$$f(x, y)=3x^2-(y-1)^2+3x^2 (y-1)+3x (y-1)^2+2(y-1)^3-2x^4-(y-1)^4$$

which of them is the right one??
 
@infinitesimal I can now understand why Kit did not want to return to that room at first.
 
2:19 PM
@ABeautifulMind yes, some of them can be mean
but this is the internet
 
@infinitesimal Kit has been the most encouraging person to me in my life.
 
@DanielFischer Could you tell me if that's what I have tries is right?
If $I,J$ ideals of $R$ show that $Rad(I \cap J)=Rad(I) \cap Rad(J)$.

Let $x \in Rad(I \cap J)$. That means that $\exists n \in \mathbb{N}$ such that $x^n \in I \cap J$. Thus: $x^n \in I$ and $x^n \in J$. That implies that $x \in Rad(I)$ and $x \in Rad(J) \Rightarrow x \in Rad(I) \cap Rad(J)$.

So $Rad(I \cap J) \subseteq Rad(I) \cap Rad(J)$.

Let $x \in Rad(I) \cap Rad(J)$. That means that $x \in Rad(I)$ and $x \in Rad(J)$, i.e. $\exists m,n \in \mathbb{N}$ such that $x^m \in I$ and $x^n \in J$. From the definition of t
 
@ABeautifulMind build on her encouragement
@ABeautifulMind if you want to start somewhere
 
2:58 PM
Hi folks
suppose $f_n(x)= x^2sin(1/(nx))$
 
Alright.
 
how can I disprove that $f_n$ is not convergent uniformly for $x\ge0$
 
@Abeautifulmind How are you doing? I think I am the most positive I have ever been right now. I might make friends!
 
any help please?
 
What is the definition of uniform convergence
Write it out please, I can solve the problem if you do
 
3:03 PM
uniform convergence is that it must be convergent for every x
 
@Answer $\forall \epsilon>0,\exists N,\forall n\ge N,\forall x,|f_n(x)-f(x)|<\epsilon$
 
Thanks
 
on other way $lim_{n \to \infty} || f_n(x) -f ||= 0$
 
I have never seen that second thing, but I can solve with the quantified version
 
as $f$ is the simple of $f_n(x)$, I know that f(x) = 0 for
ever x
 
3:05 PM
There is a flea in my room but I have no animals :(
 
@pourjour The way you say it is weird
5 mins ago, by pourjour
how can I disprove that $f_n$ is not convergent uniformly for $x\ge0$
 
Wait yeah do you want uniform convergence or not?
I am confuzzliesed
 
You want to disprove that ... is not uniformly convergent... so you wanna prove it is uniformly convergent ?
 
@Hippalectryon So can you prove the uniform convergence
 
So you want uniform convergence?
That is much more efforts
For me to do without a book - whichies I don't haves on me
 
3:08 PM
I don't know if it's unformly convergent but I have a doubt it's not
 
@Answer Nope
@Answer The fonction is not uni convergent
$\sin(1/(nx))\sim_{n\infty}\dfrac{1}{nx}$
So $x^2\sin(1/(nx))\sim_{n\infty}x/n$
Which obviously does not converge uniformly
@pourjour Done :D
 
From the few things I have written, do you guys really hate me?
 
@Hippalectryon but $x/n$ tends to zero when n tends to infinity
 
@Answer What do you mean ? I never said that !
 
isn't it
I mean it's convergent
 
3:13 PM
@pourjour That only gives you simple convergence
 
I don't know, I have the feeling that people hate me and don't say anything
 
It's convergent, but not uniformly
@Answer I don't know why you think that :-) I don't hate you at all
 
@Hippalectryon why?
 
Oh okay that is a relief @Hippal
 
@pourjour What's $\sup_{\mathbb{R}}|x/n-0|$ ?
 
3:14 PM
@hippa can you come on tox for a sec if you have time
 
Whats a tox?
Secret chat room for friends?
 
@Hippalectryon infinity?
 
@pourjour Indeed :D
Hence it does not converge uniformly
 
@Hippalectryon so I was wrong about simple convergence, the right answer for simple convergence is f(x)=x/n
 
3:16 PM
is there a tox.im math chat @Hipp
 
@Answer Tox is like Skype. I doubt there is one.
@pourjour I'm not sure what you mean by that .. Is that a question or a statement ?
 
Okay thank you very much for this informations
 
1
Q: Can we use the Nullstellensatz?

Mary StarIn $\mathbb{C}[x, y, z]$ we have that $V=\{y-x^2, z-x^3)=\{(t, t^2, t^3) | t \in \mathbb{C}\}$. To show that $$I(V(y-x^2, z-x^3))=\langle y-x^2, z-x^3\rangle $$ can we use the Nullstellensatz?? EDIT: To show that $\langle y-x^2, z-x^3\rangle$ is prime do we have to do the following?? We co...

 
@Hippalectryon I mean, is $f_n(x) \to x:f(x)=x/n$ when n tends to infinity?
 
@Answer have fun with your new friends :D
 
3:19 PM
@Answer I am happy for you. I am very unwell, and I hope for a miracle in my life.
 
@pourjour $f_n(x)$ is equivalent to $x/n$ when $n$ goes to ifinity.
 
@Hippalectryon Ok I got it, thanks
 
only you can choose @ABeautifulMind which of these^ two ways you want to live your life
 
@infinitesimal Sorry, but what Einstein is saying refers to something else.
 
@ABeautifulMind of course he never had to struggle with mental illness
but he did struggle @ABeautifulMind
in English Language & Usage, Dec 15 '14 at 10:54, by skullpatrol
Life is a struggle.
 
3:40 PM
1
Q: Intersection points of curves

user159870In my lecture notes there is the following example for intersection points of curves: $$F(x, y, z)=xz^3-y^4 \\ G(x, y, z)=xz^2-y^3$$ in $\mathbb{P}^2(\mathbb{C})$, where $\mathbb{P}^2(\mathbb{C})=U_2 \cup H$ where $U_2=\{[x, y, 1] | x, y \in \mathbb{C}\}, H=V(z)=\{[x, y, z] | x, y \in \mathbb...

 
@BalarkaSen Are you familiar with singular points of an affine curve?
 
@user159870 Please don't barge in and post a question here...
 
I wanted to know if I have calculated right the singular points:
http://math.stackexchange.com/questions/1129091/calculate-the-singular-points-of-affine-curve
 
@Hippalectryon Ok. Sorry. Do you know something about Intersection points of curves?
 
@user159870 Unfortunately I do not. I upvoted though, the question seems interesting.
 
3:46 PM
@evinda Nope.
 
@Hippalectryon Thank you!
 
@BalarkaSen A ok, no problem.. :)
 
@Hippalectryon where else would you post?
this is a math room
 
Hi @Ted
 
hi @Balarka
 
3:56 PM
Actually, I am still not convinced about chain homotopies. I mean, OK, I can see what it does but I don't think it quite explains the word "homotopy" in chain homotopies.
 
because it links the chain complexes on $X$ and $X\times I$
 
It does?
 
doesn't it?
That's what we said yesterday
salut @Hippa
 
OK, let's start again. $f, g : X \to Y$ be maps $F : X \times I \to Y$ a homotopy between $f, g$ and $f_\#, g_\#$ corresponding chain maps $C_\bullet(X) \to C_\bullet(Y)$. $F_\#$ picks up a map $\Delta^k \to X \times I$ and then sends it to $\Delta^k \to X \times I \stackrel{F}{\longrightarrow} Y$. Now we see what happens when we take $\sigma \times I : \Delta^n \times I \to X \times I$.
 
That seems muddled, @Balarka. $F$ acts on $\sigma\times I$ to start with.
$\sigma$ lives in $X$ alone.
 
4:03 PM
I seem to have completely forgotten what we did last day. That's exactly why I hate sleeping. $\sigma $ is a singular n-simplex on X, isn't it?
 
@TedShifrin Could you take a look at this? math.stackexchange.com/questions/1129091/…
 
$n$? I don't like that letter, as often $X$ is $n$-dimensional. And I'm encouraging you to think of a standard simplex, rather than singular one, just to be concrete.
@evinda: Not now.
 
Hell, it's hard to translate from topological spaces to chain complexes.
I've never tried that before, except with simplical homology.
 
Well, I'm trying to get you to think about a plain ol' $k$-dimensional simplex sitting inside $X$. It can be the image of a singular chain, fine.
 
Well, what in particular do you want me to think about? $F_\# : C_k(X) \to C_{k+1}(X)$ is an OK map, but I just don't see what it does to $f_\#$ and $g_\#$
 
4:10 PM
@infinitesimal I mean, don't advertise your MSE questions here
@TedShifrin Hello
 
and I mean
19 mins ago, by infinitesimal
this is a math room
 
@infinitesimal I know that much q_q
 
That isn't right, @Balarka, is it? $F$ is $f$ at $t=0$ and it's $g$ at $t=1$.
 
Uck I am thinking about my made-up map which I also denoted as $F_\#$
 
4:13 PM
Stop that.
 
$F_\# : C_k(X \times I) \to C_k(Y)$
 
OK, I'll be back later.
 
later
 
Byes.
 
4:47 PM
@infinitesimal "If you asked your question on the main site, please don't post it on the chat. It will get a lot of exposure without that. If you do choose to post it, [title](http://link) is a better format; it is compact and easier on the eyes."
That is from the guidelines.
 
Is there a diffrence in the result of applying algorithm of Prim and Deikstra I know the methods are different
 
@robjohn As my bounty expired without any answer, what do you think I should do about my question here ? Is it reasonable to ask those three closely related questions, or should I stick to one at a time ?
 
@Hippalectryon You are talking about this question?
 
Can you exlain to me the following:



The finite points $\left [x, y, 1\right ] \in \mathbb{P}^2(K)$ geometrically correspond to the intersection points of the lines from $(0,0,0)$ with the plane $z=1$.

The points at infinity $\left [x, y, 0\right ]$ geometrically correspond to the lines of the plane $x0y$ that pass through $(0, 0, 0)$.


???
 
@MikeMiller: Yep! I'll be coming for both days.
 
4:53 PM
@Hippalectryon There may be few tools, if any, to tackle that problem. I can see heuristically why it might be true, but I have no idea of how to compute when it would be true. It is not something that is well-known.
 

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