The Hodge Conjecture: Solved!

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The Hodge Conjecture: Solved! Anonymous Wed Jul 1 19:52:00 2015 No.7367510 [Reply] [Original]

>>	Anonymous Wed Jul 1 19:57:49 2015 No.7367517 >Polynomial equation >Non-smooth

>>	Anonymous Wed Jul 1 19:58:48 2015 No.7367518 >>7367517 Read closely.

>>	Anonymous Wed Jul 1 20:08:25 2015 No.7367524 Bump.

>>	Anonymous Wed Jul 1 20:16:33 2015 No.7367539 The solution to the hodge conjecture is ignored while topics like the gay gene are getting attention? Come on sci!

>>	Anonymous Wed Jul 1 20:41:56 2015 No.7367585 Bump.

>>	Anonymous Wed Jul 1 20:46:41 2015 No.7367595 >>7367539 Fuck off. This is a well known troll.

>>	Anonymous Wed Jul 1 20:48:50 2015 No.7367600 >>7367595 The proof is there. The question is whether it is correct. Can a grad student verify? At least find Waldo.

>>	Anonymous Wed Jul 1 20:49:29 2015 No.7367602 Don't even know Hodges conjecture but this seems fine. If anything he just needs to polish a little to make it a rigorous proof but that is to be expected from a undergrad paper.

>>	Anonymous Wed Jul 1 20:51:31 2015 No.7367603 >>7367600 Grad student here. It looks correct, but you're going to have to fill in the details. Enjoy your $1000000 OP.

>>	Anonymous Wed Jul 1 21:07:36 2015 No.7367622 Thanks.

>>	victor is schizo Wed Jul 1 21:08:44 2015 No.7367625 victor is schizo

>>	Anonymous Wed Jul 1 21:13:30 2015 No.7367634 That's irrelevant.

>>	victor is schizo Wed Jul 1 21:18:18 2015 No.7367642 is it?

>>	Anonymous Wed Jul 1 21:28:34 2015 No.7367662 It's high functioning sza as far as I can tell. Kindly get lost.

>>	Anonymous Wed Jul 1 21:29:41 2015 No.7367664 >>7367662 that's your delusions talking. you shouldn't trust them

>>	Anonymous Wed Jul 1 21:30:46 2015 No.7367665 >>7367664 This thread is about the proof. Reported.

>>	Anonymous Wed Jul 1 21:38:06 2015 No.7367679 >>7367510 Can you explain what you mean in step 2? Sorry for my ignorance but I don't see how step one implies step 2. The rest of the proof seems okay.

>>	Anonymous Wed Jul 1 21:47:33 2015 No.7367701 Certainly. A polynomial equation has a graph with turns or curves. Those of the rational are smooth and turn while retaining a curve. Those of the irrational are sharp and are not curvy or smooth.

>>	Anonymous Wed Jul 1 21:48:57 2015 No.7367706 well done, OP. mind if i take the credit?

>>	Anonymous Wed Jul 1 21:52:19 2015 No.7367715 >>7367701 Can I get an example?

>>	Anonymous Wed Jul 1 21:54:11 2015 No.7367719 Why isn't the thread blowing up? I've scored among the greats for Guatemala!

>>	Anonymous Wed Jul 1 21:59:11 2015 No.7367728 >>7367719 >implying I'm not writing it up in tex as fast as I can so I can send it to the Annals of Mathematics and get dat $$

>>	Anonymous Wed Jul 1 22:01:53 2015 No.7367735 http://m.wolframalpha.com/input/?i=%28x%5E2%29%2Bpi&x=0&y=0

>>	Anonymous Wed Jul 1 22:23:07 2015 No.7367763 >>7367735 >wolframalpha page for x^2 + pi What are you saying?

>>	Anonymous Wed Jul 1 22:32:16 2015 No.7367777 >>7367763 The parabola disproves the whole thing.

>>	Anonymous Wed Jul 1 22:37:44 2015 No.7367784 File: 99 KB, 1280x720, Applesauce Bitch.jpg [View same] [iqdb] [saucenao] [google] >>7367763 >>7367777

>>	Anonymous Wed Jul 1 22:37:54 2015 No.7367785 No. It does not. The pi is a constant in that equation. The proof remains correct.

>>	Anonymous Wed Jul 1 22:40:27 2015 No.7367787 >>7367785 http://www.wolframalpha.com/input/?i=pi%28x%5E2%29%2Be

>>	Anonymous Wed Jul 1 22:52:03 2015 No.7367804 No. The proof is correct. I don't think there's any error.

>>	Anonymous Wed Jul 1 22:55:02 2015 No.7367807 >>7367804 No it's not. Not all complex polynomials with irrational coefficients are non-smooth. Plus, learn to math in your math. >Prove this or at least provide an example of a non-smooth complex polynomial with irrational coefficients.

>>	Anonymous Wed Jul 1 23:03:11 2015 No.7367816 >>7367804 >provided counterexample >responds proclaiming there is no error >people laugh

>>	victor is schizo Wed Jul 1 23:07:14 2015 No.7367822 victor is schizo

Anonymous Wed Jul 1 23:08:20 2015 No.7367824

>>7367807
Let <span class="math"> p(x)=\pi \cdot x [/spoiler]. <span class="math"> \pi [/spoiler] takes an infinite amount of energy to compute. The only way to go from one speed to another instantly is if acceleration has a sharp point, which takes infinite energy. In other words the function isn't smooth. Therefore <span class="math">p(x) [/spoiler] has such a sharp point. QED

>>	Anonymous Wed Jul 1 23:09:35 2015 No.7367826 >>7367807 Yes they are. I'm not going to prove basic Calculus I for you. Every polynomial with irrational coefficients has corners in its graphs instead of smooth curves.

>>	Anonymous Wed Jul 1 23:11:22 2015 No.7367830 >>7367826 >at least provide an example of a non-smooth complex polynomial with irrational coefficients.

>>	Anonymous Wed Jul 1 23:12:26 2015 No.7367831 >>7367826 P(z) = zsqrt(2) is linear, smooth, and continuous.

>>	Anonymous Wed Jul 1 23:12:41 2015 No.7367833 >>7367826 http://math.stackexchange.com/questions/23534/roots-of-a-polynomial-with-irrational-coefficients

>>	Anonymous Wed Jul 1 23:16:20 2015 No.7367837 >>7367826 what then do you say about the counterexample?

>>	Anonymous Thu Jul 2 00:13:04 2015 No.7367909 Bump

>>	Anonymous Thu Jul 2 00:14:37 2015 No.7367913 File: 39 KB, 562x437, Ohwow.jpg [View same] [iqdb] [saucenao] [google] >>7367824

>>	Anonymous Thu Jul 2 00:18:26 2015 No.7367921 File: 98 KB, 894x880, comradevictor.jpg [View same] [iqdb] [saucenao] [google] >>7367913 >mfw

>>	Anonymous Thu Jul 2 00:24:02 2015 No.7367930 >>7367824 >>7367824 Uh isn't p continuous here? Its derivative exists ergo it is continuous. In fact this is just a line with slope pi. Also that is not the only way to accelerate instantly, any jump discontinuities will do that too

>>	Anonymous Thu Jul 2 00:27:16 2015 No.7367939 >>7367930 You realize you are trying to argue with /sci/'s resident schizophrenic right?

Anonymous Thu Jul 2 00:30:09 2015 No.7367946

>>7367930
Uh, no. If you instantly go from one speed to another, you have infinite acceleration, which takes an infinite amount of energy to accomplish. Also, the digits of an irrational number go on forever, so computing one takes an infinite amount of energy. Therefore, the linear function with slope <span class="math">\pi[/spoiler] must have some kind of discontinuity or sharp point.

>>	Anonymous Thu Jul 2 00:31:24 2015 No.7367953 >>7367946 How can people say things like this and not feel stupid. You don't even know if time is granular, what makes you think infinities are even necessary much less decisively exist?

>>	Anonymous Thu Jul 2 00:34:02 2015 No.7367957 File: 64 KB, 526x526, 1433920563180.jpg [View same] [iqdb] [saucenao] [google] >falling for bait of this quality keep it classy /sci/

>>	Anonymous Thu Jul 2 00:35:04 2015 No.7367961 >>7367957 It's not bait, Comrade Victor is a real schizophrenic who posts nonsense on /sci/

Anonymous Thu Jul 2 00:35:18 2015 No.7367962

>>7367953
Time isn't granular. That's why there must be discontinuities. Infinities may or may not exist, but going instantly from one velocity to another is discontinuous which is expressed in the fact that it requires infinite energy, in theory. Since calculating the digits of an irrational number also requires infinite energy, any polynomial with irrational coefficients must also be discontinuous.

Anonymous Thu Jul 2 00:48:31 2015 No.7367987

>>7367962
I have no real knowledge of deeper mathematics, which is likely why I hold the (perhaps primitive) opinion that if you have designed a system t represent reality with so many factors you don't understand, it's simply hacked together. It could seem to work at first, but be fundamentally broken. The common us of "infinity" is a symptom of this, it's a magic concept that just makes things work.

I don't know if subdivision of particles has an end, nor if time is actually done in ticks (or discrete units), but I'm apt to think things are ultimately finite. They just aren't easily measurable. When people immediately throw around the idea "you'd need infinite energy!" I don't agree with it, and it doesn't make any sense. Both intuitively and via every other bit of knowledge we've amassed.

To think people actually argued viciously with Einstein when he became attached to the idea of a hidden variable. I've yet to see anything satisfactory against it to this very day.

>>	Anonymous Thu Jul 2 00:59:59 2015 No.7368008 >>7367510 >polynomial equation >undifferentiable ebin

Anonymous Thu Jul 2 01:00:35 2015 No.7368009

>>7367987
It's a metaphor to demonstrate the proof. An easier example involves the kinetic energy required to move from one point to another which is a function of velocity squared. Ignoring relativity, it is obvious that a continuous motion from point a to b takes a finite amount of energy. However, an instantaneous (discontinuity) change in position requires "infinite speed", even reaching the finite speed of light requires infinite energy for a massive object, which is why it does not happen in physics.

However, mathematically if my polynomial also has irrational coefficients, which similarly require infinite energy to calculate, the polynomial itself must display discontinuous phenomena.

Anonymous Thu Jul 2 01:04:58 2015 No.7368017

>>7368009
It doesn't matter if it "requires infinite energy to compute". Smooth = infinitely differentiable. (πx)' = π. (π)' = 0. The shape of the manifold matters, not its specific values.
Are you going to say that y = π isn't smooth either? It's a flat line!

>>	Anonymous Thu Jul 2 01:07:03 2015 No.7368024 >>7368017 No, because I proved that any polynomial with irrational coefficients is not smooth. Therefore <span class="math"> f(x)=\pi [/spoiler] is not smooth.

>>	Anonymous Thu Jul 2 01:12:48 2015 No.7368033 >>7368024 f'(x)=d/dx(pi)=0 d/dx(0)=0 ... Hence it is smooth.

>>	Anonymous Thu Jul 2 01:15:01 2015 No.7368038 >>7368024 No you didn't. You just said it isn't. But you said wrong.

>>	Anonymous Thu Jul 2 01:15:22 2015 No.7368039 >>7368033 Wrong. <span class="math"> f(x)=0 [/spoiler] is smooth because it is a polynomial with rational coefficients. I am using smooth in the informal sense of its graph, which must necessarily have sharp points or discontinuities.

>>	Anonymous Thu Jul 2 01:16:54 2015 No.7368042 >>7368039 Where in the graph of f(x)=π is there a sharp point or discontinuity? Care to point it out?

>>	Anonymous Thu Jul 2 01:18:10 2015 No.7368044 >>7368042 You are too stupid if you can't see it

>>	Anonymous Thu Jul 2 01:21:11 2015 No.7368046 >>7368044 No, I'm just asking you to pull off an impossible feat. Specifically, pointing out a sharp point in a straight line.

>>	Anonymous Thu Jul 2 01:21:18 2015 No.7368047 >>7368044 Confirmed as 12 year old with autism or simply a tryhard troll. There is no discontinuity in a straight line, even a 9 year old can understand this.

>>	Anonymous Thu Jul 2 01:46:21 2015 No.7368084 >>7367939 No. But the more you know...

>>	Anonymous Thu Jul 2 01:50:41 2015 No.7368088 >>7368042 The number of discontinuities of <span class="math"> f [/spoiler] is dense in its graph, which look like <span class="math"> \mathbb{R} [/spoiler], so pick any open interval and there will be an infinite number of them.

>>	Anonymous Thu Jul 2 01:58:20 2015 No.7368098 File: 80 KB, 331x480, 1435678078174.jpg [View same] [iqdb] [saucenao] [google] >>7368088 Whatever you say. It's defined everywhere and its slope is zero everywhere, hence it's differentiable everywhere.

Anonymous Thu Jul 2 02:08:20 2015 No.7368108

>>7368088
1. At all discontinuities of a function, the derivative does not exist. Yes or no?

2. The derivative of f(x)=πx is f'(x)=π. Yes or no?

3. Since π exists, the derivative of f(x) exists for all x. Yes or no?

Therefore there are no discontinuities.

>>	OP Thu Jul 2 02:28:58 2015 No.7368133 So is the proof right?

>>	OP Thu Jul 2 02:36:01 2015 No.7368143 Bump

>>	Anonymous Thu Jul 2 02:37:22 2015 No.7368145 >>7368143 >bumping from page 1 I didn't ask you to prove your stupidity, that was already self-evident.

>>	Anonymous Thu Jul 2 02:39:23 2015 No.7368150 >>7368133 No, but you won't believe me anyway so why ask?

>>	OP Thu Jul 2 11:43:17 2015 No.7368729 Bump

>>	Anonymous Thu Jul 2 12:06:15 2015 No.7368768 File: 66 KB, 900x900, 1418267988946.jpg [View same] [iqdb] [saucenao] [google] >>7368088 BTFO The teachings of NJ Wildberger prevail again

>>	Anonymous Thu Jul 2 12:49:19 2015 No.7368835 >someone with barely able to grasp high school algebra attempts the hodge conjecture, one of the most abstract and complex problems in all mathematics

>>	Anonymous Thu Jul 2 13:15:42 2015 No.7368871 >>7367824 >>7367946 Why the fuck are you talking about energies and speeds?

Anonymous Thu Jul 2 15:47:05 2015 No.7369105

>>7368871
That was my first thought, I mean pi isn't even applicable

also, muh fractals
>p.p.s. muh fractals is a hint
>p.p.p.s. OP is talking about a solve for what a variable actually is

One last thing, Victor, I messaged you on facebook, do you not know what a matrice is?

>>	OP Thu Jul 2 19:23:16 2015 No.7369489 The proof is correct! I've won a million dollars!

>>	Anonymous Thu Jul 2 19:57:28 2015 No.7369543 >>7369489 so do you listen to anybody else or just delude yourself repeatedly

>>	Anonymous Thu Jul 2 20:02:40 2015 No.7369555 File: 486 KB, 1836x2448, cQTvNtS.jpg [View same] [iqdb] [saucenao] [google] fucking nonsense gobbledigook get on my level

>>	Anonymous Thu Jul 2 20:16:24 2015 No.7369570 >>7369105 > a matrice Chewy... is that you?

>>	Anonymous Thu Jul 2 20:31:22 2015 No.7369590 >>7369489 Pics or it didn't happen.

>>	OP Thu Jul 2 21:04:19 2015 No.7369643 The proof is right there perfect. The only criticisms are reiterations of already refuted "errors" claimed by trolls. Glory to God!

>>	Anonymous Thu Jul 2 21:13:13 2015 No.7369658 >>7369643 Do you have any idea how crazy you are?

Anonymous Thu Jul 2 21:18:37 2015 No.7369669

>>7368835
To be honest it's more impressive than just assuming you're not educated enough, not reading or attempting the hodge conjecture, and only doing your assignments
He will have learned more just from looking up the definitions.
And he might have even understood what Could be shown in order to satisfy the proof requirement, even if he fudged a big step.

>>	Anonymous Thu Jul 2 21:20:12 2015 No.7369671 >>7369669 Except he's got all the definitions completely wrong.

>>	Anonymous Thu Jul 2 21:22:47 2015 No.7369675 >>7369669 There's problems in almost every one of his steps. And his solution, if it was at all correct, is incredibly trivial. He doesn't even know what smooth means ffs.

>>	Anonymous Thu Jul 2 21:42:51 2015 No.7369709 >>7369675 I quite like his alternative intuition for smoothness in terms of a particles speed.

>>	OP Thu Jul 2 21:52:03 2015 No.7369719 File: 37 KB, 640x480, image.jpg [View same] [iqdb] [saucenao] [google] >>7369675 Smooth definition is used correctly. Pic related is me in a new suit I bought I might wear to the prize ceremony.

>>	Anonymous Thu Jul 2 21:55:31 2015 No.7369727 >>7369658 If he did, he wouldn't be as crazy

>>	Anonymous Fri Jul 3 04:48:34 2015 No.7370277 File: 8 KB, 184x184, kebab.jpg [View same] [iqdb] [saucenao] [google] >>7369555 REMOVE H2O

>>	Anonymous Fri Jul 3 06:19:33 2015 No.7370355 >>7367961 >>7369658 >>7369658 This dude isn't trolling. He posts on this music message board all the time and everyone is fed up with him. I'm serious. Victor please stop posting on here and there

>>	Anonymous Fri Jul 3 06:25:02 2015 No.7370360 Fuck I shouldn't have bumped this, please just let this thread die. 4chan is definitely not the best place for him.

>>	OP Fri Jul 3 08:59:47 2015 No.7370505 Shut the fuck up

>>	Anonymous Fri Jul 3 11:00:04 2015 No.7370613 >>7368042 >>7368039 >>7368033 >>7367824 >>7367777 Holy shit, all these people who don't know that there's a difference between the graph of a polynomial function and a complex projective variety. I need to stop coming here.

>>	OP Fri Jul 3 16:20:44 2015 No.7371090 I won!!!

>>	OP Fri Jul 3 18:41:54 2015 No.7371295 Where else can I share my proof? They took it gdpwn on pf

>>	Anonymous Fri Jul 3 18:42:51 2015 No.7371298 >>7371295 Send it to a professor so he can laugh at your incompetence you fucking retard.

>>	OP Fri Jul 3 18:44:33 2015 No.7371303 Most people are saying it's correct though?

>>	Anonymous Fri Jul 3 18:46:22 2015 No.7371305 >>7371303 No they aren't you delusional retard.

>>	Anonymous Fri Jul 3 21:43:31 2015 No.7371692 The sarcasm in this thread, fills me with awe and love for my fellow man.

>>	OP Fri Jul 3 22:24:24 2015 No.7371771 Irrational functions have curves that are discontinuities everywhere because all the range values are representable on a plane or number line only as approximations. So the proof is correct.

>>	OP Fri Jul 3 22:29:57 2015 No.7371782 >>7368017 Try driving your dad's car at pi miles per hour then retard.

>>	OP Fri Jul 3 22:38:10 2015 No.7371801 >>7371771 Every point is isolated to itself.

Anonymous Fri Jul 3 22:43:37 2015 No.7371808

>>7371771
>Irrational functions have curves that are discontinuities everywhere because all the range values are representable on a plane or number line only as approximations.
But the function itself is not an approximation, only our representation of it in a drawing is an approximation. This has no bearing on the function itself, which can easily be proved differentiable everywhere. Does the derivative exist for all x? Then it's differentiable.

>>7371782
Irrelevant.

>>7371801
Also irrelevant.

>>	OP Fri Jul 3 22:58:55 2015 No.7371827 >>7371808 The function must be infinitely differentiable to be smooth. Irrational functions are not so.

Anonymous Fri Jul 3 23:03:37 2015 No.7371836

>>7371827
>The function must be infinitely differentiable to be smooth. Irrational functions are not so.
They are so and this has been proven in the thread.

Take the irrational function f(x) = πx. What is it's derivative? π. Does π exist? Yes. Therefore the derivative exists everywhere. Thus it's an irrational function is smooth. Why do you keep ignoring this?

>>	OP Fri Jul 3 23:08:19 2015 No.7371852 That's not smooth because the radically different change in rate of change results in a turn in the opposite direction. At that point you start going back. Resulting in a corner.

Anonymous Fri Jul 3 23:10:41 2015 No.7371857

>>7371852
>That's not smooth because the radically different change in rate of change results in a turn in the opposite direction.
There aren't any differences in the rate of change. It's a straight line. It's derivative is constant. It's rate of change is constant. Take your meds and remember your high school math Victor.

>>	OP Fri Jul 3 23:19:54 2015 No.7371876 >>7371857 Nah the gradient is still jagged. My point holds. Plus perfectly straight lines are useless to topology.

>>	Anonymous Fri Jul 3 23:21:26 2015 No.7371879 >>7371876 >Nah the gradient is still jagged. I just proved it's not. You have no argument, therefore you're wrong.

>>	Anonymous Sat Jul 4 10:38:10 2015 No.7372639 Try driving your dumb car at pi miles per hour then retard. It's impossible.

Anonymous Sat Jul 4 16:48:41 2015 No.7373189

>>7372639
You're insane. I respect the effort but your inability to accept criticism and accept your "proof" is incorrect is what's holding you back from progressing into a more mature STUDENT and individual. You are very green and very young, don't just learn the fundamentals MASTER them. You obviously have a very limited understanding of advanced matts this you need to learn more. Grow up.

>>	Anonymous Sun Jul 5 18:48:00 2015 No.7374816 >>7373189 Answer the question: what would happen if McFly set the time machine to travel to EXACTLY pi-years?

>>	Anonymous Sun Jul 5 19:01:10 2015 No.7374840 >>7374816 exactly. By Wildberger's theorem, irrationals don't exist the hodge conjecture is simply a lemma to widberger's theorem

>>	Anonymous Sun Jul 5 19:40:34 2015 No.7374868 Finally someone with some sense!

>>	Anonymous Sun Jul 5 19:52:30 2015 No.7374883 >>7372639 Easy, drive in a circle with a radius of 1/2 mile with each circuit lasting an hour.

>>	Anonymous Sun Jul 5 20:53:17 2015 No.7374972 >>7374883 That's not speed.

>>	Anonymous Sun Jul 5 21:10:44 2015 No.7374994 As you continue differentiating an irrational function the derivative gets closer to zero, indicating a non smooth function

>>	Anonymous Sun Jul 5 21:11:25 2015 No.7374995 >>7374994 > indicating a smooth function ftfy

>>	Anonymous Sun Jul 5 21:12:56 2015 No.7374997 >>7374995 No. Indicating a NON smooth function/curve

>>	Anonymous Sun Jul 5 21:24:27 2015 No.7375011 >>7374997 If you can continue taking the derivative it's a smooth function.

>>	Anonymous Sun Jul 5 21:30:43 2015 No.7375019 >>7375011 No. If you can differentiate it infinitely it is smooth. Even then it may turn around. Otherwise it's discontinuous.

>>	Anonymous Sun Jul 5 21:31:49 2015 No.7375020 >>7375019 >Even then it may turn around. What does this mean?

>>	Anonymous Sun Jul 5 22:13:51 2015 No.7375070 Negative derivatives cause sharp turn arounds in the graph

>>	Anonymous Sun Jul 5 22:16:50 2015 No.7375079 >>7375070 No they don't. f(x) = -x

>>	Anonymous Sun Jul 5 22:19:22 2015 No.7375085 >>7374816 any irrational number described by a digital representation would be continuous. So it could be done

>>	Anonymous Sun Jul 5 22:37:30 2015 No.7375113 Can a graduate student verify the proof? I'm sick of calculus students in this thread

>>	Anonymous Mon Jul 6 10:16:17 2015 No.7376107 Bump

>>	Anonymous Mon Jul 6 10:19:38 2015 No.7376109 >>7376107 fuck off victor.

>>	Anonymous Mon Jul 6 11:43:02 2015 No.7376206 Jelly?

>>	Anonymous Mon Jul 6 11:46:44 2015 No.7376210 >>7376206 Of your schizofrenia. No, not really. Take your meds.

>>	Anonymous Mon Jul 6 11:48:41 2015 No.7376211 >>7375113 Graduate student here. You're fucking retarded.

>>	Anonymous Mon Jul 6 11:51:44 2015 No.7376219 >>7376211 You're supposed to point out specific errors in the proof.

>>	Anonymous Mon Jul 6 12:00:01 2015 No.7376230 does someone seriously think one of the greatest mysteries in math can be solved in 12 simple steps?

>>	Anonymous Mon Jul 6 12:02:38 2015 No.7376238 >>7376219 2 doesn't follow from 1. Are you really using blackboard font for a generic function?

>>	Anonymous Mon Jul 6 12:21:53 2015 No.7376258 >>7376238 Sigh. Just retake Calc 1 idiots.

>>	Anonymous Mon Jul 6 12:22:47 2015 No.7376259 >>7376258 Just take your meds Victor.

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