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/math/ - Mathematics

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25 Dec 2021Mathchan is launched into public
Catalog Archive

File: JohnvonNeumann.gif ( 187.66 KB , 490x639 , 1703776296430.gif )

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  No. 590
Reply
Happy Birthday von Neumann
>>
  No. 591
Happy Birthday to him
>>
  No. 592

File: 229f48a3e2e51980f404997bd1688f0a.jpg ( 1.79 MB , 3025x3912 , 1703801252663.jpg )

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Von Neumann helped fix the contradictions in Set Theory by rephrasing sets as classes which are members of classes.

File: logica.jpg ( 94.1 KB , 566x929 , 1703673978707.jpg )

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  No. 574
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hat is your opinion about the Logica Hamburgensis, anon?

What do you say about?
https://www.digitale-sammlungen.de/de/view/bsb11273280?page=7

File: question18.png ( 11.57 KB , 693x61 , 1690377094530.png )

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  No. 456
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>>459 >>457 >>482
Prove you are not a Midwit.
7 posts and 2 image replies omitted. Click here to view.
>>
  No. 514 >>533
>>460
Can be done in a couple of ways; summing
⌊akm⌋⌊bkm⌋\lfloor \frac{ak}{m} \rfloor \lfloor \frac{bk}{m} \rfloor
or computing
∫01{ax}{bx}\int_0^1 \{ax\}\{bx\}
works.

For(1),We will show that
λ=∫01{ax}{bx}dx\lambda = \int_{0}^{1}\left \{ ax \right \} \left \{ bx \right \}dx
satisfied. In fact,we would show that
[ab(k−1)m,abkm]∩Z=∅\left [ \frac{ab\left ( k-1 \right )}{m} ,\frac{abk}{m}\right ]\cap\mathbb{Z}= \emptyset

then we are done. This
⟺{akm}{bkm}−m∫k−1mkm{ax}{bx}dx=O(1m)\Longleftrightarrow\left \{ \frac{ak}{m} \right \} \left \{ \frac{bk}{m} \right \} -m\int_{\frac{k-1}{m} }^{\frac{k}{m} }\left \{ ax \right \}\left \{ bx \right \} dx=\mathrm {O}\left ( \frac{1}{m} \right )
and
⟺sup⁡x∈[k−1m,km]{ax}{bx}−inf⁡x∈[k−1m,km]{ax}{bx}=O(1m)\Longleftrightarrow \sup_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} -\inf_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} =\mathrm {O} \left ( \frac{1}{m} \right )
It's obviously right.

■\blacksquare
>>
  No. 533 >>534 >>536
>>458
>>514
>mathchan
>can't even render weblatex
this is sad
>>
  No. 534 >>547
>>533
You have to use 
\[ ... \]
instead of dollars.
>>
  No. 536
>>533
Exactly just what I was thinking lmao
>>
  No. 547
>>534
>he did not read the sticky

File: 20230903_200854.jpg ( 1.3 MB , 3297x1921 , 1693882765279.jpg )

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  No. 493
Reply
>>520
Has anyone read euclids elements or has ability to elucidate upon this geometry
>>
  No. 520
>>493
no
>>
  No. 524
Euclids work is just of historical interest.
I have give it a try but... I don't have the patience to follow it.

File: Gigachad.jpg ( 58.71 KB , 680x763 , 1699147008221.jpg )

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  No. 522
Reply
>There is no Eulerian path along the bridges of Konigsberg?
>What if we dropped a 5000 kg bomb on the city?

File: norman chad.jpg ( 107.98 KB , 1280x720 , 1662995967260.jpg )

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  No. 297
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>>451 >>449
He is right about everything.
1 post omitted. Click here to view.
>>
  No. 449
>>297
>>298
All me
>>
  No. 451
>>297
shes wrong tho
>>
  No. 453 >>502
Ok they're not real, but let's pretend they're real and prove all these useful theorems
>>
  No. 502
>>453
this is what happens
>>
  No. 521
you need to define real
he seems to think that math is the language of the universe or some idealism like that, and that therefore real numbers (more precisely, irrationals) can't be real since they contradict with his vision of reality
except math was never the language of the universe, it's a language that can model the universe, it can also model other things, just like other languages
math is real in the sense that it exists in our head, that we recognize collectively that it exists, it is not real in the sense that it’s reality itself
so real number aren't any less real than natural number, and, they are logically sound from the axioms we use

File: 183px-Double_torus_illustration.png ( 26.71 KB , 183x200 , 1696790827904.png )

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  No. 506
Reply
Hodge conjecture is true. How do you prove it?
>>
  No. 509 >>511
It's not true.
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  No. 511
>>509
The integral Hodge conjecture is false. Why also is Hodge conjecture false? It is proved for many special case!
But this it not prove the conjecture.
>>
  No. 513
I cant

File: TIMESAND___QDRH762aFF.jpg ( 1.25 MB , 3400x3044 , 1663048919445.jpg )

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  No. 301
Reply
The Riemann hypothesis is false. I have a few independent negations by direct counterexample and they are presented in papers with varying degrees of rigor. In this "quick" proof, I present a fully rigorous proof upon an unproven proposition. Obvioulsy, this fails to negate RH due to the unproven proposition but it concisely states the method I use. In a longer paper, I do it without that proposition's assumption and in a *much* longer paper I start with Euclid's Elements, assume nothing else, and then show that RH is consequently false. That paper is linked below and if I can post the PDF, I will post it.

Fractional Distance: The Topology of the Real Number Line with Applications to the Riemann Hypothesis
>https://vixra.org/abs/2111.0072
Recent analysis has uncovered a broad swath of rarely considered real numbers called real numbers in the neighborhood of infinity. Here we extend the catalog of the rudimentary analytical properties of all real numbers by defining a set of fractional distance functions on the real number line and studying their behavior. The main results of are (1) to prove with modest axioms that some real numbers are greater than any natural number, (2) to develop a technique for taking a limit at infinity via the ordinary Cauchy definition reliant on the classical epsilon-delta formalism, and (3) to demonstrate an infinite number of non-trivial zeros of the Riemann zeta function in the neighborhood of infinity. We define numbers in the neighborhood of infinity as Cartesian products of Cauchy equivalence classes of rationals. We axiomatize the arithmetic of such numbers, prove all the operations are well-defined, and then make comparisons to the similar axioms of a complete ordered field. After developing the many underlying foundations, we present a basis for a topology.
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  No. 302

File: TIMESAND___FractionalDistance.pdf ( 943.18 KB , 1663049229219.pdf )

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This is the one starting from Euclid.
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  No. 303

File: TIMESAND___ZetaMedium.jpg ( 3.19 MB , 3689x2457 , 1663049338048.jpg )

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This one starts with a more modern shared framework for analysis.
>>
  No. 504
Is this your work?
>>
  No. 505
I don't know but I think you're another crackpot.
>>
  No. 510
Oh boy 😔

File: PuppetWalks.gif ( 30.38 KB , 150x200 , 1694924381757.gif )

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  No. 501
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>>503
Trying to look for a more mathemathical, vectorized summarized idea of Drawing correctly all the frames involved in the walk cycle

From the idea that it is looping to maintaining constant balance between on step or cycle to another. But generally it needs to be summarized into a shape or curves or something easily read, mathemathically, that will tell where the pose go so to define the motion that is predetermined... which includes variety of things limping, running, crowling... assortments of so.

The prioritt idea is the balance and the repeat, so much less than abstract characteristic but rather the correct physics in accordance to the asked body motion
>>
  No. 503
>>501
Did you make a similar thread on hikari3.ch/aca?
Someone had a very similar question there before the thread was deleted

File: ‎image_two_grids.jpeg ( 318.43 KB , 1920x1080 , 1693178750553.jpeg )

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  No. 481
Reply
Do you know of any interesting ways to list all possible
kk
-tuples of nonnegative integers in order? Or in other words, do you know a function
f(n)f(n)
such that for every
(a,b)(a, b)
with integers
a,b≥0a, b \ge 0
, there exists an
n≥0n \ge 0
with
f(n)=(a,b)f(n) = (a, b)
? Well, I do! I'll be showing my way.

First you create a grid with two axes which you start numbering starting at 0. The square in the
pp
th row and
qq
th column will have the tuple
(p−1,q−1)=(a,b)(p - 1, q - 1) = (a, b)
. Not
(n,m)(n, m)
because we start at 0. Now, we draw a line that connects the squares like in the left grid of the image attached (translucent, red line). The line creates this one-dimensional sequence of 2-tuples:
(0,0)→(0,1)→(1,0)→(2,0)→(1,1)→(0,2)→(0,3)→(1,2)→(2,1)→(3,0)→(4,0)→…(0, 0) \rightarrow (0, 1) \rightarrow (1, 0) \rightarrow (2, 0) \rightarrow (1, 1) \rightarrow (0, 2) \rightarrow (0, 3) \rightarrow (1, 2) \rightarrow (2, 1) \rightarrow (3, 0) \rightarrow (4, 0) \rightarrow \dots
You should notice the pattern. Like that, I've listed all 2-tuples.

We define
fk(n)f_k(n)
as our function that maps
{0,1,2,…}\lbrace 0, 1, 2, \ldots \rbrace
to the set of all
kk
-tuples of nonnegative integers.

We can extend this to 3-tuples. But without using a 3-dimensional grid. We make a similar grid (right side of image), this time the upper axis is for the third element in the tuple and the left axis is just the sequence of 2-tuples I listed before for the 1st and 2nd elements of the 3-tuple.

Using that method, we get this for
k=3k = 3
:
f3(0)=(0,0,0)f3(1)=(0,0,1)f3(2)=(0,1,0)f3(3)=(1,0,0)f3(4)=(0,1,1)f3(5)=(0,0,2)   ⋮ f_3(0) = (0, 0, 0)\\ f_3(1) = (0, 0, 1)\\ f_3(2) = (0, 1, 0)\\ f_3(3) = (1, 0, 0)\\ f_3(4) = (0, 1, 1)\\ f_3(5) = (0, 0, 2)\\ \ \ \ \vdots


A slightly less intuitive pattern...

Of course, you can generalize this method to get the formula:
fk+1(n)=fk(b(n))∪s(n)f_{k+1}(n) = f_k(b(n)) \cup s(n)
. The union symbol is used here to append
s(n)s(n)
to to the
kk
-tuple
fk(b(n))f_k(b(n))
.
b(n)b(n)
refers to the first element of
f2(n)f_2(n)
and
s(n)s(n)
refers to the second. Also, it isn't particularly hard to prove that these functions indeed do return every possible tuple.

I find this formula beautiful. Correct me in case you find mistakes!
>>
  No. 483 >>484
This method is the same for demonstrating the numerability of rational numbers, but it presents a mistake.
I explain it. Natural numbers are defined by Peano's axioms, actually they are defined for recursion (second axiom), i.e. 7 does not exist without 6.
In first black line of the table you have already the naturals in order.
Now if we stop this method in some numbers, we'll see that it exist a number but the inferior number does not exist yet. e.g. If we stop in (1,1), we'll have (2,0)=(1,0,0)>(1,1)=(0,1,1)
The same is true in numerability, if we stop in 2/3, we'll have defined 1/4 without 4, but rationals are defined by naturals.
>>
  No. 484 >>485
>>483 I know that a similar method is used to prove that the set of rationals and naturals have the same cardinality except you remove all the duplicates. But I'm not quite sure what you mean by
(2,0)=(1,0,0)>(1,1)=(0,1,1)(2, 0) = (1, 0, 0) > (1, 1) = (0, 1, 1)
. Could you elaborate?
>>
  No. 485
>>484
Of course, you ask to us: "Do you know of any interesting waves to list all possible k-tuples of possible nonnegative integers in order?" In effect it exists, it is sufficient to change the base of natural numbers. e.g. For k-tuple in base two until
f(2k−1) f(2^k-1)
you have the k-tuple in order f(0))(0,...,0,0,0), f(1)=(0,...,0,0,1), f(2)=(0,...,0,1,0), f(3)=(0,...,0,1,1), f(4)=(0,...,1,0,0), f(5)=(0,...,1,0,1), etc here you can see that (1,0,0)>(0,1,1) where a>b means a comes after b.
If, otherwise, you fixed k and you need numbers bigger than
f(2k−1) f(2^k -1)
you can change base into base n for number until
f(nk−1) f(n^k -1)
. e.g. k=2, number until f(125), n≥12 you obtain: (for n=12) f(0)=(0,0), f(1)=(0,1), f(2)=(0,2), f(3)=(0,3),..., f(10)=(0, A), f(11)=(0, B), f(12)=(1,0), f(13)=(1,1), f(14)=(1,2),...,f(125)=(A,5)

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