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File: 3-adic integers with dual colorings.png ( 660.99 KB , 1200x1200 , 1641009124540.png )

Image
  No. 94
Reply
This board is for the discussion of mathematics.



Equations can be embedded in multiple ways: 
  • \eqn{...}
    or 
    \math{...}
    command substitutions
  • \[ ... \]
    or 
    \( ... \)
    block substitutions
  • $$ ... $$
    or 
    $ ... $
    special block substitutions
  • \begin{equation} ... \end{equation}
    or 
    \begin{math} ... \end{math}
    environments
  • by starting a line with 
    ,,
    or 
    ,eqn
    like one commonly would with >greentext.


Matrices can be embedded by using 
\begin{matrix} ... \end{matrix}
or 
\begin{array} ... \end{array}
environments in any of the above ways to embed equations, or by starting a line with 
,mat
, 
,pmat
, 
,smat
,bmat
, 
,Bmat
, 
,vmat
or 
,Vmat
and using 
&
and 
\\
symbols to delineate between columns and rows respectively.

Arrays can be embedded by using 
\begin{array}{c|c:c} ... \end{array}
environment in any of the above ways to embed equations, or by starting a line with 
,arr{c|c:c}
Post too long. Click here to view the full text.
>>
  No. 98

File: tikz-cd-doc.pdf ( 264.95 KB , 1641011017028.pdf )

Image
Mathchan supports embedding commutative diagrams:
A→aBb↓↑cC=D \qquad\begin{CD} A @>a>> B \\ @VbVV @AAcA \\ C @= D \end{CD}

This can be done by using the 
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environment in any of the above ways to to embed equations, or by specifying your code after starting a line with 
,cd
.

If KaTeX's support is insufficient, diagrams can rendered using the 
\tikzcd{...}
command or by using the 
\begin{tikzcd} ... \end{tikzcd}
environment.
For example, pasting the following excerpt from the attached PDF (TikZ-CD manual): 

\begin{tikzcd}[row sep=scriptsize, column sep=scriptsize]
   & f^* E_V \arrow[dl] \arrow[rr] \arrow[dd] & & E_V \arrow[dl] \arrow[dd] \\
   f^* E \arrow[rr, crossing over] \arrow[dd] & & E \\
   & U \arrow[dl] \arrow[rr] & & V \arrow[dl] \\
   M \arrow[rr] & & N \arrow[from=uu, crossing over]\\
\end{tikzcd}

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File: xiom.png ( 380.54 KB , 1080x980 , 1663002740804.png )

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  No. 299
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>>604 >>709 >>711 >>732 >>767 >>779 >>876 >>949 >>1081
When did you realize that the reals are fake?
Mathematics can do without infinities.
47 posts and 3 image replies omitted. Click here to view.
>>
  No. 954
Only one thing in the universe is infinite and that's the coping and seething of finitistcels
>>
  No. 955 >>956
In order to be mathematics, Ultrafinitist and finitists need to make a logical coherent system.
They can't. It's about the physics to deceide whether the world as a whole is infinite or finite in nature.
As far as I know, our informations doesn't allow such a inference. The topic is closed at this moment.
>>
  No. 956
>>955
I'd say finitism is moreso a philosophical perspective than anything very empirical.
>>
  No. 977
>>
  No. 1081
>>299
trve

File: Linear-Differential-Equation.png ( 667.17 KB , 2000x2068 , 1655761703189.png )

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  No. 207
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>>626
Post diff eqs and their solutions.
14 posts and 1 image reply omitted. Click here to view.
>>
  No. 626

File: FTDC9FGXwAEZ211.jpg ( 297.09 KB , 2048x1616 , 1705954986668.jpg )

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>>207
how did i not see this lol i have a wide collection

Consider
y′′−2y=ln⁡(x)y''-2y=\ln(x)
.

Let's first solve for the complementary solution, which is the general solution to the complementary equation
y′′−2y=0y''-2y=0
.

The characteristic polynomial for the complementary equation is
λ2−2=0\lambda^2-2=0
, so
λ=±2\lambda=\pm \sqrt{2}
. Hence, the complementary solution is
yc(x)=Ay1(x)+By2(x)=Ae−x2+Bex2y_c(x)=Ay_1(x)+By_2(x)=Ae^{-x\sqrt{2}}+Be^{x\sqrt2}
for some constants
AA
and
BB
. Now let's find the particular solution using variation of parameters. Using the two basis solutions of the complementary solution,
y1(x)y_1(x)
and
y2(x)y_2(x)
, we construct the Wronskian and get
W(e−x2, ex2)(x)=∣e−x2ex2(e−x2)′(ex2)′∣=22W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x) = \begin{vmatrix} e^{-x\sqrt{2}} & e^{x\sqrt{2}}\\ \left(e^{-x\sqrt{2}}\right)' & \left(e^{x\sqrt{2}}\right)' \end{vmatrix} = 2\sqrt2


Hence, for the particular solution
yp(x)=u1(x)y1(x)+u2(x)y2(x)y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)
, we have,
u1(x)=−∫ln⁡(x)y2(x) dxW(e−x2, ex2)(x)=−∫ln⁡(x)ex2 dx22=14Ei⁡(x2)−14ex2ln⁡(x) u_1(x) = - \int\frac{\ln(x)y_2(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = -\int\frac{\ln(x)e^{x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)

u2(x)=∫ln⁡(x)y1(x) dxW(e−x2, ex2)(x)=∫ln⁡(x)e−x2 dx22=14Ei⁡(−x2)−14e−x2ln⁡(x) u_2(x) = \int\frac{\ln(x)y_1(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = \int\frac{\ln(x)e^{-x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)


The particular solution then is nothing but
yp(x)=u1(x)y1(x)+u2(x)y2(x)=(14Ei⁡(x2)−14ex2ln⁡(x))e−x2+(14Ei⁡(−x2)−14e−x2ln⁡(x))ex2=14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) y_p(x) = u_1(x)y_1(x)+u_2(x)y_2(x) = \left(\frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)\right) e^{-x\sqrt2} + \left(\frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)\right) e^{x\sqrt2}\\ = \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)

Thus the complete general solution is given by
y(x)=yc(x)+yp(x)=Ae−x2+Bex2+14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) \boxed{y(x) = y_c(x) + y_p(x) = Ae^{-x\sqrt{2}}+Be^{x\sqrt2} + \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)}
>>
  No. 646
>>210
>>212
>>213

Honestly, I never understood why we defined the notion of an "elementary" function the way we did. An elementary function from what I've seen is completely restricted to polynomials/rational powers, exponentials, and logarithms (trig/hyperbolic trig and their inverses are expressible in exponentials/logs), which is extremely limited.
>>
  No. 655

File: GFv9PYsakAA1vJW.jpg ( 1.28 MB , 1400x1900 , 1708564466975.jpg )

Image
We have
y′=(1+x) y+xy2y'=(1+x)\,y+xy^2
.

Rewriting this gives
y′−(1+x) y=xy2y'-(1+x)\,y=xy^{2}
. This is a standard Bernoulli diffeq form
y′+a(x) y=b(x) yαy'+a(x)\,y=b(x)\,y^{\alpha}
, which is solved by performing the substitution
ϕ=y1−α\phi=y^{1-\alpha}
. Using this substitution, which for our specific case is
ϕ=y1−2=y−1\phi = y^{1-2} = y^{-1}
, we have
−y′y2+x+1y=−x⟺ϕ′+(x+1) ϕ=−x\displaystyle -\frac{y'}{y^2} +\frac{x+1}{y} = -x \Longleftrightarrow \phi' + (x+1)\,\phi = -x
.

To make this substitution more rigorous (since it is less obvious), differentials can be used. Also, to account for this division and ensure we miss no solutions, we make sure to check when the denominators are equal to 0, which is
y=0y=0
. Since plugging in
y=0y=0
satisfies the diffeq, this is a constant solution. Then, using the integrating factor
exp⁡(∫x+1 dx)=ex22+x\exp\left(\displaystyle\int x+1\text{ d}x\right) = e^{\frac{x^2}{2}+x}
, we get
ϕ′ex22+x+(ex22+x)′ϕ=−xex22+x∫(ϕ ex22+x)′dx=∫−xex22+x dxϕ ex22+x=π2eerfi⁡(x+12)−ex22+x+Cϕ=π2eerfi⁡(x+12)−ex22+x+Cex22+x \begin{align*} \phi'e^{\frac{x^2}{2}+x} + \left(e^{\frac{x^2}{2}+x}\right)'\phi &= -xe^{\frac{x^2}{2}+x}\\ \int\left(\phi \,e^{\frac{x^2}{2}+x}\right)'\text{d}x &= \int-xe^{\frac{x^2}{2}+x}\text{ d}x\\ \phi \, e^{\frac{x^2}{2}+x} &= \sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C\\ \phi &= \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} \end{align*}

Undoing our substitution, we have, for the positive Dawson Integral
D+(x)=π2e−x2erfi⁡(x)D_+(x) = \frac{\sqrt{\pi}}2e^{-x^2}\operatorname{erfi}(x)
that
ϕ=y−1=π2eerfi⁡(x+12)−ex22+x+Cex22+x=2 D+(x+12)−1+Ce−x22−x\displaystyle \phi = y^{-1} = \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} = \sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}

  ⟹  y(x)=0,12 D+(x+12)−1+Ce−x22−x\displaystyle\implies \boxed{y(x)=0, \qquad \frac{1}{\sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}}}
>>
  No. 1005
>>
  No. 1084
>>208
I should add to this thread, given I'm study differential equations and have to memorize how to sovle stuff like this.

File: asdasdas.png ( 2.99 KB , 605x152 , 1697767208017.png )

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  No. 512
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>>532 >>535 >>539 >>548 >>713 >>744 >>1018
What I said in the subject. It's so easy to prove. Search the problem and solve it, simple as that.
17 posts and 2 image replies omitted. Click here to view.
>>
  No. 1033
jjjjjjjjjjjj
>>
  No. 1053
test post

will dedicate my time to extrapolating a curve later
>>
  No. 1054 >>1071
>>1018
>posting about iq on a mathematics board
I just want to say that i scored 110s on the mensa when i didn't know these insane fucks were doing function identities on the object numbers and 130s after finding out
iq tests are a joke on uninformed people
>but you have got to find out during the test bro
fuck off
>>
  No. 1071
>>
  No. 1083
На этом сайте у вас есть возможность приобрести виртуальные телефонные номера различных операторов. Эти номера могут использоваться для подтверждения профилей в разных сервисах и приложениях.
В каталоге представлены как долговременные, так и одноразовые номера, которые можно использовать для получения сообщений. Это удобное решение для тех, кто не желает использовать основной номер в сети.
<a href="https://nhl-news.ru/blog/virtualnye-nomera-indii-prostoe-i-udobnoe-obshchenie-dlya-vashego-biznesa">индийские номера</a>
Оформление заказа очень простой: определяетесь с необходимый номер, оплачиваете, и он сразу будет готов к использованию. Попробуйте сервис уже сегодня!

File: Collatz.pdf ( 291.71 KB , 1738035089350.pdf )

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  No. 1064
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>>1077
Hello Mathchan. I'm not in math or anybody important, so I've found it basically impossible to communicate my ideas, but attached is a pdf outlining an idea for a proof of the Collatz conjecture I came up with about a year ago. The proof is based on a paper by Mandelbrot:
https://users.math.yale.edu/mandelbrot/web_pdfs/136multifractal.pdf

I can't attest if it's correct or not, but really I'm just posting this because I wanted it archived so that maybe some day someone will find value in it. Also you can do whatever you want with it, rewrite it or whatever.
>>
  No. 1077 >>1082
>>1064
interesting, will read later
>>
  No. 1082
>>1077
I would welcome any criticism of it.

File: Rational-and-Irrational-Numbers-768x678.jpg ( 56.32 KB , 768x678 , 1738595323081.jpg )

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  No. 1080
Reply
So what are the real numbers that neither rational or irrational.

File: bc2e95b1013a443b0ad125e8bcecb7e7.jpg ( 87.82 KB , 781x1092 , 1642824353663.jpg )

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  No. 107
Reply
>>169
Talk maths.
10 posts and 1 image reply omitted. Click here to view.
>>
  No. 657 >>658

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>>132
>>563
Wouldn't it be an improper integral over the complex plane?
>>
  No. 658
>>657
https://mathoverflow.net/questions/453862/is-the-area-of-the-mandelbrot-set-known
>>
  No. 917 >>1075
Rediscovered something cute

https://files.catbox.moe/l6ft4m.mp4
>>
  No. 1027
>>111
Trying to classify what ideals fail the weak lefschetz property
>>
  No. 1075
>>917
do u have moar?

File: Frac.png ( 116.91 KB , 256x256 , 1691684545939.png )

Image
  No. 465
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>>466 >>478 >>529 >>1017 >>1039
I don't get it. Why are so many people in fractals nowadays?

There seems to be a whole community on Youtube that makes Mandelbrotzooms and seems to appreciate the psychedelic aesthetics and the relationship of these to PC hardware. Is this simply a continuation of the graphics demo scene?

I don't quite understand the appeal. However, I don't quite understand the math behind it either.
3 posts and 1 image reply omitted. Click here to view.
>>
  No. 529
>>465
"schizos" schizing out is a good explanation
>>
  No. 1017
>>465
It looks interesting, like there could be some seret around any corner, but it isn't, so there's no pressure to find anything
>>
  No. 1032
Look:
https://youtu.be/Ed1gsyxxwM0
>>
  No. 1039 >>1046
>>465
There's a difference between being in fractals and being "in fractals". I'm sure there are plenty of people who like to look at a mandelbrot zoom, but the number of mathematicians actually studying fractals is probably quite low, even lower than in the 80s during the fractal heyday.
>>
  No. 1046
>>1039
A >>>IQ comment

File: 1705233653759.png ( 201.38 KB , 1139x804 , 1713480390587.png )

Image
  No. 724
Reply
>>727 >>731 >>940
If you had a math exam and you could use your phone with internet while attempting the paper, what service would you use to solve questions?
Topics:
-Propositional Calculus
-Methods of Proof
-Boolean Algebra and Circuits
-Sets, Relations and Functions
-Combinatorics
-Some more Counting Principles
-Partitions and Distributions
15 posts and 1 image reply omitted. Click here to view.
>>
  No. 754 >>791
>>751
No problem mate. Couldn't find anything that works on rutracker. Where exactly must I be looking?
>>
  No. 791
>>754
Mobilism
>>
  No. 940
>>724
fart
>>
  No. 1019
what i say, fuck niggers
>>
  No. 1038
Surely the answer has to be Wolfram Cloud. Mathematica + WA at your fingertips

File: 1642980098311.gif ( 3.73 MB , 406x640 , 1713832706815.gif )

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  No. 749
Reply
>>948 >>1037
Im new.
2 posts omitted. Click here to view.
>>
  No. 948

File: 100512551_p0.jpg ( 2.66 MB , 4058x2029 , 1725514201950.jpg )

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>>749
broooo i need lukyon back plsss &#128532;&#128532;&#128532;i miss him so much
>>
  No. 952
hi new
>>
  No. 1021
Congratulations
>>
  No. 1025
Hi new I'm dad
>>
  No. 1037
>>749 hey me too

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