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25 Dec 2021Mathchan is launched into public
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File: 3-adic integers with dual colorings.png ( 660.99 KB , 1200x1200 , 1641009124540.png )

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



Equations can be embedded in multiple ways: 
  • \eqn{...}
    or 
    \math{...}
    command substitutions
  • \[ ... \]
    or 
    \( ... \)
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  • $$ ... $$
    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 
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and using 
&
and 
\\
symbols to delineate between columns and rows respectively.

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\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 
\begin{CD} ... \end{CD}
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.
48 posts and 3 image replies omitted. Click here to view.
>>
  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
>>
  No. 1107
Наша частная клиника предоставляет современное лечение для всей семьи.
В нашем центре индивидуальный подход эффективные методы лечения.
Наши врачи — это высококвалифицированные специалисты, использующие передовые методики.
Наши услуги включают услуги в различных медицинских направлениях, среди которых консультации специалистов.
Мы ценим ваше доверие — основная цель нашего обслуживания.
Свяжитесь с нами, и восстановите ваше здоровье с нами.
https://brightfrenzy.com/profile.php?user=stacy-gillam.402423&com=profile&op=userinfo

  No. 136
Reply
>>137 >>173 >>667 >>371 >>372 >>403 >>414 >>448 >>474 >>550
Is thrembo real? It's an integer between 6 and 7.
46 posts and 6 image replies omitted. Click here to view.
>>
  No. 833 >>838
>>832
The final answer is 3 and in >>830 it is 2.
>>
  No. 838
>>833
no the answer is clearly 2
two integers cant represent a complex number
>>
  No. 839
sneed
>>
  No. 962
logsday is the thremboth day of the week
>>
  No. 1106

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.
16 posts and 2 image replies omitted. Click here to view.
>>
  No. 1031
Tesr
>>
  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

File: 1725321685516r.png ( 28.95 KB , 1500x1500 , 1740952559180.png )

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  No. 1099
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>>1100
It is both kinda surprising and disappointing that the easiest way to compute the derivative of
arc sinarc\ sin
is to solve the integral of
1−x2\sqrt{1-x^2}
geometrically.

Just by looking on a quarter of a circle with angle in it you know that:
∫1−x2dx=arc sinx2+x1−x22\int{\sqrt{1-x^2}}dx = \frac{arc\ sinx}{2} + \frac{x\sqrt{1-x^2}}{2}


By deriving everything we get:
1−x2=(arc sinx)′2+(x1−x2)′2\sqrt{1-x^2} = \frac{(arc\ sinx)'}{2} + \frac{(x\sqrt{1-x^2})'}{2}


And now it's easy to get
(arc sinx)′(arc\ sinx)'
out of the equation and solve it.

It is
11−x2\frac{1}{\sqrt{1-x^2}}
if anyone asked.

Then you can compute
(arc cosx)′(arc\ cosx)'
:

(arc cosx)′=(arc sin(1−x))′=(arc\ cosx)' = (arc\ sin(1-x))' =

=(1−x)′11−(1−x)2=...= (1-x)'\frac{1}{\sqrt{1-(1-x)^2}} = ...


Similarily and even easier
(arc tanx)′(arc\ tanx)'
because it's just a derivative of a fraction of functions.
>>
  No. 1100 >>1102

File: Ears.gif ( 357.71 KB , 425x599 , 1740956279555.gif )

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>>1099

the inverse function of any differentiable bijection can have
its derivative computed via the chain rule. just differentiate their composition.

gross wojack picture btw
>>
  No. 1102
>>1100
Yes, now I see. I suspected that there should be some easier way to do this because these are inverse functions.

This wojack is a homage to some Serbian who was shilling Lukyon on the soy wojak forum. He also used this "neutral feraljak".

File: mathchan.jpg ( 99.43 KB , 1200x900 , 1740499273498.jpg )

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  No. 1095
Reply
Why are you not at the math room https://depvana.com/topic/202

let's gather around mathematics there. As a start, I need examples of a real problem where complex numbers are used to derive a real solution?

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

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  No. 1080
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>>1089
So what are the real numbers that neither rational or irrational.
>>
  No. 1089
>>1080
That subset is empty and the diagram doesn't say otherwise

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

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

File: IMG_5393.jpg ( 1.42 MB , 1284x1713 , 1708976478684.jpg )

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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?

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