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/math/ - Mathematics

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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 
    \( ... \)
    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 
\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}

Will render as the following figure on Mathchan:



The drawback is that you will have to pause typing and wait until the diagram renders while KaTeX can be displayed instantaneously.
It is recommended that you compile any LaTeX code yourself - or in Overleaf - before attempting to paste it on Mathchan.

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

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  No. 512
Reply
>>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.
11 posts and 2 image replies omitted. Click here to view.
>>
  No. 800
>>706
I haven't read it yet but I'll rewrite the document
(https://www.researchgate.net/publication/316110458_SOLVING_THE_BROCARD_PROBLEM)
in
LaTeX\LaTeX
after reading it for anyone wanting to read it because...
- it is written in Word with bad formatting (like I mentioned before),
- there are some (math) typos,
- it contains unnecessary proofs of obvious facts, e.g. that any prime
p>3p > 3
is of the form
6k+16k + 1
or
6k+56k + 5
,
- it is written in a confusing way in general.
>>
  No. 835 >>1030
n==1 and m==√2
Proof
1!=1=1=1=1
1+1=2
sqrt(2)= √2
>>
  No. 1018
>>512
An IQ score is a measurement variable for intelligence, and is explicitly linked to the test given
This is like saying "if you can reach this tall shelf you are 6'4"
Pointlessly vague because the shelf is a godawful test
>>
  No. 1029
critte
>>
  No. 1030
>>835
proof by example

File: question.png ( 1.95 KB , 404x273 , 1731228993157.png )

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  No. 1012
Reply
>>1013
Can any given probability be reduced to a simple judgment, zero or one?
This would mean that we could reduce any distribution to the binomial distribution
>>
  No. 1013 >>1028
>>1012
no
>>
  No. 1028
>>1013
God, I hate it. Terrible. What a pain.

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

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  No. 107
Reply
>>169
Talk maths.
9 posts and 1 image reply omitted. Click here to view.
>>
  No. 566
>>112
same
>>
  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
Rediscovered something cute

https://files.catbox.moe/l6ft4m.mp4
>>
  No. 1027
>>111
Trying to classify what ideals fail the weak lefschetz property

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

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  No. 749
Reply
>>948
Im new.
1 post omitted. Click here to view.
>>
  No. 947
Hi im new two
>>
  No. 948

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

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>>749
broooo i need lukyon back plsss 😔😔😔i miss him so much
>>
  No. 952
hi new
>>
  No. 1021
Congratulations
>>
  No. 1025
Hi new I'm dad

File: animtourne.gif ( 3.06 KB , 16x16 , 1731881731626.gif )

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  No. 1023
Reply
test
>>
  No. 1024
I just typed in I don't know I didn't think it would work. I'm sorry

File: she's doing integrals ok.jpg ( 21.76 KB , 480x480 , 1655484028377.jpg )

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  No. 199
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Primitive function of function
f(x)f(x)
over some interval
x∈[a,b]x\in[a, b]
is a function
F(x)F(x)
whose derivative is the function
f(x)f(x)
on that interval.

∀x∈[a,b],F′(x)=f(x)\qquad \forall x\in[a, b],\quad F'(x) = f(x)


Antiderivative (aka indefinite integral) of a function
f(x)f(x)
is a family of its primitive functions which differ by a constant
C∈RC\in\mathbb{R}
:

∫f(x)dx=F(x)+C\qquad \int f(x)\mathrm{d}x = F(x) + C


In a nutshell, integration is the opposite of differentiation:

d∫f(x)dxdx=F(x)\qquad \frac{\mathrm{d}\int f(x)\mathrm{d}x}{\mathrm{d}x} = F(x)


d∫f(x)dx=f(x)dx\qquad \mathrm{d}\int f(x)\mathrm{d}x = f(x)\mathrm{d}x


∫dF(x)=F(x)+C\qquad \int\mathrm{d}F(x) = F(x) + C


Solving integrals in general is pretty hard, but there are a lot of established ways to do it. As OP I'll post some of the standard approaches, but this thread is about any kind of integration so feel free to post integrals and theirs solutions.

Most basic methods are:

  1. Using the table of integrals
  2. Using linearity property
  3. Using substitution
  4. Using partial integration
  5. Reducing quadratic to its cannonical form
  6. Partial decomposition
3 posts omitted. Click here to view.
>>
  No. 203
Partial integration

This basically allows you to swap what you integrate and what you integrate by.

∫u dv=uv−∫v du+C\qquad\int u\ \mathrm{d}v = uv - \int v\ \mathrm{d}u + C



Usually, you use it to solve an integral with polynomial multiplied by one of the non-polynomial functions.

∫Pn(x)⋅eaxsin⁡(ax)cos⁡(ax)ln⁡(ax)arcsin⁡(ax)arccos⁡(ax)dx\qquad \int P_n(x)\cdot \begin{matrix}e^{ax}\\\sin(ax)\\\cos(ax)\\\ln(ax)\\\arcsin(ax)\\\arccos(ax)\end{matrix}\mathrm{d}x


You usually choose
uu
to be the non-polynomial because calculating the derivative of it is probably going to be easier than integrating it.
On the other hand, polynomials are easy to both derive and integrate.

Example

You integrate
∫x2arccos⁡(x)dx\int x^2\arccos(x)\mathrm{d}x
by differentiating
arccos⁡(x)\arccos(x)
and by integrating
x2x^2
:

∫arccos⁡(x)⋅x2dx={u=arccos⁡(x)du=−dx1−x2dv=x2dxv=x33}=x33arccos⁡(x)+13∫x31−x2dx+C=x33arccos⁡(x)−13(1−x2+(1−x2)33)+C\begin{aligned} \qquad \int \arccos(x)\cdot x^2 \mathrm{d}x &= \begin{Bmatrix} u = \arccos(x) & \mathrm{d}u = -\frac{\mathrm{d}x}{\sqrt{1- x^2}}\\\mathrm{d}v = x^2\mathrm{d}x & v = \frac{x^3}{3}\end{Bmatrix} = \frac{x^3}{3}\arccos(x)+ \frac{1}{3}\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x +C \\&= \frac{x^3}{3}\arccos(x) - \frac{1}{3}\left(\sqrt{1-x^2} + \frac{\sqrt{(1-x^2)^3}}{3}\right) + C \end{aligned}


As for how you solve
∫x31−x2dx\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x
you do it by substituting
t=1−x2,x2=1−t2,2dx=−2tdtt = \sqrt{1-x^2},\quad x^2 = 1- t^2,\quad \cancel{2}\mathrm{d}x = -\cancel{2}t\mathrm{d}t
:

∫x31−x2dx=∫1−t2t(−t)dt=t−t33=1−x2−(1−x2)33\qquad \int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x = \int\frac{1-t^2}{t}(-t)\mathrm{d}t = t - \frac{t^3}{3} = \boxed{\sqrt{1-x^2} - \frac{\sqrt{(1-x^2)^3}}{3}}
>>
  No. 204
Quadratic trinomial

How do you solve
∫dxax2+bx+c\int\frac{\mathrm{d}x}{ax^2 + bx + c}
and
∫dxax2+bx+c\int\frac{\mathrm{d}x}{\sqrt{ax^2 + bx + c}}
? You write
ax2+bx+cax^2 + bx + c
in the following way (hint: completing the square by adding and subtracting
b24)\frac{b^2}{4})
:

ax2+bx+c=a(x2+bax+ca)=a((x2+2b2ax+b24a2)+(−b24a2+c))=a((x+b2a)2+(c−b2a)2)=a(t2+k2),t=x+b2a,k=c−b2a\begin{aligned} \qquad ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = a\left(\left(x^2 + 2\frac{b}{2a}x + \frac{b^2}{4a^2}\right) +\left(- \frac{b^2}{4a^2} + c \right)\right)\\&= a\left(\left(x + \frac{b}{2a}\right)^2 +\left(\sqrt{c - \frac{b}{2a}}\right)^2\right)\\&=a(t^2 + k^2),\quad t= x+ \frac{b}{2a},\quad k=\sqrt{c-\frac{b}{2a}} \end{aligned}


Now the integral reduces to either
1a∫dxt2+k2=1akarctan⁡tk+C\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{t^2 + k^2} = \frac{1}{ak}\arctan{\frac{t}{k}} + C}
or
1a∫dxt2+k2=1aln⁡∣t2+t2+k2∣+C\boxed{\frac{1}{a}\int\frac{\mathrm{d}x}{\sqrt{t^2 + k^2}} = \frac{1}{a}\ln\left|t^2 + \sqrt{t^2 + k^2}\right| + C}
>>
  No. 205
Partial fraction decomposition

How do you solve
∫P(x)Q(x)dx\int\frac{P(x)}{Q(x)}\mathrm{d}x
where
 deg⁡P(x)<deg⁡Q(x)\ \deg{P(x)} < \deg{Q(x)}
?


First, as a consequence of the fundamental theorem of algebra, any real polynomial can be factored into linear and quadtratic terms.
We will do that with
Q(x)Q(x)
:

Q(x)=(x−a1)A1(x−a2)A2…(x−am)Am(x2+b1x+c1)B1(x2+b2x+c2)B2…(x2+bnx+cn)Bn\qquad Q(x) = (x-a_1)^{A_1}(x-a_2)^{A_2}\dots(x - a_m)^{A_m}(x^2 + b_1x + c_1)^{B_1}(x^2 + b_2 x + c_2)^{B_2}\dots(x^2 + b_n x + c_n)^{B_n}


Now employ the partial fraction decomposition:

P(x)Q(x)=∑i=1m∑j=1Aiaij(x−ai)j+∑i=1n∑j=1Bibijx+cij(x2+bix+)j\qquad\frac{P(x)}{Q(x)} = \sum_{i=1}^m\sum_{j=1}^{A_i}\frac{a_{ij}}{(x-a_i)^j} + \sum_{i=1}^n\sum_{j=1}^{B_i}\frac{b_{ij}x + c_{ij}}{(x^2 + b_ix +)^j}


Then just use the linearity property of the integral.

For example:

∫x2+1x5−3x4+x3+7x2−6x−8dx=∫x2+1(x−2)(x+1)2(x2−3x+4)dx=∫a11x−2dx+∫a21x+1dx+∫a22(x+1)2dx+∫b11x+c11x2−3x+4dx\qquad \int\frac{x^2 + 1}{x^5 - 3x^4 + x^3 + 7x^2 - 6x - 8}\mathrm{d}x = \int\frac{x^2 + 1}{(x-2)(x +1)^2(x^2-3x + 4)}\mathrm{d}x = \int \frac{a_{11}}{x-2}\mathrm{d}x + \int\frac{a_{21}}{x+1}\mathrm{d}x + \int\frac{a_{22}}{(x+1)^2}\mathrm{d}x + \int\frac{b_{11} x + c_{11}}{x^2 - 3x + 4} \mathrm{d}x



The constants
aij,bij,cij∈Ra_{ij},b_{ij}, c_{ij}\in\mathbb{R}
have to be found e.g. using the Heaviside cover up method.
>>
  No. 464
cool nad thanks
>>
  No. 1022
If you have a function that cannot be integrated but which has a fourth
derivative, you can approximate the definite integral to a high degree of
accuracy using Simpson's rule.

Choose
Δx\Delta x
such that
[a,b][a,b]
is divided into an even number of subintervals.

∫abf(x) dx=lim⁡Δx→0+Δx3[f(x0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+...+4f(xn−1)+f(xn)]\int_{a}^b f(x)\,dx = \lim_{\Delta x \to 0^+} \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + ... + 4f(x_{n-1}) + f(x_n)]


For the error, find the maximum value
MM
of
f(4)(x)f^{(4)}(x)
on
[a,b][a,b]
.
b−a180M(Δx)4\frac{b-a}{180}M(\Delta x)^4

File: images.jpeg ( 13.87 KB , 260x194 , 1729335144595.jpeg )

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  No. 1001
Reply
>>1016
I only know mathematics upto highschool only and that too barely and I want to learn mathematics from beginning rigorously. I need your help. Where should I start, where to begin. What books, what should I do and what should I not do
>>
  No. 1016
>>1001
You should make sure to have fun!
Try mathologer, He does some fairly advanced visual proof
https://www.youtube.com/@Mathologer
Once you are done with that, any book will do, but calculus tends to be useful
>>
  No. 1020

File: 81NCPLOK4HL._SL1500_-1808559805.jpg ( 120.7 KB , 1031x1500 , 1731603827102.jpg )

Image
Start right here

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

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  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
14 posts and 1 image reply omitted. Click here to view.
>>
  No. 751 >>754
Sorry if I'm late, the captcha is hard. If you go looking, you will find free wolfram alpha for android.
>>
  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

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

Image
  No. 465
Reply
>>466 >>478 >>529 >>1017
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.
>>
  No. 466 >>467
>>465
Cauase fractals are pretty.
>>
  No. 467
>>466
Thank you. Now, I get it.
>>
  No. 478

File: Screenshot_20230819-155401_Gallery~2.jpg ( 84.25 KB , 720x541 , 1692917903705.jpg )

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>>465
A fractal
>>
  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

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