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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?

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

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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

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

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  No. 1064
Reply
>>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.

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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?

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  No. 465
Reply
>>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

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

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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
>>
  No. 1037
>>749 hey me too

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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.

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  No. 199
Reply
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

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  No. 308
Reply
>>1015
Peak oil.

It's mathematically impossible for you to not die within a year.

I posted with the alias mustang on pol, you can see in archive.
>>
  No. 1015
>>308
I lived

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  No. 523
Reply
>>528
What do people make of these numbers?
>>
  No. 528 >>1014
>>523
seems to be a relation between numbers and their cubes
>>
  No. 1014
>>528
But the cube of 2 is like eight, and the cue of eight is nowhere near 2
512?
Where the fuck did 51 come from?

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