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25 Dec 2021Mathchan is launched into public
Catalog Archive

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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
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

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  No. 749
Reply
>>948 >>1037
Im new.
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>>
  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
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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

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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

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Start right here

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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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  No. 798
Reply
>>834 >>862
Solve these problems if you're not a retard:

Let
⟨a⟩=(a1,a2,…)\langle a \rangle = (a_1, a_2, \ldots)
denote an infinite sequence of positive integers.
→\rightarrow
Prove that there is no
⟨a⟩\langle a \rangle
such that
gcd⁡(ai+j,aj+i)=1\gcd(a_i + j, a_j + i) = 1
for all
i≠ji \neq j
.
Let
p≠2p \neq 2
be a prime.
→\rightarrow
Prove that there is an
⟨a⟩\langle a \rangle
such that
gcd⁡(ai+j,aj+i)=p\gcd(a_i + j, a_j + i) = p
for all
i≠ji \neq j
.
>>
  No. 834 >>860
>>798
cool homework bro

1. if the problem didnt have the +i +j bit then the solutions would just be permutations of the primes
a things position on a list is set, so we can just subtract its number to be re added later
therefore, because the naturals will always overtake the primes, any valid list must contain negative numbers

2
prime infinity
never let rules as intended into your heart
>>
  No. 860 >>865
>>834

what the fuck
>>
  No. 862
>>798
Okay, what now.
>>
  No. 865
>>860
i wrote that at like 7 am
it had relative coherance at the time
>>
  No. 999
>> 798
(A) The problems suggests that
p=2p = 2
is the problem. If we consider
gcd⁡(a2i+2j,a2j+2i)=1\gcd(a_{2i} + 2j, a_{2j} + 2i) = 1
, we see that at least one of
a2ia_{2i}
,
a2ja_{2j}
has to be odd — say,
a2ia_{2i}
. Then
gcd⁡(a2i+2j+1,a2j+1+2i)=1\gcd(a_{2i} + 2j + 1, a_{2j + 1} + 2i) = 1
implies that
aia_i
is odd for odd
ii
. But then
gcd⁡(a2i+1+2j+1,a2j+1+2j+1)=1\gcd(a_{2i + 1} + 2j + 1, a_{2j + 1} + 2j + 1) = 1
is a contradiction since both sides are divisible by
22
.
(B) Just let
ai=pi−i+1>0a_i = pi - i + 1 > 0
, which gives
gcd⁡(pi−i+1+j,pj−j+1+i)=gcd⁡(p(i+j)+2,pj−j+1+i)\gcd(pi - i + 1 + j, pj - j + 1 + i) = \gcd(p(i + j) + 2, pj - j + 1 + i)
and
p∤p(i+j)+2p \nmid p(i + j) + 2
.

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  No. 400
Reply
>>401 >>402 >>404 >>489
Request and deliver any books or book recommendation in this thread.

OP starts:
>Algebraic Topology, Allen Hatcher
https://pi.math.cornell.edu/~hatcher/AT/AT+.pdf
You really shouldn't go with any other book. Stick with Hatcher, even when you feel lost.
9 posts and 2 image replies omitted. Click here to view.
>>
  No. 857
>>855
Thanks.
>>
  No. 858 >>873
>>852
NTA but while I really do like his treatment of analysis (i.e. chapter 3 and onwards), I think his introduction isn't that great, specifically chapter 2 (I actually think chapter 1 is great for getting used to basic set theory and learning to handle quantifiers). Especially the problems (e.g. proving Liouville's theorem) are far too difficult compared to what you got in chapter 1. For introductory stuff, I prefer Amann Escher.
>>
  No. 873 >>874
>>858
Test
>>
  No. 874
>>873
What are you testing?
>>
  No. 996
Does anyone know of any books that explain quotient vector space and filtration of nilpotent endomorphism?

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  No. 282
Reply
>>438
Hey /math/

I've been trying to understand spherical harmonics to grok an ML paper that represented points in euclidean space in a rotation-and-translation-invariant way (https://arxiv.org/pdf/1802.08219.pdf). I found a great textbook on SO(3) (https://www.diva-portal.org/smash/get/diva2:1334832/FULLTEXT01.pdf), but while I can kind of maybe sort of follow what's being done with them in this particular paper I fail to really get an intuition of what spherical harmonics are, and it feels like there's some pretty beautiful insight in there.

Do you have any advice or perspectives on how to intuitively grasp what these harmonics are and mean, beyond just group theory definitions?
>>
  No. 285
Harmonics are solutions of Laplace's equation in a given coordinate basis. Spherical harmonics are the solutions of Laplace's equation in spherical coordinates.
>>
  No. 438
>>282
https://youtu.be/Ziz7t1HHwBw
>>
  No. 893 >>983
fdsa
>>
  No. 982
fdsafas
>>
  No. 983
>>893
fdsafdsafdsa

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