/sci/ - Science & Math » Thread #14052616

21KiB, 1152x726, circle-circle intersection.png

View Same Google iqdb SauceNAO

Anonymous Mon 03 Jan 19:33:39 2022 No.14052616 View Reply Original Report

Quoted By: >>14052757 >>14054090 >>14054123 >>14056166

Post some beautiful math formulas

Anonymous

View Same Google iqdb SauceNAO soddy circle formula.png, 109KiB, 1920x1556

Anonymous Mon 03 Jan 2022 19:38:04 No.14052635 Report

Quoted By:

Anonymous

Anonymous Mon 03 Jan 2022 19:59:33 No.14052684 Report

Quoted By:

digamma
$\psi \big( \frac {p} {q} \big) = \sum_{n=1}^{q-1} \left( \log \sin \frac {n \pi} {q} \cos \frac {2 n p \pi} {q} \right) - \frac {\pi} {2} \cot \frac {p \pi} {q} - \log 2 q - \gamma$

Anonymous

Anonymous Mon 03 Jan 2022 20:08:55 No.14052705 Report

Quoted By: >>14052730 >>14052763 >>14052792 >>14054129

$ 0.999... \neq 1 $

Anonymous

View Same Google iqdb SauceNAO Chudnovsky.jpg, 41KiB, 1200x646

Anonymous Mon 03 Jan 2022 20:11:27 No.14052718 Report

Quoted By:

Anonymous

View Same Google iqdb SauceNAO CrmGtIUWIAAAoqg.jpg, 55KiB, 957x621

Anonymous Mon 03 Jan 2022 20:14:20 No.14052730 Report

Quoted By: >>14052792

>>14052705

Anonymous

View Same Google iqdb SauceNAO 1637851636071.png, 14KiB, 396x362

Anonymous Mon 03 Jan 2022 20:24:25 No.14052757 Report

Quoted By:

>>14052616
Let x be the angle from a chord's endpoints to the center of a circle for a chord that divides the area of the circle into fraction F

x - sin(x) = 2 pi F

Now for two intersecting chords with angles y and z whose intersection has angle x and fraction of area F

cos(z/2) cos(x-y/2) csc(x-y/2-z/2) (1-cos(y/2) sec(x-y/2))^2-sin(x) cos(y/2) sec(x-y/2)+x = 2 pi F

Anonymous

Anonymous Mon 03 Jan 2022 20:27:03 No.14052763 Report

Quoted By: >>14052792

>>14052705
proof?

Anonymous

View Same Google iqdb SauceNAO zeta2.png, 5KiB, 400x300

Anonymous Mon 03 Jan 2022 20:29:49 No.14052773 Report

Quoted By:

Anonymous

Anonymous Mon 03 Jan 2022 20:33:34 No.14052792 Report

Quoted By:

>>14052705
>>14052730
>>14052763
take your off topic banter to the containment thread >>14050848 or get b& faggots

Anonymous

Anonymous Tue 04 Jan 2022 02:07:50 No.14054090 Report

Quoted By: >>14054122

>>14052616
Math formulas are generally the ugliest part of math, the most beautiful are the proofs.

Anonymous

Anonymous Tue 04 Jan 2022 02:19:54 No.14054122 Report

Quoted By:

>>14054090
Hilbert class formula

Anonymous

Anonymous Tue 04 Jan 2022 02:20:12 No.14054123 Report

Quoted By:

>>14052616
$ \delta \, \epsilon \left ( 0,1 \right ) \\ \displaystyle \prod_{k=0}^{n} \left ( 1 + \delta ^{2^{k}} \right ) = (1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ (1- \delta) \prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = (1- \delta)(1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{2})(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{4})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) = (1- \delta ^{2^{n}})(1+ \delta ^{2^{n}}) = 1- \delta ^{2^{n+1}} \\ \\ \boxed{(\delta ^{2^n})^2 = \delta ^{2 \cdot 2^n}=\delta^{2^{n+1}}} \\ \displaystyle \prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = \dfrac{1- \delta^{2^{n+1}}}{1- \delta} \\ \displaystyle \lim_{n \to \infty} \prod_{k=0}^{n} \left (1+ \delta ^{2^{k}} \right ) = \lim_{n \to \infty} \dfrac{1- \delta^{2^{n+1}}}{1- \delta} = \dfrac{1}{1- \delta} $

Anonymous

Anonymous Tue 04 Jan 2022 02:22:10 No.14054129 Report

Quoted By:

>>14052705
1/inf=0

El Arcon

View Same Google iqdb SauceNAO TIMESAND___FFFkjo727658275275001 (...).png, 22KiB, 1363x165

El Arcon Tue 04 Jan 2022 02:25:36 No.14054140 Report

Quoted By: >>14054153

Anonymous

Anonymous Tue 04 Jan 2022 02:30:29 No.14054153 Report

Quoted By:

>>14054140
infinity is not a number

Anonymous

View Same Google iqdb SauceNAO 5d35ce6963bf535d6d25691ba3c235b9 (...).jpg, 89KiB, 1055x818

Anonymous Tue 04 Jan 2022 02:44:02 No.14054192 Report

Quoted By:

Assuming Goldbach, let $\langle 2n \rangle := \{ (p, q)\in \mathbb{P}^2\ |\ p+q=n\}, n\ge 2$ , where $\mathbb{P}$ is the set of all primes. Define $\sigma\colon \mathbb{P}^2 \to \mathbb{N}, (p, q)\mapsto p+q$ . Then, for any $N\in \mathbb{N}, N\ge 2$ , we have $1+2+\cdots+ N = 1+ N + \frac{ (\frac{\sigma(p, q)}{2} +1)(\frac{\sigma(p, q)}{2} -2)}{2}$ , where $(p, q) \in \langle 2N\rangle$ is an arbitrary pair. $(\ ^\circ\smile ^\circ)$

Anonymous

Anonymous Tue 04 Jan 2022 15:59:41 No.14056166 Report

Quoted By:

>>14052616
f(x) = x

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches