[OOC] Genkai Warriors Academy [OPEN TO NEW CHARACTERS]

[TheSecret]

actually this world is made of rock containing silicon, iron, magnesium, aluminum, oxygen and other minerals. the crust is made of silicon, aluminum, iron, calcium, sodium, potassium and magnesium.

@LowKey123

NEVER WONDERED WHAT KEEPS ALL THAT SH-T TOGETHER!? IT'S THE POWER OF YURI!!!

[KaiserAuto]

100 POSTS! WOOOW

GOOOOOODDDD DAMNNITTTT SEEECCREEETT!

[nichinichisou]

<Snipped quote by LowKey123>

NEVER WONDERED WHAT KEEPS ALL THAT SH-T TOGETHER!? IT'S THE POWER OF YURI!!!

@TheSecret

no

gravity does

[KaiserAuto]

TheSecret ruined m'lyfe

@LowKey123
save me.

[nichinichisou]

TheSecret ruined m'lyfe

@LowKey123
save me.

@KaiserAuto

ill math him to death

Proofs of the famous mathematical result that the rational number 22/7 is greater than π (pi) date back to antiquity. One of these proofs, more recently developed but requiring only elementary techniques from calculus, has attracted attention in modern mathematics due to its mathematical elegance and its connections to the theory of diophantine approximations. Stephen Lucas calls this proof, “One of the more beautiful results related to approximating π”.[1] Julian Havil ends a discussion of continued fraction approximations of π with the result, describing it as “impossible to resist mentioning” in that context.[2]

The purpose of the proof is not primarily to convince its readers that 22/7 (or 31⁄7) is indeed bigger than π; systematic methods of computing the value of π exist. If one knows that π is approximately 3.14159, then it trivially follows that π < 22/7, which is approximately 3.142857. But it takes much less work to show that π < 22/7 by the method used in this proof than to show that π is approximately 3.14159.

Contents [hide]
1 Background
2 The proof
3 Details of evaluation of the integral
4 Quick upper and lower bounds
5 Proof that 355/113 exceeds π
6 Extensions
7 See also
8 References
9 External links
Background[edit]
22/7 is a widely used Diophantine approximation of π. It is a convergent in the simple continued fraction expansion of π. It is greater than π, as can be readily seen in the decimal expansions of these values:

\begin{align}
\frac{22}{7} & = 3. \overline{142\,857}, \\
\pi\, & = 3.141 \,592\,65\ldots
\end{align}
The approximation has been known since antiquity. Archimedes wrote the first known proof that 22/7 is an overestimate in the 3rd century BCE, although he may not have been the first to use that approximation. His proof proceeds by showing that 22/7 is greater than the ratio of the perimeter of a circumscribed regular polygon with 96 sides to the diameter of the circle. Another rational approximation of π that is far more accurate is 355/113.

The proof[edit]
The proof can be expressed very succinctly:

0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi.
Therefore 22/7 > π.

The evaluation of this integral was the first problem in the 1968 Putnam Competition.[3] It is easier than most Putnam Competition problems, but the competition often features seemingly obscure problems that turn out to refer to something very familiar. This integral also has been used in the entrance examinations for the Indian Institutes of Technology.[4]

Details of evaluation of the integral[edit]
That the integral is positive follows from the fact that the integrand is a quotient whose numerator and denominator are both non-negative, being sums or products of powers of non-negative real numbers. Since the integrand is positive, the integral from 0 to 1 is positive because the lower limit of integration is less than the upper limit of integration.

It remains to show that the integral in fact evaluates to the desired quantity:

\begin{align}
0 & < \int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx \\[8pt]
& = \int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\,dx\quad\text{(expansion of terms in the numerator)} \\[8pt]
& = \int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right) \,dx \\
& {} \qquad \text{(polynomial long division)} \\[8pt]
& = \left.\left(\frac{x^7}{7}-\frac{2x^6}{3}+ x^5- \frac{4x^3}{3}+4x-4\arctan{x}\right)\,\right|_0^1 \quad \text{(definite integration)} \\[6pt]
& = \frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\quad (\text{since }\arctan(1) = \pi/4 \text{ and } \arctan(0) = 0) \\[8pt]
& = \frac{22}{7}-\pi. \quad \text{(addition)}
\end{align}
(See polynomial long division.)

Quick upper and lower bounds[edit]
In Dalzell (1944), it is pointed out that if 1 is substituted for x in the denominator, one gets a lower bound on the integral, and if 0 is substituted for x in the denominator, one gets an upper bound:[5]

\frac{1}{1260} = \int_0^1\frac{x^4 (1-x)^4}{2}\,dx < \int_0^1\frac{x^4 (1-x)^4}{1+x^2}\,dx < \int_0^1\frac{x^4 (1-x)^4}{1}\,dx = {1 \over 630}.
Thus we have

{22 \over 7} - {1 \over 630} < \pi < {22 \over 7} - {1 \over 1260},
hence 3.1412 < π < 3.1421 in decimal expansion. The bounds deviate by less than 0.015% from π. See also Dalzell (1971).[6]

Proof that 355/113 exceeds π[edit]
As discussed in Lucas (2005), the well-known Diophantine approximation and far better upper estimate 355/113 for π follows from the relation

0<\int_0^1\frac{x^8(1-x)^8(25+816x^2)}{3164(1+x^2)}\,dx=\frac{355}{113}-\pi.
Note that

\frac{355}{113}= 3.141\,592\,92\ldots,
where the first six digits after the period agree with those of π. Substituting 1 for x in the denominator, we get the lower bound

\int_0^1\frac{x^8(1-x)^8(25+816x^2)}{6328}\,dx =\frac{911}{5\,261\,111\,856} = 0.000\,000\,173\ldots,
substituting 0 for x in the denominator, we get twice this value as an upper bound, hence

\frac{355}{113}-\frac{911}{2\,630\,555\,928}<\pi<\frac{355}{113}-\frac{911}{5\,261\,111\,856}\,.
In decimal expansion, this means 3.141 592 57 < π < 3.141 592 74, where the bold digits of the lower and upper bound are those of π.

Extensions[edit]
The above ideas can be generalized to get better approximations of π; see also Backhouse (1995)[7] and Lucas (2005) (in both references, however, no calculations are given). For explicit calculations, consider, for every integer n ≥ 1,

\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
<\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx
<\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\,dx,
where the middle integral evaluates to

\begin{align}
&\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx\\
&\qquad=\sum_{j=0}^{2n-1}\frac{(-1)^j}{2^{2n-j-2}(8n-j-1)\binom{8n-j-2}{4n+j}}
+(-1)^n\biggl(\pi-4\sum_{j=0}^{3n-1}\frac{(-1)^j}{2j+1}\biggr)
\end{align}
involving π. The last sum also appears in Leibniz' formula for π. The correction term and error bound is given by

\begin{align}\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\,dx
&=\frac{1}{2^{2n-1}(8n+1)\binom{8n}{4n}}\\
&\sim\frac{\sqrt{\pi n}}{2^{10n-2}(8n+1)},
\end{align}
where the approximation (the tilde means that the quotient of both sides tends to one for large n) of the central binomial coefficient follows from Stirling's formula and shows the fast convergence of the integrals to π.

Calculation of these integrals[show]
The results for n = 1 are given above. For n = 2 we get

\frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\,dx=\pi -\frac{47\,171}{15\,015}
and

\frac18\int_0^1 x^8(1-x)^8\,dx=\frac1{1\,750\,320},
hence 3.141 592 31 < π < 3.141 592 89, where the bold digits of the lower and upper bound are those of π. Similarly for n = 3,

\frac1{16}\int_0^1\frac{x^{12}(1-x)^{12}}{1+x^2}\,dx= \frac{431\,302\,721}{137\,287\,920}-\pi
with correction term and error bound

\frac1{32}\int_0^1 x^{12}(1-x)^{12}\,dx=\frac1{2\,163\,324\,800},
hence 3.141 592 653 40 < π < 3.141 592 653 87. The next step for n = 4 is

\frac1{64}\int_0^1\frac{x^{16}(1-x)^{16}}{1+x^2}\,dx= \pi-\frac{741\,269\,838\,109}{235\,953\,517\,800}
with

\frac1{128}\int_0^1 x^{16}(1-x)^{16}\,dx=\frac1{2\,538\,963\,567\,360},
which gives 3.141 592 653 589 55 < π < 3.141 592 653 589 96.

See also[edit]
Approximations of π
Chronology of computation of π
Proof that π is irrational
Lindemann–Weierstrass theorem (proof that π is transcendental)
List of topics related to π
References[edit]
Jump up ^ Lucas, Stephen (2005), "Integral proofs that 355/113 > π" (PDF), Australian Mathematical Society Gazette 32 (4): 263–266, MR 2176249, Zbl 1181.11077
Jump up ^ Havil, Julian (2003), Gamma. Exploring Euler's Constant, Princeton, NJ: Princeton University Press, p. 96, ISBN 0-691-09983-9, MR 1968276, Zbl 1023.11001
Jump up ^ Alexanderson, Gerald L.; Klosinski, Leonard F.; Larson, Loren C. (editors) (1985), The William Lowell Putnam Mathematical Competition: Problems and Solutions: 1965–1984, Washington, D.C.: The Mathematical Association of America, ISBN 0-88385-463-5, Zbl 0584.00003
Jump up ^ 2010 IIT Joint Entrance Exam, question 38 on page 15 of the mathematics section.
Jump up ^ Dalzell, D. P. (1944), "On 22/7", Journal of the London Mathematical Society 19: 133–134, doi:10.1112/jlms/19.75_part_3.133, MR 0013425, Zbl 0060.15306.
Jump up ^ Dalzell, D. P. (1971), "On 22/7 and 355/113", Eureka; the Archimedeans' Journal 34: 10–13, ISSN 0071-2248.
Jump up ^ Backhouse, Nigel (July 1995), "Note 79.36, Pancake functions and approximations to π", The Mathematical Gazette 79 (485): 371–374, JSTOR 3618318

[KaiserAuto]

<Snipped quote by KaiserAuto>

ill math him to death
<Snipped quote>

@LowKey123

Ah, math is refreshing me~

This translates into a very ancient and important question:
How much smut will be allowed in the IC?!?!?!?!?!?!?!?!?

[nichinichisou]

<Snipped quote by LowKey123>

Ah, math is refreshing me~

This translates into a very ancient and important question:
How much smut will be allowed in the IC?!?!?!?!?!?!?!?!?

@KaiserAuto

let's see
kissing, hugging and other such things can be in IC, but any explicit sexual content is either fade to black or PMs

[KaiserAuto]

<Snipped quote by KaiserAuto>

let's see
kissing, hugging and other such things can be in IC, but any explicit sexual content is either fade to black or PMs

@LowKey123

Great response, as per guild regulations.

Just for the sake of asking, will you participate in the P.E. class, LowKey?

[nichinichisou]

<Snipped quote by LowKey123>

Great response, as per guild regulations.

Just for the sake of asking, will you participate in the P.E. class, LowKey?

@KaiserAuto

Of course I will. Just need a bit of inspiration to get posting. I don't really know how to jump in that well.

anti-freshman gangs

>Janette is freshman
shouldn't they be hating her by now

[Slendy]

o3o Could Slendy Join if there's enough space?
@KaiserAuto

[nichinichisou]

o3o Could Slendy Join if there's enough space?
@KaiserAuto

@Slendy

hey Slendy!

the title says [OPEN TO NEW CHARACTERS], just put a CS up and wait for Kaiser to get back on

edit: AND AS SOON AS I SAY THAT HE IS BACK ON

[KaiserAuto]

@LowKey123@Slendy

The dude has...
Well, interesting ideas in mind and since the gang name is pathetic you shouldn't expect too much from this side-gang, lol!

This is a roleplay which'll focus slightly on the comedic side of things as well.

You're greatly welcomed, just make a CS and I'll inspect it closely- after all 100% acceptance rates currently.

[nichinichisou]

@LowKey123@Slendy

interesting ideas in mind

@KaiserAuto

INTERESTING IDEAS HMM
INNNNNNNNTERESTING IDEAS
SWIGGITY SWOOGITY

[Slendy]

*face palms* ._. Duhh Slendy!!! It said open to new characters in caps and I still didn't pay it any attention. I'll have the character done in a few.

[KaiserAuto]

Take your time Slendy~

[nichinichisou]

HOW2INTRODUCEMYCHARACTERHELP
i'll just force Slendy to collab with me once he finishes his character

[karamonnom]

@LowKey123
LOL maybe you should wait for the fighting to start, rush in late, and just start randomly beating people up

Edit: or just do that.

[KaiserAuto]

HOW2INTRODUCEMYCHARACTERHELP

@LowKey123

spooky my character up

Alternatively I could create some sort of scenario for you if you give me something to work with. You could do the way that the sugoi-model-superkawaii-fighter did it but you could also discuss with me a bit and we could thought storm.

@Karamonnom I will stop there until tomorrow morning with that character as I want everyone to have the chance to enter the P.E. class.

Edit:

or just do that.

@karamonnom

[KaiserAuto]

Alternatively I'll wait for Slendy to make his character and intoruduce it and I'll advance the P.E. plot.

@LowKey123@Slendy
You guys are forced by my authority as the GM, a.k.a. the Judge-inator to attend the P.E. plot.

[KaiserAuto]

ONE HUNDREEEDDD ERMAHGERDDDD