Pi Approximations

Author

Eric W. Weisstein
July 15, 2023

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Constants/PiApproximations.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/PiApproximations.html.

©2023 Wolfram Research, Inc. except for portions noted otherwise

Pi

N[Pi,20]

Simple

Kochansky

NRoots[9x^4-240x^2+1492==0,x]

CorrectDigits[{Pi,Sqrt[40/3-Sqrt[12]]}]

Radicals

CorrectDigits[{Pi,(12/7) (-1+2 (Sqrt[2]))}]

CorrectDigits[{Pi,((24 (Sqrt[2-(Sqrt[2])]))/(4+(Sqrt[2+(Sqrt[2])])))}]

Golden ratio

CorrectDigits[{(6/5)GoldenRatio^2,Pi}]

N[(6/5)GoldenRatio^2]

N[Pi]

CorrectDigits[{3.141592653589793` ,3.141640786499874`}]

Mircea-Mugurel

Date: Wed, 30 Oct 2002 11:50:40 +0200
From: Serban Mircea-Mugurel <mugurserban@hotmail.com>

CorrectDigits[{Pi,4/Sqrt[GoldenRatio]}]

CorrectDigits[{Pi,4/Sqrt[GoldenRatio]}]

Canonical tetragonal antiwedge

Connection with ϕ noted by E. Weisstein, Apr. 19, 2022

N[(8/3)GoldenRatio(2-(GoldenRatio^(1/2))),20]

FunctionExpand[(8/3)GoldenRatio(2-(GoldenRatio^(1/2)))]//FullSimplify

Pandigital

Astle

N[3+(4/28)-(1/(790+(5/6))),20]

CorrectDigits[{Pi,3+(4/28)-(1/(790+(5/6)))}]

Barbosa

((log((((2 5!+(8-1)!))^(Sqrt[9]))+4!+(3!)!))/(Sqrt[67]))

N[Log[((8-1)!+2*5!)^Sqrt[9]+(3!)!+4!]/Sqrt[67]-π,10]

Log[((8-1)!+2*5!)^Sqrt[9]+(3!)!+4!]/Sqrt[67]

CorrectDigits[Log[((8-1)!+2*5!)^Sqrt[9]+(3!)!+4!]/Sqrt[67],π]

N[((Log[(((2 5!+(8-1)!))^(Sqrt[9]))])/(Sqrt[67]))-π,10]

CorrectDigits[((Log[(((2 5!+(8-1)!))^(Sqrt[9]))])/(Sqrt[67])),Pi]

Irvine

Plouffe e-mail 2000-04-17

irv=0+Sqrt[Sqrt[3^4+19^2/(78-56)]]

N[irv,20]

CorrectDigits[{irv,Pi}]

Pegg

((1-(1/(((9-(1/(8^5))))^6)))/(7+(1/(2^4))))+3+0

pegg=0+3+(1-(9-8^(-5))^(-6))/(7+2^(-4))

N[%,20]

CorrectDigits[pegg,Pi]

(22/17+37/47+88/83)

Dec 29, 2002

CorrectDigits[Pi,(22/17+37/47+88/83)]

Rational

Povolotsky Aug. 29, 2022

Sort[RealDigits[#,10,10][[1]]]&/@{6897253140,2195463870,FromDigits[Range[9]]}//Column

Equal@@%

N[(6897253140/2195463870),20]

N[Pi,20]

Exhaustive search

Eric Weisstein, Sep. 7, 2022

len=Length[p=FromDigits/@Permutations[Range[0,9]]]

nf=Nearest[p]

nf[Pi 2195463870]

good=DeleteCases[{#,nf[Pi #][[1]]}&/@p,{_,{}}];//Timing

Take[SortBy[{#,Abs[1./Divide@@#-Pi]}&/@good,Last],10]

N[(8405139762/2675439081),20]

N[Pi,20]

RealDigits[(8405139762/2675439081)-Pi,10,10]

Ziv

-((((((.3)^9)/7)))^((.8)^(.1)))+(2^(5^(.4)))-.6

2 ^ (5 ^ .4) - .6 - ((.3 ^ 9) / 7) ^ (.8 ^ .1)

CorrectDigits[Pi,ziv=2 ^ (5 ^ (4/10)) - 6/10 - (((3/10) ^ 9) / 7) ^ ((8/10) ^(1/10))]

N[{Pi,ziv},14]

Misc.

24 ((1/12)-(1/(28 (2^7)))-(1/(72 (2^9)))-(1/(5 (2^5))))+((3 (Sqrt[3]))/4)==π??

3Sqrt[3]/4+24(1/12-1/5/2^5-1/28/2^7-1/72/2^9)

Castellanos

castel={
	(2E^3+E^8)^(1/7),
	(553/( 311+1))^2,
	(3/14)^4(193/5)^2,
	(296/167)^2,
	((66^3+86^2)/ 55^3)^2,
	 1.09999901* 1.19999911* 1.39999931* 1.69999961, 
	(47^3+20^3)/ 30^3-1,
	 2+Sqrt[1+(413/750)^2],
	(77729/254)^(1/5),
	(31+(62^2+14)/28^4)^(1/3),
	(1700^3+82^3-10^3-9^3-6^3-3^3)/ 69^5,
	(95+(93^4+34^4+17^4+88)/ 75^4)^(1/4),
	(100-(2125^3+214^3+30^3+37^2)/82^5)^(1/4)
};
Table[CorrectDigits[{Pi,castel[[i]]}],{i,Length[castel]}]

de Jerphanion

CorrectDigits[{Pi,Log[23+1/22+2/21]}]

FullSimplify[Log[23+1/22+2/21]]

FullSimplify[Log[23+1/6-2/77]]

Deskio

From desio@katamail.com Thu Aug 17 08:46:07 2000
Date:17 Aug 2000 15:40:17+0200
From:desio@katamail.com
To:mathworld@wolfram.com

pi=(101^2+((241*8*13)^2+19*5*2+(2/3)^2)/(179^2-3))^(1/9)

FullSimplify[pi]

CorrectDigits[{pi,Pi}]

N[pi-Pi,20]

ToRadicals[%]

(Expand[(1074401915/16019)((2/3))^2])

Now I have a new formula for pi which is an approximation to the 15-th digit. And 
of pi.	      
Now let be C=B^2 + 19*5*2 + ((2/3)^2)*(793*5*2/89^2)).	      
Well	 
pi=(101^2 + (C/A))^(1/9))=3.(141592653589793)1...    
As you can see in the formula there are 793 and 89. You find these digits in pi.

c=(241*8*13)^2+19*5*2+((2/3)^2)*(793*5*2/89^2);
a=(179^2-3);
pi=(101^2+(c/a))^(1/9)

CorrectDigits[{pi,Pi}]

RootReduce[pi]

N[(34041350274878/1141978491)^(1/9)-Pi,20]

Golden Ratio

CorrectDigits[{Pi,6/5 GoldenRatio^2}]

Gosper

Gosper[n_]:=Module[{},
  f[k_]:=k*2^(-k/4)*NSum[2^-(k*Binomial[n,2]),{n,1,Infinity}]^2;
  g[0]=f[n]/f[2n];
  g[k_]:=Sqrt[1/2(g[k-1]+1/g[k-1])];
  f[n] 2 Log[2] g[1]g[2]g[3]g[4]
]

Do[Print[n,". ",N[Gosper[n],20]],{n,1,5}]

{ArcSin[x]/Sqrt[1-x^2] /. x->1/2,Sum[(i!)^2/(2i+1)!,{i,0,Infinity}]/2}

Hitchhiker's Guide Mini-Course, JMM 2010

Sin[(111111.^(-1)) Degree]/(π/2)

Sin[(222222.^(-1)) Degree]/(π/4)

Sin[(333333.^(-1)) Degree]/(π/6)

Sin[(444444.^(-1)) Degree]/(π/8)

Sin[(555555.^(-1)) Degree]/π

N[Sin[(555555^(-1)) Degree]-π (10^(-8)),20]

N[Pi,20]

Sin[(555555.^(-1)) Degree]

Sin[(666666.^(-1)) Degree]/(π/12)

Sin[(777777.^(-1)) Degree]/(π/14)

Sin[(888888.^(-1)) Degree]/(π/16)

Sin[(999999.^(-1)) Degree]/(π/18)

With[{d=5,k=6},d((10^k-1)/9)]

With[{d=2,n=4},{Sin[(π/(180 d((10^n-1)/9)))],(π/(2d))(10^(-n-1))}]//N

Series[Sin[(π/(180 d((10^n-1)/9)))],{n,∞,2}]

Hoffman

David W. Hoffman

CorrectDigits[(((((10^100)/11222.11122)))^(1/193)),π]

Irvine

Plouffe e-mail 2000-04-17

9^2+19^2/22

102-2222/22^2

97+1/2-1/11

97+9/22

Iuliano

iul=1/((3 123449)^(-1/2)+57/180)

N[{iul,Pi},17]

CorrectDigits[{iul,Pi}]

Joseph

π==(log^((log^((log^((log^(log(2)))(3)))(4)))(5)))(6)

CorrectDigits[π,Log[6]^Log[5]^Log[4]^Log[3]^Log[2]]

May 3, 2006

N[((Log[6])^((Log[5])^((Log[4])^((Log[3])^(Log[2]))))),10]

(31^(1/3))

CorrectDigits[(31^(1/3)),π]

Leppänen

Jul 15, 2023

4488*^-4

With[{x=(561/1250)},(((345666+x))^(x/5))]//FullSimplify

N[%,20]

CorrectDigits[With[{x=(561/1250)},(((345666+x))^(x/5))],π]

Lindborg

(314+142)/(2*3*5*7)*1373/(13*73)

FactorInteger[(314+142)]

FactorInteger[(104348/33215)]

Through[{Numerator,Denominator}[%]]

CorrectDigits[(314+142)/(2*3*5*7)*1373/(13*73),π]

Fernando Mancebo

Apr 29, 2012

With[{p=8},(((p(((p+2))^(2p))))^(1/(4p+2)))]

N[%,20]

N[Pi,20]

CorrectDigits[{Pi,(2^(19/34)) (5^(8/17))}]

Misc

CorrectDigits[{Pi,Log[20+Pi]}]

Plouffe

plouffe={
43^(7/23),
Log[2198]/Sqrt[6],
(13/4)^(1181/1216),
689/396/Log[689/396],
(2143/22)^(1/4),
Log[5280]/Sqrt[67/9],
(63023/30510)^(1/3)+1/4+1/2(Sqrt[5]+1),
Log[60318/13387]48/23,
(228+16/1329)^(1/41)+2,
Log[28102/1277]*125/123,
(3/(Sqrt[163]))Log[640320],
(276694819753963/226588)^(1/158)+2,
Log[640320^3+744]/Sqrt[163]
};
Table[CorrectDigits[{Pi,plouffe[[i]]}],{i,Length[plouffe]}]

640320^3+744

Povolotsky

(Sqrt[4 e-1])

CorrectDigits[Sqrt[4E-1],Pi]

N[Sqrt[4E-1],10]

Solve[(E-1)/((pi^2)/3-1)==3/4,pi]

Ramanujan

99^2/1103/2/Sqrt[2]

N[%,20]

N[Pi,20]

CorrectDigits[{%,Pi}]

ramanujan={
	19Sqrt[7]/ 16,
	7/3(1+1/5Sqrt[3]) ,
	9/5+Sqrt[9/5],
	(9^2+19^2/ 22)^(1/4),
	(97+1/2 -1/11)^(1/4),
	(63/25)(17+15Sqrt[5])/( 7+15Sqrt[5]),
	355/113(1-0.0003/ 3533),
	12/Sqrt[130] Log[(3+Sqrt[13])(Sqrt[8]+Sqrt[10])/ 2],
	24/Sqrt[142]Log[(Sqrt[10+11Sqrt[2]]+Sqrt[10+7Sqrt[2]])/ 2],
	12/Sqrt[190]Log[(3+Sqrt[10])(Sqrt[8]+Sqrt[10])],
	12/Sqrt[310]Log[1/4 (3+Sqrt[5])(2+Sqrt[2])(5+2Sqrt[10]+Sqrt[61+20Sqrt[10]])],
	4/Sqrt[522]Log[((5+Sqrt[29])/Sqrt[2])^3(5Sqrt[29]+11Sqrt[6])*
		(Sqrt[(9+3Sqrt[6])/4]+Sqrt[(5+3Sqrt[6])/ 4])^6]
};
Table[CorrectDigits[{Pi,ramanujan[[i]]}],{i,Length[ramanujan]}]

Rashid

CorrectDigits[(((((802GoldenRatio-801)/(602GoldenRatio-601))))^4),π]

Schneider

π==(Sqrt[7+(Sqrt[6+(Sqrt[5])])])

CorrectDigits[π,(Sqrt[7+(Sqrt[6+(Sqrt[5])])])]

von Hippel

(e^(e^(e^(-2))))

CorrectDigits[Exp[Exp[Exp[-2]]],π]

Exp[Exp[Exp[-2]]]//N

(e^(e^(-2)))+2

CorrectDigits[2+Exp[Exp[-2]],π]

2+Exp[Exp[-2]]//N

Voormanns

12 Dec 2003

(1/7) (((13 y-6 7)/(13 y))+3 7)

Function[y,N[(3 7+(13y-6 7)/(13 y))/7,20]]/@{365,365.2425`7,365.242190`9}

CorrectDigits[{Pi,#}]&/@Function[y,N[(3 7+(13y-6 7)/(13 y))/7,20]]/@{365,365.2425`7,365.242190`9}

Warda

15 Nov 2004

Formula 1

CorrectDigits[{Pi,Log[(640320^3+744)^2 - 2 196884]/(2Sqrt[163])}]

Log[(640320^3+744)^2 - 393768]

With[{qbar=-Exp[-pi(Sqrt[163])]},{(1/qbar)+744+196884qbar,(-640320)^3}]

(E^((Sqrt[163]) pi))(1+196884 (E^(-2pi(Sqrt[163])))) ≈ 640320^3+744

(E^(2(Sqrt[163]) pi))(1+196884 (E^(-2pi(Sqrt[163]))))^2≈(640320^3+744)^2
(E^(2(Sqrt[163]) pi))+2 196884 ≈(640320^3+744)^2

(E^(2(Sqrt[163]) pi)) ≈(640320^3+744)^2-2 196884

π ≈Log[(640320^3+744)^2 - 2 196884]/(2Sqrt[163])

Solve[Equal@@With[{qbar=-Exp[-pi(Sqrt[163])]},{(1/qbar)+744+196884qbar,(-640320)^3}],pi]

((Log[6 (21878117720064062+25 (Sqrt[765843255956769776567618214295]))])/(Sqrt[163]))//InputForm

Block[{$MaxExtraPrecision=200},N[Subtract@@{((Log[6 (21878117720064062+25 (Sqrt[765843255956769776567618214295]))])/(Sqrt[163])),Log[(640320^3+744)^2 - 393768]/(2Sqrt[163])},100]]

$Version

expr={Log[6(21878117720064062+25Sqrt[765843255956769776567618214295])]/Sqrt[163],Pi};

RealDigits[Subtract@@{Log[6*(21878117720064062+25*Sqrt[765843255956769776567618214295])]/Sqrt[163],Pi},10,1000]

CorrectDigits[{Log[6*(21878117720064062+25*Sqrt[765843255956769776567618214295])]/Sqrt[163],Pi},100]

N[Log[6*(21878117720064062+25*Sqrt[765843255956769776567618214295])]/Sqrt[163]-Pi,200]

N[Subtract@@{Log[Ceiling[E^(Pi*Sqrt[163])]^2 - 393768]/(Sqrt[163]*2),((Log[6 (21878117720064062+25 (Sqrt[765843255956769776567618214295]))])/(Sqrt[163]))},100]

Formula 1

Block[{$MaxExtraPrecision=10^4},
CorrectDigits[{#,Pi},2000]&/@Table[((Log[Round[Exp[Pi Sqrt[163n]]]])/(Sqrt[163n])),{n,10}]
]

The -2000s are a bug.

Block[{$MaxExtraPrecision=1000},
Position[Table[RealDigits[((Log[Round[Exp[Pi Sqrt[163n]]]])/(Sqrt[163n]))-Pi,10,10],{n,100}],_RealDigits]//Flatten
]

Bug

CorrectDigits/@Block[{$MaxExtraPrecision=1000},
Table[((Log[Round[Exp[Pi Sqrt[163n]]]])/(Sqrt[163n]))-Pi,{n,100}]//N[#,1000]&
]

SetAttributes[f1,NumericFunction]

f1[n_?InexactNumberQ]:=Log[Abs[((Log[Round[Exp[Pi Sqrt[163n]]]])/(Sqrt[163n]))-Pi]]
f2[n_]:=N[f1[SetPrecision[n,Infinity]],$MachinePrecision+1]

Block[{$MaxExtraPrecision=1000},
Plot[f2[n],{n,1,10},PlotStyle->Red,PlotPoints->100,PlotRange->All]
]//Timing

<<Utilities`Plot`

IntegerListPlot[%158,AxesLabel->TraditionalForm/@{n,d[n]}]