A pi and e equation challenge.
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From: Danny73 on 24 Jul 2010 23:44 On Jul 24, 5:35 pm, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote: > On 24 July, 17:35, Danny73 <fasttrac...(a)att.net> wrote: > > > > > > > On Jul 24, 10:30 am, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> > > wrote: > > > > On 24 July, 14:19, Danny73 <fasttrac...(a)att.net> wrote: > > > > > (3) separate equations using pi and e where the > > > > results are close too the value of pi > > > > (correct too 6 decimal digits) > > > > > (((e^3/pi^2)^2)-1) = 3.141592835... > > > > (slightly greater than pi) > > > > > (1/((e^6/pi^5)-1)) = 3.1415920835... > > > > (slightly less than pi) > > > > > (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912... > > > > (slightly greater than pi) > > > > > The challenge -- > > > > > Are more results possible that give pi > > > > correct too 6 decimal places or more > > > > using only pi and e and their powers in each > > > > equation but with slightly different results > > > > from above. > > > > > The criteria for each equation involves pi and e > > > > and using any or all of these operators (-,+,*,/ and ^n) > > > > giving 6 or more correct decimal places of pi in the > > > > result. > > > > You seem to be finding solutions to > > > say > > > > A1*pi^5 +A2*pi^4 +A3*e^6 a.e. 0 > > > > A1*pi = -A2 + (A3/A2)(e^3/pi^2)^2) > > > > with A1 =A2=A3 =1 > > > > would pi = (-A2/A1) +(A3/(A2*A1))e^3/pi^2)^2) > > > > also be acceptable ?- Hide quoted text - > > > > - Show quoted text - > > > Sure if you have the actual integer values for a1,a2 and a3 > > depicted for each equation. > > > Thanks for your input. > > A simple method of finding integer relations between > irrationals is to write the irrationals to a large > number of decimal places in an nx1 column vector with the identity > matrix beside it. > > 1 0 0 306.0196848 = pi^5 = r1 > 0 1 0 97.40909103 = pi^4 = r2 > 0 0 1 403.4287935 = e^5 = r3 > > As r1 >r3 -r1 = 97.4091087 >0 take row 1 from row 3 > of the matrix > > 1 0 0 306.0196848 = r1 > 0 1 0 97.40909103 = r2 > -1 0 1 97.4091087 = r3 > > As r3 is only slightly greater than r2 > take row2 from row 3 > > 1 0 0 306.0196848 = r1 > 0 1 0 97.40909103 = r2 > -1 -1 1 0.00001717 = r3 > > This means that > > (-1,-1,1).(pi^5,pi^4, e^6) = 0.0000172 > > -pi^5 -pi^4 + e^6 = 0.0000172 > > pi +1 -(e^6/pi^4) = -0.0000172/pi^4 > > pi + 0.0000172/pi^4 = (e^6/pi^4)-1 > > Obviously, you can reduce the values in the column > vector as close as you want to 0, but the absolute vales > in the matrix rows will grow in size correspondingly. > > There is no guarantee that a1,a2,a3 will be "small" > integers.- Hide quoted text - > > - Show quoted text - Also another interesting observation. Am I wrong in assuming that it will greatly improve decimal digit accuracy for pi with large a1,a2 and a3? Dan
From: sttscitrans on 25 Jul 2010 03:42 On 25 July, 04:44, Danny73 <fasttrac...(a)att.net> wrote: > On Jul 24, 5:35 pm, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> > wrote: > > > > > > > On 24 July, 17:35, Danny73 <fasttrac...(a)att.net> wrote: > > > > On Jul 24, 10:30 am, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> > > > wrote: > > > > > On 24 July, 14:19, Danny73 <fasttrac...(a)att.net> wrote: > > > > > > (3) separate equations using pi and e where the > > > > > results are close too the value of pi > > > > > (correct too 6 decimal digits) > > > > > > (((e^3/pi^2)^2)-1) = 3.141592835... > > > > > (slightly greater than pi) > > > > > > (1/((e^6/pi^5)-1)) = 3.1415920835... > > > > > (slightly less than pi) > > > > > > (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912... > > > > > (slightly greater than pi) > > > > > > The challenge -- > > > > > > Are more results possible that give pi > > > > > correct too 6 decimal places or more > > > > > using only pi and e and their powers in each > > > > > equation but with slightly different results > > > > > from above. > > > > > > The criteria for each equation involves pi and e > > > > > and using any or all of these operators (-,+,*,/ and ^n) > > > > > giving 6 or more correct decimal places of pi in the > > > > > result. > > > > > You seem to be finding solutions to > > > > say > > > > > A1*pi^5 +A2*pi^4 +A3*e^6 a.e. 0 > > > > > A1*pi = -A2 + (A3/A2)(e^3/pi^2)^2) > > > > > with A1 =A2=A3 =1 > > > > > would pi = (-A2/A1) +(A3/(A2*A1))e^3/pi^2)^2) > > > > > also be acceptable ?- Hide quoted text - > > > > > - Show quoted text - > > > > Sure if you have the actual integer values for a1,a2 and a3 > > > depicted for each equation. > > > > Thanks for your input. > > > A simple method of finding integer relations between > > irrationals is to write the irrationals to a large > > number of decimal places in an nx1 column vector with the identity > > matrix beside it. > > > 1 0 0 306.0196848 = pi^5 = r1 > > 0 1 0 97.40909103 = pi^4 = r2 > > 0 0 1 403.4287935 = e^5 = r3 > > > As r1 >r3 -r1 = 97.4091087 >0 take row 1 from row 3 > > of the matrix > > > 1 0 0 306.0196848 = r1 > > 0 1 0 97.40909103 = r2 > > -1 0 1 97.4091087 = r3 > > > As r3 is only slightly greater than r2 > > take row2 from row 3 > > > 1 0 0 306.0196848 = r1 > > 0 1 0 97.40909103 = r2 > > -1 -1 1 0.00001717 = r3 > > > This means that > > > (-1,-1,1).(pi^5,pi^4, e^6) = 0.0000172 > > > -pi^5 -pi^4 + e^6 = 0.0000172 > > > pi +1 -(e^6/pi^4) = -0.0000172/pi^4 > > > pi + 0.0000172/pi^4 = (e^6/pi^4)-1 > > > Obviously, you can reduce the values in the column > > vector as close as you want to 0, but the absolute vales > > in the matrix rows will grow in size correspondingly. > > > There is no guarantee that a1,a2,a3 will be "small" > > integers.- Hide quoted text - > > > - Show quoted text - > > Also another interesting observation. > Am I wrong in assuming that it will greatly improve > decimal digit accuracy for pi with large a1,a2 and a3? > Once you have reduced the other column elements mod 0.00001717 ( = r3), you will have at least 6 digit approximation for a1,a2,a3 of about 300/0.0002. As you continue to take row nultiples avway from each other, you could get lucky and find another pair or triplet of values very close together, which means a relatively small increase in a1,a2,a3. Even though you can approximate pi to 100 decimal places or more, the formulas might contain integers with 1,000,000 decimal places and that is not as pleasing as a "surpisingly good" approximation with "significant" small integers, such as primes or powers.
From: sttscitrans on 25 Jul 2010 07:30 On 24 July, 14:19, Danny73 <fasttrac...(a)att.net> wrote: > (3) separate equations using pi and e where the > results are close too the value of pi > (correct too 6 decimal digits) > > (((e^3/pi^2)^2)-1) = 3.141592835... > (slightly greater than pi) > > (1/((e^6/pi^5)-1)) = 3.1415920835... > (slightly less than pi) > > (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912... > (slightly greater than pi) > > The challenge -- > > Are more results possible that give pi > correct too 6 decimal places or more > using only pi and e and their powers in each > equation but with slightly different results > from above. > > The criteria for each equation involves pi and e > and using any or all of these operators (-,+,*,/ and ^n) > giving 6 or more correct decimal places of pi in the > result. If the sqrt(2) is acceptable ((17^2)p^2 - 95e^2)/((22^2)sqrt(2)) = 3.14159265//42.. Gives pi to 8 places and has two relatively small integer squares.
From: Danny73 on 25 Jul 2010 10:25 On Jul 25, 7:30 am, "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote: > On 24 July, 14:19, Danny73 <fasttrac...(a)att.net> wrote: > > > > > > > (3) separate equations using pi and e where the > > results are close too the value of pi > > (correct too 6 decimal digits) > > > (((e^3/pi^2)^2)-1) = 3.141592835... > > (slightly greater than pi) > > > (1/((e^6/pi^5)-1)) = 3.1415920835... > > (slightly less than pi) > > > (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912... > > (slightly greater than pi) > > > The challenge -- > > > Are more results possible that give pi > > correct too 6 decimal places or more > > using only pi and e and their powers in each > > equation but with slightly different results > > from above. > > > The criteria for each equation involves pi and e > > and using any or all of these operators (-,+,*,/ and ^n) > > giving 6 or more correct decimal places of pi in the > > result. > > If the sqrt(2) is acceptable > > ((17^2)p^2 - 95e^2)/((22^2)sqrt(2)) = 3.14159265//42.. > > Gives pi to 8 places and has two relatively small integer squares.- Hide quoted text - > > - Show quoted text - Interesting, and yes it should be ecceptable where I overlooked the (sqrt) operator. I should also include sqrt(n) as an operator as long as (n) is an integer. The new criteria --(-,+,*,/,(sqrt),^n) and having pi and (e) within the equation and either pi or (e) as the result. The other criteria is the same as above but having an integer as the result as with the case of e^(pi*(sqrt(163))) ~ 262537412640768744 or using the natural log of (e) along with pi- ((log(1.92932559972567304972024812259531566433349783047577528040155793074025543587927999223373032998386289025375167228752542895869119788592042536004022640601542769951592967417765008384707347928761185634055546471521383604751331331401391405583924e +236))/pi)^2 ~ 29992. followed by 234 zeros Thanks again for your interesting observations.
From: Raymond Manzoni on 25 Jul 2010 10:43 Danny73 a �crit : > On Jul 24, 2:37 pm, Raymond Manzoni <raym...(a)free.fr> wrote: (snip) >> Since there is no 'pi' smell in that one either ;-) I shouldn't >> propose it anyway : >> e + (69 - e^(-8) - e^(-17))/163 = 3.1415926535877.. >> >> Of course the first one may be ameliorated too : >> e^pi - 20 + e/pi^7 + e^(-2)/pi^14 = 3.141592653549... and so on... >> >> You may too combine two results from the link to get : >> e^pi-(67*pi^9/e^8-430)/12 = 3.141592678... >> with the variant (e/pi)^8*((e^pi-pi)*12+430)/67 = 3.14159265498.. >> >> or combine with pi^2 + pi/24 ~= 10 to get >> e^(-1863/422)-(pi^2-10)*24 = 3.14159265436... >> or >> e^(-12/e)-(pi^2-10)*24 = 3.14159433... >> e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563... >> >> and so on... >> >> You may of course approximate pi with a predefined precision if you >> accept the formula to be long enough (by adding each time a >> e^(-fraction) term)! >> A better 'quality criteria' could be obtained by dividing the number >> of correct digits by the length of the formula >> <http://en.wikipedia.org/wiki/Kolmogorov_complexity>. >> (snip) > > Thanks again Raymond, > > It is interesting on how you come up with these. (entering in the kitchen... :-)) Start for example with : pi^3 - 2 pi/10^3 ~= 31 and replace one of the terms appearing (pi, 2, pi, 10^3, 31) by another term so that the equality 'become more exact'. For example 10^3 should be replaced by 2*pi/(pi^3-31) = 1001.036377... so that 1001 should be better but not good enough... subtract 1000, take the log, invert and you get nearly 28 so that pi^3 - 2 pi/(10^3+e^(1/28)) ~= 31 with more precision and you got another approximation : 31/(Pi^2-2/(10^3+e^(1/28))) = 3.141592653600... Of course I used other recipes : - integer relations algorithms (extension of continued fractions) <http://en.wikipedia.org/wiki/Integer_relation_algorithm> (I use lindep() in the excellent pari/gp <http://pari.math.u-bordeaux.fr/download.html>) - combine two (or more) relations in one for example by dividing the errors in the two relations and replace the result by a fraction - add more error terms using integer relations applied to the logarithm of the |error term|, log(e), log(pi), log(2)... (trick to get a multiplicative answer using an additive algorithm!) - of course many other tricks may (and should :-)) be tried! Let's use this with pi^2 + pi/24 ~= 10 : - search a better fraction than 24 => 2337/97 - get a new result (without 'e' terms) : 10/pi-97/2337 = 3.14159265730... - compute the log of the error term : a= log(10/Pi-97/2337-Pi) - use the integer relation algorithm (changing precision...) lindep([a,1,log(pi),log(2)],8) -> [-1, 72, 14, -155]~ so that |-log(10/Pi-Pi-97/2337)+72*log(e)+14*log(pi)-155*log(2)| << 1 - conclude that the error term was near e^72*pi^14/2^155 - continue these iterations to get : 10/pi-97/2337 = 3.14159265730... 10/pi-97/2337-e^72*pi^14/2^155 = 3.141592653589804... 10/pi-97/2337-e^72*pi^14/2^155-e*(e/pi)^114/2^24 = 3.141592653589793238465869... 10/pi-97/2337-e^72*pi^14/2^155-e*(e/pi)^114/2^24-2^84/e^31/pi^65 = 3.1415926535897932384626433800... I'll stop here and let you play with all that too! Enjoy! Raymond > Although that last one looks most interesting -- > e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563... > I fail to get it to come out right. > I am probably getting the order of preference wrong > for one or more of the operators. Perhaps that this output will clarify my notations : / 12 \ 2 exp(6) exp| - ------ | - 24 PI - ---------- + 240 \ exp(1) / 21/2 (2 PI) > > So using my criteria for these equations it appears > there are many more equations then just (4) and > with a higher degree of accuracy also.
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