A pi and e equation challenge.
in [Math]
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From: Danny73 on 24 Jul 2010 12:29 On Jul 24, 12:22 pm, Danny73 <fasttrac...(a)att.net> wrote: > On Jul 24, 10:38 am, Raymond Manzoni <raym...(a)free.fr> wrote: > > > > > > > Danny73 a écrit : > > > > (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. > > > > BTW > > > I know (e^(pi*i)) * pi * -1 = pi > > > So don't try to pull any fast ones! ;-) > > > Also imaginary (i) is not one of the operators > > > that is allowed. > > > > Dan > > > Your three approximations are variants of pi^4 + pi^5 ~= e^6 given > > here : <http://en.wikipedia.org/wiki/Mathematical_coincidence> > > > Other results may be found using this link like : > > > e^pi - 20 + e/pi^7 = 3.1415926387... > > > e + (69 - e^(-8))/163 = 3.1415926538... > > > 355/113 = 3.14159292..... > > > 103993/33102 = 3.1415926530... :-) > > > Hoping this helped, > > Raymond- Hide quoted text - > > > - Show quoted text - > > Thanks Raymond, > > Your first equation fits the criteria for my (3) equations > with a better approximation to boot. > > Your next one does not fit the criteria but is also interesting > because pi is involved indirectly with 163 along with (e). > > Where e^((sqrt(163)) * pi) ~ 262537412640768744. > > The next (2) do not fit the criteria. > > So now there are a total of (4) equations that fit the criteria. > > Are there anymore? > > That link brought me someplace else. > > Thanks for your input. > > Dan- Hide quoted text - > > - Show quoted text - Sorry I clicked on link because it was highlighted. The actual link (wiki) was very informative. Thanks again.
From: Danny73 on 24 Jul 2010 12:35 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. Dan
From: Raymond Manzoni on 24 Jul 2010 14:37 Danny73 a �crit : >> Your three approximations are variants of pi^4 + pi^5 ~= e^6 given >> here : <http://en.wikipedia.org/wiki/Mathematical_coincidence> >> >> Other results may be found using this link like : >> >> e^pi - 20 + e/pi^7 = 3.1415926387... >> >> e + (69 - e^(-8))/163 = 3.1415926538... >> >> 355/113 = 3.14159292..... >> >> 103993/33102 = 3.1415926530... :-) >> >> Hoping this helped, >> Raymond- Hide quoted text - >> >> - Show quoted text - > > Thanks Raymond, > > Your first equation fits the criteria for my (3) equations > with a better approximation to boot. > > Your next one does not fit the criteria but is also interesting > because pi is involved indirectly with 163 along with (e). > > Where e^((sqrt(163)) * pi) ~ 262537412640768744. > > The next (2) do not fit the criteria. > > So now there are a total of (4) equations that fit the criteria. > > Are there anymore? 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>. Wishing you fun, Raymond > > That link brought me someplace else. > > Thanks for your input. > > Dan
From: sttscitrans on 24 Jul 2010 17:35 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.
From: Danny73 on 24 Jul 2010 23:19 On Jul 24, 2:37 pm, Raymond Manzoni <raym...(a)free.fr> wrote: > Danny73 a écrit : > > > > > > >> Your three approximations are variants of pi^4 + pi^5 ~= e^6 given > >> here : <http://en.wikipedia.org/wiki/Mathematical_coincidence> > > >> Other results may be found using this link like : > > >> e^pi - 20 + e/pi^7 = 3.1415926387... > > >> e + (69 - e^(-8))/163 = 3.1415926538... > > >> 355/113 = 3.14159292..... > > >> 103993/33102 = 3.1415926530... :-) > > >> Hoping this helped, > >> Raymond- Hide quoted text - > > >> - Show quoted text - > > > Thanks Raymond, > > > Your first equation fits the criteria for my (3) equations > > with a better approximation to boot. > > > Your next one does not fit the criteria but is also interesting > > because pi is involved indirectly with 163 along with (e). > > > Where e^((sqrt(163)) * pi) ~ 262537412640768744. > > > The next (2) do not fit the criteria. > > > So now there are a total of (4) equations that fit the criteria. > > > Are there anymore? > > 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>. > > Wishing you fun, > Raymond > > > > > > > That link brought me someplace else. > > > Thanks for your input. > > > Dan- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Thanks again Raymond, It is interesting on how you come up with these. 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. 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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