A pi and e equation challenge.
in [Math]
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From: Danny73 on 24 Jul 2010 09:19 (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
From: sttscitrans on 24 Jul 2010 10:30 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. 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 ?
From: Mike Terry on 24 Jul 2010 10:35 "Danny73" <fasttrack2a(a)att.net> wrote in message news:5410028f-c562-4749-ab86-fb8849d50130(a)s9g2000yqd.googlegroups.com... > > (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. pi = 3.141592653589793... (If this is not acceptable because e "does not appear", replace lhs with (pi+e-e)) Mike. > > 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
From: Raymond Manzoni on 24 Jul 2010 10:38 Danny73 a �crit : > (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
From: Danny73 on 24 Jul 2010 12:22 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
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