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1 decimal integer ring cycle of many
Quantum Field Fractal Polarization Math Constants
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ᐱ Y φ Θ P Q Ψ
condensed matter
Complex sets of Y Phi Theta Prime Q Psi Quotient Based Numerals
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Given F=∈2⅄(Θ/Q)cn
The definition of Q variable should be accurate to its defining base path where 1⅄Q and 2⅄Q, differ from prime base just as φ and Θ differ in paths 1⅄ and 2⅄ of Y base.
so
F=∈2⅄(Θn1/1⅄Qn2)cn
and
F=∈2⅄(Θn1/2⅄Qn2)cn
then
Fn1 of 2⅄(Θn1/1⅄Qn2)=(0/1.^6)=0
Fn2 of 2⅄(Θn2/1⅄Qn3)=(1/1.4)=0.^714285
Fn3 of 2⅄(Θn3/1⅄Qn4)=(0.5/1.^571428)=0.31818193388433959430530702011164367696133707684984612721677353337219395352507400 with an extended shell of decimal having a potential of 1,571,427 total digits in the final quotient based on probability.
Fn4 of 2⅄(Θn4/1⅄Qn5)=(0.^6/1.^18)=^0.508474576271186440677966101694915254237288135593220338983 for c1 of both variables in equation set Fn4 of 2⅄(Θn4c1/1⅄Qn5c1)
Fn5 of 2⅄(Θn5/1⅄Qn6)=(0.6/1.^307692)=0.4588236373702676165335568314251368059145425681276630888 with an extended shell of decimal having a potential of 1,307,691 total digits in the final quotient based on probability.
Fn6 of 2⅄(Θn6/1⅄Qn7)=(0.625/1.^1176470588235294)=0.559210526315789479570637119113573469164601253827089149101065829758833148432271892198243667708125181034143870611844010885724953808884325112894250 with an extended shell of decimal having a potential of 11,176,470,588,235,293 total digits in the final quotient based on probability.
Fn7 of 2⅄(Θn7/1⅄Qn8)=(0.^615384/1.^210526315789473684)=0.5083606956521739131318888166351606805446763159365496835729871853802695101866064670226555669889750377430705333893869630857513971111977327105654603671648230800983409334199692313214505971165163880559044516724376327053746872473804578 with an extended shell of decimal having a potential of 1,210,526,315,789,473,684 total digits in the final quotient based on probability.
if F=∈2⅄(Θn1/1⅄Qn2)cn
and
F=∈2⅄(Θn1/2⅄Qn2)cn
then
Fn1 of 2⅄(Θn1/2⅄Qn2)=(0/0.6)=0
Fn2 of 2⅄(Θn2/2⅄Qn3)=(1/0.^714285)=^1.40000 or 1.40000140000 and so on...
Fn3 of 2⅄(Θn3/2⅄Qn4)=(0.5/0.^63)=^0.79365 or 0.79365079365 and so on...
Fn4 of 2⅄(Θn4/2⅄Qn5)=(0.^6/0.^846153)=^0.70909161818252727343636434545525454616363707272798181889090980000
Fn5 of 2⅄(Θn5/2⅄Qn6)=(0.6/0.^7647058823529411)=^0.78461538461538469384615384615385400000000000000
Fn6 of 2⅄(Θn6/2⅄Qn7)=(0.625/0.^894736842105263157)=^0.69852941176470588305147058823529411834558823529411764775735294117647058893382352941176470658088235294117647128676470588235294187500000000000000
Fn7 of 2⅄(Θn7/2⅄Qn8)=(0.^615384/0.^8260869565217391304347)=^0.74493852631578947368428502016842105263157895481780631578947368421127125431578947368421060080964210526315789474429149052631578947368495546484210526315789481133595789473684210527060728000000000000000