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Number Systems
Background: Number Systems is a post to explore number systems in general and for use in the
physical and computational sciences.
Post 8.2
Natural Events in Fibonacci Number Space
Energy Production
Posts 1 – 8 have established:
1 𝐷 = (1 +
𝛾∞
𝑓
𝑇𝐷
)
−1
(1 +
𝛾 𝐷
𝑓
𝑇𝐷
)
+1
For natural events, this definition should correlate to the Bernoulli base of natural logarithms:
∫
1
𝑥
𝑑𝑥
𝑒
1
= 1 where lim
𝑛→∞
(1 +
1
𝑛
)
𝑛
= 𝑒
A mathematical description of nature should not be accurate unless the number system complies
with both natural conditions of the number one shown above.
Natural examples:
1
𝑐3
2 =
1
35
2 𝑥 10−16
meter-2 sec+2
h = 6.6260700 E-34 = 6.6260700 x (1∞ − 𝑅 𝐸
𝑓{3}
) x 10-34
meter+2 kg+1 sec-1
𝒘𝒉𝒆𝒓𝒆 𝒂 𝒈 = 𝒈
when g = gEarthSurface <g units: acceleration+1 second+2>
It has been shown
𝐸
𝐸 𝐵
= 𝑚𝑉𝐵
To be rigorous, energy can be defined as a ratio:
𝐸
𝐸 𝐵
= 𝑚𝑉𝐵
Kilogram+1 Meter+3
Define
𝐸
𝐸 𝐵𝐷
=
(𝑘𝑎𝑝𝑝𝑎) 𝐷
𝑛
𝑥 𝑐 𝐷
1 𝐷 = (1 +
𝛾∞
𝑓
𝑇𝐷
)
−1
(1 +
𝛾 𝐷
𝑓
𝑇𝐷
)
+1
𝑒1 = 0
𝑒2 = 1
𝑒3 = 𝑒
𝑒5 = 𝑒 𝑓{3}
𝒆 𝑫 = 𝒆 𝑫−𝟏𝑫
𝒇{𝑫−𝟏𝑫}
𝑚 𝐷 =
(1 − 𝑅 𝐸
𝑓{𝐷}
)
𝑒 𝐷
𝐷
𝐷+1𝐷
𝒉 𝟓 = 𝒎 𝟑 𝒉 𝟑 + 𝒃 𝟑
And so on for D = F(n).
Post 8.2.1 is intended to clarify energy production in Fibonacci energy space.

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Post_Number Systems_8.2

  • 1. Number Systems Background: Number Systems is a post to explore number systems in general and for use in the physical and computational sciences. Post 8.2 Natural Events in Fibonacci Number Space Energy Production Posts 1 – 8 have established: 1 𝐷 = (1 + 𝛾∞ 𝑓 𝑇𝐷 ) −1 (1 + 𝛾 𝐷 𝑓 𝑇𝐷 ) +1 For natural events, this definition should correlate to the Bernoulli base of natural logarithms: ∫ 1 𝑥 𝑑𝑥 𝑒 1 = 1 where lim 𝑛→∞ (1 + 1 𝑛 ) 𝑛 = 𝑒 A mathematical description of nature should not be accurate unless the number system complies with both natural conditions of the number one shown above. Natural examples: 1 𝑐3 2 = 1 35 2 𝑥 10−16 meter-2 sec+2 h = 6.6260700 E-34 = 6.6260700 x (1∞ − 𝑅 𝐸 𝑓{3} ) x 10-34 meter+2 kg+1 sec-1 𝒘𝒉𝒆𝒓𝒆 𝒂 𝒈 = 𝒈 when g = gEarthSurface <g units: acceleration+1 second+2> It has been shown 𝐸 𝐸 𝐵 = 𝑚𝑉𝐵 To be rigorous, energy can be defined as a ratio: 𝐸 𝐸 𝐵 = 𝑚𝑉𝐵 Kilogram+1 Meter+3
  • 2. Define 𝐸 𝐸 𝐵𝐷 = (𝑘𝑎𝑝𝑝𝑎) 𝐷 𝑛 𝑥 𝑐 𝐷 1 𝐷 = (1 + 𝛾∞ 𝑓 𝑇𝐷 ) −1 (1 + 𝛾 𝐷 𝑓 𝑇𝐷 ) +1 𝑒1 = 0 𝑒2 = 1 𝑒3 = 𝑒 𝑒5 = 𝑒 𝑓{3} 𝒆 𝑫 = 𝒆 𝑫−𝟏𝑫 𝒇{𝑫−𝟏𝑫} 𝑚 𝐷 = (1 − 𝑅 𝐸 𝑓{𝐷} ) 𝑒 𝐷 𝐷 𝐷+1𝐷 𝒉 𝟓 = 𝒎 𝟑 𝒉 𝟑 + 𝒃 𝟑 And so on for D = F(n). Post 8.2.1 is intended to clarify energy production in Fibonacci energy space.