Relationship between Napier number e and Pi without i April 17, 2019 Yuji Masuda Bachelor of Mechanical Systems Engineering in Tokyo University of Agriculture and Technology

I am Yuji Masuda from Japan. I have a job of lending rooms to companies, which called “share office or share w ork place”. And I am working there as an hourly-employee.

The length √x, that x is a prime number, can be set in a geometric pattern composed of circles with a diameter 1. And in many lines or curves through the center of circles of prime, some limited conditions, which is on the center of the circles of x=3 and 7(Fig.1), shows some graphs. Especially, graph of Gauss’ prime number theorem. which meets the limited condition, shows relationship between Napier number e and π without imaginary number i.

Key Words : Napier number e, π, imaginary unit i

1. Introduction 4．Formula about Napier number e and pi without i. There is unknown area in prime numbers. Therefore, this I thought about b of the equation below. research aimed to deepen knowledge on unknown areas through a 1 basic research on prime numbers. e = (∵ a = e 3 + b) log( a)

2. Geometric pattern and limited condition And as a result , 1  14  a = e 3 + b∵ b =  The center of a circle with a close-packed structure of a circle  15  with a radius of 1/2 is set as the O, is focused on x at a distance a ≅ e √ x from O. log( a) The limited the condition is go through the centers of two And I thought two things below. circles of x=3&7. 2  1   3  = ⋯ ≅ 10 − 1 1 ：  e  .3 17307306 x   π 100 x 1 1 2 2   O  e 3    1   2   = .9 96850302 ⋯ ≅ π π ×  3  ≅ π 2 ： 1  e  ,   π 1 log ()π ≅ 3 1 1 Here, I noticed below. 4 3 4 1 = .1 74713705 ⋯ ≅ 14 − 2 9 7 7 9 log ()π 16 13 12 13 16 14 × log ()π = .2 14159352 ⋯ ≅ π −1 25 21 19 19 21 25

36 31 28 27 28 31 36 And after trial & error, I could finally get the 49 43 39 37 37 39 43 49 following equation.

π 64 57 52 49 48 49 52 57 64     81 73 67 63 61 61 63 67 73 81            B  A ⋅   π                  B   π ()()−+ a1 −+ a1 ⋅⋅    1 1  A π           Fig.1 two curves which one meets the limited conditions  a2 a2  B    π ()()−+ 1 −+ 1 A ⋅⋅  π       a3 a3         π ()()()−+ 1 −+ 1 ()A ⋅ ⋯     and another doesn’t meet it.         −= 1 (1) 3．Curve of Gauss’ prime number theorem graph First, I thought of some quadratic curves. However, it did (∵ ⋅ln14 ()π π =− A ⋅ ln14, ()π 1 =+ B) (∵ a p ±= ,1 n p ()()22 q +⋅⋅= ,1 ∵ , qp = int eger ) not give a desired result. n Second, I considered a curve of Gauss’ prime number 5．Conclusion theorem graph. The equation (1) that I found shows the relationship between x=1.78131217...on the curve which meets the limited Napier number e and π . Furthermore, the equation (1) x condition x : = 3:1 doesn’t include imaginary unit i of e^(iπ)= −1.[1] log( x) 1 log( x) = 6．References 3 1 [1]. 新装版 オイラーの贈物ー人類の至宝 e^i π=−1 を学ぶ x = e 3 = 1.78131217 ⋯ 吉田武（著） 2010 年 1 月 1 日 Next, I considered more deeply about the curve of the Gauss (New edition Euler's gift-The treasure of humanity. Learn e ^ i π = 1. prime theorem graph. Takeshi Yoshida. 2010 Jan 1)(by google translation)