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Relationship Between Napier Number E and Pi Without I

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Relationship Between Napier Number E and Pi Without I 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 (∵ 14⋅ ln()π −π = A, 14⋅ ln()π +1 = B) (∵ a =p ± ,1 n =2p ⋅()() 2 ⋅q + ,1 ∵ p, q = 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 : = 1: 3 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).
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