Generalizing Trigonometry with Adjoining roots

Generalized Trignometry and its Applications Mohammed Farhaan, [email protected] 18 May 2025 Introduction We begin with an algebraic construction of trigonometric functions, stemming from a differential equation (D.E.) structure proposed in this paper. This approach aims to provide a novel perspective on these fundamental functions. We will explore the orthogonality of this generalized class of trigonometric functions, investigating their properties and relationships within this new framework. 1 Differential Equation Graphs 1.1 Zero-Order D.E Graph Let the edges of the graph be represented by the differential operator d dx and the vertices by {t}. The graph can be visualized as: t d dx The solution to this differential equation is of the form: t= exp(x) 1 1.2 First-Order D.E Graph Let the edges of the graph be d dx and the vertices be t, dt dx . The graph can be represented as: tdt d dx d dx In this graph, if we consider a specific solution, t= cosh(x), which can be expressed as: t= cosh(x) = ex+e−x 2 This demonstrates how specific functions can be represented and analyzed within this differential equation graph framework. 1.3 Second-Order D.E Graph Let the edges be {d dx }and the vertices be {t, dt dx ,−t, −dt dx }. The graph can be represented as: t dt −t −dt d dx d dx d dx d dx Note that d2t dx2=−t, which implies the solution is of the form t=eix +e−ix 2. 2 1.4 Third Order D.E Graph? If the edges/arrows be d dx  If the vertices be t, dt dx, it, i dt dx,−t, −dt dx,−it, −idt dx Find tsuch that it satisfies the graph represented as: t dt it idt −t −dt −it −idt d dx d dx d dx d dx d dx d dx d dx d dx Let’s assume t:A→A, where A=C[a3]/(a2 3−i). Note: 1 = a0,−1 = a1,i=a2 These are the first case of generalized trigonometric function. So tcan be satisfied by t=ea3x−e−a3x 2a3 We also impose the restriction that a3/∈Cso that we avoid adjoining a unit that already exists. Let’s go ahead and prove that tsatistfies our D.E with the following steps: t=ea3x−e−a3x 2a3 dt dx =ea3x+e−a3x 2 3 d2t dx2=a3 ea3x−e−a3x 2=a3t d3t dx3=a3 dt dx d4t dx4=a2 3t=it =−t d5t dx5=−dt dx d6t dx5=−it d7t dx7=−idt dx d8t dx8=t hence it proves that sina3(x) is the proved solution to this problem 1.5 The general case We say, let us recursively adjoin units to Rin the following way: Let R=A1. C=A2=A1[a2]/(a2 2+ 1) A3=A2[a3]/(a2 3+a2) where a3/∈A2. Or recursively, we can say: An=An−1[an]/(a2 n+an−1) where an/∈An−1. We need to prove that the same equivalent (if this graph) can be solved by sinan(x) in An. We know this is true for A1,A2, and A3. Let’s assume it’s true for Ak. Is it true for Ak+1? For Ak, let’s see: In Ak+1, we know that a2 k+1 =ak and supposedly d2t dx2=akt 4 Now, let t= sinak(x). dt dx =akcosak(x) d2t dx2=ak eakx−e−akx 2 =a2 ksinak(x) =ak+1 sinak(x) d4t dx4=ak+2 sinakx=−d(23) dx(23)t d23t dx23=ak−3sinak(x) d2kt dx2k=ak−ksinak(x) =a0sinak(x) = sinak(x) We can also show that d2k+1 t dx2k+1 = sinak+1 (x) Hence we have shown that this structure is solved by sinan(x) for all n∈N. Now comes the hard part: Prove that there exist nfor which sinak(x) and cosak(x) are suitable for Fourier analysis. 1.6 Finding What these functions parameterize 1.6.1 Find what each trigonometry parameterizes. 1. sina1(x) and cosa1(x) are sinh(x) and cosh(x), parametrizing a real hyperbola. 2. sina2(x) and cosa2(x) are sin(x) and cos(x) respectively, and parametrize a real circle. 5 3. What do sina3(x) and cosa3(x) parametrize? For that, we must find out: sina3(x)2−icosa3(x)2=? sina3(x) = ea3x−e−a3x 2a3 cosa3(x) = ea3x+e−a3x 2 So, sina3(x)2−icosa3(x)2=ea3x−e−a3x 2a32 −iea3x+e−a3x 22 =(ea3x−e−a3x)2 4a2 3 −i(ea3x+e−a3x)2 4 =e2a3x−2 + e−2a3x 4a2 3 −ie2a3x+2+e−2a3x 4 =1 4ie2a3x+e−2a3x−2 −i 4(4) =? sina3(x)2−icosa3(x)2=−i This parametrizes a curve: z1(x)2−iz2(x)2=−i or equivalently, z2(x)2+iz1(x)2= 1 So we can say that sina3(x) and cosa3(x) parameterize a complex sphere It’s not unreasonable to assume it’s unbounded as well. Note: a2it is z2(x)2+z1(x)2= 1 4. Now we see what sina4(x) and cosa4(x) parametrize. Let a2 4=−a3. Checking if cosa4(x)2+a3sina4(x)2= 1 6 cosa4(x)2+a3sina4(x)2=ea4x+e−a4x 2 2 +a3ea4x−e−a4x 2a42 =e2a4x+2+e−2a4x 4+a3·e2a4x−2 + e−2a4x 4a2 4 =1 4e2a4x+2+e−2a4x+a3 4a2 4e2a4x−2 + e−2a4x =1 41 + a3 a2 4(e2a4x+e−2a4x) + (2 −2a3 a2 4 ) Since a2 4=−a3, so a3 a2 4 =−1: =1 4(1 −1)(e2a4x+e−2a4x) + (2 + 2)=1 4(0 + 4) = 1 Therefore, cosa4(x)2+a3sina4(x)2= 1 So cosa4(x) and sina4(x) (suspiciously) parametrize some algebraic version of a sphere. The whole basis of doing Fourier analysis is that these functions should have a period. However, sinan(x) : R→An+1 cosan(x) : R→An+1 So by this we have to assume that our period is a Real number If they have a period T, we should find it. To make our lives easier we can focus on just cosak(x) for k= 2,3... and so on 1. What is the period of cosa2(x)? 2. What is the period of cosa3(x)? 1.6.2 Explicit Power Series Notation for sinan(x)and cosan(x) 1) sina3(x): Let a2 3= +i, sina3(x) = ∞ X n=0 (−1)nx4n+1 (4n+ 1)! +i ∞ X n=0 (−1)nx4n+3 (4n+ 3)! cosa3(x) = ∞ X n=0 (−1)nx4n (4n)! +i ∞ X n=0 (−1)nx4n+2 (4n+ 2)! 7 2) Let a2 4=a3 sina4(x) = ∞ X n=0 (−1)nx8n+1 (8n+ 1)!+ ∞ X n=0 (−1)nx8n+3 (8n+ 3)!+ ∞ X n=0 (−1)nx8n+5 (8n+ 5)!+ ∞ X n=0 (−1)nx8n+7 (8n+ 7)! cosa4(x) = ∞ X n=0 (−1)nx8n (8n)! + ∞ X n=0 (−1)nx8n+2 (8n+ 2)! +ia3 ∞ X n=0 (−1)nx8n+4 (8n+ 4)! +ia3 ∞ X n=0 (−1)nx8n+6 (8n+ 6)! 1.6.3 General Explicit formula for sinak(x)and cosak(x) sinak(x) = 2k−1−1 X j=0 "(ak)j ∞ X n=0 (−1)nx2kn+2j+1 (2kn+ 2j+ 1)!# cosak(x) = 2k−1−1 X j=0 "(ak)j ∞ X n=0 (−1)nx2kn+2j (2kn+ 2j)!# tanak(x) will always be sinak(x) cosak(x) 2 Possible Generalizations: 2.1 Change in Recursive Structure of Adjoining To generalize sinh and cosh, we had the structure arise with a2 k=ak−1, where ak/∈Ak−1. This results in a graph with elements of a set of vertices that has 2kvertices. Similarly, if we impose the recursive relation of a2 k=ak−1 where ak/∈Ak−1, then, after nsteps, we must get ak−n. So I’m guessing that we can impose an arbitrary condition of an k=ak−n 8 2.2 b) Non-homogeneous change in recursive structure: Instead of your adjunctions change, we have: A1=R, A2=A1[a2]/(a2 2+ 1), A3=A2[a3]/(a2 3+a1) where a2/∈A1,a3/∈A2. 2.3 c) Change in differential operator: Instead of D=d dx being the structure, we can have the Caputo fractional derivative: RDα t(f(t)) = Dn(In−αy(t)) When we introduce the fractional differential operator, it is useful to consider the fractional analog of exor exp(x), which is the Mittag-Leffler function. The Mittag-Leffler function Eαmay preserve the differential equation structure; however, the nature of the adjunction of algebras is very different. Eα(x) = ∞ X k=0 xk Γ(αk + 1) Within these possible generalizations, we have room to explore different useful cases. 2.4 d) Applying Our Findings to Fourier Analysis A good place to understand Fourier series is to think of it as a weighted sum of exponentials. f(x+a) = ∞ X n=0 Anenx For eix, f(t) = ∞ X n=−∞ Cnei2πnt T Now, f(t) = ∞ X n=−∞ CnaM N where N=Tis the period of f(t). My guess is that if sinak(x) and cosak(x) are periodic functions, and if solutions exist to sinak(x) = sinak(x+T) 9