The Existence of the Exponential Function

Paperlets Navigation Panel   Algebraic Knot Theory Algebraic Knot Theory - A Call for Action. The Kontsevich Integral for Knotted Trivalent Graphs. Knotted Trivalent Graphs are Finitely Presented. VS, TS and TG Algebras. The Alexander Polynomial of a Knotted Trivalent Graph. Some Questions About Trinions. Finite Type Invariants The Envelope of a Finite Type Invariant. The Envelope of The Alexander Polynomial. Some Quotients of A. Associators with Frozen Feet. The HOMFLY Braidor Algebra. Coloured Chord Diagrams. Khovanov Homology Mutation Invariance of Khovanov Homology. Khovanov Homology of Alternating Tangles. Other The Existence of the Exponential Function. The Many Faces of Alexander.

Introduction

The purpose of this paperlet is to use some homological algebra in order to prove the existence of a power series [math]\displaystyle{ e(x) }[/math] (with coefficients in [math]\displaystyle{ {\mathbb Q} }[/math]) which satisfies the non-linear equation

[Main]

[math]\displaystyle{ e(x+y)=e(x)e(y) }[/math]

as well as the initial condition

[Init]

[math]\displaystyle{ e(x)=1+x+ }[/math](higher order terms).

Alternative proofs of the existence of [math]\displaystyle{ e(x) }[/math] are of course available, including the explicit formula [math]\displaystyle{ e(x)=\sum_{k=0}^\infty\frac{x^k}{k!} }[/math]. Thus the value of this paperlet is not in the result it proves but rather in the story it tells: that there is a technique to solve functional equations such as [Main] using homology. There are plenty of other examples for the use of that technique, in which the equation replacing [Main] isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation [math]\displaystyle{ e'=e }[/math].

The Scheme

We aim to construct [math]\displaystyle{ e(x) }[/math] and solve [Main] inductively, degree by degree. Equation [Init] gives [math]\displaystyle{ e(x) }[/math] in degrees 0 and 1, and the given formula for [math]\displaystyle{ e(x) }[/math] indeed solves [Main] in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial [math]\displaystyle{ e_7(x) }[/math] which solves [Main] up to and including degree 7, but at this stage of the construction, it may well fail to solve [Main] in degree 8. Thus modulo degrees 9 and up, we have

[M]

[math]\displaystyle{ e_7(x+y)-e_7(x)e_7(y)=M_8(x,y) }[/math],

where [math]\displaystyle{ M_8(x,y) }[/math] is the "mistake for [math]\displaystyle{ e_7 }[/math]", a certain homogeneous polynomial of degree 8 in the variables [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math].

Our hope is to "fix" the mistake [math]\displaystyle{ M_8 }[/math] by replacing [math]\displaystyle{ e_7(x) }[/math] with [math]\displaystyle{ e_8(x)=e_7(x)+\epsilon_8(x) }[/math], where [math]\displaystyle{ \epsilon_8(x) }[/math] is a degree 8 "correction", a homogeneous polynomial of degree 8 in [math]\displaystyle{ x }[/math] (well, in this simple case, just a multiple of [math]\displaystyle{ x^8 }[/math]).

So we substitute [math]\displaystyle{ e_8(x)=e_7(x)+\epsilon_8(x) }[/math] into [math]\displaystyle{ e(x+y)-e(x)e(y) }[/math] (a version of [Main]), expand, and consider only the low degree terms - those below and including degree 8. The terms containing no [math]\displaystyle{ \epsilon_8 }[/math]'s make a copy of the left hand side of [M]. The terms linear in [math]\displaystyle{ \epsilon_8 }[/math]