The Existence of the Exponential Function

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Introduction

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

[Main]

$e(x+y)=e(x)e(y)$

as well as the initial condition

[Init]

$e(x)=1+x+$(higher order terms).

Alternative proofs of the existence of $e(x)$ are of course available, including the explicit formula $e(x)=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}$. Thus the value of this paperlet is not in the result it proves but rather in the allegorical 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 $e'=e$.

The Scheme

We aim to construct $e(x)$ and solve [Main] inductively, degree by degree. Equation [Init] gives $e(x)$ in degrees 0 and 1, and the given formula for $e(x)$ indeed solves [Main] in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial $e_{7}(x)$ 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]

$e_{7}(x+y)-e_{7}(x)e_{7}(y)=M(x,y)$,

where $M(x,y)$ is the "mistake for $e_{7}$", a certain homogeneous polynomial of degree 8 in the variables $x$ and $y$.

Our hope is to "fix" the mistake $M$ by replacing $e_{7}(x)$ with $e_{8}(x)=e_{7}(x)+\epsilon (x)$, where $\epsilon _{8}(x)$ is a degree 8 "correction", a homogeneous polynomial of degree 8 in $x$ (well, in this simple case, just a multiple of $x^{8}$).

*1 The terms containing no $\epsilon$'s make a copy of the left hand side of [M]. The terms linear in $\epsilon$ are $\epsilon (x+y)$, $-e_{7}(x)\epsilon (y)$ and $-\epsilon (x)e_{7}(y)$. Note that since the constant term of $e_{7}$ is 1 and since we only care about degree 8, the last two terms can be replaced by $-\epsilon (y)$ and $-\epsilon (x)$, respectively. Finally, we don't even need to look at terms higher than linear in $\epsilon$, for these have degree 16 or more, high in the stratosphere.

So we substitute $e_{8}(x)=e_{7}(x)+\epsilon (x)$ into $e(x+y)-e(x)e(y)$ (a version of [Main]), expand, and consider only the low degree terms - those below and including degree 8:^*1

$e_{8}(x+y)-e_{8}(x)e_{8}(y)=M(x,y)-\epsilon (y)+\epsilon (x+y)-\epsilon (x)$.

We define a "differential" $d:{\mathbb {Q} }[x]\to {\mathbb {Q} }[x,y]$ by $(df)(x,y)=f(y)-f(x+y)+f(x)$, and the above equation becomes

$e_{8}(x+y)-e_{8}(x)e_{8}(y)=M(x,y)-(d\epsilon )(x,y)$.

*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of $e(x+y)=e(x)e(y)$, is quite unlikely. For $e$ must be an element of the relatively small space ${\mathbb {Q} }[[x]]$ of power series in one variable, but the equation it is required to satisfy lives in the much bigger space ${\mathbb {Q} }[[x,y]]$. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!

To continue with our inductive construction we need to have that $e_{8}(x+y)-e_{8}(x)e_{8}(y)=0$. Hence the existence of the exponential function hinges upon our ability to find an $\epsilon$ for which $M=d\epsilon$. In other words, we must show that $M$ is in the image of $d$. This appears hopeless unless we learn more about $M$, for the domain space of $d$ is much smaller than its target space and thus $d$ cannot be surjective, and if $M$ was in any sense "random", we simply wouldn't be able to find our correction term $\epsilon$.^*2

As we shall see momentarily by "finding syzygies", $\epsilon$ and $M$ fit within the 0th and 1st chain groups of a rather short complex

$\left(\epsilon \in C_{1}={\mathbb {Q} }[[x]]\right)\longrightarrow \left(M\in C_{2}={\mathbb {Q} }[[x,y]]\right)\longrightarrow \left(C_{3}={\mathbb {Q} }[[x,y,z]]\right)$,

whose first differential was already written and whose second differential is given by $(d^{2}m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)$ for any $m\in {\mathbb {Q} }[[x,y]]$. We shall further see that for "our" $M$, we have $d^{2}M=0$. Therefore in order to show that $M$ is in the image of $d^{1}$, it suffices to show that the kernel of $d^{2}$ is equal to the image of $d^{1}$, or simply that $H^{2}=0$.

Finding a Syzygy

So what kind of relations can we get for $M$? Well, it measures how close $e_{7}$ is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have $M(x,y)=M(y,x)$, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute $e_{7}(x+y+z)$ associating first as $(x+y)+z$ and then as $x+(y+z)$. In the first way we get

$e_{7}(x+y+z)=M(x+y,z)+e_{7}(x+y)e_{7}(z)=M(x+y,z)+\left(M(x,y)+e_{7}(x)e_{7}(y)\right)e_{7}(z).$

In the second we get

$e_{7}(x+y+z)=M(x,y+z)+e_{7}(x)e_{7}(y+z)=M(x+y,z)+e_{7}(x)\left(M(y,z)+e_{7}(y)e_{7}(z)\right)$.

Comparing these two we get an interesting relation for $M$: $M(x+y,z)+M(x,y)e_{7}(z)=M(x,y+z)+e_{7}(x)M(y,z)$. Since we'll only use $M$ to find the next highest term, we can be sloppy about all but the first term of $M$. This means that in the relation we just found we can replace $e_{7}$ by its constant term, namely 1. Upon rearranging, we get the relation promised for $M$: $d^{2}M=M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)=0$.

Computing the Homology

Now let's prove that $H^{2}=0$ for our (piece of) chain complex. That is, letting $M(x,y)\in \mathbb {Q} [[x,y]]$ be such that $d^{2}M=0$, we'll prove that for some $E(x)\in \mathbb {Q} [[x]]$ we have $d^{1}E=M$.

Write the two power series as $E(x)=\sum {{\frac {e_{i}}{i!}}x^{i}}$ and $M(x,y)=\sum {{\frac {m_{ij}}{i!j!}}x^{i}y^{j}}$, where the $e_{i}$ are the unknowns we wish to solve for.

The coefficient of $x^{i}y^{j}z^{k}$ in $M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)$ is

${\frac {\delta _{i0}m_{jk}}{j!k!}}-{{i+j} \choose i}{\frac {m_{i+j,k}}{(i+j)!k!}}+{{j+k} \choose j}{\frac {m_{i,j+k}}{i!(j+k)!}}-{\frac {\delta _{k0}m_{ij}}{i!j!}}.$

Here, $\delta _{i0}$ is a Kronecker delta: 1 if $i=0$ and 0 otherwise. Since $d^{2}M=0$, this coefficient should be zero. Multiplying by $i!j!k!$ (and noting that, for example, the first term doesn't need an $i!$ since the delta is only nonzero when $i=0$) we get:

$\delta _{i0}m_{jk}-m_{i+j,k}+m_{i,j+k}+\delta _{k0}m_{ij}=0$.

An entirely analogous procedure tells us that the equations we must solve boil down to $\delta _{i0}e_{j}-e_{i+j}+\delta _{j0}e_{i}=m_{ij}$.

By setting $i=j=0$ in this last equation we see that $e_{0}=m_{00}$. Now let $i$ and $j$ be arbitrary positive integers. This solves for most of the coefficients: $e_{i+j}=-m_{ij}$. Any integer at least two can be written as $i+j$, so this determines all of the $e_{m}$ for $m\geq 2$. We just need to prove that $e_{m}$ is well defined, that is, that $m_{ij}$ doesn't depend on $i$ and $j$ but only on their sum.

But when $i$ and $k$ are strictly positive, the relation for the $m_{ij}$ reads $m_{i+j,k}=m_{i,j+k}$, which show that we can "transfer" $j$ from one index to the other, which is what we wanted.

It only remains to find $e_{1}$ but it's easy to see this is impossible: if $E$ satisfies $d^{1}E=M$, then so does $E(x)+kx$ for any $k$, so $e_{1}$ is abritrary. How do our coefficient equations tell us this?

Well, we can't find a single equation for $e_{1}$! We've already tried taking both $i$ and $j$ to be zero, and also taking them both positive. We only have taking one zero and one positive left. Doing so gives two neccessary conditions for the existence of the $e_{m}$: $m_{0r}=m_{r0}=0$ for $r>0$. So no $e_{1}$ comes up, but we're still not done. Fortunately setting one of $i$ and $k$ to be zero and one positive in the realtion for the $m_{r}s$ does the trick.