Link btw. exponential and derivatives from an algebraic perspective [closed]

Question

I have been attempting to understand my math education (as a bachelor in electrical engineering) from a more algebraic perspective recently. I would like to understand more about the link between exponentiation and taking the derivative.

(From Wikipedia) given an algebra $A$ over a ring or field $K(+,\cdot,0_K,1_K)$, consider a $K$-linear map $D : A \rightarrow A$ that satisfies

$$D(a\cdot b) = a\cdot D(b) + D(a)\cdot b$$

Consider also (from Wikipedia) a homomorphism $E$ from the additive group of $K$ to the multiplicative group of $K$ such that:

$$ E(0_K) = 1_K $$ $$ \forall a, b \in K_+ \mid E(a + b) = E(a) \cdot E(b) $$

It seems that the only homomorphism $E$ that satisfies the above for the ring of integers is $ n \mapsto ( -1 )^n $. Let's consider the ring on $\mathbb{Q}$ instead.

Is there any nice way to understand the idea that

$$ \exists E \mid \exists c \in K \mid D ( E(x) ) = c \cdot E(x)$$

and that that homomorphism is the natural exponential function?

It seems to me that the homomorphism $E$ does a good job of capturing the concept of exponentiation. I have a bit of trouble seeing how the map $D$ encodes the idea of a derivative, but that's okay; I think I can work it out with some more reading and thought.

I am mainly curious if we can have this relation between the natural exponential $e$ and the derivative without invoking the idea of a power series. I'll admit, I'm not very opposed to the idea of using power series over the field of $\mathbb{Q}$, but I would like to invoke the reals in my thinking as little as possible.

Specifically I would like to understand functional analysis and signals from as much an algebraic point of view as possible, relying as little as possible on topological/real notions. Most of the spaces I think with have a metric which induces a measure and topology, but for my own sanity and symbol pushing, I would like to understand these things from as algebraic a position as possible.

Also, if there are any resources for understanding signals and electromagnetism from an algebraic perspective, that would be greatly appreciated.

Answer 1

Take a look at the Heaviside operational calculus and its relation to the Laplace transform solns. of differential eqns. by transforming them into algebraic equations.

As far as the derivation formula goes, just look at action on $1 = 1 \cdot 1$ and $x = x \cdot 1$ then $x^2 = xx$, etc. to develop a formula for derivation of each term of the Taylor series for the exponential.

Answer 2

I think I see now. Assume we want a function that is its own derivative.

$$ DF(x) = F(x) $$

With the assumption that the integers $\mathbb{Z}$ are invertible, construct the Taylor series using the above definition to get the familiar power series representation of the exponential function

$$ F(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$

We can check if this function satisfies the properties of the homomorphism $E$ in the question

\begin{align} F(a + b) &\stackrel{?}{=} F(a)\cdot F(b) \\ \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} &\stackrel{?}{=} \sum_{n=0}^{\infty} \frac{a^n}{n!} \cdot \sum_{n=0}^{\infty} \frac{b^n}{n!} \end{align}

For the left side, the binomial theorem gives

$$ \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{1}{n!} \cdot \frac{n!}{k!(n-k)!} \cdot a^k \cdot b^{n-k} \right) $$

For the right side, we can use the multiplication of power series to get

$$ \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \frac{1}{i!}\cdot \frac{1}{j!}\cdot a^i\cdot b^j = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{1}{k!}\cdot \frac{1}{(n - k )!}\cdot a^k\cdot b^{n - k} \right) $$

So we see that this function $F$ also satisfies the desired properties of $E$.

In this way, with any ring and acknowledging the invertibility of $\mathbb{Z}$, we can construct a Taylor series for a function which satisfies these two properties. This function is both its own derivative and is a homomorphism from the ring's additive group to its multiplicative group.

I believe that

$$ \exists F \mid \exists c \in K \mid D ( F(c\cdot x) ) = c \cdot F(x)$$

follows from the product rule on the above power series.