This post is essentially an adaptation to an answer I gave to a question on MathSE. In there, I gave another approach that may be worth checking: response in MathSE.
Initial observation: I'll assume that it is well-understood what is a number raised to a rational power.
Firstly, there are a lot of "motivational" approaches to the study of the exponential function. I'll adopt a purely mathematical motivation (specifically, algebraic) that is essentially in the nucleus of almost all others.
We know that there are two canonical algebraic structures in $ \mathbb{R}$: addition and multiplication. Formally, we consider the groups: $ (\mathbb{R},+)$ and $(\mathbb{R}_{> 0},\times)$. The question we ask is:
Now, let's try to extract some properties of what a function of this type, which we will call $\xi : (\mathbb{R},+) \rightarrow (\mathbb{R}_{> 0},\times)$ should have. Firstly, she should satisfy, by its definition, the following equation:
$\xi(x+y)=\xi(x).\xi(y)$
Since $\xi(0)=\xi(0+0)=\xi(0)\xi(0)$, $\xi(0)=1$.
Now, define $k:= \xi(1)$. What we will show now is that, with the value of $k$, we have the value of $\xi$ in all rationals.
We want to find a $\displaystyle \xi_1 (x)=\sum_{n=0}^{\infty}a_nx^n$ that is of the "type" of $ \xi$, meaning: $\xi_1 (x+y)=\xi_1(x)\xi_1(y)$. Do:
$ \displaystyle \xi_1(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the Binomial Theorem.
Notice that the last equality is BEGGING for the Cauchy Product Formula. If $ a_nn!$ was $1$... the last side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. But we have the power to choose who the $a_n$ are! We are searching a candidate to the series, after all. So, take $a_nn!=1$, which implies $ \displaystyle a_n=\frac{1}{n!}$. Now, stare at our luck! Since $\displaystyle a_n=\frac{1}{n!}$, what we have in the last equality is precisely the product $ \xi_1(x).\xi_1(y)$.
So, the function $\displaystyle \xi_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ satisfies all we want. (The series is well defined for all x real, which is easily verifiable by the ratio test. The series is obviously greater than zero is x is greater than 0, and the fact that $\xi_1(x).\xi_1(-x)=\xi_1(0)=1$ implies the series is greater than zero for all x real. Other observations also imply that it is bijective, since it is surjective (follows from the fact that it goes to infinity when x goes to infinity, to zero when x goes to -infinity, and it is continuous) and injective (because it is strictly increasing)
As we saw, it is enough knowing $\xi_1$ in $1$ to know its value in all rationals. But we know what it is in $1$:
$\displaystyle \xi_1(1)=\sum_{n=0}^{\infty}\frac{1}{n!}$
We will call this amazing number $e$.
So, $\xi_1(r)= e^r$ if $r$ is rational, in the usual sense we have for "a number raised to a rational power" (this follows from (1), (2) and (3)). Note that we already have an answer to our initial problem: we found a function $\xi_1$ which is our isomorphism between the groups. But it may be worth a while to try and say a few more things about it.
Since $\xi_1(r)=e^r$ if $r$ is rational, we define :
$e^x:=\xi_1(x)$
when $x$ is real, and this satisfies what we would wait of $e^x$ by previous observations. OK then, we found one function that extends what we call "exponentiation" of rational powers to real powers. But this function is analytic. Is there another less smooth function, for example $ C^{\infty}$ or even only continuous that could extend exponentiation? The answer is not, since this function would coincide in rational points, and the rationals are dense in $\mathbb{R} $ (Two continuous functions that coincide in a dense subset coincide in all $\mathbb{R}$). Then, this extension is unique if we require it to be continuous. This is enough to call $\xi_1(x)$ "the exponential", and have no regrets for calling it $e^x$.
Now, since $e^x$ is bijective, there is an inverse, which we will call $ln(x)$. This allows us to define, for an arbitraty $ a >0 $ :
$a^x:=e^{ln(a).x}$
Since, by definition $ a^1=a$ e $ a^{(x+y)}=a^x.a^y$, it follows from the same considerations as before that this function is what deserves to be called "raising $a$ to a real power".
Thus, we have:
$ (e^x) ^y=e^{(ln(e^x)).y}=e^{(x.y)}$
But going back to our previous question: what if we forget the constraint of a continuous extension: is there another function that coincides with $e^x$ on the rationals?
Yes, but we will need the following theorem to "show" a function of this kind:
Theorem: Every vector space has a basis.
Now, consider $ \mathbb{R}$ as a vector space over $ \mathbb{Q}$, and take a basis $ \{x_i\}_{i \in \Lambda}$ that has $1$. Define:
$f(1):=e$
$ f(x_i):=e^{x_i/2}$ if $ x_i \neq 1$
Since we defined the function on the basis and we want that $f(x+y)=f(x).f(y)$, we define:
$ f(x):=f(x_1)^{\lambda_1}f(x_2)^{\lambda_2}...f(x_n)^{\lambda_n}$,
where $ x=\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n$ is the representation of $x$ as a linear combination of elements of the basis (keep in mind that $\lambda_i$ are rationals!), and everything is well defined since$\{x_i\}$ is a basis. The fact that this function satisfies $f(x+y)=f(x).f(y)$ follows from definition, and the non-continuity of the function follows from the fact that it coincides with the exponentiation on the rationals but is not the exponentiation on the reals (she takes different values on $ x_i$, for example).
We see that requiring continuity is not only very reasonable, but also necessary. Otherwise, we would have a lot of "non-constructive exponentials" to consider.
Initial observation: I'll assume that it is well-understood what is a number raised to a rational power.
Firstly, there are a lot of "motivational" approaches to the study of the exponential function. I'll adopt a purely mathematical motivation (specifically, algebraic) that is essentially in the nucleus of almost all others.
We know that there are two canonical algebraic structures in $ \mathbb{R}$: addition and multiplication. Formally, we consider the groups: $ (\mathbb{R},+)$ and $(\mathbb{R}_{> 0},\times)$. The question we ask is:
- Is there an isomorphism between these two structures?
Now, let's try to extract some properties of what a function of this type, which we will call $\xi : (\mathbb{R},+) \rightarrow (\mathbb{R}_{> 0},\times)$ should have. Firstly, she should satisfy, by its definition, the following equation:
$\xi(x+y)=\xi(x).\xi(y)$
Since $\xi(0)=\xi(0+0)=\xi(0)\xi(0)$, $\xi(0)=1$.
Now, define $k:= \xi(1)$. What we will show now is that, with the value of $k$, we have the value of $\xi$ in all rationals.
- Firstly, if $n \in \mathbb{N}$, we have$\xi(n)=\xi(1)...\xi(1)$ $ n$ times=$\xi(1)^n$
- If $-n$ is a negative integer, we have $ 1=\xi(0)=\xi(n-n)=\xi(n)\xi(-n)$, which implies: $\displaystyle \xi(-n)=\frac{1}{\xi(n)}=\frac{1}{\xi(1)^n}=\xi(1)^{-n}$
- If $a=\frac{p}{q} \in \mathbb{Q}$, $ \xi(p)=\xi(q.\frac{p}{q})=\xi(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $ q$ times $ =\xi(p/q)^q$. So, $ \xi(\frac{p}{q})=\xi(p)^{1/q}=(\xi(1)^p)^{1/q}$
We want to find a $\displaystyle \xi_1 (x)=\sum_{n=0}^{\infty}a_nx^n$ that is of the "type" of $ \xi$, meaning: $\xi_1 (x+y)=\xi_1(x)\xi_1(y)$. Do:
$ \displaystyle \xi_1(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the Binomial Theorem.
Notice that the last equality is BEGGING for the Cauchy Product Formula. If $ a_nn!$ was $1$... the last side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. But we have the power to choose who the $a_n$ are! We are searching a candidate to the series, after all. So, take $a_nn!=1$, which implies $ \displaystyle a_n=\frac{1}{n!}$. Now, stare at our luck! Since $\displaystyle a_n=\frac{1}{n!}$, what we have in the last equality is precisely the product $ \xi_1(x).\xi_1(y)$.
So, the function $\displaystyle \xi_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ satisfies all we want. (The series is well defined for all x real, which is easily verifiable by the ratio test. The series is obviously greater than zero is x is greater than 0, and the fact that $\xi_1(x).\xi_1(-x)=\xi_1(0)=1$ implies the series is greater than zero for all x real. Other observations also imply that it is bijective, since it is surjective (follows from the fact that it goes to infinity when x goes to infinity, to zero when x goes to -infinity, and it is continuous) and injective (because it is strictly increasing)
As we saw, it is enough knowing $\xi_1$ in $1$ to know its value in all rationals. But we know what it is in $1$:
$\displaystyle \xi_1(1)=\sum_{n=0}^{\infty}\frac{1}{n!}$
We will call this amazing number $e$.
So, $\xi_1(r)= e^r$ if $r$ is rational, in the usual sense we have for "a number raised to a rational power" (this follows from (1), (2) and (3)). Note that we already have an answer to our initial problem: we found a function $\xi_1$ which is our isomorphism between the groups. But it may be worth a while to try and say a few more things about it.
Since $\xi_1(r)=e^r$ if $r$ is rational, we define :
$e^x:=\xi_1(x)$
when $x$ is real, and this satisfies what we would wait of $e^x$ by previous observations. OK then, we found one function that extends what we call "exponentiation" of rational powers to real powers. But this function is analytic. Is there another less smooth function, for example $ C^{\infty}$ or even only continuous that could extend exponentiation? The answer is not, since this function would coincide in rational points, and the rationals are dense in $\mathbb{R} $ (Two continuous functions that coincide in a dense subset coincide in all $\mathbb{R}$). Then, this extension is unique if we require it to be continuous. This is enough to call $\xi_1(x)$ "the exponential", and have no regrets for calling it $e^x$.
Now, since $e^x$ is bijective, there is an inverse, which we will call $ln(x)$. This allows us to define, for an arbitraty $ a >0 $ :
$a^x:=e^{ln(a).x}$
Since, by definition $ a^1=a$ e $ a^{(x+y)}=a^x.a^y$, it follows from the same considerations as before that this function is what deserves to be called "raising $a$ to a real power".
Thus, we have:
$ (e^x) ^y=e^{(ln(e^x)).y}=e^{(x.y)}$
But going back to our previous question: what if we forget the constraint of a continuous extension: is there another function that coincides with $e^x$ on the rationals?
Yes, but we will need the following theorem to "show" a function of this kind:
Theorem: Every vector space has a basis.
Now, consider $ \mathbb{R}$ as a vector space over $ \mathbb{Q}$, and take a basis $ \{x_i\}_{i \in \Lambda}$ that has $1$. Define:
$f(1):=e$
$ f(x_i):=e^{x_i/2}$ if $ x_i \neq 1$
Since we defined the function on the basis and we want that $f(x+y)=f(x).f(y)$, we define:
$ f(x):=f(x_1)^{\lambda_1}f(x_2)^{\lambda_2}...f(x_n)^{\lambda_n}$,
where $ x=\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n$ is the representation of $x$ as a linear combination of elements of the basis (keep in mind that $\lambda_i$ are rationals!), and everything is well defined since$\{x_i\}$ is a basis. The fact that this function satisfies $f(x+y)=f(x).f(y)$ follows from definition, and the non-continuity of the function follows from the fact that it coincides with the exponentiation on the rationals but is not the exponentiation on the reals (she takes different values on $ x_i$, for example).
We see that requiring continuity is not only very reasonable, but also necessary. Otherwise, we would have a lot of "non-constructive exponentials" to consider.