Characterizations of $e^x$

Question

I've been thinking about the following problem for a while:

(AFAIK) the 'exponential function', $e^x$ can be characterized as the unique solution to the following differential equation with initial conditions specified:

$f'(x) = f(x)$ , $f(0) = 1$.

Prior to learning this, I thought of $e^x$ in the following way:

I first thought of $e$ to be the supremum of the sequence $u_n =(1+ \frac{1}{n})^n$, and since $u_n$ can be shown to be monotone increasing and bounded above, this is the same as defining it to be $e = \lim_{n \to \infty} (1+\frac{1}{n})^n$

I then showed that via this characterization, we can also show that $e= lim_{h \to 0} (1+h)^{\frac{1}{h}}$, which allows us to show that the function $e^x$ has a nice property, namely, that its own derivative is itself, which immediately makes it $C^{\infty}$.

Using this, as well as the Lagrange form of the remainder for the Taylor polynomial, I was then able to show that $\displaystyle e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$

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I then wanted to try and show that the characterization of $e^x$ as the unique function that satisfies:

$f'(x) = f(x)$ , $f(0) = 1$.

and the characterization I previously thought about were equivalent.

So I started with showing that $f'(x) = f(x)$ , $f(0) = 1$ implied my previous characterization, as my previous characterization is a solution to $f'(x) = f(x)$ , $f(0) = 1$

I noted that as $f$ was defined at $0$, and that $f$ had a first derivative at $0$, it can be inferred that $f$ is continuous at $0$, and exists in some neighborhood, lets say $(-k,k)$ of $0$.

Now I also realized if $f$ is a solution to this differential equation, it is $C^{\infty}$, and all of its derivatives must exist in the neighborhood $(-k,k)$ of $0$.

Then I applied the Lagrange form of the remainder to this function to show that the error as this function is approximated with $\sum_{i=0}^n \frac{x^i}{i!}$ tends to $0$ as $n \to \infty$.

So, in $(-k,k)$, the function must be $e^x$, as $e^x$ can also be made arbitrarily close to $\sum_{i=0}^n \frac{x^i}{i!}$ given $n$ large enough.

Okay, so I understand that $f(x) = e^x$ in $(-k,k)$, and I realize that $\sum_{i=0}^{\infty} \frac{x^i}{i!}$ has an interval of convergence of $\mathbb{R}$, where it is uniformly convergent, and so $f$ even has a smooth uniformly convergent continuation outside $(-k,k)$, which is also just $e^x$.

But this is what I am struggling to understand, why does $f$ need to be defined on all of $\mathbb{R}$, as in, why does $f$ need to actually be the same function as $e^x$, same domain and all?

Surely $f$ could be defined on any arbitrarily small interval $(-k,k)$ around zero, and still fulfill its defining differential equation, but not be defined outside of $(-k,k)$, so while within $(-k,k)$ it is $e^x$, it may not exist outside $(-k,k)$ as itself, but only as a smooth unique continuation of itself (namely, $e^x$).

Another problem I have related to this is when talking about solutions to the following differential equation (again with specified initial conditions):

$f''(x) = -kf(x)$, $f(0) = A$, $f'(0) = 0$, I realize that $f(x)$ is essentially the cosine function, and that this is in fact the characterizing equation of simple harmonic motion, which a pendulum approximates when it swings back and forth with small incline angle.

I could show this in a similar way, by talking about how $f$ is again $C^{\infty}$ and then showing that within whatever interval its defined in its equal to a function that is basically cosine, but now my question is, what makes us believe that $f(x)$, which can be thought of as the displacement of a particle doing simple harmonic motion, is defined outside a small open interval containing $t=0$? How can we model these particles with cosine in real life if you can't directly conclude all solutions to $f''(x) = -kf(x)$, $f(0) = A$ have to be defined themselves on all of $\mathbb{R}$ (and not as themselves on $(-k,k)$ and then as a unique smooth continuation (cosine) on $\mathbb{R} \setminus (-k,k)$).

I would really appreciate it if someone were to help me understand this better, and I am sorry if I have been unclear

Answer 1

If for some $a<0<b$ we have $f\colon(a,b)\to\Bbb R$ with $f'(x)=f(x)$ for all $x\in(a,b)$ and $f(0)=1$, note that for any fixed $c\in(a,b)$ with $f(c)\ne 0$, we can define $$g_c(x)=\frac{f(x+c)}{f(c)} $$ which gives us a function $g_c\colon (a-c,b-c)\to \Bbb R$ with $$g_c'(x)=\frac{f'(x+c)}{f(c)}=\frac{f(x+c)}{f(c)}=g_c(x)$$ for all $x\in(a-c,b-c)$. In particular, if $0\in(a-c,b-c)$ we have $g_c(0)=\frac{f(c)}{f(c)}=1$ and by uniqueness, $f(x)=g_c(x)$ for all $x\in(a,b)\cap(a-c,b-c)$. This allows us to glue the solutions together to a solution of the differential equation that is defined on $(a,b)\cup(a-c,b-c)$. As $f(c)\ne 0$ certainly holds for $c$ in some neighbourhood of $0$, we conclude that our solution can in fact be extended to all of $\Bbb R$.

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Remark. By the way, we have shown in passing that $f(x+y)=g_y(x)f(y)=f(x)f(y) $ whenever $f(y)\ne 0$. It follows that the continuous function $h(x)=f(x)f(-x)$ can take on only the values $1$ and $0$. As $h$ is defined on all of $\Bbb R$, we conclude that $h(x)=1$ for all $x$ and so $f(x)\ne 0$ for all $x$ (in fact, $$ \tag1f(x)>0$$ per Intermediate Value Theorem) and so $$\tag2f(x+y)=f(x)f(y) $$ for all $x,y\in\Bbb R$. Finally, we have another important property of $f$, namely $$ \tag3f(x)\ge 1+x.$$ Indeed, by $(1)$ we have $f''(x)=f'(x)=f(x)>0$ so that $f$ is a convex function. Therefore it stays above each of its tangents, and $y=1+x$ is the tangent at $(0,f(0))$. It is remarkable how many properties about the exponential function can be shown by using only $(1)$, $(2)$, $(3)$.

Answer 2

Reading your long post, I come to the conclusion that you have some working definition of $a^{b}$ for $a, b \in \mathbb{R}$ and $a > 0$ (that's why you are talking about limit $e = \lim_{h \to 0}(1 + h)^{1/h}$). Although a working definition of $a^{b}$ without using any knowledge of $e^{x}$ is hard, I will assume that you have already managed to solve this hard problem.

I am bit surprised as to why you want to limit your solution to the differential equation $y' = y, y(0) = 1$ in the interval $(-k, k)$. It is rather better to establish the following theorem:

Theorem: There is a unique function $f:\mathbb{R}\to\mathbb{R}^{+}$ such that $f'(x) = f(x), f(0) = 1$. Moreover any solution to the differential equation $$\frac{dy}{dx} = y$$ is given by $y(x) = y(0)f(x)$ where $f$ is the unique function mentioned earlier.

For a proof see this answer.

Further it is easy to show that the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all $x, y$ and with some more effort we can show that $f(1) = \lim_{n \to \infty}(1 + (1/n))^{n}$. The part regarding $f(1)$ as a limit is done by defining another function $$g(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ and showing that $g'(x) = g(x), g(0) = 1$ so that $g$ is the same unique function $f$ mentioned in the theorem above. See this answer for more details.

The solution to $y'' = -ky$ where $k > 0$ is essentially based on theory of circular functions. It is easy to show that if $f$ is a function satisfying $f''(x) = -f(x)$ for all $x \in \mathbb{R}$ then $$f(x) = f(0)\cos x + f'(0)\sin x$$ for all $x$. For a proof see later part of this blog post. The equation $y''= -ky$ is then solved by putting $t = \sqrt{k}x$.