Why do the antiderivatives of $y=1$ follow $e^{x}$?

Question

If you integrate $y=1$, you get $x$. If you integrate that, you get $\frac{x^{2}}{2}$, following the power rule. If you continue this, integrating over and over, the antiderivatives are: $1$, $x$, $\frac{x^{2}}{2}$, $\frac{x^{3}}{6}$, $\frac{x^{4}}{24}$, and so on. It follows the pattern of $\frac{x^{n}}{n!}$, which happens to be the infinite polynomial of $e^{x}$. Just by integrating $y=1$, you follow the $exp()$ function. Is there any intuitive way why this is?

Answer 1

As a heuristic explanation, let consider

$$f(x)=1+\int_0^x I_0(t) dt+\int_0^x I_1(t)dt+\int_0^x I_2(t)dt+\cdots$$

with $I_0(t)=1$, $I_1(t)=\int_0^t I_0(u) du=t$, $I_2(t)=\int_0^t I_1(u) du=\frac{t^2}2$, $\ldots$, $I_n(t)=\int_0^t I_{n-1}(u) du=\frac{t^n}{n!}$ such that

$$\frac{d(f(x))}{dx}=0+I_0(x)+I_1(x)+I_2(x)+\ldots=f(x)$$

that is

$$f'(x)=f(x) \quad f(0)=1 \implies f(x)=e^x$$

Refer to the related

• Possible definitions of exponential function
• Definition of $\exp(x)$

As noticed in the comments more care is needed to actually prove not only that the sum converges but that we can differentiate the sum term by term.

Answer 2

Here's a less technical, more intuitive sense: what you're building up is the series of antiderivatives of any constant $C$. By doing it an infinite amount of times, and summing all the terms, you'll get $Ce^x$. Note that $1$ is just the $C$ that you won't notice, because it can be ignored, being the identity for multiplication.

If you do this process an infinite number of times, in a "Hilbert hotel" kind of way:

• integrating each term and throwing away the potential constant is like "sliding the term to the right"
• similarly, deriving each term only "slides it to the left".

What's interesting is that, if you run this process to infinity, there's always one more term to pick from "at the end of infinity" (on the right of your sum). This "infinite source" of terms makes it so that the derivative of this power series (infinite polynomial) is itself.

The reason this "resolves to the exponential" specifically, is because the exponential (and more generally, functions of the form $y_0e^{x - x_0}$) are solutions to the equation $f(x) = f'(x)$ for initial conditions $(x_0, f(x_0) = y_0)$.