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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?

e4494s
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    Note that the derivative of $e^x$ is $e^x$ – J. W. Tanner Oct 10 '21 at 17:02
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    Take any function $g$ and define $f = \sum_{n\in\Bbb Z} g^{(n)}$. Then $f' = f$ so $f=\exp$. But in the case of $g=1$, the derivatives stop when $n≥ 0$ since $g'=0$. – LL 3.14 Oct 10 '21 at 17:14

2 Answers2

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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

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.

user
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  • @ThomasAndrews Yes I think you are right! I fix that. Thanks – user Oct 10 '21 at 17:10
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    All you’ve shown is that if $e^x$ is that sum, then it’s derivative is that sum. But the sum alone doesn’t obviously converge. – Thomas Andrews Oct 10 '21 at 17:11
  • @ThomasAndrews I'm assuming $f'(x)=f(x) \quad f(0)=1 \implies f(x)=e^x$ as a known result/definition. – user Oct 10 '21 at 17:13
  • It does not change the fact that $f$ could be $+\infty$ ... but I do not think it is relevant to the question anyway – LL 3.14 Oct 10 '21 at 17:15
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    This is a nice heuristic explanation but more care is needed to actually prove not only that the sum converges but that you can differentiate the sum term by term. – Eric Wofsey Oct 10 '21 at 17:17
  • @EricWofsey Thanks I add that! – user Oct 10 '21 at 17:18
  • @ThomasAndrews and others, thanks a lot for the comments. I would like also to see your answers to have a complete discussion for these issues. – user Oct 10 '21 at 17:20
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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 $(x_0, f(x_0) = y_0)$.

Tristan Duquesne
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