Calculus I · Building a Function From Its Derivatives

The flagship: $e^x$ rebuilds itself

Now the example the whole chapter was waiting for. Center at $a = 0$. We need all the derivatives of $e^x$ at $0$ — and here is where Chapter 5 pays an enormous dividend. Since $e^x$ is its own derivative, every derivative of $e^x$ is again $e^x$, and at $x = 0$ every one of them equals $e^0 = 1$. So every coefficient is $\tfrac{1}{k!}$, and

$$\boxed{\;e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\;}$$

Look at how inevitable that was. We didn’t grind through a table of messy derivatives; the self-similarity of $e^x$ handed us every coefficient at once. The most beautiful series in mathematics exists because its function refuses to change under differentiation. Test it: $e^{0.1} \approx 1 + 0.1 + \tfrac{0.01}{2} + \tfrac{0.001}{6} = 1.10517\ldots$, and a calculator gives $e^{0.1} = 1.10517\ldots$ — agreement to five places from four terms, because the factorials in the denominators crush the later terms fast. (Figure 7.3: $e^x$ with $T_1, T_2, T_3, T_4$ layered on top, each clinging to the curve over a visibly wider stretch.)

Two more, and an old friend returns

$\dfrac{1}{1-x}$ at $0$. Its derivatives are $f^{(k)}(x) = \dfrac{k!}{(1-x)^{k+1}}$, so $f^{(k)}(0) = k!$, and the coefficient is $\tfrac{k!}{k!} = 1$. Every coefficient is $1$:

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$

That is the geometric series from Chapter 1, now revealed as a Taylor series. And it carries a warning you must respect: this equality only holds for $|x| < 1$. Plug in $x = 2$ and the left side is $\tfrac{1}{1-2} = -1$ while the right side is $1 + 2 + 4 + \cdots = +\infty$. A Taylor series is a local promise; it is true only where it converges. We’ll make that precise with the remainder in a moment.

$\ln(1+x)$ at $0$. Here $f'(x) = \tfrac{1}{1+x}$, and the derivatives give $f^{(k)}(0) = (-1)^{k-1}(k-1)!$, so the coefficient is $\tfrac{(-1)^{k-1}(k-1)!}{k!} = \tfrac{(-1)^{k-1}}{k}$:

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$

Notice the leading term is just $x$. That is the exact reason behind a limit we borrowed back in Chapter 1: $\tfrac{\ln(1+t)}{t} \to 1$ as $t\to 0$, the fact that made $\left(1+\tfrac{x}{n}\right)^n \to e^x$ work. The “loan” was that $\ln(1+t) \approx t$ near $0$ — and there it is, the first Taylor term, finally explained.

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