Taylor Series Expansion of the Exponential Function near $0$ simple question

Question

We all know that $1+x$ is approximately equal to $e^x$ for small $x$ near $0$. But can one actually claim that $1+x=e^x$ for a sufficiently small $x$ near $0$? Thanks in advance.

Answer 1

Over the Reals

The equation $e^x = 1 + x$ considered over the real numbers holds only for $x = 0$.

We can see that for $f(x) = e^x - x - 1$, we have a root at $x = 0$: $f(0) = e^0 - 0 - 1 = 1-1 = 0$.

We can also see that $f'(0) = e^x - 1\rvert_{x=0} = e^0 - 1 = 1-1 = 0$, so $x=0$ is also a minimum, and as $f''(x) = e^x$ is everywhere-positive, $x=0$ is a global minimum.

Thus, $x = 0$ is the only root of $f(x)$.

In Other Contexts

In other contexts, we can indeed have $\exp(x) = 1 + x$ for nonzero $x$:

Dual Numbers

The dual numbers are a hypercomplex number system $\mathbb{R}[\varepsilon]$, where $\varepsilon \neq 0$ is a unit such that by definition $\varepsilon^2 := 0$. A dual number is a number $a + b\varepsilon$, with real part $a$ and dual part $b$. It is plain to see that in the dual numbers, for any $x = b\varepsilon$, we have

$$\exp(x) = \exp(b\varepsilon) = \sum_{n = 0}^\infty{\frac{(b\varepsilon)^n}{n!}} = 1 + b\varepsilon + (b\varepsilon)^2 \sum_{n = 2}^\infty{\frac{(b\varepsilon)^{n - 2}}{n!}} = 1 + b\varepsilon = 1 + x\text{ .}$$

$\varepsilon$ may be thought of as an infinitesimal, and in that case you could say that for all points $x$ infinitesimally close to $0$, we have $\exp(x) = 1 + x$.

Square Matrices

For a nilsquare $n \times n$ matrix $\mathbf{X}$ (i.e. a matrix such that $\mathbf{X}^2 = \mathbf{0}_n$), we have

$$\exp(\mathbf{X}) = \sum_{n = 0}^\infty{\frac{\mathbf{X}^n}{n!}} = \mathbf{I}_n + \mathbf{X} + \mathbf{X}^2 \sum_{n = 2}^\infty{\frac{\mathbf{X}^{n - 2}}{n!}} = \mathbf{I}_n + \mathbf{X}\text{ .}$$ In fact, this is a generalization of the case of the dual numbers, as the dual numbers are isomorphic to a certain ring of $2 \times 2$ matrices; $\mathbb{R}[\varepsilon] \cong \left\langle \mathbf{I}_2, \mathbf{E}\right\rangle$, where $$\mathbf{I}_2 = \begin{bmatrix}1 &0\\ 0 &1 \end{bmatrix}, \qquad \mathbf{E} = \begin{bmatrix}0 &1\\ 0 &0\end{bmatrix}\text{ ,}$$ under the isomorphism $\varphi \colon a + b \varepsilon \mapsto a\mathbf{I}_2 + b \mathbf{E}$.

Rings

In fact, this is true for any ring $R$. For any nilsquare element $a \in R$, i.e. any element such that $a^2 = 0$, we have $$\exp(a) = \sum_{n = 0}^\infty{\frac{(a)^n}{n!}} = 1 + a + a^2 \sum_{n = 2}^\infty{\frac{a^{n - 2}}{n!}} = 1 + a\text{ .}$$ This generalizes the matrix result, as $M_n(\mathbb{K})$ is a ring. We can see that over any ring, $x = 0$ satisfies $\exp(x) = 1 + x$, as definitionally zero is nilsquare.