Prove $e^x = 1+x+\frac{x^2}{2!}+O(x^3)$ as $x\to 0^{\pm}$

Question

Context is characteristic functions (or also moment generating functions) in probability.

From Wiki:

To prove $$e^x = 1+x+\frac{x^2}{2!}+O(x^3) \ \text{as} \ x\to 0 \tag{*}$$, can we avoid splitting into cases, i.e. $x\to 0^{\pm}$?

For $x\to 0^{+}$, I guess $$\delta = 1, M = \sum_{n=3}^{\infty} \frac{1}{n!}$$ would work. For $x\to 0^{-}$, I think the same $\delta$ and $M$ could work but if we instead do: $e^x = 1+x+\frac{x^2}{2!}+O(-x^3)$ as $x\to 0^{-}$. I mean, if it's $O(x^3)$ instead, then how could we prove $\underbrace{|\frac{x^3}{3!} + \dots + \frac{x^4}{4!} + \dots |}_{+} \le \underbrace{M}_{+} \underbrace{x^3}_{-}$?

I think $(*)$ should instead be

$$e^x = 1+x+\frac{x^2}{2!}+O(\pm x^3) \ \text{as} \ x\to 0$$

or split into two cases:

$$e^x = 1+x+\frac{x^2}{2!}+O(\pm x^3) \ \text{as} \ x\to 0^{\pm}$$

Answer 1

It is wrong, and it possibly a matter of different definitions. In the case where the term instead big $O$ has to be strictly positive, we may use $O(|x^3|)$ instead of $O(x^3)$ as $x \to 0$. Likely, a proposition that uses $O(x^3)$ has a different definition of big $O$, namely one that does not require strictly positive term instead big $O$ or some other condition.

Answer 2

It's really great that you're thinking this through in such detail. I believe the issue you're encountering is with how big-O notation is defined. Check out https://en.wikipedia.org/wiki/Big_O_notation for details. They even mention this example on that page that "The second expression (the one with $O(x^3)$) means the absolute-value of the error $e^x − (1 + x + x^2/2)$ is at most some constant times $|x^3|$ when $x$ is close enough to $0$."

Answer 3

$$e^x- \left(1+x+\frac{x^2}{2!}\right) = \frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots = x^3\left(\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+\cdots\right)$$

so $|x| \le 1$ then $$\left|e^x- \left(1+x+\frac{x^2}{2!}\right)\right| \le \left|x^3\right|\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots\right)=\left|x^3\right|\left(e-\frac52\right)$$

and thus $ e^x- \left(1+x+\dfrac{x^2}{2!}\right)$ is between $x^3\left(e-\frac52\right)$ and $x^3\left(\frac52 -e\right)$, noting that which of those bounds is greater or less than $0$ depends on the sign of $x$ - either way they are constants times $x^3$

So the difference is $O(x^3)$ for $-1 \le x \le 1$ and you can say $$e^x= 1+x+\dfrac{x^2}{2!} +O(x^3)$$