In Probability Theory by Athreya and Lahiri, they give a very elegant proof of Central Limit Theorem (The Lindeberg one) wherein they use a lemma:
For $x \in \mathbb{R}$ and $r \geq 1$, ($i=\sqrt{-1}$ here)
$$\left|e^{ix} – \sum_{k=0}^{r-1} \frac{(ix)^k}{k!}\right| \leq \min\left\{\frac{|x|^r}{r!},\frac{2|x|^{r-1}}{(r-1)!}\right\}\quad (1)$$
In the proof of the lemma, they give the following expansion of $e^{ix}$ arguing that it follows from Taylor series.
$$e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^k}{k!}\right] + \frac{(ix)^r}{(r-1)!}\int_0^1(1-u)^re^{iux}du \quad (2)$$
By taking absolute values, they would get $(1)$. What I don't get is how did they arrive at $(2)$ in the first place. I have seen Taylor series (and Taylor's theorem) for real valued functions but not for complex valued functions. Moreover I don't recall an integral being used to approximate the remainder.
Hence my question is twofold:
1) Does anyone know the equivalent of a Taylor Theorem for complex functions as above?
2) Is there an alternative way to prove $(1)$ without resorting to Taylor Series? I tried to dominate the remainder terms by a geometric series but that failed.
Good references and proofs/hints for this are welcome. Let me know if you need more details.
Update1: In my second question, what I meant is "Is there an alternate way to prove (1)". You can use Taylor series and find a clever way to bound the higher order terms. As I mentioned, I used a geometric series bound but that didn't work.
Best Answer
By the fundamental theorem of calculus, $$ e^{ix} = 1 + (ix) \int_{0}^{1} e^{iux}\, du, $$ which is the case $r = 1$ of $$ e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^{k}}{k!}\right] + \frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r-1} e^{iux}\, du. \tag*{$P(r)$} $$ (N.B. $(1 - u)^{r-1}$ in the integrand, not $(1 - u)^{r}$.)
Assume inductively that $P(r)$ is true for some $r \geq 1$. Integrating by parts with \begin{align*} U &= e^{iux}, & V &= -\frac{(1 - u)^{r}}{r}, \\ dU &= ix e^{iux}\, du, & dV &= (1 - u)^{r-1}\, du, \\ \end{align*} gives \begin{align*} \frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du &= \frac{(ix)^{r}}{(r - 1)!} \left[-e^{iux}\frac{(1 - u)^{r}}{r}\bigg|_{u=0}^{u=1} + \frac{(ix)}{r} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du\right] \\ &= \frac{(ix)^{r}}{r!} + \frac{(ix)^{r+1}}{r!} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du. \end{align*} Substituting this into $P(r)$ gives $P(r + 1)$.