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.

  • $\begingroup$ Are you asking how to get a Taylor series for $e^{ix}$? $\endgroup$
    – Aleksandar
    Jun 28, 2015 at 12:26
  • $\begingroup$ No. What I am asking is if anyone knows how to derive (2). And note that I mentioned "Taylor Theorem" i.e. I am looking for something like this: $f(z) = f(0) + zf'(0) +... + \frac{z^n}{n!}f^{(n)}(s)$ for some $s \in \mathbb{C}$. I'm not sure how to write it for complex valued functions or if it is true at all. $\endgroup$ Jun 28, 2015 at 12:53

2 Answers 2


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 $P(r + 1)$.

  • 1
    $\begingroup$ Oh, so there was an error in the book. Didn't realize it. Thanks. Good use of induction. $\endgroup$ Jun 28, 2015 at 13:30

There are various forms for the remainder term of a finite Taylor expansion. One of them is $$f(x)=\sum_{k=0}^nf^{(k)}(a){(x-a)^k\over k!} +\int_a^xf^{(n+1)}(t){(x-t)^n\over n!}\ dt\ ,\tag{1}$$ whereby $f$ is assumed sufficiently differentiable in the neighborhood of $x=a$. This is of course valid as well if $f$ is complex-valued.

For the proof of $(1)$ begin with $$f(x)=f(a)+\int_a^xf'(t)\cdot 1\ dt$$ and set up an induction, using repeated partial integration.


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