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$h\in \mathbb{R}$, because we have defined the Trigonometric Functions only on $\mathbb{R}$ so far.

I have a look at $e^{ih}=\sum_{k=0}^{\infty}\frac{(ih)^k}{k!}=1+ih-\frac{h^2}{2}+....$

How can one describe the nth term of the sum?

Then I look at $\frac{e^{ih}-1}{h}=\frac{(1-1)}{h}+i-\frac{h}{2}+...=i-\frac{h}{2}+....$

Again how can I describe that the nth term of the sum?

Because $\frac{e^{ih}-1}{h}=\sum_{k=1}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}<\sum_{k=0}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}=\sum_{k=1}^{\infty}\frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$

and $ih^{-1}$ is a complex number and the exponential-series converges absolutely for all Elements in $\mathbb{C}$, I have found a convergent majorant. And I can apply the properties of Limits on $\frac{e^{ih}-1}{h}\forall, h\in \mathbb{R}$.

How can I now prove formally (i.e by chosing an explicit $\delta$) that

$$\forall_{\epsilon>0}\exists_{\delta>0}\forall_{h\in\mathbb{R}}|h-0|=|h|<\delta\Longrightarrow |(\frac{e^{ih}-1}{h}=i-\frac{h}{2}+...)-i|<\epsilon$$

I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $\delta$ ).

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    $\begingroup$ (1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $\mathbb C$ $\endgroup$ – Calvin Khor Jan 23 at 13:42
  • $\begingroup$ @CalvinKhor On (2), If I would put $|\cdot |$ around the sums would that work? $\endgroup$ – RM777 Jan 23 at 13:47
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    $\begingroup$ You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|\cdot|$s everywhere) $\endgroup$ – Calvin Khor Jan 23 at 13:49
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Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:

Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $\mathbb C$ are such that $\sum_{n=0}^\infty f_n(h) := \lim_{N\to\infty} \sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $h\in [-a,a]$ (i.e. $\sum_{n=0}^\infty \|f_n\|_{\infty} < \infty $). Then $$ \lim_{h\to 0} \sum_{n=0}^\infty f_n(h) = \sum_{n=0}^\infty f_n(0)$$

Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := \sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=\sum_{n=0}^\infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.


Application: $$\frac{e^{ih}-1}{h}\\= \frac{(\sum_{n=0}^\infty (ih)^n/n! )- 1} {h} \\= \frac{\sum_{n=1}^\infty (ih)^n/n! } {h} \\= \sum_{n=1}^\infty \frac{i(ih)^{n-1}}{n!} \\= \sum_{m=0}^\infty \frac{i(ih)^{m}}{(m+1)!} $$ After you verify we can apply the theorem, this yields $\lim_{h\to 0 } \frac{e^{ih}-1}{h} = i+0+0+\dots = i$.


PS make sure you realise that all infinite sums are defined in terms of limits in $\mathbb C$ (e.g. $a_n \to a \in\mathbb C$ iff $|a_n - a| \to 0$)

PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.

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  • $\begingroup$ Nice answer, Calvin. I was wondering if you had a source or proof of the theorem? $\endgroup$ – Jam Jan 23 at 14:15
  • $\begingroup$ @Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help? $\endgroup$ – Calvin Khor Jan 23 at 14:16
  • $\begingroup$ It does. Thank you $\endgroup$ – Jam Jan 23 at 14:17
  • $\begingroup$ @Jam you're welcome $\endgroup$ – Calvin Khor Jan 23 at 14:21
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    $\begingroup$ @RM777 I have added the proof to the answer, as in the comment to Jam $\endgroup$ – Calvin Khor Jan 23 at 22:11
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Hint: Use Euler's formula and split the limit into well known trigonometric limits.

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Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.

To reiterate the question for clarity we have $h\in\mathbb{R} $ and the symbol $e^{ih} $ is defined by the series $\sum_{n=0}^{\infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that \begin{align*} \left|\frac {e^{ih} - 1}{h}-i\right|&=\left|\frac{i^2h}{2!}+\frac{i^3h^2}{3!}+\dots\right|\\ &\leq\frac{|h|}{2!}+\frac{|h|^2}{3!}+\dots\\ &\leq\frac{|h|} {2}+\frac{|h|^2}{2^2}+\frac{|h|^3}{2^3}+\dots\\ &=\frac{|h|}{2-|h|}\text{ (sum of infinite GP)} \\ &<|h| \end{align*} Thus given any $\epsilon>0$ if we choose $\delta=\min(1,\epsilon) $ then we have $$\left|\frac{e^{ih} - 1}{h}-i\right|<\epsilon $$ whenever $0<|h|<\delta$. Therefore by definition of limit we have $$\lim_{h\to 0}\dfrac{e^{ih}-1}{h}=i$$


Hardy uses the above limit to prove the formula $$e^{ix} =\cos x+i\sin x \,\forall x\in\mathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(\cos x-i\sin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =\cos x+i\sin x$.

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Show that $\lim_{h\rightarrow 0}\frac{e^{ih}-1}{h}=i$:

As we know: $ \cos \theta + i \sin \theta = e^{i \theta}$

So, $$\lim \limits_{h\rightarrow 0} [\frac { \cos h + i \sin h - 1 } {h} ] = \lim \limits_{h \rightarrow 0} [ \frac {1 - \frac{h^2}{2!} + \phi_1 (h^4) + i( h - \frac{h^3}{3!}... \phi_2(h^5) )-1} {h} ] = \lim \limits_{h \rightarrow0} \frac { i(1 - higher \space powers\space of\space h) }{1} = i$$

Alternatively,

$$e^{ih} = 1 + ih + \frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!

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