# Taylor series expansion for $e^{-x}$

could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how

$e^{-i\theta}$=$\cos\theta-i\sin\theta$.

More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.

• Just plug $-x$ into the expansion of $e^x$. – Mark McClure Nov 4 '14 at 11:14
• @Mark McClure:are you saying that the Taylor series expansion becomes $e^{-x}$=$1 - x + \frac{x^2} {2!} - \frac{x^3} {3!} + \frac{x^4} {4!} - \frac{x^5} {5!} +\cdots$ – justin Nov 4 '14 at 11:25
• Yes! This type of manipulation with power series is very common and has some nice applications. Also, note that if you differentiate your power series, you end up with exactly -(what you started with), just as if you differentiated $e^{-x}$. – Mark McClure Nov 4 '14 at 11:29
• @Mark McClure:while if you do $\cos x-i\sin x$ what I got is $e^{-x}$=$1 - x - \frac{x^2} {2!} + \frac{x^3} {3!} + \frac{x^4} {4!} - \frac{x^5} {5!} +\cdots$.which isn't which I think we should obtain. – justin Nov 4 '14 at 11:37
• @Mark McClure:that answer really helped me. – justin Nov 4 '14 at 12:32

$$e^{-x} = \sum\limits_{k=0}^{\infty}\frac{(-x)^k}{k!} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$$ So plugging in $x=i\theta$ we have that \begin{align} e^{-i\theta} &=1-i\theta+\frac{(i\theta)^2}{2!}-\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\dots\\ &=1-i\theta-\frac{\theta^2}{2!}+i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+\dots \end{align} using the fact that $i^2=-1$, $i^3=-i$ and $i^4=1$, etc. The taylor expansions of $\sin$ and $\cos$ are $$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots\hspace{10px}\text{and}\hspace{10px}\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$$ So \begin{align} e^{-i\theta} &=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\dots\right)-i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\dots\right)\\ &=\cos\theta-i\sin\theta. \end{align} As a side-note if you already know that $e^{i\theta}=\cos\theta+i\sin\theta$, then it is easy to show that $e^{-i\theta}$ without using taylor-series using the fact that cosine is even and sine is odd. That is $$e^{-i\theta} = \cos(-\theta)+i\sin(-\theta)=\cos\theta-i\sin\theta,$$ because $\cos(x)=\cos(-x)$ and $\sin(-x)=-\sin(x).$

• that's what I expected.could you show it in terms of $\cos\theta$ and $\sin\theta$ so I could reach at $\cos\theta-i\sin\theta$. – justin Nov 4 '14 at 12:04

$$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\cdots$$ So, taking into account the fact that $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, etc., we get \begin{align} e^{i\theta} &= 1+(i\theta)+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+\frac{(i\theta)^6}{6!}+\frac{(i\theta)^7}{7!}+\cdots \\ &= 1+i\theta+i^2\frac{\theta^2}{2!}+i^3\frac{\theta^3}{3!}+i^4\frac{\theta^4}{4!}+i^5\frac{\theta^5}{5!}+i^6\frac{\theta^6}{6!}+i^7\frac{\theta^7}{7!}+\cdots \\ &= 1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+i\frac{\theta^5}{5!}-\frac{\theta^6}{6!}-i\frac{\theta^7}{7!}+\cdots \\ &= \left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} + \cdots\right) +i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!} \cdots\right) \\ &= \cos(\theta) + i\sin(\theta). \end{align}

At this point, you can plug $-\theta$ in for $\theta$ to get to $$e^{-i\theta} = \cos(\theta) - i\sin(\theta).$$

• :is that because $\cos({-x})=\cos{x}$ and $\sin({-x})=-\sin{x}$. – justin Nov 4 '14 at 12:13
• You're missing an $i$ in your last equation. – Axoren Nov 4 '14 at 12:13
• @justin I think it's more because $(-\theta)^{2n}=\theta^{2n}$ and $(-\theta)^{2n+1}=-\theta^{2n+1}$. – Mark McClure Nov 4 '14 at 12:15
• @Axoren Thanks! I'm surprised (if) that's all I missed. – Mark McClure Nov 4 '14 at 12:16
• @Mark McClure:could you explain the relation of $(\theta^{2n})$ with $\cos\theta$ and $\sin\theta$ – justin Nov 4 '14 at 12:18