Taylor expansion of $e^{\cos x}$ I have to find the 5th order Taylor expansion of $e ^{\cos x}$. I know how to do it by computing the derivatives of the function, but the 5th derivative is about a mile long, so I was wondering if there is an easier way to do it.
I'd appreciate any help.
 A: Hint
$$e^{\cos x}=\sum_{n=0}^\infty \frac{(\cos x)^n}{n!}$$
A: You can start with the Taylor expansion of $\cos x$ and when you expand 
$\exp(\cos x)$, you just throw away terms you know won't affect the final result.
For lower order Taylor expansion, the derivation is actually pretty short and straight forward.
$$\begin{align}
\exp(\cos x)
&= \exp\left(1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)\right)\\
&= e\left[1 + \left(-\frac{x^2}{2} + \frac{x^4}{24} + O(x^6)\right) + \frac{1}{2}\left( -\frac{x^2}{2} + O(x^4)\right)^2 + O(x^6) \right]\\
&= e\left[ 1 - \frac{x^2}{2} + \left(\frac{1}{24} + \frac{1}{8}\right) x^4 \right] + O(x^6)\\
&= e\left[ 1 - \frac{x^2}{2} + \frac{x^4}{6} \right] + O(x^6)
\end{align}
$$
A: Use 
$$e^{\cos x}=e\cdot e^{\cos x-1}.$$
Then substitute the power series expansion of $\cos x-1$ for $t$ in the power series expansion of $e^t$. What makes this work is that the series for $\cos x-1$ has $0$ constant term. For terms in powers of $x$ up to $x^5$, all we need is the part $1+t+\frac{t^2}{2!}$ of the power series expansion of $e^t$, and only the part $-\frac{x^2}{2!}+\frac{x^4}{4!}$ of the series expansion of $\cos x -1$. 
A: This article explains how to get the power series for $e^{f(x)}$ if you know the power series for $f(x)$.
A: One needs to keep in mind the point about which we are expanding. 
The question posed doesn't specify, but I guess we are talking about the taylor series about x=0. 
firstly, instead of computing the derivatives (which is rarely the best approach), you can note that the function is a composition of two functions. both of which you probably know the taylor series for. 
What one does (this is elementary, and if you don't know this by the time you are given the above question, you might need to go back to your textbook and revise...) is compute the taylor series of cos(x) about x=0 substitute it into the taylor series of e^x 
the taylor series are: 
$$ \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+H.O.T. \\  
 e^x = 1 + () + \frac{()^2}{2!}+\frac{()^3}{3!}+H.O.T$$
substituting in: 
$$ e^{\cos(x)} = 1 + (1-\frac{x^2}{2!}+H.O.T.)+ \frac{(1-\frac{x^2}{2!}+H.O.T.)^2}{2!}+\frac{(1-\frac{x^2}{2!}+H.O.T.)^3}{3!}\\+H.O.T$$
Now there is a problem: 
we are expanding cos(x) about x = 0
but, we are not expanding e^x about zero but rather about cos(0), which = 1
Therefore you would have to use the formula for e^x about a, where a=1
which is increasingly complicated and computationally heavy. The best way of approaching this problem would be to use the suggestion given by André Nicolas. 
$$e^{\cos x}=e\cdot e^{\cos x-1}.$$ 
A: Hint: the function $f(x) = e^{\cos x}$ is even (that is, $f(x) = f(-x)$).
