I have to do the following task that goes like this:
Calculate the first 5 terms of Taylor expansion of function $f(x) = e^x \cos x$ centered in point $x_0 = 0$ by:
a) forming Cauchy product of Taylor series from $e^x$ and $\cos x$ at point $0$,
b) by finding the derivatives of $f(x)$ and inserting them into the general formula for the Taylor series.
I will now write how I think the first part of the assignment should be done, please correct me if I'm wrong.
Taylor series formula:
$$\sum_{n=0}^{\infty}{\frac{f^{(n)}(x_0)}{n!} (x-x_0)^n}$$
First 5 terms when $x_0 = 0$: $$\begin{align} &f(x) = e^x\cos x = 1\\ &f'(x) = e^x(\cos x - \sin x) = 1\\ &f''(x) = e^x(-2\sin x) = 0\\ &f'''(x) = 2e^x(-\sin x - \cos x) = -2\\ &f''''(x) = 2e^x(-2\cos x) = -4 \end{align} $$
a) We can represent $e^x$ and $\cos x$ as Taylor series:
$$e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}} \quad\text{and}\quad\cos x = \sum_{n=0}^{\infty}{(-1)^n \frac{x^{2n}}{(2n)!}}$$
Now, we also have formula for Cauchy product, which is: $$(\sum_{i=0}^{\infty}{a_i}) (\sum_{j=0}^{\infty}{b_j}) = \sum_{k=0}^{\infty}{c_k}$$, which gives us: $$c_k = \sum_{k=0}^{n}{a_kb_{n-k}}$$
Now I'm starting to be confused, since I couldn't find anywhere solution for the Cauchy product of these to series, so I just tried this (again, please correct me if it's wrong):
$$\sum_{k=0}^{n}{\frac{x^k}{k!}(-1)^{n-k}\frac{x^{2n-2k}}{(2n-2k)!}} = \sum_{k=0}^{n}{(-1)^{n-k}\frac{x^{2n-k}}{k!(2n-2k)!}}$$
If this is correct, is this the final solution? Also, why it was said "at point $0$", should I substitute something with $0$?
b) I don't have a clue what are we suppose to do here. Please help :)
Lastly, we are asked to establish the principle (general formula?) of $n$-th derivative ($n \in \Bbb{N}$) of $f(x)$ and give the complete Taylor expansion of $f(x)$ in point $0$.
P.S. I was translating from German to English, which both aren't my mother languages, so maybe I have mistranslated something.
Sorry for the long post.
Thank you in advance
Hint for (a). Note that $a_n=1/n!$ and $b_{2n}=(-1)^n/(2n)!$ (and zero otherwise). Hence $$\frac{f^{(n)}(0)}{n!}=c_n = \sum_{k=0}^{n}{a_kb_{n-k}}=\sum_{k=0}^{n}a_{n-k}b_{k}=\sum_{k=0}^{\lfloor n/2\rfloor}a_{n-2k}b_{2k}=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{1}{(n-2k)!}\cdot \frac{(-1)^k}{(2k)!}.$$
Hint for (b). By the general Leibniz rule we have that $$f^{(n)}(0)=\sum_{k=0}^n \binom{n}{k} D^{(n-k)}(e^x)_{x=0}\cdot D^{(k)}(\cos(x))_{x=0}\\=\sum_{k=0}^n \binom{n}{k} \cdot D^{(k)}(\cos(x))_{x=0} =\sum_{k=0}^n \binom{n}{2k} (-1)^k.$$ By the way note that $f(x)=\mbox{Re}(e^{x(1+i)})$ and therefore $$f^{(n)}(0)=\mbox{Re}( (1+i)^n).$$