Finding the $n$-th derivative of $f(x) =e^x \sin x$, solving the recurrence relation I am trying to find a closed solution for the nth derivative of the function:
$f(x) = e^x \sin x$
So far I have been able to obtain the derivative as:
$f^{(n)}(x) = e^x S_n \sin x + e^x C_n \cos x$
The sequences S and C are defined as below:
$S_n = S_{n-1} - C_{n-1}$
$C_n = S_{n-1} + C_{n-1}$
$S_0 = 1$,
$C_0 = 0$
I have been able to further simplify this by combining the two equations and obtaining:
$C_n = 2S_{n-2}$
$S_n = S_{n-1} - 2 S_{n-3}$
However, I have no idea what to do now. Can anyone help me find the closed form solution?
 A: Note that $S_{n+4} = -4 S_n$ and $C_{n+4} =-4 C_n$.  So $S_{4k+j} = (-4)^k S_j$ and 
$C_{4k+j} = (-4)^k C_j$.  List the cases $j = 0$ to $3$ and you're done.
A: \begin{align}
\left( \begin{array}{r} S_n\\C_n\\ \end{array} \right) &=
\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)
\left(\begin{array}{c} S_{n-1}\\C_{n-1}\\ \end{array} \right)\\
&=\left( \begin{array}{rr}
1 &-1\\
1 &1\\
\end{array}
\right)^n
\left(\begin{array}{c} S_{0}\\C_{0}\\ \end{array} \right)
\end{align}
Approach 1: Compute eigen-decomposition of the matrix and continue.
Approach 2: Eigenvalues are $(1+i)$ and $(1-i)$. Therefore, $S_n$ must be of the form $A(1+i)^n+B(1-i)^n$. Now, use values of $S_0$ and $S_1$ to compute $A$ and $B$. $A=B=1/2$.
After further simplification, we get
$$S_n = \sqrt{2^n} \cos \frac{n\pi}{4}, C_n = \sqrt{2^n} \sin \frac{n\pi}{4}$$
A: The recurrence $S_n=S_{n-1}-2S_{n-3}$ can be solved mechanically. Its auxiliary equation is $x^3-x^2+2=0$. By inspection $-1$ is a solution, so $x+1$ is a factor of the cubic: $$x^3-x^2+2=(x+1)(x^2-2x+2)\;.$$ The other roots are $$\frac{2\pm\sqrt{-4}}2=1\pm i\;,$$ so the solution is of the form $S_n=A(-1)^n+B(1+i)^n+C(1-i)^n$. From the initial values $S_0=S_1=1$ and $S_2=0$ we get
$$\left\{\begin{align*}
&A+B+C=1\\
&-A+B+C+Bi-Ci=1\\
&A+2Bi-2Ci=0
\end{align*}\right.$$
This system has the solution $A=0,B=C=\frac12$, so $$S_n=\frac12\left((1+i)^n+(1-i)^n\right)$$ and $$C_n=(1+i)^{n-2}+(1-i)^{n-2}\;.$$
Now $1+i=\sqrt2 e^{i\pi/4}$ and $1-i=\sqrt2 e^{-i\pi/4}$, so 
$$\left\{\begin{align*}
&S_n=2^{(n-2)/2}\left(e^{in\pi/4}+e^{-in\pi/4}\right)=2^{n/2}\cos\frac{n\pi}4\\
&C_n=2^{(n-2)/2}\left(e^{i(n-2)\pi/4}+e^{-i(n-2)\pi/4}\right)=2^{n/2}\cos\frac{(n-2)\pi}4=2^{n/2}\sin\frac{n\pi}4\;.
\end{align*}\right.$$
A: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
so $$f(x)=\frac{e^{(1+i)x}-e^{(1-i)x}}{2i}$$
Thus $$f'(x)=\frac{(1+i)e^{(1+i)x}-(1-i)e^{(1-i)x}}{2i}$$
$$f''(x)=\frac{(1+i)^2e^{(1+i)x}-(1-i)^ne^{(1-i)x}}{2i}$$
Following in similar manner gives, $$f^n(x)=\frac{(1+i)^ne^{(1+i)x}-(1-i)^ne^{(1-i)x}}{2i}$$
A: Here's a nice way that makes use of complex exponentials and involves no nasty recurrence relations.
Start by noticing that $\sin x$ is the imaginary part of $e^{ix}$.
Define a function $F$ as follows:
$F(x)=e^xe^{ix}=e^x\cos x+i e^x \sin x$
The function $f(x)=e^x\sin x$ can now be expressed as the imaginary part of $F$:
$f(x)=Im(F(x))$
If you write $F$ as $F(x)=e^{(1+i)x}$ you can easily find its $nth$ derivative:
$F^{(n)}(x)=(1+i)^ne^{(1+i)x}=e^x \cdot(1+i)^ne^{ix}$
To make the extraction of the imaginary part of the above expression easier we can rewrite $1+i$ in polar form:
$1+i=\sqrt2 e^{i\pi /4}$
With this we  finally write $F^{(n)}(x)$ as:
$F^{(n)}(x)=e^x\cdot 2^{n/2}e^{in\pi /4}e^{ix}=2^{n/2}e^x e^{i(x+n\pi /4)}$
If you now take the imaginary part you find:
$f^{(n)}(x)=Im(F^{(n)}(x))=2^{n/2}e^x \sin (x+n\pi /4)$
A: I introduced a complete solution to the problem of finding the $n$th derivative and the $n$th anti derivative of many classes of elementary and special functions. The problem you are trying to solve is covered under the power exponential class. See this reference (section 6) or Ph.D. thesis, 2004  (page 85), where you find the general method.  
Here is a formula found based on the general method
$$ f^{(n)}(x)={{\rm e}^{x}}{2}^{\frac{n}{2}}\sin \left( x+\frac{n\pi}{4}  \right)\,. $$
