How to resolve multiply differentiation function algorithms?

Question

My simple function is $f(x)=\frac{1}{2}e^{-x}\sin(2x)$; Can I resolve for multiply differentiation $f^{(n)}=?$ algorithm?

Thx for answer.

Answer 1

Write $f(x)=g(x)h(x)$, where $g(x)=\frac{1}{2}e^{-x}$ and $h(x)=\sin(2x)$.

Then apply the general Leibniz rule: $$ (g \cdot h)^{(n)}=\sum_{k=0}^n {n \choose k} g^{(k)} h^{(n-k)} $$

noting that the high-order derivatives of $g$ and $h$ are easy to find.

You may also enjoy proving that $f^{(n)}$ is of the form $$ \frac{1}{2}e^{-x} (A_n \sin(2 x)+B_n \cos(2 x)) $$ for integers $A_n$ and $B_n$ which can be found recursively.

Answer 2

Edit: A much simpler solution can be obtained writing: $$f(x)=\frac{1}{4i}e^{-x}\left(e^{2ix}-e^{-2ix}\right)=\frac{1}{4i}\left(e^{(-1+2i)x}-e^{(-1-2i)x}\right)\ ,$$ hence $$f^{(n)}(x)=\frac{1}{4i}\left((-1+2i)^{n}e^{(-1+2i)x}-(-1-2i)^{n}e^{(-1-2i)x}\right)\ .$$

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Longer procedure (using recurrence relations): $$f(x)=\frac{1}{2}e^{-x}\sin(2x)$$ $$f'(x)=-\frac{1}{2}e^{-x}\sin(2x)+e^{-x}\cos(2x)$$ $$f''(x)=-\frac{3}{2}e^{-x}\sin(2x)-2e^{-x}\cos(2x)=-5f(x)-2f'(x)$$ You thus reduce to the following linear recurrence relation (http://en.wikipedia.org/wiki/Recurrence_relation).

Solving $\lambda^2+2\lambda+5=0$ you get $\lambda_1=-1+2i$, $\lambda_2=-1-2i$, hence the general solution of your problem is: $$f^{(n)}(x)=A\lambda_1^n+B\lambda_2^n$$ where $A$ and $B$ can be determined by solving $$\left\{\begin{array}{l} f(x)=A+B\\ f'(x)=A(-1+2i)+B(-1-2i) \end{array}\right. \ .$$ Hence $$A=\frac{1}{2}\left[f(x)+\frac{f'(x)+f(x)}{2i}\right]\qquad B=\frac{1}{2}\left[f(x)-\frac{f'(x)+f(x)}{2i}\right]\ .$$

Finally:

$$f^{(n)}(x)=\frac{1}{2}\left[f(x)+\frac{f'(x)+f(x)}{2i}\right](-1+2i)^n+\frac{1}{2}\left[f(x)-\frac{f'(x)+f(x)}{2i}\right](-1-2i)^n\ .$$