Finding the 19th derivative of $\frac{x-1}{e^x}$ using taylor series

Question

Full context: The problem is multiple choice and originally asks to find the 19th derivative which I prefer to do by Taylor Series. My method get's the answer (after plugging in $x=0$) but it is a bit unsatisfying since it relies on there being multiple choice options and I am wondering if there is a less multiple choice-y way to do the question under time pressure.

I took $$f(x)=e^{-x}(x-1)=\sum_{n=0}^{\infty}\frac{1}{n!}(-x)^{n+1}-\sum_{n=0}^{\infty}\frac{1}{n!}(-x)^{n}$$ A taylor series centered at zero. Then we can equate $$ \frac{f^{(n)}(0)}{n!}=a_n\Rightarrow \frac{f^{(19)}(0)}{19!}=\frac{1}{18!}+\frac{1}{19!}\Rightarrow f^{(19)}(0)=20 $$ Then plugging in, only one solution satisfies $f^{(19)}(0)=20$: $(20-x)e^{-x}$.

I am wondering: Is there a more sound way to find a closed form solution? I tend to not be great at looking for a pattern, especially under pressure, so if that is your method please explain how you go about seeing a pattern quickly after a couple of computations.

Answer 1

I think this maybe useful for you $$y=e^{-x}(x-1)$$ $$y'=-e^{-x}(x-1)+e^{-x}=-y+e^{-x}$$ $$y''=-y'-e^{-x}=y-2e^{-x}$$ if you continue, you will get the following equation for nth derivative $$y^n=(-1)^{n}y+(-1)^{n+1}*n*e^{-x}$$

Answer 2

Hint: By the quotient rule \begin{align*} f'(x) &= \frac{(1)e^{x} - e^x(x-1)}{e^{2x}}\\ &=e^{-x} - \frac{(x-1)}{e^x}\\ &=e^{-x} - f(x) \end{align*} If you want to be rigorous, you can use induction to finish this off! However, its clear from this point that

$$f^n(x) = (-1)^{n+1}ne^{-x} + (-1)^nf(x)$$ (which gives your answer for $n=19$).

Hope this helps.