Proving that nth derivate of $x e^{-x}$ is $(-1)^n (e^{-x})(x-n)$ by induction.

Question

I'm quite stuck on this. How would you prove that the $n^{th}$ derivative of $x e^{-x}$ if the $(-1)^n (e^{-x})(x-n)$ by induction?

I did:

$\frac{d}{dx}(x e^{-x})=(e^{-x}) - x(e^{-x})$

Now I have no idea how to proceed and prove this by induction. I would really appreciate if anyone could guide me. A clear answer would be highly appreciated.

Thanks

Answer 1

Carrying out a proof by induction usually (probably always?) involves three steps:

1. The base step: Proof the assumption for a specific $n$. That is what you already did, you showed that the assumption holds for $n=1$. You have no choice but to simply try it out.

2. Induction step: now you try to show, that, if the assumption holds for a given $n$, it is also valid for the successor of $n$, namely $n+1$. Let me demonstrate: You assume (and write down on your scrap sheet), that the relationship \begin{eqnarray} \frac{d^{n}}{dx^{n}}(xe^{-x})&=&(-1)^{n}(e^{-x})(x-n) \end{eqnarray} holds for a given, but yet unknown $n$. "You assume" means, that for the moment you don't proof it, but rather accept the idea that there might be such a $n$. The question is: Does it also work for $n+1$? Again, you find this out by trying. You want to know what the $(n+1)$th derivative is, so just try it out: \begin{eqnarray} \frac{d^{n+1}}{dx^{n+1}}(xe^{-x})&=&\frac{d}{dx}\left(\frac{d^{n}}{dx^{n}}(xe^{-x})\right)\\ &=& \frac{d}{dx}\left((-1)^{n}(e^{-x})(x-n)\right)\\ &=& (-1)^n(-1)(e^{-x})(x-n)+(-1)^n(e^{-x}), \end{eqnarray} by application of chain rule. Now, what you obviously need is the result: \begin{eqnarray} \frac{d^{n+1}}{dx^{n+1}}(xe^{-x})=(-1)^{n+1}(e^{-x})(x-(n+1)). \end{eqnarray} It means you have to manipulate the result of your differentiation in a way that it resembles this one: \begin{eqnarray} \frac{d^{n+1}}{dx^{n+1}}(xe^{-x})&=& (-1)^n(-1)(e^{-x})(x-n)+(-1)^n(e^{-x})\\ &=& (-1)^{n+1}(e^{-x})(x-n)+(-1)^{n+1}(-1)(e^{-x}) \quad\text{because $(-1)(-1)=1$}\\ &=& (-1)^{n+1}(e^{-x})\left(x-n-1\right)\quad\text{by factoring out $(-1)^{n+1}(e^{-x})$}\\ &=& (-1)^{n+1}(e^{-x})(x-(n+1)). \end{eqnarray}

3. The last step is a rather formal one, but maybe the intelectually most demanding. You have to bring your two results together. In the induction step you showed, that, if there is any $n$ for which the assumption holds, then it also holds for $n+1$. In your base step you showed indeed, that it holds for $n=1$. With the induction step it follows that it is also valid for $n+1=2$. If you set $k=n+1$ you see that it is also valid for $k+1=n+2=3$, and so on. Thus, the assumption is valid for any $n$, because you can reach any be adding $1$ often enough.

According to your task it may be neccessary to show, that it also holds for $n=0$. This is the general scheme to solve such problems. The second step is the most difficult. You should write down what you expect at the end (as I did) and then simply try different algebraic manipulations until you reach the goal. With some practice you get more skilled in that.

Hope that helped you. Kind regards,

Christoph