Tricky exponentialccomposite function

Question

I have a composite function problem. Given $f(x) =e^{-x}$ Find $f^{215} (x)$ I use graphing calculator and I realise $f^4(x$) is the same as $f^2(x)$. With this is mind I'd say $f^{215}$ is the same as $f^3$.

But how could I show it.

Answer 1

We have that

• $f'(x)=\frac{d}{dx}e^{g(x)}=e^{g(x)}\cdot g'(x)=-f(x)$
• $f''(x)=f(x)$
• $f'''(x)=f'(x)=-f(x)$
• $\dots$

therefore your guess is correct.

To prove it rigorously, from $f'(x)=-f(x)$, we have that for any $n\in N$

$$f^{2n-1}(x)=-f(x)\implies f^{2n}(x)=-\frac{d}{dx}f^{2n-1}(x)=-\frac{d}{dx}f(x) =-f'(x)=f(x)$$

$$\implies f^{2n+1}(x)=\frac{d}{dx}f^{2n}(x)=\frac{d}{dx}f(x)=-f(x)$$

and we are looking for $n$ from $1$ to $107$.