# If $f'(x) = f(x)$ for all $x \in \mathbb R$, show that $f^{(n)}(x) = f(x)$ for all natural $n$.

If $f'(x) = f(x)$ for all $x \in \mathbb R$, show that $f^{(n)}(x) = f(x)$ for all natural $n$.

This problem seems to be best approached via induction and, as far as I can see, using Taylor's theorem; we are given the fact that for $n = 1$, the case holds. However, I can't quite grasp how to properly establish the truth of the general equality via induction.

• Just differentiate both sides again and again; you can do this because derivative is also differentiable, being equal to $f$. For instance $f''(x)=f'(x)=f(x)$ differentiating both sides of $f'=f$ – user160738 Dec 8 '16 at 3:38
• Huh? If $f ^n (x)=f (x)$ then $f^{n+1}=(f^n (x))'=f '(x)=f (x)$. Surely that's a no brainer. – fleablood Dec 8 '16 at 3:41

Note that if $$f'=f$$, then $$f$$ is differentiable (of course), but also , $$f'$$ is differentiable, because it is equal to a differentiable function ($$f$$) !

So, we can differentiate this equation again, that gives $$f''=f'$$, but then we already know that $$f'=f$$, so that $$f''=f$$.

I think you can perform the induction by yourself. A more interesting question would be( if you don't know the answer yet) to find such a function that satisfies $$f'=f$$.

$f^{'}(x)=f(x)\implies \dfrac{f^{'}(x)}{f(x)}=1$.

Integrating $\ln f(x)=x+C\implies f(x)=e^{x+C}\implies f^{n}(x)=e^{x+C}=f(x)$.

The base case, $n=1$, follows by assumption. Next suppose that for some fixed $n\geq 1$, $f^{(n)}(x)=f(x)$. Then, $$f^{(n+1)}(x)=\left(f^{(n)}(x)\right)'(x)=f'(x)=f(x),$$ as desired.

Sure, induction works.

The induction assumption is that $f^{(n-1)}(x) = f(x).$ You must use this assumption to prove that $f^{(n)}(x) = f(x).$ You can do this using the fact that the n-th derivative is the derivative of the $(n-1)$-th derivative: $$f^{(n)}(x) = \frac{d}{dx}\left(f^{(n-1)}(x)\right) = \frac{d}{dx}\left(f(x)\right) = f'(x) = f(x)$$

Let $n$ be a fixed natural number. By linearity of the derivative, $$f^{(k)}-f^{(k-1)}=(f'-f)^{(k-1)}=0,\quad\forall\ k\in\mathbb{N}\tag{\#}$$ and thus $$f^{(k)}=f^{(k-1)},\quad\forall\ k\in\mathbb{N}.\tag{*}$$ Applying $(*)$ for $k=1$, $k=2$, ..., $k=n$ we get $$f^{(n)}=f^{(n-1)}=\cdots=f''=f'=f.$$ As $n$ is arbitrary, the desired result follows:

$$f^{(n)}=f,\quad\forall \ n\in\mathbb N.$$

Remark. This solution doesn't use induction on $n$ directly. However, the first equality in $(\#)$ (which is a general result independent of your particular problem) is proved by induction on $k$. In other words, this solution uses a general result (which is proved by induction) to solve your particular problem without induction.