Show that if $$f: \mathbb{R} \to \mathbb{R}$$ is differentiable in second order which satisfies the equation $$f'' =f'+f$$ then $f$ is indefinitely differentiable.

I was thinking to write the $n$the derivative as $$f^{(n)}=a_nf+b_nf', a_n=b_{n-1}, b_n=a_{n-1}+b_{n-1} $$ I calculated a few derivatives so I think this must be the form of the sequence, but I don't know how to solve it and I cannot see any other pattern to write the general term.


3 Answers 3


Since $f$ is twice differentiable, $f''=f'+f$ is the sum of two differentiable functions, so $f'''$ exists and is equal to $f'''=f''+f'$. By induction you can now show that it's infinitely differentiable, as $f^{(n)}=f^{(n-1)}+f^{(n-2)}$, implying that $f^{(n+1)}$ exists.

  • $\begingroup$ Can we simply neglect the coefficients of f and f' ? $\endgroup$
    – Andrei
    May 31, 2019 at 20:48
  • $\begingroup$ @Andrei What coefficients are you talking about? $\endgroup$
    – DonAntonio
    May 31, 2019 at 20:52
  • 6
    $\begingroup$ @Andrei I see what you are trying to do, but you are overthinking the problem. $$\dfrac{d}{dx}f'' = \dfrac{d}{dx}(f'+f)$$ Then $$f''' = f''+f'$$ You don't need to then convert $f''$ into $f'+f$. You can immediately apply induction. This is the base case. $\endgroup$ May 31, 2019 at 20:52

You wanted to know the general form for $f^{(n)}$ in terms of $f$ and $f'$. The solution Alex R. provided is the correct solution. But, if you just want to see what form the $n$-th derivative takes (in terms of $f$ and $f'$), you have:

$$f^{(n)} = a_nf' + b_nf$$

Take the derivatives of both sides:

$$f^{(n+1)} = a_nf''+b_nf' = a_n(f'+f)+b_nf' = (a_n+b_n)f' + a_nf$$

From this, we get:

$$b_{n+1} = a_n$$

$$a_{n+1} = a_n+b_n$$

So, we have:

$$a_{n+2} = a_{n+1}+b_{n+1} = a_{n+1}+a_n$$

This is the Fibonacci recurrence relation. From this, we obtain:

$$f^{(n)} = F_{n+1}f'+F_nf, n\ge 0$$

where $$F_0=1, F_1=0$$ and $$F_{n+2} = F_{n+1}+F_n, n\ge 0$$ is the Fibonacci sequence.

  • $\begingroup$ a very interesting approach to acquire some extra information $\endgroup$ May 31, 2019 at 21:34

Another way (different way) to look at this is to think of it as an ODE: $$ \frac{d^2 f}{dx^2} - \frac{df}{dx} - f = 0 $$ This is a pretty simple ODE which has the solution of: $$ f(x) = a_1 e^{\frac{\sqrt{5} + 1 }{2}x } + a_2e^{-\frac{\sqrt{5} + 1 }{2}x } $$ Where $a_1$ and $a_2$ are some constant. Since we know that exponential functions are infinite differentiable, this imply that $f(x)$ is infinite differentiable since it is just linear combination of exponential functions.


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