# Prove that a function is indefinitely differentiable

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.

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.

• Can we simply neglect the coefficients of f and f' ? May 31, 2019 at 20:48
• @Andrei What coefficients are you talking about? May 31, 2019 at 20:52
• @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. 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.

• a very interesting approach to acquire some extra information 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.