4
$\begingroup$

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.

$\endgroup$

3 Answers 3

19
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ a very interesting approach to acquire some extra information $\endgroup$ May 31, 2019 at 21:34
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.