1
$\begingroup$

Let $f(x)=\log(x)$ (the natural logarithm). I'm asked to find a system in $f', f'', f^{(3)}, f^{(4)}$ and use induction to prove my system is correct.

Edit: After the comments I now have the system $f^{(n)}=(-1)^{n+1}\cdot (n-1)! \cdot x^{-n}$

I proved the induction start (obviously since I found a system being true in the beginning), but can't do the induction step - is the guess correct? Could you provide me with some tip to do the induction step?

Thanks

  • 0
    $f'(x)=x^{-1}$, and now the derivatives of higher order are computed byt the elementary rule for (negative) powers. Hence $f''(x)=-x^{-2}$, $f'''(x)=2 x^{-3}$, etc2012-09-18

2 Answers 2

1

Hint: $f^{(n+1)}(x)=\frac{d}{dx}\bigl[f^{(n)}(x)\bigr]$.

Also, you can give an explicit form for $f^{(n)}(x)$ in $x$ only. I think you'd have an easier time of it if you figured that out.

  • 0
    Thanks. I rewrote it to just depend on$x$and it got really easy! I should have just kept looking in the first place instead of assuming it had to be as messy as my first guess, but it's easy to be wise after the event :)2012-09-18
0

Assuming the inductive hypothesis (that the statement holds for $n$), you would verify such a formula by showing that $f^{(n+1)}=(-1)^{(n)}\cdot (n)\cdot f' \cdot f^{(n)}$ holds.

To do that, you would take your formula for $f^{(n)}$ and show that its derivative (you would just need the product rule, for your formula) turns out to fit what your formula predicts for $f^{(n+1)}$.

  • 0
    @Henrik That's OK! It makes a good motivational example of why it's better to append changes you make to the end of your question, in the future.2012-09-18