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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?


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$f'(x)=x^{-1}$, and now the derivatives of higher order are computed byt the elementary rule for (negative) powers. Hence $f''(x)=-x^{-2}$, $f'''(x)=2 x^{-3}$, etc – Siminore Sep 18 '12 at 12:40
up vote 1 down vote accepted

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.

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Yeah, the formula in $x$ is definitely more convenient to verify. And if the poster really does need it in terms of $f^{(i)}$ then it wouldn't be too hard to translate. – rschwieb Sep 18 '12 at 12:44
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 :) – Henrik Sep 18 '12 at 14:29

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 $f^{(n+1)}$.

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How did the $f\prime$ get in there? – Ben Millwood Sep 18 '12 at 13:31
@BenMillwood Aside from substituting $n$ for $n-1$, that is verbatim what the OP wrote in the original post. Everything was originally in terms of $f$ and its derivatives. I should also clarify that I made no attempt to review the correctness of what is there: I just gave the necessary strategy (which is what the OP asked for). – rschwieb Sep 18 '12 at 13:43
Sorry rschwieb I didn't need to undermine your answer - it was most appreciated. I chose to write it in terms of x instead since I agreed it looked nicer - the product rule you proposed I had already tried but without luck (probably because my expression was so ugly). – Henrik Sep 18 '12 at 14:27
@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. – rschwieb Sep 18 '12 at 16:36

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