1

While solving another problem I stumbled upon this fact that seems to be true but I do not how to prove.

For a positive integer $ n $, we have $$ \left[ \frac{d^n}{dx^n} \left((x - a)^{n+1} (x - b)^{n+1}\right) \right]_{x=a} = 0. $$

Is this true? If so, how can I prove this? I tried proving it by induction but I failed to reduce the case of ( n + 1 ) to that of ( n ) in a way that would make the induction work. A proof of this statement (if it is true) would be very appreciated by me.

A proof that does not rely on induction is fine too.

Lone Learner
  • 1,076
  • Use the product formula and induction but first replace $n+1$ by $m>n$ and formulate the statement of induction correctly. – markvs Nov 25 '21 at 18:14
  • 3
    Think of it this way. With each derivative you take, you get an extra term involving a lower power of $x-a$. But even after $n$ derivatives, the final extra term you get will still have a factor of $(x-a)^{n+1-n}=x-a$. So at the end, you have a sum in which every term has a factor of $x-a$. – Will R Nov 25 '21 at 18:19
  • 2
    It might be easier to prove by induction that for any polynomial $P(x)$, $\left[\frac{d^n}{dx^n} (x-a)^{n+1}P(x)\right]_{x=a}=0$ – Alan Abraham Nov 25 '21 at 18:26

0 Answers0