What's this problem getting at? (sum and derivative)

Question

Let $$P(x) = \sum_{m=1}^n b_m (x-a)^m $$ Then $$P^{(k)}(x) = \sum_{m=1} ^n b_m(m)(m-1)\cdots(m-k+1)(x-a)^{m-k}$$ Is above correct? But what about when $m < k$, that seems to ruin the formula?

Then, a side question to above assignment is posed, and it asks somewhat mysteriously: "why do we say that P is unambiguously determined by the first $n$ derivatives $P^k(a)$, $k = 1,\ldots,n$?", and I honestly have no idea what's going on. Do they mean that all other derivatives are zero, so there's no more information to get?

When $m < k$ one of the terms, $m-i$, in the product will be zero. — Joel, Apr 06 '15 at 17:33
Just note that you are differentiating a polynomial of degree $n$. — science, Apr 06 '15 at 17:53

science · Answer 1 · 2015-04-08T13:18:54.860

0

You can write it as

$$P^{(k)}(x) = k! \sum_{m=0}^{n} {m\choose k}b_m (x-a)^{m-k}. $$

edited Apr 08 '15 at 13:18

answered Apr 06 '15 at 18:03

science

2,900

1

You've forgotten the coefficients $b_m$ inside the sum. – Alex M. Apr 06 '15 at 19:54
@AlexM.: It's fixed. Thank you. – science Apr 09 '15 at 10:29

What's this problem getting at? (sum and derivative)

1 Answers1