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How would you show that $p(x)= \sum\limits_{i=0}^n b_i(x-c)^i$ is equivalent to $p(x)=\sum\limits_{i=0}^n a_ix^i$ by expressing the $a_i$ in terms of $b_i$ and $c$?

Also we know that the polynomial $p$ in $P_n$ that interpolates $n+1$ distinct points is unique.

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I really don't understand what are you asking –  Belgi Apr 7 '12 at 21:21

3 Answers 3

We can write $$p(x)=\sum_{i=0}^nb_i\sum_{k=0}^i\binom ikx^k(-c)^{k-i}=\sum_{0\leq k\leq i\leq n}\binom ikx^k(-c)^{k-i}=\sum_{k=0}^n\sum_{i=k}^n\binom ikx^k(-c)^{k-i}$$ hence $p(x)=\sum_{k=0}^n\left(\sum_{i=k}^n\binom ik(-c)^{k-i}\right)x^k$. You get what you want putting $a_k:=\sum_{i=k}^n\binom ik(-c)^{k-i}$.

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A mosquito nuking solution relies on Taylor (Maclaurin) expansion:

$$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

and the fact that $\dfrac{\mathrm d^k}{\mathrm du^k}(u-c)^n=\begin{cases}\frac{n}{(n-k)!}(u-c)^{n-k}&k\leq n\\0&k>n\end{cases}$. Take $f(x)=p(x)$ and $x_0=0$, so that

$$p^{(k)}(0)=\left.\dfrac{\mathrm d^k}{\mathrm dx^k}\sum_{i=0}^n b_i (x-c)^i\right|_{x=0}=\sum_{i=k}^n \frac{b_i i!}{(i-k)!} (-c)^{i-k}$$

and use the fact that $\dbinom{i}{k}=\dfrac{i!}{k!(i-k)!}$ to obtain the answer sought. (Remember also that $\dbinom{i}{k}=0$ if $i < k$.)

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One can exploit the fact that $\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)x^k=(x+c)^k$ to get $$ \begin{align} p(x) &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)p(x)\\ &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)x^i\\ &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^\infty\frac{c^k}{k!}\frac{\mathrm{d}^k}{\mathrm{d}x^k}x^i\\ &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^ic^k\binom{i}{k}x^{i-k}\\ &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^ic^{i-k}\binom{i}{k}x^k\\ &=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{k=0}^n\left(\sum_{i=k}^na_ic^{i-k}\binom{i}{k}\right)x^k\\ &=\sum_{k=0}^n\left(\sum_{i=k}^na_ic^{i-k}\binom{i}{k}\right)(x-c)^k\\ &=\sum_{k=0}^nb_k(x-c)^k\\\end{align} $$

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In essence, this derivation is half-way between those by Davide and JM. –  robjohn Apr 14 '12 at 15:41
    
Neat, operator manipulations... :) –  J. M. Apr 15 '12 at 15:25

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