Polynomial interpolation $n+1$ distinct points 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. 
 A: 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 $a_k:=\sum_{i=k}^n\binom ik(-c)^{k-i}$.
A: 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 $i &lt k$.)
A: One can exploit the fact that $\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)x^k=(x+c)^k$ to get
$$
\begin{align}
p(x)
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)p(x)\\
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\exp\left(c\frac{\mathrm{d}}{\mathrm{d}x}\right)x^i\\
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^\infty\frac{c^k}{k!}\frac{\mathrm{d}^k}{\mathrm{d}x^k}x^i\\
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^ic^k\binom{i}{k}x^{i-k}\\
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{i=0}^na_i\sum_{k=0}^ic^{i-k}\binom{i}{k}x^k\\
&=\exp\left(-c\frac{\mathrm{d}}{\mathrm{d}x}\right)\sum_{k=0}^n\left(\sum_{i=k}^na_ic^{i-k}\binom{i}{k}\right)x^k\\
&=\sum_{k=0}^n\left(\sum_{i=k}^na_ic^{i-k}\binom{i}{k}\right)(x-c)^k\\
&=\sum_{k=0}^nb_k(x-c)^k\\\end{align}
$$
