Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I really don't understand what are you asking – Belgi Apr 7 '12 at 21:21

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 $a_k:=\sum_{i=k}^n\binom ik(-c)^{k-i}$.

share|cite|improve this answer

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 $i < k$.)

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.