$$\begin{align*}f(x)=x^4-2x^3+x-1\end{align*}$$
express $f(x)$ as the following form
$$\begin{align*}c_0+c_1x+c_2x(x-1)+c_3x(x-1)(x-2)+c_4x(x-1)(x-2)(x-3)\end{align*}$$
I know how to express as power series by Ruffini's rule.
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Hint: Let $x=0$. Note that $f(0)=-1$. See what that tells you about $c_0$.
Let $x=1$, and use the information to find $c_1$.
Continue, finding $c_2$ and $c_3$. Finding $c_4$ uses a somewhat different idea.
André Nicolas
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Not intentional, and fixed. – André Nicolas Aug 08 '13 at 15:34
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Got it, is there any relation with Binomial? or other methods?$(x-1)(x-2)(x-3)$ – forlorn Aug 08 '13 at 15:43
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1Yes, there are a lot of relationships. It turns out that expressions of the shape $x(x-1)(x-2)\dots(x-k+1)$ in many ways behave better than powers. Lots of nice identities. – André Nicolas Aug 08 '13 at 15:44
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Wow, glad to know about that, any references? – forlorn Aug 08 '13 at 15:52
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Sorry, cannot think of any. But note that repeated differentiation of $t^\alpha$ yields products of the type in your expression. – André Nicolas Aug 08 '13 at 16:08