Let $n$ be a natural number and $f$ be a polynomial with $\deg(f)\leq n$. We consider the basis $$ \mathcal{B} = (1, x-1, (x-1)^2, \dots, (x-1)^n) $$ of the vector space with all polynomials with degree less or equal $n$.
Question: How does the coordinate vector of $f$ look like?
Approach:
- Let $f = \sum_{k=0}^n a_k x^k$, i.e. the coefficients of $f$ are the $a_k$'s. Then $(f)_\mathcal{E} = (a_0,\dots,a_n)$ where $\mathcal{E} = (1,x, \dots, x^n)$ is the monomial basis. I also know that $(f(t-1))_\mathcal{B} = (a_0,\dots,a_n)$.
- Since we are considering the terms $(x-1)^k$ for $k=0,\dots,n$, it might be useful to use the binomial theorem to obtain $$ (x-1)^k = \sum_{\ell=0}^k \binom{k}{\ell}x^{k-\ell}(-1)^\ell. $$ It would be nice if I could obtain a closed formula.
Could you please help me with this problem? Thank you!
Use $$ f(x)=f(1+(x-1))=\sum_{k=0}^n a_k\sum_{j=0}^k\binom{k}{j}(x-1)^j\\ =\sum_{0\le j\le k\le n} a_k\binom{k}{j}(x-1)^j =\sum_{j=0}^n(x-1)^j\sum_{k=j}^na_k\binom{k}{j} $$