Proposition 4.1.2
The sets \((M)\), \((\gamma)\), \((\Delta)\), and \((h^*)\) are bases for the vector space
\[\mathbb{C}[x]_{\le d}
:= \{\,f \in \mathbb{C}[x] : \deg(f)\le d\,\}.\]
Proof: Recall that
$$
\begin{align*}
(M)\text{-basis}: &\{ x^m \mid 0 \leq m \leq d\} \\
(\gamma)\text{-basis}: &\{ x^m(1-x)^{d-m} \mid 0 \leq m \leq d\} \\
(\Delta)\text{-basis}: &\{ \tbinom{x}{m} \mid 0 \leq m \leq d\} \\
(h^{*})\text{-basis}: &\{ \tbinom{x+m}{d} \mid 0 \leq m \leq d\}
\end{align*}
$$
1. \((\gamma) \rightarrow (M) \)
\((M)\) is the canonical basis and the dimension of the vector space is exactly \(d+1\).. To give an explicit change of bases from \((\gamma)\) to \((M)\), we want to take each basis vector of \((\gamma)\) and write it as a linear combination of the basis vectors of \((M)\). So take
$$
\begin{align*}
\gamma_m(x) = x^m(1-x)^{d-m}.
\end{align*}
$$
We want to find coefficients \(a_0,a_1,\cdots,a_d\) such that
$$
\begin{align*}
x^m(1-x)^{d-m} = a_0x^0 + a_1x^1 + \cdots + a_dx^d.
\end{align*}
$$
We can expand \(x^m(1-x)^{d-m}\) using the Binomial Theorem and then multiply by \(x^m\)
$$
\begin{align*}
x^m(1-x)^{d-m} &= x^m\sum_{i=0}^{d-m} \binom{d-m}{i} (-x)^{i} \\
&= \sum_{i=0}^{d-m} \binom{d-m}{i} (-1)^i x^{m+i}.
\end{align*}
$$
Let \(r = m+i\), then
$$
\begin{align*}
x^m(1-x)^{d-m} &= \sum_{r=m}^{d} \binom{d-m}{r - m} (-1)^{r-m} x^{r}.
\end{align*}
$$
2. \((M) \rightarrow (\gamma) \)
Conversely, we can we can express the canonical basis \((M)\) in terms of \((\gamma)\)-basis vectors as follows
$$
\begin{align*}
x^j &= x^j \cdot 1^{d-j} \\
&= x^j \cdot (x+ (1-x))^{d-j} \\
&= x^j \sum_{i=0}^{d-j} \binom{d-j}{i} x^i (1-x)^{(d-j)-i} \\
&= \sum_{i=0}^{d-j} \binom{d-j}{i} x^{i+j} (1-x)^{d-(i+j)}
\end{align*}
$$
Let \(k = i+j\) to get
$$
\begin{align*}
x^j &= \sum_{k=j}^{d} \binom{d-j}{k-j} x^{k} (1-x)^{d-k}
\end{align*}
$$
When \(j = 0\), this reduces to
$$
\begin{align*}
1 &= \sum_{k=0}^{d} \binom{d}{k} x^{k} (1-x)^{d-k}
\end{align*}
$$
which is equation 4.1.1 in the textbook. From (1) and (2), every \((\gamma)\)-basis vector can be expressed in terms of \((M)\) vice versa. Hence both span the same set of vectors. Since each contains \(d+1\) vectors, then they are bases.
Next, we’ll verify that \(\Delta\) is a basis as follows:
1. \((\Delta) \rightarrow (M) \)
Recall that \((\Delta)\text{-basis} = \{ \tbinom{x}{m} \mid 0 \leq m \leq d\}\). Hence we want to express \(\binom{x}{m}\) in terms of the \((M)\)-basis vectors. $\tbinom{x}{m}$$ is a polynomial in $x$ where
$$
\begin{align*}
\binom{x}{0} &= 1 \\
\binom{x}{1} &= x \\
\binom{x}{2} &= \frac{x(x-1)}{2} = \frac{x^2-x}{2} \\
\binom{x}{3} &= \frac{x(x-1)(x-2)}{6} = \frac{x^3-3x^2+2x}{6} \\
\binom{x}{4} &= \frac{x(x-1)(x-2)(x-3)}{24} = \frac{x^4-6x^3+11x^2-6x}{24} \\
\cdots \\
\binom{x}{m} & = \frac{x(x-1)(x-2)\cdots(x-m+1)}{m!}.
\end{align*}
$$
Note that the coefficients in the numerator are also the Stirling numbers of the first kind where
$$
\begin{align*}
\binom{x}{m}
=
\frac{1}{m!}
\sum_{k=0}^{m}
s(m,k)x^k.
\end{align*}
$$
2. \((M) \rightarrow (\Delta) \)
We want to express a monomial \(x^m\) in terms of the \((\Delta)\)-basis vectors so
$$
\begin{align*}
x^m = \sum_{i=0}^m a_i \binom{x}{i}.
\end{align*}
$$
The first few examples are
$$
\begin{align*}
x &= \binom{x}{1}\\
x^2 &= \binom{x}{1} + 2\binom{x}{2}\\
x^3 &= \binom{x}{1} + 6\binom{x}{2} + 6\binom{x}{3}.
\end{align*}
$$
This reduces to
$$
\begin{align*}
x^m
=
\sum_{k=0}^{m}
S(m,k)\,k!\binom{x}{k}.
\end{align*}
$$
where \(S(m,k)\) are the Stirling numbers of the second kind.
1. \((h^*) \rightarrow (M) \)
TODO
References