Exercise 1.4.12* (Pseudo derivatives)

Question

Show that $\binom{x+1}{k} - \binom{x}{k} = \binom{x}{k-1}$ when $k$ is a positive integer and $x$ is a real number. Show that if $P(x)$ is given by
$$P(x) = \sum_{k=0}^n c_k \binom{x}{k} \qquad (Ex. 1.4.11),$$
then
$$\Delta P (x) = \sum_{k=1}^n c_k \binom{x}{k-1}.$$
Note the similarity to the formula for the derivative of a polynomial.

Show that if $P(x)$ is a polynomial with real coefficients and of degree $n$, then $\Delta P$ is a polynomial of degree $n-1$.

Richard Ganaye · Accepted Answer

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\begin{proof} We know from Pascal's formula (1.14) that, for all $n \in \N$
$$\binom{n+1}{k} - \binom{n}{k} = \binom{n}{k-1}.$$
Therefore the polynomial $P(X) = \binom{X+1}{k} - \binom{X}{k} - \binom{X}{k-1}$, where $X$ is a variable (indeterminate) has infinitely many roots (all the natural integers), therefore is identically zero, so for all $x\in \R$,
$$\binom{x+1}{k} - \binom{x}{k} = \binom{x}{k-1}.$$
If $P_k(x) = \binom{x}{k}$, where $k\geq 1$, then, for all $x \in \R$,
\begin{align*}
\Delta P_k(x) &= P_k(x+1) - P_k(x) \
&= \binom{x+1}{k} - \binom{x}{k}\
&= \binom{x}{k-1}\
&= P_{k-1}(x).
\end{align*}
Note that $\Delta P_0 = 0$.

Moreover, $\Delta$ is a linear operator: if $f,g$ are some functions, and $\alpha, \beta$ are real numbers, then for all $x \in \R$
\begin{align*}
\Delta(\alpha f + \beta g)(x) &= (\alpha f + \beta g)(x+1) - (\alpha f + \beta g)(x)\
&= \alpha (f(x+1) -f(x)) + \beta g(x+1) -g(x))\
&= (\alpha \Delta f + \beta \Delta g)(x),
\end{align*}
thus $$\Delta(\alpha f + \beta g) = \alpha \Delta f + \beta \Delta g.$$
Using the linearity of $\Delta$, if $P(x) = \sum_{k=0}^n c_k P_k(x) =\sum_{k=0}^n c_k \binom{x}{k}$, then
\begin{align*}
\Delta P (x) &= \sum_{k=0}^n c_k \Delta P_k(x) \
&=\sum_{k=1}^n c_k  P_{k-1}(x)\
&= \sum_{k=1}^n c_k \binom{x}{k-1},
\end{align*}
so
$$\Delta P (x) =  \sum_{k=1}^n c_k \binom{x}{k-1}.$$
If $\deg(P) = n$, then 
$$\deg(\Delta P) = \deg\left ( \sum_{k=1}^n c_k P_{k-1} ight) \leq  \max(\deg(P_0),\ldots, \deg(P_{n-1})) = n-1.$$
Since $\deg(P) = n$, $c_n 
e 0$, thus $\deg(c_n P_{n-1}) = n-1$. 
Moreover ${\deg\left ( \sum_{k=1}^{n-1} c_k P_{k-1} ight) < n-1}$, so we obtain
$$\deg(\Delta P) = n-1.$$
\end{proof}

Exercise 1.4.12* (Pseudo derivatives)

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Exercise 1.4.12* (Pseudo derivatives)

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