Exercise 1.4.13* (Pseudo primitives)

Answers

Proof.

a)

By Exercise 12, for all

x \in ℝ

(\binom{x + 1}{k}) - (\binom{x}{k}) = (\binom{x}{k - 1}) .

If we replace $x$ by $x + m$ , and $k$ by $k + 1$ , we obtain

(\binom{x + m + 1}{k + 1}) - (\binom{x + m}{k + 1}) = (\binom{x + m}{k}) .

Therefore

\begin{array}{l} \sum_{m = 0}^{M} (\binom{x + m}{k}) & = \sum_{m = 0}^{M} (\binom{x + m + 1}{k + 1}) - \sum_{m = 0}^{M} (\binom{x + m}{k + 1}) \\ = \sum_{m = 1}^{M + 1} (\binom{x + m}{k + 1}) - \sum_{m = 0}^{M} (\binom{x + m}{k + 1}) \\ = (\binom{x + M + 1}{k + 1}) + \sum_{m = 1}^{M} (\binom{x + m}{k + 1}) - \sum_{m = 1}^{M} (\binom{x + m}{k + 1}) - (\binom{x}{k + 1}) \\ = (\binom{x + M + 1}{k + 1}) - (\binom{x}{k + 1}), \end{array}

\sum_{m = 0}^{M} (\binom{x + m}{k}) = (\binom{x + M + 1}{k + 1}) - (\binom{x}{k + 1}) .

b)

Let

P (x) = \sum_{k = 0}^{n} c_{k} (\binom{x}{k})

be any polynomial, in the form (1.16). Then, by part (a),

\begin{array}{l} \sum_{m = 0}^{M} P (x + m) & = \sum_{m = 0}^{M} \sum_{k = 0}^{n} c_{k} (\binom{x + m}{k}) \\ = \sum_{k = 0}^{n} c_{k} \sum_{m = 0}^{M} (\binom{x + m}{k}) \\ = \sum_{k = 0}^{n} c_{k} [(\binom{x + M + 1}{k + 1}) - (\binom{x}{k + 1})] \\ = \sum_{k = 0}^{n} c_{k} (\binom{x + M + 1}{k + 1}) - \sum_{k = 0}^{n} c_{k} (\binom{x}{k + 1}) \\ = Q (x + M + 1) - Q (x), \end{array}

where

Q (x) = \sum_{k = 0}^{n} c_{k} (\binom{x}{k + 1}) .

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2024-07-27 14:44