 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/p/p4/a/solution

Subtract the two finite sums. In the shifted sum replace the summation index

k

k - 1

. For every interior index the new coefficient is

(- 1)^{k} (k r) - (- 1)^{k - 1} (k - 1 r) = (- 1)^{k} (k r + 1)

by Pascal's identity; the two endpoint coefficients agree as well. Therefore

Q_{(r)} (x) - Q_{(r)} (x - h) = Q_{(r + 1)} (x) .

For a degree-

d

c \neq = 0

, the backward difference has leading term

c (x^{d} - (x - h)^{d}) = c d h x^{d - 1} + lower-degree terms .

Since

h > 0

, this coefficient is nonzero for

d \geq 1

. Starting with

Q_{(0)} = P

n - r

c h^{r} n! / (n - r)!

for

0 \leq r \leq n

. The next difference of the resulting constant is zero:

de g Q_{(r)} = n - r (0 \leq r \leq n), Q_{(n + 1)} \equiv 0.

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