Inspired by the fact that 0! = 1 and 1! = 1, let h(x) satisfy(i) h(x) = 1 for all 0 ≤ x ≤ 1, and(ii) h(x) = xh(x − 1) for all x ∈ R.(a) Find a formula for h(x) on [1, 2], [2, 3], and [n, n + 1] for arbitrary n ∈ N.(b) Now do the same for [−1, 0], [−2,−1], and [−n,−n + 1].(c) Sketch h over the domain [−4, 4]. Notice that h(x) satisfies h(n) = n! and it is at least continuous for x ≥ 0, but its piecewise definition and its many non-differentiable corners disqualify it from being our sought after factorial function. One legitimate conclusion that arises out of this exercise is that x!, when we find it, will exhibit the same asymptotic behavior as h at x = −1,−2,−3, . . . , and thus won’t be defined on the negative integers.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
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Inspired by the fact that 0! = 1 and 1! = 1, let h(x) satisfy
(i) h(x) = 1 for all 0 ≤ x ≤ 1, and
(ii) h(x) = xh(x − 1) for all x ∈ R.
(a) Find a formula for h(x) on [1, 2], [2, 3], and [n, n + 1] for arbitrary n ∈ N.
(b) Now do the same for [−1, 0], [−2,−1], and [−n,−n + 1].
(c) Sketch h over the domain [−4, 4].

Notice that h(x) satisfies h(n) = n! and it is at least continuous for x ≥ 0, but its piecewise definition and its many non-differentiable corners disqualify it from being our sought after factorial function. One legitimate conclusion that arises out of this exercise is that x!, when we find it, will exhibit the same asymptotic behavior as h at x = −1,−2,−3, . . . , and thus won’t be defined on the negative integers.

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