Exercise 15.2.3

Here are some useful properties of φ ′ .

(a)
φ has period 2 ϖ . Explain why this implies that the same is true for φ ′ .
(b)
φ is an odd function by (15.9). Explain why this implies that φ ′ is even.
(c)
Use (15.9) to prove that φ ′ ( ϖ − s ) = − φ ′ ( s ) .
(d)
Use Proposition 15.2.1 to prove that φ ″ ( s ) = − 2 φ 3 ( s ) .

Answers

Proof. (a) For all s ∈ ℝ , φ ( s + 2 ϖ ) = φ ( s ) . By differentiation, and the chain rule, we obtain

φ ′ ( s + 2 φ ) ( s ) = φ ( s ) .

φ ′ has period 2 ϖ . (b) Since φ ( − s ) = − φ ( s ) for all s ∈ ℝ , the chain rule gives

− φ ′ ( − s ) = − φ ′ ( s ) ,

thus φ ′ is even. (c) By (15.9), φ ( ϖ − s ) = φ ( s ) for all s ∈ ℝ . Then the chain rule gives − φ ′ ( ϖ − s ) = φ ′ ( s ) , thus

φ ′ ( ϖ − s ) = − φ ′ ( s ) , s ∈ ℝ .

(d) By differentiation of φ ′ 2 ( s ) = 1 − φ 4 ( s ) ( s ∈ ℝ ) , we obtain

2 φ ′ ( s ) φ ″ ( s ) = − 4 φ 3 ( s ) φ ′ ( s ) .

If s ≠ ϖ 2 + nϖ , n ∈ ℤ , then φ ′ ( s ) ≠ 0 , so that

φ ″ ( s ) = − 2 φ 3 ( s ) , s ≠ ϖ 2 + nϖ , n ∈ ℤ .

If s = ϖ 2 + nϖ for some integer n ∈ ℤ , since φ is infinitely differentiable, φ ″ is continuous, therefore

φ ″ ( s ) = lim ⁡ t → s , t ≠ s φ ″ ⁡ ( t ) = lim ⁡ t → s , t ≠ s ( − 2 φ 3 ( t ) ) = − 2 φ 3 ( s ) .

Therefore

φ ″ ( s ) = − 2 φ 3 ( s ) , s ∈ ℝ .

□

User profile picture
2022-07-19 00:00
Comments