Trigonometric version of second difference

Question

I am reading an article Hall P, Turlach BA. On the estimation of a convex set with corners. IEEE Transactions on Pattern Analysis and Machine Intelligence. 1999 Mar;21(3):225-34 and stuck at an identity used there (p. 228). It can be written as follows: $$ p(\theta_i)\sin(\theta_j - \theta_k) + p(\theta_j)\sin(\theta_k - \theta_i) + p(\theta_k)\sin(\theta_i - \theta_j) = -\frac{1}{2}(\theta_i - \theta_j)(\theta_j - \theta_k)(\theta_k - \theta_i)\{p(\theta_j) + p''(\theta_j)\} + O\{(\theta_k - \theta_i)^4\}, $$ where $-\pi < \theta_i < \theta_j < \theta_k < \pi$ and $p:(-\pi, \pi) \to \mathbb{R}$ is a function with three bounded derivatives on $[\theta_i, \theta_k]$.

Authors call this idenitity as trigonometric version of second difference. In the paper $p(\theta)$ is a support function of the convex figure on the plane, but the idenitity does not require this property.

I have tried to prove this idenitity, but failed. Here are my steps:

1. We can write the identity in the equivalent form, by dividing both parts by $(\theta_i - \theta_j)(\theta_j - \theta_k)(\theta_k - \theta_i)$ and multiplying by $-1$: $$ \frac{1}{2}\{p(\theta_j) + p''(\theta_j)\} = \frac{p(\theta_i)\sin(\theta_k - \theta_j) - p(\theta_j)\sin(\theta_k - \theta_i) + p(\theta_k)\sin(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i), $$

2. Substitute sines with first two parts of Taylor series as $\sin x = x - \dfrac{x^3}{6} + O(x^5)$:

$$ \frac{1}{2}\{p(\theta_j) + p''(\theta_j)\} = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} - \frac{1}{6} \frac{p(\theta_i)(\theta_k - \theta_j)^3 - p(\theta_j)(\theta_k - \theta_i)^3 + p(\theta_k)(\theta_j - \theta_i)^3}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i), $$

3. We can write a second divided difference in classic form without sines: $$ \frac{1}{2}p''(\zeta) = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} $$ for some $\zeta \in [\theta_i, \theta_k]$ (see e.g. Bakhvalov N.S. Numerical methods: analysis, algebra, ordinary differential equations. MIR Publishers, 1977. P. 50). Hence, $$ \frac{1}{2}p''(\theta_j) = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i), $$ because $p''(\theta_j) = p''(\zeta) + O(\theta_k - \theta_i)$.

4. After the subtraction of the second identity from the first one, we obtain: $$ \frac{p(\theta_j)}{2} = - \frac{1}{6} \frac{p(\theta_i)(\theta_k - \theta_j)^3 - p(\theta_j)(\theta_k - \theta_i)^3 + p(\theta_k)(\theta_j - \theta_i)^3}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i) $$

The last identity does not seem to be true for any arbitrary function $p(\theta)$. If we try to take $\theta_i = \phi, \theta_j = 2 \phi, \theta_k = 3 \phi$, then

$$ \frac{p(\theta_j)}{2} = - \frac{1}{6} \frac{p(\theta_i)\phi^3 - 8 p(\theta_j)\phi^3 + p(\theta_k)\phi^3}{2 \phi^3} + O(\phi), $$

or, equivalently,

$$ p(\theta_j) = \frac{p(\theta_i) + p(\theta_k)}{2} + O(\phi), $$ and this is not true in the general case.

I need to prove the initial identity. Could someone find a mistake in my calculations?

Answer 1

$\def\peq{\mathrel{\phantom{=}}{}}$Since $p'''$ is bounded, then\begin{align*} p(θ_i) &= p(θ_j) + (θ_i - θ_j) p'(θ_j) + \frac{1}{2} (θ_i - θ_j)^2 p''(θ_j) + O((θ_i - θ_j)^3), \\ p(θ_k) &= p(θ_j) + (θ_k - θ_j) p'(θ_j) + \frac{1}{2} (θ_k - θ_j)^2 p''(θ_j) + O((θ_k - θ_j)^3). \end{align*} Also,\begin{align*} \sin(θ_i - θ_j) &= (θ_i - θ_j) - \frac{1}{6} (θ_i - θ_j)^3 + O((θ_i - θ_j)^4), \\ \sin(θ_j - θ_k) &= (θ_j - θ_k) - \frac{1}{6} (θ_j - θ_k)^3 + O((θ_j - θ_k)^4), \\ \sin(θ_k - θ_i) &= (θ_k - θ_i) - \frac{1}{6} (θ_k - θ_i)^3 + O((θ_k - θ_i)^4). \end{align*} Note that $|θ_i - θ_j| \leqslant |θ_k - θ_i|$, $|θ_j - θ_k| \leqslant |θ_k - θ_i|$. After combining terms with respect to $p(θ_j)$, $p'(θ_j)$, $p''(θ_j)$,\begin{align*} &\peq p(θ_i) \sin(θ_j - θ_k) + p(θ_j) \sin(θ_k - θ_i) + p(θ_k) \sin(θ_i - θ_j) \\ &= A_0 p(θ_j) + A_1 p'(θ_j) + A_2 p''(θ_j) + O((θ_k - θ_i)^4), \end{align*} where\begin{align*} A_0 &= (θ_j - θ_k) + (θ_k - θ_i) + (θ_i - θ_j) \\ &\peq - \frac{1}{6} ((θ_j - θ_k)^3 + (θ_k - θ_i)^3 + (θ_i - θ_j)^3) \\ &= -\frac{1}{6} ((θ_j^3 - 3θ_j^2 θ_k + 3θ_j θ_k^2 - θ_k^3) + (θ_k^3 - 3θ_k^2 θ_i + 3θ_k θ_i^2 - θ_i^3) \\ &\peq + (θ_i^3 - 3θ_i^2 θ_j + 3θ_i θ_j^2 - θ_j^3)) \\ &= -\frac{1}{2} (-θ_i^2 θ_j + θ_i θ_j^2 - θ_j^2 θ_k + θ_j θ_k^2 - θ_k^2 θ_i + θ_k θ_i^2) \\ &= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i), \end{align*}$$ A_1 = (θ_i - θ_j)(θ_j - θ_k) + (θ_k - θ_j)(θ_i - θ_j) = 0, $$\begin{align*} A_2 &= \frac{1}{2} (θ_i - θ_j)^2 (θ_j - θ_k) + \frac{1}{2} (θ_k - θ_j)^2 (θ_i - θ_j) \\ &= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i). \end{align*} Therefore,\begin{align*} &\peq p(θ_i) \sin(θ_j - θ_k) + p(θ_j) \sin(θ_k - θ_i) + p(θ_k) \sin(θ_i - θ_j) \\ &= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i) (p(θ_j) + p''(θ_j)) + O((θ_k - θ_i)^4). \end{align*}