Need to prove trigonometric equation using definition of forward difference operator

Question

I am trying to prove

$ \triangle^{2}\sin(\log x)=2\cos(\log \sqrt{x^{2}+xh})\sin(\log\sqrt{1+\frac{h}{x}}) $

Using definition of forward difference operator

$\triangle f(x)=f(x+h)−f(x) $

But I can't get my head around it, I have basic knowledge of interpolation which I think is not enough for proving this, can anybody help

Answer 1

NOTE: I think instead of the second difference ($\Delta^2$) the question is about the first difference ($\Delta^1$). You may want to recheck your problem.

Using $\sin A -\sin B=2 \sin\frac{(A-B)}{2} \cos\frac{(A+B)}{2}$, we get

\begin{align*} \Delta \sin(\log x) & =\sin(\log(x+h)) - \sin (\log x)\\ & = 2 \sin\left(\frac{\log(x+h)-\log x}{2}\right) \, \cos\left(\frac{\log(x+h)+\log x}{2}\right)\\ & = 2 \sin\left(\frac{1}{2}\log\left(\frac{x+h}{x}\right)\right) \, \cos\left(\frac{1}{2}\log\left((x+h)x\right)\right)\\ & = 2 \sin\left(\log\sqrt{1+\frac{h}{x}}\right) \, \cos\left(\log\sqrt{x^2+xh}\right)\\ \end{align*}

Answer 2

Hint: Let $D$ be the linear operation mapping $f(x)$ to $f(x+h)$. Let $I$ be the linear operation mapping $f(x)$ to itself. Then,

$$\Delta=D-I$$

$$\Delta^2=(D-I)^2=D^2-2D+I^2$$

$$=D^2-2D+I$$

This means $\Delta^2$ sends $f(x)$ to $f(x+2h)-2f(x+h)+f(x)$.