Does this approximation always overestimate?

Question

So, I'm working on a method that calculates $\cos\theta : \theta \in [0, \frac{\pi}{2})$ using geometric methods, and I'm trying to work out the error of the approximation. I have a wonderful estimate for the approximation if I know that the following always produces an overestimate.

We have a $h_0 := \frac{\theta}{2^k}$, where $k$ is an arbitrary fixed positive integer, which produces the sequence $h_n := h_{n-1}\sqrt{4 - h_{n-1}^2} \quad \forall n \in \mathbb{N}$. These values are approximating $2\sin\frac{\theta}{2^{k+1-n}}$, which is the length of the chord of an arc of a circle with angle $\frac{\theta}{2^{k-n}}$. It is obvious, from observation that $h_0$ is the arc length of the circular arc with angle $\frac{\theta}{2^k}$, and thus is an overestimate for $2\sin\frac{\theta}{2^{k+1}}$.

My dilemma is that empirical evidence shows that $h_n$ overestimates $2\sin\frac{\theta}{2^{k+1-n}} \quad \forall n \in [0, k-1] \cap \mathbb{Z}$, and in some cases for $h_k$ as well. However despite varied attempts I have been unable to prove that this is the case. Any help would be most appreciated.

Answer 1

As per your considerations, and introducing a distinction between the exact and approximate angle, $$ h_0=\fracθ{2^k}=2\sin\fracϕ{2^{k+1}} $$ you get for the exact angle of the iteration \begin{align} ϕ&=2^{k+1}\arcsin\frac{θ}{2^{k+1}} \\ &= 2^{k+1}\left(\frac{θ}{2^{k+1}}+\frac16\frac{θ^3}{2^{3k+3}}+…+\binom{2n}{n}\frac{θ^{2n+1}}{2^{nk+n}4^n(2n+1)}+…\right) \\ &= θ+\frac{θ^3}{6·2^{2k+2}}+O(2^{-4k}) \end{align}

Which means that $$ h_{k+1}=2\sinϕ=2\sin\left( θ+\frac{θ^3}{6·2^{2k+2}}+…\right) $$ will be slightly larger than $2\sin θ$.

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Of course, $$ h_k=2\sin\fracϕ2=\frac{\sinϕ}{\cos\fracϕ2} $$ will be systematically larger than $\sinϕ$ as $\cos\fracϕ2<1$.

Answer 2

You need the fact that for $x \gt 0, \sin x \lt x$. You can prove this by using the Taylor series and alternating series theorem, which says that (once the terms are decreasing) the error is the same sign as and smaller than the first neglected term. As the Taylor series for $\sin x$ is $x-\frac {x^3}6 + \dots$ we have $\sin x \lt x$ for $0 \lt x \le 1$. For larger $x$ (not your issue here) you can just use $\sin x \le 1$