Approxiamating Arc Length with Chords and Taylor Series ($S=\frac{1}{3}(8c-C)(1+\epsilon)$) The problem I am working on comes directly from Chapter 3 of Advanced Calculus by Edwin Bidwell Wilson. A chapter that deals with Taylor series.
The problem in full is :



*If S is  a circular arc and C its chord and v is the chord of half the arc, prove $$S= \frac{1}{3}(8v-C)(1+\epsilon)$$ where $$\epsilon \approx S^4/7680R^4$$

Here is a picture that helps visualize how I see the problem.

Clearly this problem begs the question of where do I start. I can tell you I started by working with the fact that
$$\theta = 2\arcsin(C/2R)$$
thus
$$S = R\theta = 2R\arcsin(C/2R)$$
using the taylor series of $\arcsin$ gives us
$$S = 2R \sum_{n=0}^{\infty} \frac{(C/2R)^{2n+1}}{2n+1}{2n \choose n}\frac{1}{4^n}$$
You could also derive two pieces of information using the equations presented:
$$R = \frac{4h^2+c^2}{8h}$$
and
$$ h = \sqrt{v^2 - C^2/4}$$
to get
$$R = \frac{v^2}{2\sqrt{v^2-C^2/4}}$$
which gets us
$$S = C \sum_{n=0}^{\infty} \frac{(\frac{C^2}{v^2}(v^2-C^2/4))^{n}}{2n+1}{2n \choose n}\frac{1}{4^n}$$
might be stuck .

$v=2R\sin\frac{\theta}{4}=2R\left(\frac{\theta}{4}-\frac{\theta^3}{384}+\frac{\theta^5}{122880}+o(\theta^5)\right)$
$C=2R\sin\frac{\theta}{2}=2R\left(\frac{\theta}{2}-\frac{\theta^3}{48}+\frac{\theta^5}{3840}+o(\theta^5)\right)$
$8v-C=2R\left(\frac{3\theta}{2}-\frac{\theta^5}{5120}+o(\theta^5)\right)$
$\frac{1}{1+\epsilon}=\frac{8v-C}{3S}=\frac{2R\left(\frac{3\theta}{2}-\frac{\theta^5}{5120}+o(\theta^5)\right)}{3R\theta}=1-\frac{\theta^4}{7680}+o(\theta^4)$
$1+\epsilon=\frac{1}{1-\frac{\theta^4}{7680}+o(\theta^4)}=1+\frac{\theta^4}{7680}+o(\theta^4)$
$\epsilon=\frac{\theta^4}{7680}+o(\theta^4)\approx \frac{\theta^4}{7680}=\frac{S^4}{7680R^4}$
