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We are currently covering arc length in the calculus class I'm teaching, and since most of the integrals involved are impossible to solve analytically, I'd like to have my students do some approximations instead. Of course we could use the usual rectangle-based methods on the integral $$ \int_a^b \sqrt{1 + f'(x)^2}\,dx $$ but this is somewhat unnatural compared to the "sum-of-chords" approximation using evenly-spaced points $(x_i,f(x_i))$ along the curve:

$$ \sum_{i=1}^n \sqrt{(x_{i+1} - x_i)^2 + (f(x_{i+1}) - f(x_i))^2} $$

(Not to mention that computing the maximum value of the second derivative of $\sqrt{1 + f'(x)^2}$ along $[a,b]$ would be incredibly time-consuming for my students.)

The problem with this approximation is that I don't know a bound on the error. This question (and the papers referenced there) suggests that the order of the error is $o(1/n^2)$, but the exact form of the bound isn't mentioned, and my (rudimentary) calculations only gave an error bound on the order of $o(1/n)$.

So: is there an order-$o(1/n^2)$ bound on the error from the sum-of-chords approximation? What form does it take?

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Paul McKenney
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1 Answers1

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\begin{align*} e_{h} &= \int_{a}^{a+h} \sqrt{1+f'(x)^{2}} \, dx-\sqrt{h^{2}+[f(a+h)-f(a)]^{2}} \\ &= \frac{h^{3}}{24} \frac{f''(a)^{2}}{\left[ 1+f'(a)^{2} \right]^{\frac{3}{2}}}+O(h^{4}) \\ E &=\frac{(b-a)^{3}}{24n^{2}} \frac{f''(\xi)^{2}}{\left[ 1+f'(\xi)^{2} \right]^{\frac{3}{2}}} \end{align*}

Alternatively

Using local canonical form of a curve parametrized by arc length \begin{align*} \mathbf{r}(s) &= \mathbf{r}(0)+ \left( s-\frac{\kappa^{2} s^{3}}{6} \right) \mathbf{T}+ \left( \frac{\kappa s^{2}}{2}+\frac{\kappa' s^{3}}{6} \right) \mathbf{N}+ \left( \frac{\kappa \tau s^{3}}{6} \right) \mathbf{B}+O(s^{4}) \\ |\mathbf{r}(s)-\mathbf{r}(0)| & \approx \sqrt{\left( s-\frac{\kappa^{2} s^{3}}{6} \right)^{2}+ \left( \frac{\kappa s^{2}}{2} \right)^{2}} \\ &= s-\frac{\kappa^{2}}{24}s^{3}+O(s^{4}) \end{align*}

Ng Chung Tak
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