Sixth Derivative for Series Expansion

Question

Per Roark's Formulas for Stress and Strain (as well as other resources), the distance from the perimeter to the centroid of a circle segment is:

$$R(1-\frac{2}{3}\frac{sin^3\theta}{\theta -sin\theta\cos\theta})$$

Roark's also gives this in polynomial form, ($\theta\lt\pi/4$)

$$.3R\theta^2(1-0.0976\theta^2+0.0028\theta^4)$$

I'm assuming the polynomial form is a Taylor series. My goal in this question is to be able to verify this.

The differentiation however, becomes very complicated very quickly. Even with some simplifications (just start with $\frac{sin^3\theta}{\theta -sin\theta\cos\theta}$), this seems to be way too complicated to do by hand. The first derivative of this simplification is:

$$\frac{3\theta\sin^2\theta\cos\theta-2sin^3\theta\cos^2\theta-sin^3\theta-sin^5\theta}{(\theta -sin\theta\cos\theta)^2}$$

Considering that the numerator of the first derivative has four terms, by the sixth derivative, I expect at least a thousand terms, possibly far more.

Are there shortcuts that would assist? The series given is sixth-order - am I wrong that I'd need to take the derivative six times? The only shortcut I can see is if I can predict that a term will always have sine in it for up to the required number of derivatives, then I can discard that term as the sine will always be zero due to choosing $\theta=0$. Not sure if I can do that effectively.

Answer 1

Here are some tools that should be handy:

• If you know the series expansions of $f(x)$ and $g(x)$, then you can compute the series expansion of $f(x)g(x)$ to any order desired—just multiply the two series as if they were polynomials.
• If $h(0)=1$ and you know the series expansion of $h(x)$, then you can compute the series expansion of $1/h(x)$ using the geometric series $1/(1-u) = 1+u+u^2+\cdots$ with $u=1-h(x)$. (Note this requires the series multiplication technique above to find the powers of $1-h(x)$.) If $h(0)$ is some other nonzero constant, just factor it out first; if $h(0)=0$, just factor out the appropriate power of $x$ first.

With those two techniques, you can compute the series for a quotient of two functions using just the series for the functions themselves—which can save a lot of differentiation pain.