# What is the arc length of f(x)=x^2/(4-x^2)  on x in [-1,1]?

Jun 20, 2018

$L = 2 + 2 {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 4 n \\ m\end{matrix}\right) {\left(- 1\right)}^{m} / \left(\left(2 n + 2 m + 1\right) \cdot {4}^{n + m}\right)$ units.

#### Explanation:

$f \left(x\right) = {x}^{2} / \left(4 - {x}^{2}\right) = 1 - \frac{4}{4 - {x}^{2}}$

$f ' \left(x\right) = \frac{8 x}{4 - {x}^{2}} ^ 2$

Arc length is given by:

$L = {\int}_{-} {1}^{1} \sqrt{1 + \frac{64 {x}^{2}}{4 - {x}^{2}} ^ 4} \mathrm{dx}$

For $x \in \left[- 1 , 1\right]$, $\frac{64 {x}^{2}}{4 - {x}^{2}} ^ 4 < 1$. Take the series expansion of the square root:

$L = {\int}_{-} {1}^{1} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\frac{64 {x}^{2}}{4 - {x}^{2}} ^ 4\right)}^{n} \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{-} {1}^{1} \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {4}^{3 n} {\int}_{-} {1}^{1} {x}^{2 n} / {\left(4 - {x}^{2}\right)}^{4 n} \mathrm{dx}$

Rearrange:

$L = 2 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\int}_{-} {1}^{1} {x}^{2 n} {\left(1 - \frac{1}{4} {x}^{2}\right)}^{- 4 n} \mathrm{dx}$

For $x \in \left[- 1 , 1\right]$, $\frac{1}{4} {x}^{2} \le 1$. Take another series expansion:

$L = 2 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\int}_{-} {1}^{1} {x}^{2 n} \left\{{\sum}_{m = 0}^{\infty} \left(\begin{matrix}- 4 n \\ m\end{matrix}\right) {\left(- \frac{1}{4} {x}^{2}\right)}^{m}\right\} \mathrm{dx}$

Simplify:

$L = 2 + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 4 n \\ m\end{matrix}\right) {\left(- 1\right)}^{m} / {4}^{n + m} {\int}_{-} {1}^{1} {x}^{2 n + 2 m} \mathrm{dx}$

Integrate directly:

$L = 2 + {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 4 n \\ m\end{matrix}\right) {\left(- 1\right)}^{m} / {4}^{n + m} {\left[{x}^{2 n + 2 m + 1}\right]}_{-} {1}^{1} / \left(2 n + 2 m + 1\right)$

Insert the limits of integration and simplify:

$L = 2 + 2 {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}- 4 n \\ m\end{matrix}\right) {\left(- 1\right)}^{m} / \left(\left(2 n + 2 m + 1\right) \cdot {4}^{n + m}\right)$