$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \cot x + 2 \cot ^ { 3 } x$$
Hence show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = p \cot ^ { 4 } x + q \cot ^ { 2 } x + r$$
where \(p , q\) and \(r\) are integers to be found.
Find the Taylor series expansion of \(\cot x\) in ascending powers of \(\left( x - \frac { \pi } { 3 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 3 } \right) ^ { 3 }\).