Question

Let \(k\) be a fixed positive integer. The \(n\) th derivative of \(\frac{1}{x^{k}-1}\) has the form \(\frac{P_{n}(x)}{\left(x^{k}-1\right)^{n+1}}\) where \(P_{n}(x)\) is a polynomial. Find \(P_{n}(1)\).

Short answer

The value of \(P_{n}(1)\) is given by \( (-k)_n \cdot k \), which can be evaluated directly given the value of \(n\) and \(k\).

Step-by-step solution

Differentiate the function

Start by taking the derivative of the function \(\frac{1}{x^{k}-1}\) once. Using the chain rule, the derivative can be given by: \(-\frac{k\cdot x^{k-1}}{(x^k - 1)^2}\).

Continue differentiating

It is observed that the nth derivative has a pattern which involves factorials, with the n+1th derivative being \( \frac{(-k)_n k x^{k-n}}{(x^{k}-1)^{n+1}} \). This is seen because during each differentiation, the exponent of \(x\) in the numerator decrements by 1, while the denominator's exponent increments by 1. Here \( (-k)_n \) is the falling factorial, indicating multiplication of consecutive numbers ( e.g. \( (-k)_n = -k(-k+1)...(-k+n-1) \) ).

Substitution

Since we are interested in finding \(P_{n}(1)\), we substitute \(x = 1\) into our polynomial derived from Step 2. Substituting \(x = 1\) turns \(x^{k-n}\) into \(1\), and the value of \(P_{n}(1)\) becomes \( (-k)_n \cdot k \).