decompose-inside-C-x-F-1-x-1-x-n-1-

Question Number 30599 by abdo imad last updated on 23/Feb/18

d e c o m p o s e i n s i d e C (x) F = \frac{1}{(x - 1) (x^{n} - 1)} .

Commented by abdo imad last updated on 25/Feb/18

t h e r o o t s o f z^{n} - 1 = 0 a r e t h e c o m p l e x z_{k} = e^{i \frac{2 k π}{k}} w i t h

k \in [[0, n - 1]] \Rightarrow F w i l l b e d e c o m p o s e d i n s i d e C (x) a t

f o r m F (x) = \frac{a}{x - 1} + \frac{b}{{(x - 1)}^{2}} + \sum_{k = 1}^{n - 1} \frac{λ_{k}}{x - z_{k}} w e h a v e

(x - 1) (x^{n} - 1) = x^{n + 1} - x - x^{n} + 1 \Rightarrow

\frac{d}{d x} ((x - 1) (x^{n} - 1)) = (n + 1) x^{n} - {n x}^{n - 1} - 1 \Rightarrow

λ_{k} = \frac{1}{(n + 1) z_{k}^{n} - n z_{k}^{n - 1} - 1} = \frac{1}{n + 1 - \frac{n}{z_{k}} - 1} = \frac{z_{k}}{{n z}_{k} - n}

t h e c h . x - 1 = y g i v e F (x) = \frac{1}{{(x - 1)}^{2} (1 + x + x^{2} + \dots + x^{n - 1})}

= g (y) = \frac{1}{y^{2} (1 + (1 + y) + {(1 + y)}^{2} + \dots . {(1 + y)}^{n - 1})} a f t e r

d i v i d e 1 p e r 2 + y + {(1 + y)}^{2} + \dots {(1 + y)}^{n - 1} f o l o w i n g t h e

i n c r e a s i n g p o w e r s a f t e r a l l c a l c u l u s w e f i n d

a = \frac{1 - n}{2 n} a n d b = \frac{1}{n} l o o k a l s o t h a t

b = {l i m}_{x \to 1} {(x - 1)}^{2} F (x) = {l i m}_{x \to 1} \frac{1}{1 + x + x^{2} + \dots + x^{n - 1}} = \frac{1}{n}