let-f-x-1-x-2-2cos-x-1-find-f-n-

Question Number 30595 by abdo imad last updated on 23/Feb/18

l e t f (x) = \frac{1}{x^{2} - 2 c o s α x + 1} f i n d f^{(n)} .

Commented by abdo imad last updated on 24/Feb/18

r o o t s o f p (x) = x^{2} - 2 c o s α x + 1 = 0

Δ^{,} = {(- c o s α)}^{2} - 1 = - {s i n}^{2} α = {(i s i n α)}^{2} \Rightarrow

z_{1} = c o s α + i s i n α = e^{i α} a n d z_{2} = c o s α - i s i n α = e^{- i α} \Rightarrow

f (x) = \frac{1}{(x - e^{i α}) (x - e^{- i α})} = \frac{a}{x - e^{i α}} + \frac{b}{x - e^{- i α}}

a = {l i m}_{x \to e^{i α}} (x - e^{i α}) f (x) = \frac{1}{2 i s i n (α)}

b = {l i m}_{x \to e^{- i α}} (x - e^{- i α}) f (x) = \frac{- 1}{2 i s i n α} \Rightarrow

f (x) = \frac{1}{2 i s i n α} (\frac{1}{x - e^{i α}} - \frac{1}{x - e^{- i α}}) \Rightarrow

f^{(n)} (x) = \frac{1}{2 i s i n α} ({(\frac{1}{x - e^{i α}})}^{(n)} - {(\frac{1}{x - e^{- i α}})}^{(n)})

f^{(n)} (x) = \frac{1}{2 i s i n α} (\frac{{(- 1)}^{n} n!}{{(x - e^{i α})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - e^{- i α})}^{n + 1}})

= \frac{{(- 1)}^{n} n!}{2 i s i n α} (\frac{{(x - e^{- i α})}^{n + 1} - {(x - e^{i α})}^{n + 1}}{{(x^{2} - 2 c o s α x + 1)}^{2}}) f o r α \neq k π

i f α = k π f (x) = \frac{1}{{(x - 1)}^{2}} o r f (x) = \frac{1}{{(x + 1)}^{2}} l e t f i n d f^{(n)}

i n c a s e f (x) = \frac{1}{{(x + 1)}^{2}} w e h a v e f^{(1)} (x) = \frac{- 2 (x + 1)}{{(x + 1)}^{4}} = \frac{- 2}{{(x + 1)}^{3}}

f^{(2)} (x) = \frac{3 \times 2 {(x + 1)}^{2}}{{(x + 1)}^{6}} = \frac{3 \times 2}{{(x + 1)}^{4}} l e t s u p p o s e

f^{(n)} (x) = \frac{(n + 1)! {(- 1)}^{n}}{{(x + 1)}^{n + 2}} \Rightarrow f^{(n + 1)} (x) = \frac{{(- 1)}^{n + 1} (n + 1)! (n + 2) {(x + 1)}^{n + 1}}{{(x + 1)}^{2 n + 4}}

= \frac{{(- 1)}^{n + 1} (n + 2)!}{{(x + 1)}^{n + 3}} \Rightarrow f^{(n)} (x) = \frac{{(- 1)}^{n} (n + 1)!}{{(x + 1)}^{n + 2}} .