let-f-x-1-1-x-2-calculate-f-n-x-

Question Number 30752 by abdo imad last updated on 25/Feb/18

l e t f (x) = \frac{1}{1 + x^{2}} c a l c u l a t e f^{(n)} (x) .

Commented by abdo imad last updated on 28/Feb/18

w e h a v e f (x) = \frac{1}{(x + i) (x - i)} = \frac{1}{2 i} (\frac{1}{x - i} - \frac{1}{x + i}) \Rightarrow

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

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

= \frac{n! {(- 1)}^{n}}{2 i} (\frac{{(x + i)}^{n + 1} - {(x - i)}^{n + 1}}{{(1 + x^{2})}^{n + 1}}) b u t

{(x + i)}^{n + 1} - {(x - i)}^{n + 1} = \sum_{k = 0}^{n + 1} C_{n + 1}^{k} i^{k} x^{n + 1 - k}

- \sum_{k = 0}^{n + 1} {(- i)}^{k} x^{n + 1 - k} = \sum_{k = 0}^{n + 1} C_{n + 1}^{k} (i^{k} - {(- i)}^{k}) x^{n + 1 - k}

= \sum_{p = 0}^{[\frac{n}{2}]} C_{n + 1}^{2 p + 1} (2 i) x^{n + 1 - 2 p - 1} \Rightarrow

f^{(n)} (x) = n! {(- 1)}^{n} \frac{\sum_{p = 0}^{[\frac{n}{2}]} C_{n + 1}^{2 p + 1} x^{n - 2 p}}{{(1 + x^{2})}^{n + 1}} . a l s o w e c a n g i v e

f^{(n)} a t f o r m o f a r c t a n \dots .