decompose-inside-C-x-the-fraction-F-x-1-x-2-1-n-with-n-integr-natural-and-n-1-

Question Number 57922 by maxmathsup by imad last updated on 14/Apr/19

d e c o m p o s e i n s i d e C (x) t h e f r a c t i o n F (x) = \frac{1}{{(x^{2} + 1)}^{n}} w i t h n i n t e g r n a t u r a l

a n d n ⩾ 1

Commented by maxmathsup by imad last updated on 30/Apr/19

w e h a v e F (x) = \frac{1}{{(x - i)}^{n} {(x + i)}^{n}} = \sum_{k = 1}^{n} \frac{a_{k}}{{(x - i)}^{k}} + \sum_{k = 1}^{n} \frac{b_{k}}{{(x + i)}^{k}}

f r o m a n o t h e r s i d e c h . x - i = t g i v e F (x) = G (t) = \frac{1}{t^{n} {(t + 2 i)}^{n}}

l e t d e t e r m i n e D_{n - 1} (0) f o r u (t) = \frac{1}{{(t + 2 i)}^{n}} \Rightarrow u (t) = \sum_{k = 0}^{n - 1} \frac{u^{(k)} (0)}{k!} t^{k} + \frac{t^{n}}{n!} ξ (t^{n})

w e h a v e u (t) = {(t + 2 i)}^{- n} \Rightarrow u^{(1)} (t) = (- n) {(t + 2 i)}^{- n - 1} \Rightarrow

u^{(2)} (t) = {(- 1)}^{2} n (n + 1) {(t + 2 i)}^{- n - 2} \Rightarrow u^{(p)} (t) = {(- 1)}^{p} n (n + 1) \dots (n + p - 1) {(t + 2 i)}^{- n - p}

\Rightarrow u^{(k)} (0) = {(- 1)}^{k} n (n + 1) \dots (n + k - 1) {(2 i)}^{- n - k} \Rightarrow

u (t) = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots . (n + k - 1)}{k! {(2 i)}^{n + k}} t^{k} + \frac{t^{n}}{n!} ξ (t^{n}) \Rightarrow

\frac{u (t)}{t^{n}} = \sum_{k = 0}^{n - 1} \frac{{(- 1)}^{k} n (n + 1) \dots . (n + k - 1)}{{(2 i)}^{n + k} k! t^{n - k}} + \frac{1}{n!} ξ (1) (n - k = p)

= \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) \dots . (n + n - p)}{(n - p)! t^{p} {(2 i)}^{2 n - p}} + \frac{1}{n!} ξ (1)

= \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) \dots . (2 n - p)}{(n - p)! {(x - i)}^{p} {(2 i)}^{2 n - p}} + \frac{1}{n!} ξ (1) \Rightarrow

a_{k} = \frac{{(- 1)}^{n - k} n (n + 1) \dots . (2 n - k)}{(n - k)! {(2 i)}^{2 n - k}} n o w l e t f i n d b_{k} w e u s e c h . x + i = t \Rightarrow

F (x) = H (t) = \frac{1}{t^{n} {(t - 2 i)}^{n}} l e t d e t e r m i n e D_{n - 1} (0) f o r v (t) = {(t - 2 i)}^{- n} \Rightarrow

v (t) = \sum_{k = 0}^{n - 1} \frac{v^{(k)} (0)}{k!} t^{k} + t^{n} ξ (t^{n}) b u t v^{(k)} (0) = \frac{{(- 1)}^{k} n (n + 1) \dots . (n + k - 1)}{{(- 2 i)}^{n + k}} \Rightarrow

\frac{v (t)}{t^{n}} = \sum_{p = 1}^{n} \frac{{(- 1)}^{n - p} n (n + 1) \dots (2 n - p)}{(n - p)! t^{p} {(- 2 i)}^{2 n - p}} + \frac{1}{n!} ξ (1) \Rightarrow

b_{k} = \frac{{(- 1)}^{n - k} n (n + 1) \dots . (2 n - k)}{(n - k)! {(- 2 i)}^{2 n - k}} \Rightarrow

F (x) = \sum_{k = 1}^{n} \frac{{(- 1)}^{n - k} n (n + 1) \dots . (2 n - k)}{(n - k)! {(2 i)}^{2 n - k} {(x - i)}^{k}} + \sum_{k = 1}^{n} \frac{{(- 1)}^{n - k} n (n + 1) \dots (2 n - k)}{(n - k)! {(- 2 i)}^{2 n - k} {(x + i)}^{k}}

a n d w e s e e t h a t b_{k} = c o n j (a_{k}) s o w e c a n s e a r c h o n l y a_{k} \dots

e r r o r o f t y p i n g c h a n g e ξ (t^{n}) b y ξ (t) \dots

Commented by maxmathsup by imad last updated on 30/Apr/19

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