let-f-x-x-1-n-arctan-nx-calculate-f-n-0-

Question Number 62210 by maxmathsup by imad last updated on 17/Jun/19

l e t f (x) = {(x + 1)}^{n} a r c t a n (n x)

c a l c u l a t e f^{(n)} (0) .

Commented by mathmax by abdo last updated on 24/Jun/19

l e t d e t e r m i n e f^{(p)} (x) l e i b n i z f o r m u l s e g i v e

f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {(a r c t a n (n x))}^{(p - k)}

w e h a v e {(X^{n})}^{(k)} = 0 s i k > n

{(X^{n})}^{(1)} = {n X}^{n - 1} \Rightarrow {(X^{n})}^{(k)} = n (n - 1) . . (n - k + 1) X^{n - k} = \frac{n!}{(n - k)!} X^{n - k} i f k ⩽ n a n d

{(X^{n})}^{(n)} = n! s o f o r p ⩾ n w e g e t

f^{(p)} (x) = \sum_{k = 0}^{n} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {a r c t a n (n x)}^{(p - k)} + \sum_{k = n + 1}^{p} C_{p}^{k} {{(x + 1)}^{n}}^{(k)} {a r c t a n (n x)}^{(p - k)}

= \sum_{k = 0}^{n} C_{p}^{k} \frac{n!}{(n - k)!} {(x + 1)}^{n - k} {a r c t a n (n x)}^{(p - k)} \Rightarrow

f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} \frac{n!}{(n - k)!} {a r c t a n (n x)}^{(n - k)}

= \sum_{k = 0}^{n} C_{n}^{k} k! \frac{n!}{k! (n - k)!} {a r c t a n (n x)}^{(n - k)} = \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {a r c t a n (n x)}^{(n - k)}

w e h a v e {(a r c t a n (n x))}^{(1)} = \frac{n}{1 + n^{2} x^{2}} = \frac{n}{n^{2} (x^{2} + \frac{1}{n^{2}})} = \frac{1}{n (x - \frac{i}{n}) (x + \frac{i}{n})}

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

{a r c t a n (n x)}^{(p)} = \frac{1}{2 i} {{(\frac{1}{x - \frac{i}{n}})}^{(p - 1)} - {(\frac{1}{x + \frac{i}{n}})}^{(p - 1)}}

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

{a r c t a n (n x)}^{(n - k)} = \frac{1}{2 i} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(x - \frac{i}{n})}^{n - k}} - \frac{1}{{(x + \frac{i}{n})}^{n - k}}} \Rightarrow

f^{(n)} (x) = \frac{1}{2 i} \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(x - \frac{i}{n})}^{n - k}} - \frac{1}{{(x + \frac{i}{n})}^{n - k}}}

Commented by mathmax by abdo last updated on 24/Jun/19

f^{(n)} (0) = \frac{1}{2 i} \sum_{k = 0}^{n} k! {(C_{n}^{k})}^{2} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(- \frac{i}{n})}^{n - k}} - \frac{1}{{(\frac{i}{n})}^{n - k}}} .