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Question Number 62809 by mathmax by abdo last updated on 25/Jun/19

l e t f (x) = a r c t a n (n x) w i t h n i n t e g r n a t u r a l

1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)

2) d e v e l o p p f a t i n t e g r s e r i e .

Commented by mathmax by abdo last updated on 02/Jul/19

1) w e h a v e f^{^{'}} (x) = \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} \frac{n}{2 i} {\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

f^{(p)} (x) = \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}}}

= \frac{{(- 1)}^{p - 1} (p - 1)!}{2 i} {\frac{{(x + \frac{i}{n})}^{p} - {(x - \frac{i}{n})}^{p}}{{(x^{2} + \frac{1}{n^{2}})}^{p}}} \Rightarrow f^{(p)} (0) = \frac{{(- 1)}^{p - 1} (p - 1)!}{2 i} \frac{n^{2 p}}{n^{p}} \frac{{(n x + i)}^{p} - {(n x - i)}^{p}}{{(x^{2} n^{2} + 1)}^{p}}

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

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

f^{(n)} (x) = \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} \frac{{(n x + i)}^{n} - {(n x - i)}^{n}}{{(n^{2} x^{2} + 1)}^{n}} w i t h n ⩾ 1

x = 0 \Rightarrow f^{(n)} (0) = \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} {i^{n} - {(- i)}^{n}}

= \frac{n^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} \times 2 i I m (i^{n}) = n^{n} {(- 1)}^{n - 1} (n - 1)! s i n (\frac{n π}{2})

2) f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (0)}{p!} x^{p}

= \sum_{p = 0}^{\infty} \frac{n^{p} {(- 1)}^{p - 1}}{2 i} \frac{{(n x + i)}^{p} - {(n x - i)}^{p}}{{(n^{2} x^{2} + 1)}^{p}} \frac{x^{p}}{p}