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Question Number 57486 by Abdo msup. last updated on 05/Apr/19

l e t f (x) = \frac{l n (1 + x)}{2 - x^{2}}

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

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

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

Commented by maxmathsup by imad last updated on 07/Apr/19

1) w e h a v e f (x) = - \frac{l n (1 + x)}{(x - \sqrt{2}) (x + \sqrt{2})} = - \frac{1}{2 \sqrt{2}} {\frac{1}{x - \sqrt{2}} - \frac{1}{x + \sqrt{2}}) l n (1 + x)

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

l e t d e t e r m i n e {(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} l e i b n i z f o r m u l a g i v e

{(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(l n (1 + x))}^{(k)} {(\frac{1}{x - \sqrt{2}})}^{(n - k)}

{(l n (1 + x))}^{(1)} = \frac{1}{1 + x} \Rightarrow {(l n (1 + x))}^{) k)} = {(\frac{1}{1 + x})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}} \Rightarrow

{(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} = l n (1 + x) \frac{{(- 1)}^{n} n!}{{(x - \sqrt{2})}^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}} \frac{{(- 1)}^{n - k} (n - k)!}{{(x - \sqrt{2})}^{n - k + 1}}

= {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x - \sqrt{2})}^{n + 1}} + \sum_{k = 1}^{n} {(- 1)}^{n - 1} (k - 1)! (n - k)! \frac{n!}{k! (n - k)! {(x + 1)}^{k} {(x - \sqrt{2})}^{n - k + 1}}

= {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x - \sqrt{2})}^{n + 1}} + {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(x + 1)}^{k} {(x - \sqrt{2})}^{n - k + 1}} a l s o w e h a v e

{(\frac{l n (1 + x)}{x + \sqrt{2}})}^{(n)} = {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x + \sqrt{2})}^{n + 1}} + {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(x + 1)}^{k} {(x + \sqrt{2})}^{n - k + 1}}

s o t h e v a l u e o f f^{(n)} (x) i s d e t e r m i n e d .

Commented by maxmathsup by imad last updated on 07/Apr/19

2) w e h a v e f^{(n)} (0) = \frac{1}{2 \sqrt{2}} {{(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(- \sqrt{2})}^{n - k + 1}}

- {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(\sqrt{2})}^{n - k + 1}}}

= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2}} {\sum_{k = 1}^{n} \frac{1}{k} (\frac{1}{{(- \sqrt{2})}^{n - k + 1}} - \frac{1}{{(\sqrt{2})}^{n - k + 1}})}

= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2}} \sum_{k = 1}^{n} \frac{1}{k} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{{(- 2)}^{n - k + 1}}

= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2} {(- 1)}^{n + 1}} \sum_{k = 1}^{n} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{2^{n - k + 1}} {(- 1)}^{k} \Rightarrow

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

Commented by maxmathsup by imad last updated on 07/Apr/19

3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} (f (0) = 0)

f (x) = \frac{1}{2 \sqrt{2}} \sum_{n = 1}^{\infty} {\sum_{k = 1}^{n} {(- 1)}^{k} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{2^{n - k + 1}}} x^{n} .

Commented by maxmathsup by imad last updated on 07/Apr/19

f o r g i v e e r r o r o f s i n e f (x) = \frac{1}{2 \sqrt{2}} {\frac{l n (1 + x)}{x + \sqrt{2}} - \frac{l n (1 + x)}{x - \sqrt{2}}}