let f(x) =((ln(1+x))/(2−x^2 )) 1)calculate f^((n)) (x) 2) calculate f^((n)) (0) 3)developp f(x) at integr serie.


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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
$1) we have f(x)=−((ln(1+x))/((x−(√2))(x+(√2)))) =−(1/(2(√2))){(1/(x−(√2))) −(1/(x+(√2))))ln(1+x) = (1/(2(√2))){ ((ln(1+x))/(x−(√2))) −((ln(1+x))/(x+(√2)))} ⇒f^((n)) (x) =(1/(2(√2))){ (((ln(1+x))/(x−(√2))))^n −(((ln(1+x))/(x+(√2))))^((n)) } let determine (((ln(1+x))/(x−(√2))))^((n)) leibniz formula give (((ln(1+x))/(x−(√2))))^((n)) =Σ_(k=0) ^n C_n ^k (ln(1+x))^((k)) ((1/(x−(√2))))^((n−k)) (ln(1+x))^((1) ) =(1/(1+x)) ⇒(ln(1+x))^()k)) =((1/(1+x)))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+1)^k )) ⇒ (((ln(1+x))/(x−(√2))))^((n)) =ln(1+x)(((−1)^n n!)/((x−(√2))^(n+1) )) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/((x+1)^k )) (((−1)^(n−k) (n−k)!)/((x−(√2))^(n−k+1) )) =(−1)^n n!((ln(1+x))/((x−(√2))^(n+1) )) +Σ_(k=1) ^n (−1)^(n−1) (k−1)!(n−k)! ((n!)/(k!(n−k)!(x+1)^k (x−(√2))^(n−k+1) )) =(−1)^n n! ((ln(1+x))/((x−(√2))^(n+1) )) +(−1)^(n−1) n! Σ_(k=1) ^n (1/(k(x+1)^k (x−(√2))^(n−k +1) )) also we have (((ln(1+x))/(x+(√2))))^((n)) =(−1)^n n!((ln(1+x))/((x+(√2))^(n+1) )) +(−1)^(n−1) n! Σ_(k=1) ^n (1/(k(x+1)^k (x+(√2))^(n−k+1) )) so the value of f^((n)) (x) isdetermined .$
$1) w e h a v e f (x) = - \frac{l n (1 + x)}{(x - \sqrt{2}) (x + \sqrt{2})} = - \frac{1}{2 \sqrt{2}} {\frac{1}{x - \sqrt{2}} - \frac{1}{x + \sqrt{2}}) l n (1 + x)$ $= \frac{1}{2 \sqrt{2}} {\frac{l n (1 + x)}{x - \sqrt{2}} - \frac{l n (1 + x)}{x + \sqrt{2}}} \Rightarrow f^{(n)} (x) = \frac{1}{2 \sqrt{2}} {{(\frac{l n (1 + x)}{x - \sqrt{2}})}^{n} - {(\frac{l n (1 + x)}{x + \sqrt{2}})}^{(n)}}$ $l e t d e t e r m i n e {(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} l e i b n i z f o r m u l a g i v e$ ${(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(l n (1 + x))}^{(k)} {(\frac{1}{x - \sqrt{2}})}^{(n - k)}$ ${(l n (1 + x))}^{(1)} = \frac{1}{1 + x} \Rightarrow {(l n (1 + x))}^{) k)} = {(\frac{1}{1 + x})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}} \Rightarrow$ ${(\frac{l n (1 + x)}{x - \sqrt{2}})}^{(n)} = l n (1 + x) \frac{{(- 1)}^{n} n!}{{(x - \sqrt{2})}^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}} \frac{{(- 1)}^{n - k} (n - k)!}{{(x - \sqrt{2})}^{n - k + 1}}$ $= {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x - \sqrt{2})}^{n + 1}} + \sum_{k = 1}^{n} {(- 1)}^{n - 1} (k - 1)! (n - k)! \frac{n!}{k! (n - k)! {(x + 1)}^{k} {(x - \sqrt{2})}^{n - k + 1}}$ $= {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x - \sqrt{2})}^{n + 1}} + {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(x + 1)}^{k} {(x - \sqrt{2})}^{n - k + 1}} a l s o w e h a v e$ ${(\frac{l n (1 + x)}{x + \sqrt{2}})}^{(n)} = {(- 1)}^{n} n! \frac{l n (1 + x)}{{(x + \sqrt{2})}^{n + 1}} + {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(x + 1)}^{k} {(x + \sqrt{2})}^{n - k + 1}}$ $s o t h e v a l u e o f f^{(n)} (x) i s d e t e r m i n e d .$
Commented by maxmathsup by imad last updated on 07/Apr/19
$2) we have f^((n)) (0) =(1/(2(√2))){ (−1)^(n−1) n! Σ_(k=1) ^n (1/(k(−(√2))^(n−k+1) )) −(−1)^(n−1) n! Σ_(k=1) ^n (1/(k((√2))^(n−k +1) ))} =(((−1)^(n−1) n!)/(2(√2))){ Σ_(k=1) ^n (1/k)( (1/((−(√2))^(n−k +1) )) −(1/(((√2))^(n−k +1) )))} =(((−1)^(n−1) n!)/(2(√2))) Σ_(k=1) ^n (1/k)((((√2))^(n−k +1) −(−(√2))^(n−k+1) )/((−2)^(n−k +1) )) =(((−1)^(n−1) n!)/(2(√2)(−1)^(n+1) )) Σ_(k=1) ^n ((((√2))^(n−k +1) −(−(√2))^(n−k +1) )/2^(n−k +1) )(−1)^k ⇒ f^((n)) (0) =((n!)/(2(√2))) Σ_(k=1) ^n (−1)^k ((((√2))^(n−k+1) −(−(√2))^(n−k +1) )/2^(n−k +1) ) .$
$2) w e h a v e f^{(n)} (0) = \frac{1}{2 \sqrt{2}} {{(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(- \sqrt{2})}^{n - k + 1}}$ $- {(- 1)}^{n - 1} n! \sum_{k = 1}^{n} \frac{1}{k {(\sqrt{2})}^{n - k + 1}}}$ $= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2}} {\sum_{k = 1}^{n} \frac{1}{k} (\frac{1}{{(- \sqrt{2})}^{n - k + 1}} - \frac{1}{{(\sqrt{2})}^{n - k + 1}})}$ $= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2}} \sum_{k = 1}^{n} \frac{1}{k} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{{(- 2)}^{n - k + 1}}$ $= \frac{{(- 1)}^{n - 1} n!}{2 \sqrt{2} {(- 1)}^{n + 1}} \sum_{k = 1}^{n} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{2^{n - k + 1}} {(- 1)}^{k} \Rightarrow$ $f^{(n)} (0) = \frac{n!}{2 \sqrt{2}} \sum_{k = 1}^{n} {(- 1)}^{k} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{2^{n - k + 1}} .$
Commented by maxmathsup by imad last updated on 07/Apr/19
$3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n (f(0)=0) f(x) =(1/(2(√2))) Σ_(n=1) ^∞ { Σ_(k=1) ^n (−1)^k ((((√2))^(n−k+1) −(−(√2))^(n−k+1) )/2^(n−k+1) )} x^n .$
$3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} (f (0) = 0)$ $f (x) = \frac{1}{2 \sqrt{2}} \sum_{n = 1}^{\infty} {\sum_{k = 1}^{n} {(- 1)}^{k} \frac{{(\sqrt{2})}^{n - k + 1} - {(- \sqrt{2})}^{n - k + 1}}{2^{n - k + 1}}} x^{n} .$
Commented by maxmathsup by imad last updated on 07/Apr/19

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