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


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Question Number 55269 by maxmathsup by imad last updated on 20/Feb/19

$l e t f (x) = (2 x + 1) l n (1 - x^{2})$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$ $3) c a l c u l a t e \int_{0}^{1} f (x) d x .$
Commented by turbo msup by abdo last updated on 02/Mar/19

$1) l e i b n i z g o r m u l a g i v e$ $f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(2 x + 1)}^{(k)} (l n {(1 - x^{2})}^{(n - k)}$ $= (2 x + 1) (l n {(1 - x^{2})}^{(n)} + 2 n (l n {(1 - x^{2})}^{(n - 1)}$ $w e h a v e l n {(1 - x^{2})}^{(1)} = \frac{- 2 x}{1 - x^{2}} \Rightarrow$ $(l n {(1 - x^{2})}^{(n)} = {(- \frac{2 x}{1 - x^{2}})}^{(n - 1)}$ $= {(\frac{2 x}{x^{2} - 1})}^{(n - 1)} = {(\frac{1}{x + 1} + \frac{1}{x - 1})}^{(n - 1)}$ $= {(\frac{1}{x + 1})}^{(n - 1)} + {(\frac{1}{x - 1})}^{(n - 1)}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + 1)}^{n}} + \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - 1)}^{n}}$
Commented by turbo msup by abdo last updated on 02/Mar/19
$also (ln(1−x^2 ))^((n−1)) =(((−1)^(n−2) (n−2)!)/((x+1)^(n−1) )) +(((−1)^(n−2) (n−2)!)/((x−1)^(n−1) )) ⇒ f^((n)) (x) =(2x+1)(−1)^(n−1) (n−1)!{(1/((x+1)^n )) +(1/((x−1)^n ))} +2n(−1)^(n−2) (n−2)!{(1/((x+1)^(n−1) )) +(1/((x−1)^(n−1) ))}$
$a l s o {(l n (1 - x^{2}))}^{(n - 1)}$ $= \frac{{(- 1)}^{n - 2} (n - 2)!}{{(x + 1)}^{n - 1}} + \frac{{(- 1)}^{n - 2} (n - 2)!}{{(x - 1)}^{n - 1}} \Rightarrow$ $f^{(n)} (x) = (2 x + 1) {(- 1)}^{n - 1} (n - 1)! {\frac{1}{{(x + 1)}^{n}} + \frac{1}{{(x - 1)}^{n}}}$ $+ 2 n {(- 1)}^{n - 2} (n - 2)! {\frac{1}{{(x + 1)}^{n - 1}} + \frac{1}{{(x - 1)}^{n - 1}}}$
Commented by maxmathsup by imad last updated on 02/Mar/19
$⇒f^((n)) (0) =(−1)^(n−1) (n−1)!{1+(−1)^n } +2n(−1)^n (n−2)!{1+(−1)^(n−1) } =−(−1)^n (n−1)!{1+(−1)^n } +2n(−1)^n (n−2)!{1−(−1)^n } ⇒ f^((n)) (0) =−(−1)^n (n−1)(n−2)!{1+(−1)^n } +2n(−1)^n (n−2)!{1−(−1)^n } =(−1)^n (n−2)!{(1−n)(1+(−1)^n ) +2n{1−(−1)^n }} =(−1)^n (n−2)!{1+(−1)^n −n−n(−1)^n +2n−2n(−1)^n } =(−1)^n (n−2)!{ 1+(1−3n)(−1)^n +n}$
$\Rightarrow f^{(n)} (0) = {(- 1)}^{n - 1} (n - 1)! {1 + {(- 1)}^{n}} + 2 n {(- 1)}^{n} (n - 2)! {1 + {(- 1)}^{n - 1}}$ $= - {(- 1)}^{n} (n - 1)! {1 + {(- 1)}^{n}} + 2 n {(- 1)}^{n} (n - 2)! {1 - {(- 1)}^{n}} \Rightarrow$ $f^{(n)} (0) = - {(- 1)}^{n} (n - 1) (n - 2)! {1 + {(- 1)}^{n}} + 2 n {(- 1)}^{n} (n - 2)! {1 - {(- 1)}^{n}}$ $= {(- 1)}^{n} (n - 2)! {(1 - n) (1 + {(- 1)}^{n}) + 2 n {1 - {(- 1)}^{n}}}$ $= {(- 1)}^{n} (n - 2)! {1 + {(- 1)}^{n} - n - n {(- 1)}^{n} + 2 n - 2 n {(- 1)}^{n}}$ $= {(- 1)}^{n} (n - 2)! {1 + (1 - 3 n) {(- 1)}^{n} + n}$
Commented by maxmathsup by imad last updated on 02/Mar/19
$f^′ (x)=2ln(1−x^2 ) +(2x+1)((−2x)/(1−x^2 )) ⇒f^′ (0)=0 and f(0)=0 f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=2) ^∞ (((−1)^n )/(n(n−1))){n+1+(1−3n)(−1)^n } x^n =Σ_(n=2) ^∞ (((n+1)(−1)^n +1−3n)/(n(n−1)))x^n$
$f^{^{'}} (x) = 2 l n (1 - x^{2}) + (2 x + 1) \frac{- 2 x}{1 - x^{2}} \Rightarrow f^{^{'}} (0) = 0 a n d f (0) = 0$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n (n - 1)} {n + 1 + (1 - 3 n) {(- 1)}^{n}} x^{n}$ $= \sum_{n = 2}^{\infty} \frac{(n + 1) {(- 1)}^{n} + 1 - 3 n}{n (n - 1)} x^{n}$
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