let f(x)=ln∣((x−1)/(x+1))∣ 1)determine D_f 2) calculatef^((n)) (x) and f^((n)) (0) 3) developp f at integr serie 4) calculate ∫


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Question Number 62882 by mathmax by abdo last updated on 26/Jun/19

$l e t f (x) = l n ∣ \frac{x - 1}{x + 1} ∣$ $1) d e t e r m i n e D_{f}$ $2) {c a l c u l a t e f}^{(n)} (x) a n d f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e$ $4) c a l c u l a t e \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x .$
Commented by mathmax by abdo last updated on 27/Jun/19

$4) \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ \frac{x - 1}{x + 1} ∣ d x = \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ 1 - x ∣ d x - \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ 1 + x ∣ d x$ $= \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 - x) d x - \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 + x) d x = H - K$ $H =_{1 - x = t} \int_{\frac{3}{2}}^{\frac{1}{2}} l n (t) (- d t) = \int_{\frac{1}{2}}^{\frac{3}{2}} l n (t) d t = {[t l n (t) - t]}_{\frac{1}{2}}^{\frac{3}{2}}$ $= \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2}$ $K = \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 + x) d x =_{1 + x = t} \int_{\frac{1}{2}}^{\frac{3}{2}} l n (t) d t = {[t l n (t) - t]}_{\frac{1}{2}}^{\frac{3}{2}}$ $= \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2} \Rightarrow$ $A = \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2} - \frac{3}{2} l n (\frac{3}{2}) + \frac{3}{2} + \frac{1}{2} l n (\frac{1}{2}) - \frac{1}{2} \Rightarrow$ $A = 0$ $a n o t h e r [w a y l e t p r o v e t h a t f i s o d d w e h a v e$ $f (- x) = l n ∣ \frac{- x - 1}{- x + 1} ∣ = l n ∣ \frac{x + 1}{x - 1} ∣ = - l n ∣ \frac{x - 1}{x + 1} ∣= - f (x) \Rightarrow \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x = 0 .$
Commented by mathmax by abdo last updated on 27/Jun/19
$1) D_f =R−{−1,1} 2)we have f(x) =ln∣x−1∣−ln∣x+1∣ ⇒ f^′ (x) =(1/(x−1)) −(1/(x+1)) ⇒f^((n)) (x) =((1/(x−1)))^((n−1)) −((1/(x+1)))^((n−1)) ⇒ f^((n)) (x) =(((−1)^(n−1) (n−1)!)/((x−1)^n )) −(((−1)^(n−1) (n−1)!)/((x+1)^n )) =(−1)^(n−1) (n−1)!{(1/((x−1)^n ))−(1/((x+1)^n ))} f^((n)) (x)=(−1)^(n−1) (n−1)!(((x+1)^n −(x−1)^n )/((x^2 −1)^n )) with x≥1 f^((n)) (0) =(−1)^(n−1) (n−1)! ((1−(−1)^n )/((−1)^n )) =−(n−1)!(1−(−1)^n ) ={(−1)^n −1}(n−1)! ⇒f^((2p)) (0) =0 and f^((2p+1)) (0) =−2(2p)! 3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(p=0) ^∞ ((−2 (2p)!)/((2p+1)!)) x^(2p+1) ⇒ f(x)=−2Σ_(p=0) ^∞ (x^(2p+1) /(2p+1))$
$1) D_{f} = R - {- 1, 1}$ $2) w e h a v e f (x) = l n ∣ x - 1 ∣ - l n ∣ x + 1 ∣ \Rightarrow$ $f^{^{'}} (x) = \frac{1}{x - 1} - \frac{1}{x + 1} \Rightarrow f^{(n)} (x) = {(\frac{1}{x - 1})}^{(n - 1)} - {(\frac{1}{x + 1})}^{(n - 1)} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - 1)}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + 1)}^{n}} = {(- 1)}^{n - 1} (n - 1)! {\frac{1}{{(x - 1)}^{n}} - \frac{1}{{(x + 1)}^{n}}}$ $f^{(n)} (x) = {(- 1)}^{n - 1} (n - 1)! \frac{{(x + 1)}^{n} - {(x - 1)}^{n}}{{(x^{2} - 1)}^{n}} w i t h x ⩾ 1$ $f^{(n)} (0) = {(- 1)}^{n - 1} (n - 1)! \frac{1 - {(- 1)}^{n}}{{(- 1)}^{n}} = - (n - 1)! (1 - {(- 1)}^{n}) = {{(- 1)}^{n} - 1} (n - 1)!$ $\Rightarrow f^{(2 p)} (0) = 0 a n d f^{(2 p + 1)} (0) = - 2 (2 p)!$ $3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{p = 0}^{\infty} \frac{- 2 (2 p)!}{(2 p + 1)!} x^{2 p + 1} \Rightarrow$ $f (x) = - 2 \sum_{p = 0}^{\infty} \frac{x^{2 p + 1}}{2 p + 1}$
Commented by mathmax by abdo last updated on 27/Jun/19

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