let g(x)=((ln(1+x))/(3+x^2 )) 1) find g^((n)) (x)and g^((n)) (0) 2)developp g at integr serie


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Question Number 72394 by mathmax by abdo last updated on 28/Oct/19

$l e t g (x) = \frac{l n (1 + x)}{3 + x^{2}}$ $1) f i n d g^{(n)} (x) a n d g^{(n)} (0)$ $2) d e v e l o p p g a t i n t e g r s e r i e$
Commented by mathmax by abdo last updated on 29/Oct/19
$1) we have g^((n)) (x)=Σ_(k=0) ^n C_n ^k (ln(1+x))^((k)) ((1/(x^2 +3)))^((n−k)) but ln(x+1))^((1)) =(1/(x+1)) ⇒(ln(x+1))^((k)) =((1/(x+1)))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+1)^k )) we have (1/(x^2 +3)) =(1/((x−i(√3))(x+i(√3)))) =(1/(2i(√3)))((1/(x−i(√3)))−(1/(x+i(√3)))) ⇒ ((1/(x^2 +3)))^((n−k)) =(1/(2i(√3))){ ((1/(x−i(√3))))^((n−k)) −((1/(x+i(√3))))^((n−k)) } =(1/(2i(√3))){(((−1)^(n−k) (n−k)!)/((x−i(√3))^(n−k+1) ))−(((−1)^(n−k) (n−k)!)/((x+i(√3))^(n−k+1) ))} =(((−1)^(n−k) (n−k)!)/(2i(√3))){(((x+i(√3))^(n−k+1) −(x−i(√3))^(n−k+1) )/((x^2 +3)^(n−k+1) ))} =(((−1)^(n−k) (n−k)!)/(√3))×((Im(x+i(√3))^(n−k+1) )/((x^2 +3)^(n−k+1) )) ⇒ g^((n)) (x)=ln(1+x)((1/(x^2 +3)))^((n)) +Σ_(k=1) ^n ((n!)/(k!(n−k)!)) (((−1)^(k−1) (k−1)!)/((x+1)^k ))×(((−1)^(n−k) (n−k)!×Im(x+i(√3))^(n−k+1) )/((√3)(x^2 +3)^(n−k+1) )) g^((n)) (x)=ln(x+1)((1/(x^2 +3)))^((n)) +Σ_(k=1) ^n ((n!(−1)^(n−1) .Im(x+i(√3))^(n−k+1) )/(k(√3)(x+1)^k (x^2 +3)^(n−k+1) )) also we have ((1/(x^2 +3)))^((n)) =(1/(2i(√3))){ ((1/(x−i(√3))))^((n)) −((1/(x+i(√3))))^((n)) } =(1/(2i(√3))){ (((−1)^n n!)/((x−i(√3))^(n+1) ))−(((−1)^n n!)/((x+i(√3))^(n+1) ))} =(((−1)^n n!)/(2i(√3))){((2iIm(x+i(√3))^(n+1) )/((x^2 +3)^(n+1) ))} =(((−1)^n n!)/(√3))×((Im(x+i(√3))^(n+1) )/((x^2 +3)^(n+1) )) ⇒ g^((n)) (x)=(((−1)^n n!)/(√3))×((Im(x+i(√3))^(n+1) )/((x^2 +3)^(n+1) ))ln(x+1) +Σ_(k=1) ^n ((n!(−1)^(n−1) ×Im(x+i(√3))^(n−k+1) )/(k(√3)(x+1)^k (x^2 +3)^(n−k+1) )).$
$1) w e h a v e g^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(l n (1 + x))}^{(k)} {(\frac{1}{x^{2} + 3})}^{(n - k)}$ ${b u t l n (x + 1))}^{(1)} = \frac{1}{x + 1} \Rightarrow {(l n (x + 1))}^{(k)} = {(\frac{1}{x + 1})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}}$ $w e h a v e \frac{1}{x^{2} + 3} = \frac{1}{(x - i \sqrt{3}) (x + i \sqrt{3})} = \frac{1}{2 i \sqrt{3}} (\frac{1}{x - i \sqrt{3}} - \frac{1}{x + i \sqrt{3}}) \Rightarrow$ ${(\frac{1}{x^{2} + 3})}^{(n - k)} = \frac{1}{2 i \sqrt{3}} {{(\frac{1}{x - i \sqrt{3}})}^{(n - k)} - {(\frac{1}{x + i \sqrt{3}})}^{(n - k)}}$ $= \frac{1}{2 i \sqrt{3}} {\frac{{(- 1)}^{n - k} (n - k)!}{{(x - i \sqrt{3})}^{n - k + 1}} - \frac{{(- 1)}^{n - k} (n - k)!}{{(x + i \sqrt{3})}^{n - k + 1}}}$ $= \frac{{(- 1)}^{n - k} (n - k)!}{2 i \sqrt{3}} {\frac{{(x + i \sqrt{3})}^{n - k + 1} - {(x - i \sqrt{3})}^{n - k + 1}}{{(x^{2} + 3)}^{n - k + 1}}}$ $= \frac{{(- 1)}^{n - k} (n - k)!}{\sqrt{3}} \times \frac{I m {(x + i \sqrt{3})}^{n - k + 1}}{{(x^{2} + 3)}^{n - k + 1}} \Rightarrow$ $g^{(n)} (x) = l n (1 + x) {(\frac{1}{x^{2} + 3})}^{(n)}$ $+ \sum_{k = 1}^{n} \frac{n!}{k! (n - k)!} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + 1)}^{k}} \times \frac{{(- 1)}^{n - k} (n - k)! \times I m {(x + i \sqrt{3})}^{n - k + 1}}{\sqrt{3} {(x^{2} + 3)}^{n - k + 1}}$ $g^{(n)} (x) = l n (x + 1) {(\frac{1}{x^{2} + 3})}^{(n)}$ $+ \sum_{k = 1}^{n} \frac{n! {(- 1)}^{n - 1} . I m {(x + i \sqrt{3})}^{n - k + 1}}{k \sqrt{3} {(x + 1)}^{k} {(x^{2} + 3)}^{n - k + 1}} a l s o w e h a v e$ ${(\frac{1}{x^{2} + 3})}^{(n)} = \frac{1}{2 i \sqrt{3}} {{(\frac{1}{x - i \sqrt{3}})}^{(n)} - {(\frac{1}{x + i \sqrt{3}})}^{(n)}}$ $= \frac{1}{2 i \sqrt{3}} {\frac{{(- 1)}^{n} n!}{{(x - i \sqrt{3})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x + i \sqrt{3})}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{2 i \sqrt{3}} {\frac{2 i I m {(x + i \sqrt{3})}^{n + 1}}{{(x^{2} + 3)}^{n + 1}}} = \frac{{(- 1)}^{n} n!}{\sqrt{3}} \times \frac{I m {(x + i \sqrt{3})}^{n + 1}}{{(x^{2} + 3)}^{n + 1}} \Rightarrow$ $g^{(n)} (x) = \frac{{(- 1)}^{n} n!}{\sqrt{3}} \times \frac{I m {(x + i \sqrt{3})}^{n + 1}}{{(x^{2} + 3)}^{n + 1}} l n (x + 1)$ $+ \sum_{k = 1}^{n} \frac{n! {(- 1)}^{n - 1} \times I m {(x + i \sqrt{3})}^{n - k + 1}}{k \sqrt{3} {(x + 1)}^{k} {(x^{2} + 3)}^{n - k + 1}} .$
Commented by mathmax by abdo last updated on 29/Oct/19

$g^{(n)} (0) = \sum_{k = 1}^{n} \frac{n! {(- 1)}^{n - 1} \times I m {(i \sqrt{3})}^{n - k + 1}}{k \sqrt{3} \times 3^{n - k + 1}} b u t$ ${(i \sqrt{3})}^{n - k + 1} = {(\sqrt{3})}^{n - k + 1} {(e^{\frac{i π}{2}})}^{(n - k + 1)}$ $= {(\sqrt{3})}^{n - k + 1} e^{\frac{i (n - k + 1 π}{2}} \Rightarrow I m {(i \sqrt{3})}^{n - k + 1} = {(\sqrt{3})}^{n - k + 1} s i n (\frac{(n - k + 1) π}{2})$ $\Rightarrow g^{(n)} (0) = \sum_{k = 1}^{n} \frac{n! {(- 1)}^{n - 1}}{k \sqrt{3} {(\sqrt{3})}^{n - k + 1}} s i n (\frac{(n - k + 1) π}{2})$
Commented by mathmax by abdo last updated on 29/Oct/19

$2) g (x) = \sum_{n = 0}^{\infty} \frac{g^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 1}^{\infty} (\sum_{k = 1}^{n} \frac{{(- 1)}^{n - 1}}{k {(\sqrt{3})}^{n - k + 2}} s i n (\frac{(n - k + 1) π}{2})) x^{n} (g (0) = 0)$
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