let f(x)=e^(−nx) ln(1+x^2 ) 1)determine f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie (n integr natural)


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Question Number 74502 by mathmax by abdo last updated on 25/Nov/19

$l e t f (x) = e^{- n x} l n (1 + x^{2})$ $1) d e t e r m i n e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e (n i n t e g r n a t u r a l)$
Commented by mathmax by abdo last updated on 25/Nov/19

$1) f^{(p)} (x) = \sum_{k = 0}^{p} C_{p}^{k} {(l n (1 + x^{2}))}^{(k)} {(e^{- n x})}^{(p - k)}$ $= l n (1 + x^{2}) {(- n)}^{p} e^{- n x} + \sum_{k = 1}^{p} C_{p}^{k} {(l n (x^{2} + 1))}^{(k)} {(- n)}^{p - k} e^{- n x}$ $w e h a v e {l n}^{^{'}} (x^{2} + 1) = \frac{2 x}{x^{2} + 1} = \frac{1}{x + i} + \frac{1}{x - i} \Rightarrow$ ${(l n (x^{2} + 1))}^{(k)} = {(\frac{1}{x + i} + \frac{1}{x - i})}^{(k - 1)} = \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{k}} + \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - i)}^{k}}$ $= {(- 1)}^{k - 1} (k - 1)! (\frac{1}{{(x + i)}^{k}} + \frac{1}{{(x - i)}^{k}}) = {(- 1)}^{k - 1} (k - 1)! \times \frac{{(x + i)}^{k} + {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}$ $f^{(p)} (x) = {(- n)}^{p} e^{- n x} l n (x^{2} + 1)$ $+ \sum_{k = 1}^{p} C_{p}^{k} {(- 1)}^{k - 1} (k - 1)! \times \frac{{(x + i)}^{k} + {(x - i)}^{k}}{{(x^{2} + 1)}^{k}} {(- n)}^{p - k} e^{- n x}$
Commented by mathmax by abdo last updated on 25/Nov/19

$f^{(n)} (x) = {(- n)}^{n} e^{- n x} l n (x^{2} + 1)$ $+ \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k - 1} (k - 1)! \times \frac{{(x + i)}^{k} + {(x - i)}^{k}}{{(x^{2} + 1)}^{k}} {(- n)}^{n - k} e^{- n x}$ $a n d$ $f^{(n)} (0) = \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k - 1} (k - 1)! \times (i^{k} + {(- i)}^{k}) {(- n)}^{n - k} e^{- n x}$ $i^{k} + {(- i)}^{k} = 2 R e (i^{k}) = 2 c o s (\frac{k π}{2}) \Rightarrow$ $f^{(n)} (0) = 2 {(- 1)}^{n - 1} n^{n} e^{- n x} \sum_{k = 1}^{n} (k - 1)! C_{n}^{k} c o s (\frac{k π}{2})$
Commented by mathmax by abdo last updated on 25/Nov/19
$2)f(x)=Σ_(p=0) ^∞ ((f^((p)) (0))/(p!)) x^p we have f^((p)) (0) =Σ_(k=1) ^p C_p ^k (−1)^(p−1) (k−1)!((2cos(((kπ)/2)))/1)(n)^(p−k) =2Σ_(k=1) ^p ((p!)/(k!(p−k)!))(−1)^(p−1) (k−1)! cos(((kπ)/2)) n^(p−k) =2(−1)^(p−1) n^p Σ_(k=1) ^p ((p!)/(k(p−k)!n^k )) cos(((kπ)/2)) ⇒ f(x)=2Σ_(p=0) ^∞ (−1)^(p−1) n^p { Σ_(k=1) ^p ((cos(((kπ)/2)))/(k n^k (p−k)!))}x^p$
$2) f (x) = \sum_{p = 0}^{\infty} \frac{f^{(p)} (0)}{p!} x^{p} w e h a v e$ $f^{(p)} (0) = \sum_{k = 1}^{p} C_{p}^{k} {(- 1)}^{p - 1} (k - 1)! \frac{2 c o s (\frac{k π}{2})}{1} {(n)}^{p - k}$ $= 2 \sum_{k = 1}^{p} \frac{p!}{k! (p - k)!} {(- 1)}^{p - 1} (k - 1)! c o s (\frac{k π}{2}) n^{p - k}$ $= 2 {(- 1)}^{p - 1} n^{p} \sum_{k = 1}^{p} \frac{p!}{k (p - k)! n^{k}} c o s (\frac{k π}{2}) \Rightarrow$ $f (x) = 2 \sum_{p = 0}^{\infty} {(- 1)}^{p - 1} n^{p} {\sum_{k = 1}^{p} \frac{c o s (\frac{k π}{2})}{k n^{k} (p - k)!}} x^{p}$
Commented by mathmax by abdo last updated on 25/Nov/19

$f^{(n)} (0) = 2 {(- 1)}^{n - 1} n^{n} \sum_{k = 1}^{n} (k - 1)! C_{n}^{k} c o s (\frac{k π}{2}) .$
	Answered by mind is power last updated on 25/Nov/19
	$ln^((k)) (1+x^2 )=ln(1+x^2 ),k=0 =((1/(x+i))+(1/(x−i)))^((k−1)) ,k≥1 =((((−1)^(k−1) (k−1)!)/((x+i)^((k)) ))+(((−1)^(k−1) (k−1)!)/((x−i)^((k)) ))) =(−1)^(k−1) (k−1)!.((((x−i)^((k−1)) +(x+i)^((k−1)) )/((x^2 +1)^k ))) =(((−1)^(k−1) (k−1)!.2Re((x+i)^((k−1)) ))/((x^2 +1)^k )) x+i= { (((√(x^2 +1)).e^(iarctan((1/x))) ,x>0)),(((√(x^2 +1))e^(i((π/2)+arctan((1/x)))) ,x<0)) :} x≥0 (d/dx^k ){ln(1+x^2 )}=(((−1)^(k−1) (k−1)!(x^2 +1)^((k−1)/2) 2Re{e^(i(k−1)arctan((1/x))) })/((x^2 +1)^k )),k≥1 f(x)=e^(−nx) ln(1+x^2 ) f^((n)) (x)=Σ_(k=0) ^n C_n ^k .(e^(−nx) )^((n−k)) .(ln^((k)) (1+x^2 )) =Σ_(k=1) ^n C_n ^k (−n)^((n−k)) .e^(−nx) .(((−1)^(k−1) (k−1)!.cos((k−1)(arctan((1/x))))/((x^2 +1)^((k+1)/2) ))+(−n)^n e^(−nx) ln(1+x^2 ) f^((n)) (0) =Σ_(k=1) ^n C_n ^k (−n)^(n−k) .(−1)^(k−1) .(k−1)!.cos((k−1)(π/2)) 2) f(x)=Σ_(j=0) ^(+∞) ((f^j (0)x^j )/(j!)) f(x)=Σ_(j≥1) ((f^j (0)x^j )/(j!))=Σ_(j≥1) .(Σ_(k=1) ^j C_j ^k .(−n)^(j−k) (−1)^(k−1) cos(((k−1)/2)π))x^j .(1/(j!))$
	${l n}^{(k)} (1 + x^{2}) = l n (1 + x^{2}), k = 0$ $= {(\frac{1}{x + i} + \frac{1}{x - i})}^{(k - 1)}, k ⩾ 1$ $= (\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{(k)}} + \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - i)}^{(k)}})$ $= {(- 1)}^{k - 1} (k - 1)! . (\frac{{(x - i)}^{(k - 1)} + {(x + i)}^{(k - 1)}}{{(x^{2} + 1)}^{k}})$ $= \frac{{(- 1)}^{k - 1} (k - 1)! . 2 R e ({(x + i)}^{(k - 1)})}{{(x^{2} + 1)}^{k}}$ $x + i = {\begin{cases} \sqrt{x^{2} + 1} . e^{i a r c t a n (\frac{1}{x})}, x > 0 \\ \sqrt{x^{2} + 1} e^{i (\frac{π}{2} + a r c t a n (\frac{1}{x}))}, x < 0 \end{cases}$ $x ⩾ 0$ $\frac{d}{{d x}^{k}} {l n (1 + x^{2})} = \frac{{(- 1)}^{k - 1} (k - 1)! {(x^{2} + 1)}^{\frac{k - 1}{2}} 2 R e {e^{i (k - 1) a r c t a n (\frac{1}{x})}}}{{(x^{2} + 1)}^{k}}, k ⩾ 1$ $f (x) = e^{- n x} l n (1 + x^{2})$ $f^{(n)} (x) = \underset{k = 0}{\sum^{n}} C_{n}^{k} . {(e^{- n x})}^{(n - k)} . ({l n}^{(k)} (1 + x^{2}))$ $= \underset{k = 1}{\sum^{n}} C_{n}^{k} {(- n)}^{(n - k)} . e^{- n x} . \frac{{(- 1)}^{k - 1} (k - 1)! . c o s ((k - 1) (a r c t a n (\frac{1}{x}))}{{(x^{2} + 1)}^{\frac{k + 1}{2}}} + {(- n)}^{n} e^{- n x} l n (1 + x^{2})$ $f^{(n)} (0)$ $= \underset{k = 1}{\sum^{n}} C_{n}^{k} {(- n)}^{n - k} . {(- 1)}^{k - 1} . (k - 1)! . c o s ((k - 1) \frac{π}{2})$ $2)$ $f (x) = \underset{j = 0}{\sum^{+ \infty}} \frac{f^{j} (0) x^{j}}{j!}$ $f (x) = \sum_{j ⩾ 1} \frac{f^{j} (0) x^{j}}{j!} = \sum_{j ⩾ 1} . (\underset{k = 1}{\sum^{j}} C_{j}^{k} . {(- n)}^{j - k} {(- 1)}^{k - 1} c o s (\frac{k - 1}{2} π)) x^{j} . \frac{1}{j!}$
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