let ϕ(x) =((arctan(2x))/(1−x^2 )) 1) calculate ϕ^((n)) (x) 2) calculate ϕ^((n)) (0) anddevelpp ϕ at integr serie


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Question Number 53961 by maxmathsup by imad last updated on 27/Jan/19

$l e t φ (x) = \frac{a r c t a n (2 x)}{1 - x^{2}}$ $1) c a l c u l a t e φ^{(n)} (x)$ $2) c a l c u l a t e φ^{(n)} (0) a n d d e v e l p p φ a t i n t e g r s e r i e$
Commented by maxmathsup by imad last updated on 06/Feb/19
$1) we have ϕ(x)=(1/2)arctan(2x){(1/(1−x)) +(1/(1+x))) =(1/2) ((arctan(2x))/(1−x)) +(1/2) ((arctan(2x))/(1+x)) =W(x) −H(x) with W(x)=(1/2) ((arctan(2x))/(x+1)) and H(x)=(1/2) ((arctan(2x))/(x−1)) ⇒ ϕ^((n)) (x)=W^((n)) (x)−H^((n)) (x) leibniz formula give W^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) ((1/(x+1)))^((n−k)) but ((1/(x+1)))^((n−k)) =(((−1)^(n−k) (n−k)!)/((x+1)^(n−k+1) )) we have (arctan(2x))^′ =(2/(1+4x^2 )) ⇒ (arctan(2x))^((k)) =2((1/(4x^2 +1)))^((k−1)) =−i { (1/(2x−i)) −(1/(2x+i))}^((k−1)) =i{(1/(2(x+(i/2)))) −(1/(2(x−(i/2))))}^((k−1)) =(i/2){ (((−1)^(k−1) (k−1)!)/((x+(i/2))^k )) −(((−1)^(k−1) (k−1)!)/((x−(i/2))^k ))} =(i/2)(−1)^(k−1) (k−1)!{ (((x−(i/2))^k −(x+(i/2))^k )/((x^2 +(1/4))^k ))} ⇒ W^((n)) (x) = arctan(2x)(((−1)^n n!)/((x+1)^(n+1) )) +Σ_(k=1) ^n C_n ^k (i/2)(−1)^(k−1) (k−1)!{(((x−(i/2))^k −(x+(i/2))^k )/((x^2 +(1/4))^k ))}(((−1)^(n−k) (n−k)!)/((x+1)^(n−k+1) )) . =(((−1)^n n!)/((x+1)^(n+1) )) arctan(2x) −Σ_(k=1) ^n ((n!)/k) (−1)^k ( (( Im(x+(i/2))^k )/((x^2 +(1/4))^k ))) (1/((x+1)^(n−k+1) )) .$
$1) w e h a v e φ (x) = \frac{1}{2} a r c t a n (2 x) {\frac{1}{1 - x} + \frac{1}{1 + x}) = \frac{1}{2} \frac{a r c t a n (2 x)}{1 - x} + \frac{1}{2} \frac{a r c t a n (2 x)}{1 + x}$ $= W (x) - H (x) w i t h W (x) = \frac{1}{2} \frac{a r c t a n (2 x)}{x + 1} a n d H (x) = \frac{1}{2} \frac{a r c t a n (2 x)}{x - 1}$ $\Rightarrow φ^{(n)} (x) = W^{(n)} (x) - H^{(n)} (x) l e i b n i z f o r m u l a g i v e$ $W^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n (2 x))}^{(k)} {(\frac{1}{x + 1})}^{(n - k)} b u t$ ${(\frac{1}{x + 1})}^{(n - k)} = \frac{{(- 1)}^{n - k} (n - k)!}{{(x + 1)}^{n - k + 1}}$ $w e h a v e {(a r c t a n (2 x))}^{^{'}} = \frac{2}{1 + 4 x^{2}} \Rightarrow {(a r c t a n (2 x))}^{(k)} = 2 {(\frac{1}{4 x^{2} + 1})}^{(k - 1)}$ $= - i {\frac{1}{2 x - i} - \frac{1}{2 x + i}}^{(k - 1)} = i {\frac{1}{2 (x + \frac{i}{2})} - \frac{1}{2 (x - \frac{i}{2})}}^{(k - 1)}$ $= \frac{i}{2} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{i}{2})}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - \frac{i}{2})}^{k}}}$ $= \frac{i}{2} {(- 1)}^{k - 1} (k - 1)! {\frac{{(x - \frac{i}{2})}^{k} - {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}}} \Rightarrow$ $W^{(n)} (x) = a r c t a n (2 x) \frac{{(- 1)}^{n} n!}{{(x + 1)}^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} \frac{i}{2} {(- 1)}^{k - 1} (k - 1)! {\frac{{(x - \frac{i}{2})}^{k} - {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}}} \frac{{(- 1)}^{n - k} (n - k)!}{{(x + 1)}^{n - k + 1}} .$ $= \frac{{(- 1)}^{n} n!}{{(x + 1)}^{n + 1}} a r c t a n (2 x) - \sum_{k = 1}^{n} \frac{n!}{k} {(- 1)}^{k} (\frac{I m {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}}) \frac{1}{{(x + 1)}^{n - k + 1}} .$
Commented by maxmathsup by imad last updated on 06/Feb/19
$also we have H^((n)) (x) =(((−1)^n n!)/((x−1)!)) arctan(2x)−Σ_(k=1) ^n n!(((−1)^k )/k) ((Im(x+(i/2))^k )/((x^2 +(1/4))^k (x−1)^(n−k +1) )) . ϕ^((n)) (0) =W^((n)) (0)−H^((n)) (0) =n! Σ_(k=1) ^n (((−1)^(k−1) )/k) ((Im((i^k /2^k )))/(((1/4))^k )) −(n! Σ_(k=1) ^n (((−1)^(k−1) )/k) ((Im((i^k /2^k )))/(((1/4))^k ))(−1)^(n−k +1) =n!{Σ_(k=1) ^n (((−1)^(k−1) )/k) 4^(−k) (1/2^k )sin(((kπ)/2))−Σ_(k=1) ^n (−1)^n 4^(−k) (1/2^k )sin(((kπ)/2))} .$
$a l s o w e h a v e$ $H^{(n)} (x) = \frac{{(- 1)}^{n} n!}{(x - 1)!} a r c t a n (2 x) - \sum_{k = 1}^{n} n! \frac{{(- 1)}^{k}}{k} \frac{I m {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k} {(x - 1)}^{n - k + 1}} .$ $φ^{(n)} (0) = W^{(n)} (0) - H^{(n)} (0)$ $= n! \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} \frac{I m (\frac{i^{k}}{2^{k}})}{{(\frac{1}{4})}^{k}} - (n! \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} \frac{I m (\frac{i^{k}}{2^{k}})}{{(\frac{1}{4})}^{k}} {(- 1)}^{n - k + 1}$ $= n! {\sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} 4^{- k} \frac{1}{2^{k}} s i n (\frac{k π}{2}) - \sum_{k = 1}^{n} {(- 1)}^{n} 4^{- k} \frac{1}{2^{k}} s i n (\frac{k π}{2})} .$
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