let f(x)=((arctan(2x))/(1+x)) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie


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Question Number 81431 by abdomathmax last updated on 13/Feb/20

$l e t f (x) = \frac{a r c t a n (2 x)}{1 + x}$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e$
Commented by abdomathmax last updated on 20/Feb/20
$1) we have f(x)=((arctan(2x))/(1+x)) ⇒ f^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) ((1/(1+x)))^((n−k)) =arctan(2x)×(((−1)^n )/((x+1)^(n+1) )) +Σ_(k=1) ^n C_n ^k (arctan(2x))^((k)) ×(((−1)^(n−k) )/((x+1)^(n−k+1) )) we have (arctan(2x))^((1)) =(2/(1+4x^2 )) ⇒ (arctan(2x))^((k)) =((2/(1+4x^2 )))^((k−1)) =(1/2){ (1/(x^2 +(1/4)))}^()k−1)) =(1/2){ (1/((x−(i/2))(x+(i/2))))}^((k−1)) =(1/(2i)){(1/(x−(i/2)))−(1/(x+(i/2)))}^((k−1)) =(1/(2i)){ (((−1)^(k−1) )/((x−(i/2))^k ))−(((−1)^(k−1) )/((x+(i/2))^k ))} =(((−1)^(k−1) )/(2i)){(((x+(i/2))^k −(x−(i/2))^k )/((x^2 +(1/4))^k ))} =(−1)^(k−1) ×((Im(x+(i/2))^k )/((x^2 +(1/4))^k )) ⇒ f^((n)) (x)=(((−1)^n arctan(2x))/((x+1)^n )) +(−1)^(n−1) Σ_(k=1) ^n C_n ^k ×((Im(x+(i/2))^k )/((x^2 +(1/4))^k )) ×(1/((x+1)^(n−k+1) ))$
$1) w e h a v e f (x) = \frac{a r c t a n (2 x)}{1 + x} \Rightarrow$ $f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n (2 x))}^{(k)} {(\frac{1}{1 + x})}^{(n - k)}$ $= a r c t a n (2 x) \times \frac{{(- 1)}^{n}}{{(x + 1)}^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} {(a r c t a n (2 x))}^{(k)} \times \frac{{(- 1)}^{n - k}}{{(x + 1)}^{n - k + 1}}$ $w e h a v e {(a r c t a n (2 x))}^{(1)} = \frac{2}{1 + 4 x^{2}} \Rightarrow$ ${(a r c t a n (2 x))}^{(k)} = {(\frac{2}{1 + 4 x^{2}})}^{(k - 1)}$ $= \frac{1}{2} {\frac{1}{x^{2} + \frac{1}{4}}}^{) k - 1)} = \frac{1}{2} {\frac{1}{(x - \frac{i}{2}) (x + \frac{i}{2})}}^{(k - 1)}$ $= \frac{1}{2 i} {\frac{1}{x - \frac{i}{2}} - \frac{1}{x + \frac{i}{2}}}^{(k - 1)}$ $= \frac{1}{2 i} {\frac{{(- 1)}^{k - 1}}{{(x - \frac{i}{2})}^{k}} - \frac{{(- 1)}^{k - 1}}{{(x + \frac{i}{2})}^{k}}}$ $= \frac{{(- 1)}^{k - 1}}{2 i} {\frac{{(x + \frac{i}{2})}^{k} - {(x - \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}}}$ $= {(- 1)}^{k - 1} \times \frac{I m {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n} a r c t a n (2 x)}{{(x + 1)}^{n}}$ $+ {(- 1)}^{n - 1} \sum_{k = 1}^{n} C_{n}^{k} \times \frac{I m {(x + \frac{i}{2})}^{k}}{{(x^{2} + \frac{1}{4})}^{k}} \times \frac{1}{{(x + 1)}^{n - k + 1}}$
Commented by abdomathmax last updated on 20/Feb/20
$f^((n)) (0) =(−1)^(n−1) Σ_(k=1) ^n C_n ^k ((sin(((kπ)/2)))/(2^k ((1/4))^k )) =(−1)^(n−1) Σ_(k=1) ^n C_n ^k 2^k sin(((kπ)/2)) 2) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =f(0) +Σ_(n=1) ^∞ (−1)^(n−1) ×(1/(n!)){Σ_(k=1) ^n C_n ^k 2^k sin(((kπ)/2))}x^n =Σ_(n=1) ^∞ (((−1)^(n−1) )/(n!)) Σ_(k=1) ^n ((n!)/(k!(n−k)!)) 2^k sin(((kπ)/2)) x^n f(x) =Σ_(n=1) ^∞ (−1)^(n−1) (Σ_(k=1) ^n ((2^k sin(((kπ)/2)))/(k!(n−k)!)))x^n$
$f^{(n)} (0) = {(- 1)}^{n - 1} \sum_{k = 1}^{n} C_{n}^{k} \frac{s i n (\frac{k π}{2})}{2^{k} {(\frac{1}{4})}^{k}}$ $= {(- 1)}^{n - 1} \sum_{k = 1}^{n} C_{n}^{k} 2^{k} s i n (\frac{k π}{2})$ $2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= f (0) + \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \times \frac{1}{n!} {\sum_{k = 1}^{n} C_{n}^{k} 2^{k} s i n (\frac{k π}{2})} x^{n}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n!} \sum_{k = 1}^{n} \frac{n!}{k! (n - k)!} 2^{k} s i n (\frac{k π}{2}) x^{n}$ $f (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\sum_{k = 1}^{n} \frac{2^{k} s i n (\frac{k π}{2})}{k! (n - k)!}) x^{n}$
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