let f(x)= ((artan(x+1))/(1+2x)) developp f at integr serie .


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Question Number 34863 by a.i msup by abdo last updated on 12/May/18

$l e t f (x) = \frac{a r t a n (x + 1)}{1 + 2 x}$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by math khazana by abdo last updated on 13/May/18
$for ∣x∣<(1/2) we have (1/(1+2x)) = Σ_(n=0) ^∞ (−2x)^n =Σ_(n=0) ^∞ (−2)^n x^n let put w(x)=arctan(x+1) w^′ (x) = (1/(1+(x+1)^2 )) = (1/((x+1)^2 −i^2 )) = (1/((x+1−i)(x+1+i))) =(1/(2i)) ( (1/(x+1−i)) −(1/(x+1+i))) ⇒ ⇒w^((n+1)) (x) =(1/(2i)){ ((1/(x+1−i)))^((n)) −((1/(x+1+i)))^((n)) } =(1/(2i)) (((−1)^n n!)/((x+1−i)^(n+1) )) −(1/(2i)) (((−1)^n n!)/((x+1+i)^(n+1) )) ⇒ w^((n+1)) (0) = (((−1)^n n!)/(2i(1−i)^(n+1) )) −(((−1)^n n!)/(2i(1+i)^(n+1) )) ⇒w^((n)) (0)= (((−1)^(n−1) (n−1)!)/(2i)){ (((1+i)^n −(1−i)^n )/2^n )} = (((−1)^(n−1) (n−1)!)/(2i .2^n )) 2iIm(1+i)^n =(((−1)^(n−1) (n−1)!)/2^n ) ((√2))^n sin(n(π/4)) we have w(x)= Σ_(n=0) ^∞ ((w^((n)) (0))/(n!)) x^n = 1+Σ_(n=1) ^∞ (((−1)^(n−1) (n−1)! ((√2))^n )/(2^n n!)) sin(n(π/4))x^n f(x)=( Σ_(n=0) ^∞ a_n x^n ) (Σ_(n=0) ^∞ b_n x^n ) =Σ c_n x^n with c_n = Σ_(i+j=n) a_i bj so tbe developpement of f(x) is known .$
$f o r ∣ x ∣< \frac{1}{2} w e h a v e \frac{1}{1 + 2 x} = \sum_{n = 0}^{\infty} {(- 2 x)}^{n}$ $= \sum_{n = 0}^{\infty} {(- 2)}^{n} x^{n} l e t p u t w (x) = a r c t a n (x + 1)$ $w^{^{'}} (x) = \frac{1}{1 + {(x + 1)}^{2}} = \frac{1}{{(x + 1)}^{2} - i^{2}}$ $= \frac{1}{(x + 1 - i) (x + 1 + i)} = \frac{1}{2 i} (\frac{1}{x + 1 - i} - \frac{1}{x + 1 + i}) \Rightarrow$ $\Rightarrow w^{(n + 1)} (x) = \frac{1}{2 i} {{(\frac{1}{x + 1 - i})}^{(n)} - {(\frac{1}{x + 1 + i})}^{(n)}}$ $= \frac{1}{2 i} \frac{{(- 1)}^{n} n!}{{(x + 1 - i)}^{n + 1}} - \frac{1}{2 i} \frac{{(- 1)}^{n} n!}{{(x + 1 + i)}^{n + 1}} \Rightarrow$ $w^{(n + 1)} (0) = \frac{{(- 1)}^{n} n!}{2 i {(1 - i)}^{n + 1}} - \frac{{(- 1)}^{n} n!}{2 i {(1 + i)}^{n + 1}}$ $\Rightarrow w^{(n)} (0) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{{(1 + i)}^{n} - {(1 - i)}^{n}}{2^{n}}}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i . 2^{n}} 2 i I m {(1 + i)}^{n}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2^{n}} {(\sqrt{2})}^{n} s i n (n \frac{π}{4}) w e h a v e$ $w (x) = \sum_{n = 0}^{\infty} \frac{w^{(n)} (0)}{n!} x^{n}$ $= 1 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} (n - 1)! {(\sqrt{2})}^{n}}{2^{n} n!} s i n (n \frac{π}{4}) x^{n}$ $f (x) = (\sum_{n = 0}^{\infty} a_{n} x^{n}) (\sum_{n = 0}^{\infty} b_{n} x^{n}) = Σ c_{n} x^{n} w i t h$ $c_{n} = \sum_{i + j = n} a_{i} b j s o t b e d e v e l o p p e m e n t o f f (x) i s$ $k n o w n .$
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