let f(x) =((cosx)/(x^2 +1)) 1) calculate f^((n)) (x) then f^((n)) (0) 2)calculste f^((n)) (0) 3) developp f at integr serie .


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Question Number 56583 by maxmathsup by imad last updated on 18/Mar/19

$l e t f (x) = \frac{c o s x}{x^{2} + 1}$ $1) c a l c u l a t e f^{(n)} (x) t h e n f^{(n)} (0)$ $2) c a l c u l s t e f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 20/Mar/19
$1) we have f(x) =((cosx)/((x−i)(x+i))) =(1/(2i)){((cosx)/(x−i)) −((cosx)/(x+i))} ⇒ f^((n)) (x)=(1/(2i)){ (((cosx)/(x−i)))^((n)) −(((cosx)/(x+i)))^((n)) } leibniz formula give (((cosx)/(x+i)))^((n)) =Σ_(k=0) ^n C_n ^k ((1/(x+i)))^((k)) (cosx)^((n−k)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+i)^(k+1) )) cos(x +(n−k)(π/2)) also for the same raison (((cosx)/(x−i)))^((n)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x−i)^(k+1) )) cos(x+(n−k)(π/2)) ⇒f^((n)) (x) =(1/(2i)) Σ_(k=0) ^n C_n ^k (−1)^k k!cos(x+(n−k)(π/2)){(1/((x−i)^(k+1) )) −(1/((x+i)^(k+1) ))} f^((n)) (x)=(1/(2i)) Σ_(k=0) ^n C_n ^k (−1)^k k! cos(x+(n−k)(π/2))(((x+i)^(k+1) −(x−i)^(k+1) )/((x^2 +1)^(k+1) )) 2)x=0 ⇒f^((n)) (0) =(1/(2i)) Σ_(k=0) ^n C_n ^k (−1)^k k! cos((n−k)(π/2)) ((2i Im(i^(k+1) ))/1) ⇒ f^((n)) (0) =Σ_(k=0) ^n C_n ^k (−1)^k k! cos((n−k)(π/2))sin((k+1)(π/2)) . 3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ (1/(n!)){Σ_(k=0) ^n ((n!)/(k!(n−k)!))(−1)^k k! cos((n−k)(π/2))sin((k+1)(π/2)}x^n ⇒ f(x) =Σ_(n=0) ^∞ {Σ_(k=0) ^n (((−1)^k )/((n−k)!)) cos((n−k)(π/2))sin((k+1)(π/2))}x^n .$
$1) w e h a v e f (x) = \frac{c o s x}{(x - i) (x + i)} = \frac{1}{2 i} {\frac{c o s x}{x - i} - \frac{c o s x}{x + i}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{2 i} {{(\frac{c o s x}{x - i})}^{(n)} - {(\frac{c o s x}{x + i})}^{(n)}} l e i b n i z f o r m u l a g i v e$ ${(\frac{c o s x}{x + i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{x + i})}^{(k)} {(c o s x)}^{(n - k)} = \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x + i)}^{k + 1}} c o s (x + (n - k) \frac{π}{2})$ $a l s o f o r t h e s a m e r a i s o n {(\frac{c o s x}{x - i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x - i)}^{k + 1}} c o s (x + (n - k) \frac{π}{2})$ $\Rightarrow f^{(n)} (x) = \frac{1}{2 i} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} k! c o s (x + (n - k) \frac{π}{2}) {\frac{1}{{(x - i)}^{k + 1}} - \frac{1}{{(x + i)}^{k + 1}}}$ $f^{(n)} (x) = \frac{1}{2 i} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} k! c o s (x + (n - k) \frac{π}{2}) \frac{{(x + i)}^{k + 1} - {(x - i)}^{k + 1}}{{(x^{2} + 1)}^{k + 1}}$ $2) x = 0 \Rightarrow f^{(n)} (0) = \frac{1}{2 i} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} k! c o s ((n - k) \frac{π}{2}) \frac{2 i I m (i^{k + 1})}{1} \Rightarrow$ $f^{(n)} (0) = \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} k! c o s ((n - k) \frac{π}{2}) s i n ((k + 1) \frac{π}{2}) .$ $3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 0}^{\infty} \frac{1}{n!} {\sum_{k = 0}^{n} \frac{n!}{k! (n - k)!} {(- 1)}^{k} k! c o s ((n - k) \frac{π}{2}) s i n ((k + 1) \frac{π}{2}} x^{n} \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} {\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{(n - k)!} c o s ((n - k) \frac{π}{2}) s i n ((k + 1) \frac{π}{2})} x^{n} .$
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