let f(x)=arctan(2x+1) 1) calculate f^((n)) (x) 2) calculate f^((n)) (0) 3) developp f at integr serie 4) calculate ∫_0 ^1 f(x)dx 5) calculate ∫


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 39891 by math khazana by abdo last updated on 13/Jul/18

$l e t f (x) = a r c t a n (2 x + 1)$ $1) c a l c u l a t e f^{(n)} (x)$ $2) c a l c u l a t e f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e$ $4) c a l c u l a t e \int_{0}^{1} f (x) d x$ $5) c a l c u l a t e \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{4 x^{2} + 4 x + 2} d x$
Commented by maxmathsup by imad last updated on 13/Jul/18
$1) we have f^′ (x)= (2/(1+(2x+1)^2 )) ⇒f^((n)) (x)=2 ((1/((2x+1)^2 +1)))^((n−1)) let w(x)= (1/((2x+1)^2 +1)) .let decompose w inside C(x) w(x)= (1/((2x+1−i)(2x+1 +i))) = (1/(4(x +((1−i)/2))(x +((1+i)/2)))) =(1/(4(x +(1/(√2))e^(−((iπ)/4)) )(x +(1/(√2))e^(i(π/4)) ))) = (a/(x+(1/(√2))e^(−((iπ)/4)) )) +(b/(x+(1/(√2))e^((iπ)/4) )) a =lim_(x→−(1/(√2))e^(−((iπ)/4)) ) (x +(1/(√2))e^(−((iπ)/4)) )w(x)= (1/(4(1/(√2))(e^((iπ)/4) −e^(−((iπ)/4)) ))) = ((√2)/(4(2i(1/(√2))))) = (1/(4i)) b =lim_(x→−(1/(√2)) e^(i(π/4)) ) (x+ (1/(√2)) e^((iπ)/4) )w(x)= (1/(4(1/((√2) ))(−e^((iπ)/4) +e^(−((iπ)/4)) ))) =((√2)/(−4(2i(1/(√2))))) = ((−2)/(8i)) = −(1/(4i)) ⇒w(x)= (1/(4i)){ (1/(x+(1/(√2))e^(−((iπ)/4)) )) −(1/(x +(1/(√2))e^((iπ)/4) ))} ⇒ w^((n−1)) = (1/(4i)){ (((−1)^(n−1) (n−1)!)/((x +(1/(√2))e^(−((iπ)/4)) )^n )) −(((−1)^(n−1) (n−1)!)/((x +(1/(√2))e^((iπ)/4) )^n ))} ⇒ f^((n)) (x) =2 w^((n−1)) (x) = (((−1)^(n−1) (n−1)!)/(2i)){ (1/((x +(1/(√2))e^(−((iπ)/4)) )^n )) −(1/((x +(1/(√2))e^((iπ)/4) )^n ))}.$
$1) w e h a v e f^{^{'}} (x) = \frac{2}{1 + {(2 x + 1)}^{2}} \Rightarrow f^{(n)} (x) = 2 {(\frac{1}{{(2 x + 1)}^{2} + 1})}^{(n - 1)}$ $l e t w (x) = \frac{1}{{(2 x + 1)}^{2} + 1} . l e t d e c o m p o s e w i n s i d e C (x)$ $w (x) = \frac{1}{(2 x + 1 - i) (2 x + 1 + i)} = \frac{1}{4 (x + \frac{1 - i}{2}) (x + \frac{1 + i}{2})}$ $= \frac{1}{4 (x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) (x + \frac{1}{\sqrt{2}} e^{i \frac{π}{4}})} = \frac{a}{x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} + \frac{b}{x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}}$ $a = {l i m}_{x \to - \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} (x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) w (x) = \frac{1}{4 \frac{1}{\sqrt{2}} (e^{\frac{i π}{4}} - e^{- \frac{i π}{4}})} = \frac{\sqrt{2}}{4 (2 i \frac{1}{\sqrt{2}})} = \frac{1}{4 i}$ $b = {l i m}_{x \to - \frac{1}{\sqrt{2}} e^{i \frac{π}{4}}} (x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}) w (x) = \frac{1}{4 \frac{1}{\sqrt{2}} (- e^{\frac{i π}{4}} + e^{- \frac{i π}{4}})} = \frac{\sqrt{2}}{- 4 (2 i \frac{1}{\sqrt{2}})}$ $= \frac{- 2}{8 i} = - \frac{1}{4 i} \Rightarrow w (x) = \frac{1}{4 i} {\frac{1}{x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} - \frac{1}{x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}}} \Rightarrow$ $w^{(n - 1)} = \frac{1}{4 i} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}} \Rightarrow$ $f^{(n)} (x) = 2 w^{(n - 1)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{1}{{(x + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{1}{{(x + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}} .$
Commented by maxmathsup by imad last updated on 13/Jul/18
$2) for x=0 we get f^((n)) (0) = (((−1)^(n−1) (n−1)!)/(2i)) { (1/(((1/(√2))e^(−((iπ)/4)) )^n )) −(1/(((1/(√2))e^((iπ)/4) )^n ))} =(((−1)^(n−1) (n−1)!)/(2i)){ ((√2))^n e^((inπ)/4) −((√2))^n e^(−((inπ)/4)) } =((((√2))^n (−1)^(n−1) (n−1)!)/(2i)) (2i sin(((nπ)/4))) ⇒ f^((n)) (0) =(−1)^n (n−1)! ((√2))^n sin(((nπ)/4)) .$
$2) f o r x = 0 w e g e t$ $f^{(n)} (0) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{1}{{(\frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}^{n}} - \frac{1}{{(\frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}^{n}}}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {{(\sqrt{2})}^{n} e^{\frac{i n π}{4}} - {(\sqrt{2})}^{n} e^{- \frac{i n π}{4}}}$ $= \frac{{(\sqrt{2})}^{n} {(- 1)}^{n - 1} (n - 1)!}{2 i} (2 i s i n (\frac{n π}{4})) \Rightarrow$ $f^{(n)} (0) = {(- 1)}^{n} (n - 1)! {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) .$
Commented by maxmathsup by imad last updated on 14/Jul/18

$3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $f (x) = \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} (n - 1)! {(\sqrt{2})}^{n}}{n!} x^{n}$ $= \frac{π}{4} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) x^{n} .$
Commented by maxmathsup by imad last updated on 14/Jul/18

$4) \int_{0}^{1} f (x) d x = \int_{0}^{1} a r c t a n (2 x + 1) d x b y p a r t s u^{^{'}} = 1 a n d v = a r c t a n (2 x + 1)$ $\int_{0}^{1} a r c t a n (2 x + 1) d x = {[x a r c t a n (2 x + 1)]}_{0}^{1} - \int_{0}^{1} x \frac{2}{1 + {(2 x + 1)}^{2}} d x$ $= a r c t a n (3) - \int_{0}^{1} \frac{2 x}{4 x^{2} + 4 x + 2} d x b u t$ $\int_{0}^{1} \frac{2 x}{4 x^{2} + 4 x + 2} d x = \int_{0}^{1} \frac{x}{2 x^{2} + 2 x + 1} d x = \frac{1}{4} \int_{0}^{1} \frac{4 x + 2 - 2}{2 x^{2} + 2 x + 1} d x$ $= \frac{1}{4} \int_{0}^{1} \frac{4 x + 2}{2 x^{2} + 2 x + 1} d x - \frac{1}{2} \int_{0}^{1} \frac{d x}{2 (x^{2} + x + \frac{1}{2})}$ $= \frac{1}{4} {[l n ∣ 2 x^{2} + 2 x + 1 ∣]}_{0}^{1} - \frac{1}{4} \int_{0}^{1} \frac{d x}{x^{2} + 2 \frac{1}{2} x + \frac{1}{4} + \frac{1}{4}}$ $= \frac{1}{4} l n (5) - \frac{1}{4} \int_{0}^{1} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{1}{4}} (c h . x + \frac{1}{2} = \frac{1}{2} t)$ $= \frac{1}{4} l n (5) - \frac{1}{4} \int_{0}^{1} \frac{1}{\frac{1}{4} (1 + t^{2})} \frac{d t}{2} = \frac{l n (5)}{4} - \frac{1}{2} \frac{π}{4} = \frac{l n (5)}{4} - \frac{π}{8} \Rightarrow$ $\int_{0}^{1} f (x) d x = a r c t a n (3) - \frac{l n (5)}{4} + \frac{π}{8} .$
Commented by maxmathsup by imad last updated on 14/Jul/18

$5) w e h a v e \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{4 x^{2} + 4 x + 2} d x = \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} a n d b y p a r t s$ $u^{^{'}} = \frac{1}{{(2 x + 1)}^{2} + 1} a n d v = a r c t a n (2 x + 1) w e g e t$ $\int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} d x = {[\frac{1}{2} a r c t a n (2 x + 1) . a r c t a n (2 x + 1)]}_{0}^{1}$ $- \int_{0}^{1} \frac{1}{2} a r c t a n (2 x + 1) . \frac{2}{{(2 x + 1)}^{2} + 1} d x$ $= \frac{1}{2} {(a r c t a n (3))}^{2} - \frac{π^{2}}{32} - \int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} \Rightarrow$ $\int_{0}^{1} \frac{a r c t a n (2 x + 1)}{{(2 x + 1)}^{2} + 1} d x = \frac{1}{4} {(a r c t a n (3))}^{2} - \frac{π^{2}}{64} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum