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Question Number 39699 by maxmathsup by imad last updated on 09/Jul/18

l e t f (x) = a r c t a n (2 x^{2} + 1)

1) c a l c u l a t e f^{(n)} (x)

2) d e t e r m i n e f^{(n)} (n)

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{x a r c t a n (2 x^{2} + 1)}{2 x^{2} + 2 x + 1} d x

Commented by math khazana by abdo last updated on 10/Jul/18

2) d e t e r m i n e f^{(n)} (0)

Commented by maxmathsup by imad last updated on 13/Jul/18

4) l e t i n t e g r a t e b y p a r t s u^{^{'}} = 1 a n d v = a r c t a n (2 x^{2} + 1)

\int_{0}^{1} f (x) d x = {[x a r c t a n (2 x^{2} + 1)]}_{0}^{1} - \int_{0}^{1} x \frac{4 x}{1 + {(2 x^{2} + 1)}^{2}} d x

= a r c t a n (3) - 4 \int_{0}^{1} \frac{x^{2}}{1 + {(2 x^{2} + 1)}^{2}} d x c h a n g e m e n t x = \frac{1}{\sqrt{2}} t a n θ g i v e

\int_{0}^{1} \frac{x^{2}}{1 + {(2 x^{2} + 1)}^{2}} d x = \int_{0}^{a r c t a n (\sqrt{2})} \frac{1}{2} \frac{{t a n}^{2} θ}{1 + {(1 + {t a n}^{2} θ)}^{2}} \frac{1}{\sqrt{2}} (1 + {t a n}^{2} θ) d θ

= \frac{1}{2 \sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{{t a n}^{2} θ (1 + {t a n}^{2} θ)}{1 + {(1 + {t a n}^{2} θ)}^{2}} d θ

= \frac{1}{2 \sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{(\frac{1}{{c o s}^{2} θ} - 1) (\frac{1}{{c o s}^{2} θ})}{1 + \frac{1}{{c o s}^{4} θ}} d θ

= \frac{1}{2 \sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{1 - {c o s}^{2} θ}{1 + {c o s}^{4} θ} d θ = \frac{1}{2 \sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{1 - \frac{1 + c o s (2 θ)}{2}}{1 + {(\frac{1 + c o s (2 θ)}{2})}^{2}} d θ

= \frac{1}{2 \sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{1 - c o s (2 θ)}{4 + {(1 + c o s (2 θ))}^{2}} 2 d θ

= \frac{1}{\sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{1 - c o s (2 θ)}{4 + 1 + 2 c o s (2 θ) + \frac{1 + c o s (4 θ)}{4}} d θ

= \frac{4}{\sqrt{2}} \int_{0}^{a r c t a n (\sqrt{2})} \frac{1 - c o s (2 θ)}{21 + 8 c o s (2 θ) + c o s (4 θ)} d θ \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 13/Jul/18

1) w e h a v e f^{^{'}} (x) = \frac{4 x}{1 + {(2 x^{2} + 1)}^{2}} \Rightarrow f^{(n)} (x) = 4 {\frac{x}{1 + {(2 x^{2} + 1)}^{2}}}^{(n - 1)}

l e t w (x) = \frac{x}{1 + {(2 x^{2} + 1)}^{2}} l e t d e c o m p o s e w (x)

w (x) = \frac{x}{{(2 x^{2} + 1)}^{2} - i^{2}} = \frac{x}{(2 x^{2} + 1 - i) (2 x^{2} + 1 + i)}

= \frac{x}{4 (x^{2} + \frac{1 - i}{2}) (x^{2} + \frac{1 + i}{2})} = \frac{x}{4 (x^{2} + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) (x^{2} + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}

= \frac{x}{4 (x - \frac{i}{\sqrt{\sqrt{2}}} e^{- \frac{i π}{8}}) (x + \frac{i}{\sqrt{\sqrt{2}}} e^{- \frac{i π}{8}}) (x - \frac{i}{\sqrt{\sqrt{2}}} e^{\frac{i π}{8}}) (x + \frac{1}{\sqrt{\sqrt{2}}} e^{\frac{i π}{8}})} l e t

z_{0} = \frac{i}{\sqrt{\sqrt{2}}} e^{\frac{i π}{8}}, z_{1} = - \frac{i}{\sqrt{\sqrt{2}}} e^{\frac{i π}{8}}, z_{2} = \frac{i}{\sqrt{\sqrt{2}}} e^{- \frac{i π}{8}}, z_{3} = - \frac{i}{\sqrt{\sqrt{2}}} e^{- \frac{i π}{8}} \Rightarrow

f^{^{'}} (x) = 4 w (x) = \frac{a}{x - z_{0}} + \frac{b}{x - z_{1}} + \frac{c}{x - z_{2}} + \frac{d}{x - z_{3}} = \frac{x}{(x - z_{0}) (x - z_{1}) (x - z_{2}) (x - z_{3})}

w e h a v e a = \frac{z_{0}}{(z_{0} - z_{1}) (z_{0}^{2} + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}

b = \frac{z_{1}}{(z_{1} - z_{0}) (z_{1}^{2} + \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}})}

c = \frac{z_{2}}{(z_{2} - z_{3}) (z_{2}^{2} + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})}

d = \frac{z_{3}}{(z_{3} - z_{2}) (z_{3}^{2} + \frac{1}{\sqrt{2}} e^{\frac{i π}{4}})} a n d

f^{(n)} (x) = a \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{0})}^{n}} + b \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{1})}^{n}} + c \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{2})}^{n}}

+ d \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{3})}^{n}}

2) f^{(n)} (0) = {(- 1)}^{n - 1} (n - 1)! {\frac{a}{{(- z_{0})}^{n}} + \frac{b}{{(- z_{1})}^{n}} + \frac{c}{{(- z_{2})}^{n}} + \frac{d}{{(- z_{3})}^{n}}}

= - (n - 1)! {\frac{a}{z_{0}^{n}} + \frac{b}{z_{1}^{n}} + \frac{c}{z_{2}^{n}} + \frac{d}{z_{3}^{n}}}

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{1}{n} {\frac{a}{z_{0}^{n}} + \frac{b}{z_{1}^{n}} + \frac{c}{z_{2}^{n}} + \frac{d}{z_{3}^{n}}} x^{n}

Commented by maxmathsup by imad last updated on 13/Jul/18

5) w e h a v e f^{^{'}} (x) = \frac{4 x}{1 + {(2 x^{2} + 1)}^{2}} = \frac{4 x}{1 + 4 x^{4} + 4 x^{2} + 1} = \frac{4 x}{4 x^{4} + 4 x^{2} + 2}

= \frac{2 x}{2 x^{4} + 2 x^{2} + 1} \Rightarrow \frac{x}{2 x^{4} + 2 x^{2} + 1} = \frac{1}{2} f^{^{'}} (x) \Rightarrow l e t i n t e g r a t e b y p a r t s

\int_{0}^{1} \frac{x a r c t a n (2 x^{2} + 1)}{2 x^{4} + 2 x^{2} + 1} d x = \frac{1}{2} \int_{0}^{1} f . f^{^{'}} d x = {[\frac{1}{4} f^{2} (x)]}_{0}^{1}

= \frac{1}{4} {f^{2} (1) - f^{(2)} (0)} = \frac{1}{4} {{a r c t a n}^{2} (3) - \frac{π^{2}}{16}} .

Commented by maxmathsup by imad last updated on 13/Jul/18

5) t h e Q i s [f i n d \int_{0}^{1} \frac{x a r c t a n (2 x^{2} + 1)}{2 x^{4} + 2 x^{2} + 1} d x .