let-f-x-ln-1-ix-2-1-extrsct-Re-f-x-and-Im-f-x-2-developp-f-at-integr-serie-3-calculate-f-x-by-two-methods-

Question Number 34770 by abdo mathsup 649 cc last updated on 10/May/18

l e t f (x) = l n (1 + {i x}^{2})

1) e x t r s c t R e (f (x)) a n d I m (f (x))

2) d e v e l o p p f a t i n t e g r s e r i e

3) c a l c u l a t e f^{^{'}} (x) b y t w o m e t h o d s

Commented by abdo mathsup 649 cc last updated on 13/May/18

1) w e h a v e p r o v e d t h a t

f (x) = \frac{1}{2} l n (1 + x^{4}) + i a r c t a n (x^{2}) \Rightarrow

R e (f (x)) = \frac{1}{2} l n (1 + x^{4}) a n d I m (f (x)) = a r c t a n (x^{2})

2) f o r ∣ u ∣< 1 l n (1 + u) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow

\frac{1}{2} l n (1 + x^{4}) = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{4 n}

\frac{d}{d x} (a r c t a n u) = \frac{1}{1 + u^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n} \Rightarrow

a r c t a n u = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} u^{2 n + 1} \Rightarrow

a r c t a n (x^{2}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{4 n + 2}

s o

f (x) = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{4 n} + i \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{4 n + 2}

t h e r a d i u s o f c o n v e r g e n c e i s R = 1

Commented by abdo mathsup 649 cc last updated on 13/May/18

3) w e h a v e f (x) = \frac{1}{2} l n (1 + x^{4}) + i a r c t a n (x^{2}) \Rightarrow

f^{^{'}} (x) = \frac{2 x^{3}}{1 + x^{4}} + i \frac{2 x}{1 + x^{4}} = \frac{2 x}{1 + x^{4}} (x^{2} + i)

a n o t h e r m e t h o d

f (x) = l n (1 + {i x}^{2}) \Rightarrow f^{^{'}} (x) = \frac{2 i x}{1 + {i x}^{2}}

= \frac{2 i x (1 - {i x}^{2})}{1 + x^{4}} = \frac{2 i x + 2 x^{3}}{1 + x^{4}} = \frac{2 x^{3}}{1 + x^{4}} + i \frac{2 x}{1 + x^{4}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/May/18

l n (1 + {i x}^{2}) = \frac{1}{2} l n {1^{2} + {(x^{2})}^{2}} + {i t a n}^{- 1} (x^{2} / 1)

= \frac{1}{2} (x^{4} - \frac{{(x^{4})}^{2}}{2} + \frac{{(x^{4})}^{3}}{3} \dots .) + i (x^{2} - \frac{{(x^{2})}^{3}}{3} + \frac{{(x^{2})}^{5}}{5} \dots)

= \frac{1}{2} (x^{4} - \frac{x^{8}}{2} + \frac{x^{12}}{3} \dots .) + i (x^{2} - \frac{x^{6}}{3} + \frac{x^{10}}{5} - \dots .)

Commented by tanmay.chaudhury50@gmail.com last updated on 11/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/May/18