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 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 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
	$ln(1+ix^2 )=(1/2)ln{1^2 +(x^2 )^2 }+itan^(−1) (x^2 /1) =(1/2)(x^4 −(((x^4 )^2 )/2)+(((x^4 )^3 )/3)....)+i(x^2 −(((x^2 )^3 )/3)+(((x^2 )^5 )/5)...) =(1/2)(x^4 −(x^8 /2)+(x^(12) /3)....)+i(x^2 −(x^6 /3)+(x^(10) /5)−....)$
	$l n (1 + {i x}^{2}) = \frac{1}{2} l n {1^{2} + {(x^{2})}^{2}} + {i t a n}^{- 1} (x^{2} / 1)$ $= \frac{1}{2} (x^{4} - \frac{{(x^{4})}^{2}}{2} + \frac{{(x^{4})}^{3}}{3} . . . .) + i (x^{2} - \frac{{(x^{2})}^{3}}{3} + \frac{{(x^{2})}^{5}}{5} . . .)$ $= \frac{1}{2} (x^{4} - \frac{x^{8}}{2} + \frac{x^{12}}{3} . . . .) + i (x^{2} - \frac{x^{6}}{3} + \frac{x^{10}}{5} - . . . .)$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/May/18

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

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum