let f(x)=arctan((2/x)) calculate f^((n)) (x) and f^((n)) (1) find f^((7)) ((1/7))


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 124066 by Bird last updated on 30/Nov/20

$l e t f (x) = a r c t a n (\frac{2}{x})$ $c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (1)$ $f i n d f^{(7)} (\frac{1}{7})$
	Answered by Olaf last updated on 30/Nov/20

	$f (x) = \arctan (\frac{2}{x})$ $f^{'} (x) = - \frac{2}{x^{2}} \times \frac{1}{1 + \frac{4}{x^{2}}} = - \frac{2}{x^{2} + 4}$ $f^{'} (x) = \frac{i}{2} [\frac{1}{x - 2 i} - \frac{1}{x + 2 i}]$ $f^{(n)} (x) = \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! [\frac{1}{{(x - 2 i)}^{n}} - \frac{1}{{(x + 2 i)}^{n}}]$ $f^{(n)} (x) = \frac{i}{2} {(- 1)}^{n - 1} (n - 1)! [\frac{1}{{(\sqrt{x^{2} + 4} e^{- i \arctan \frac{2}{x}})}^{n}} - \frac{1}{{(\sqrt{x^{2} + 4} e^{i \arctan \frac{2}{x}})}^{n}}]$ $f^{(n)} (x) = \frac{i {(- 1)}^{n - 1} (n - 1)!}{2 {(x^{2} + 4)}^{\frac{n}{2}}} [\frac{1}{e^{- i n \arctan \frac{2}{x}}} - \frac{1}{e^{i n \arctan \frac{2}{x}}}]$ $f^{(n)} (x) = \frac{i {(- 1)}^{n - 1} (n - 1)!}{2 {(x^{2} + 4)}^{\frac{n}{2}}} [e^{i n \arctan \frac{2}{x}} - e^{- i n \arctan \frac{2}{x}}]$ $f^{(n)} (x) = - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x^{2} + 4)}^{\frac{n}{2}}} \sin (n \arctan \frac{2}{x})$ $f^{(n)} (x) = \frac{{(- 1)}^{n} (n - 1)!}{{(x^{2} + 4)}^{\frac{n}{2}}} \sin (n \arctan \frac{2}{x})$ $f^{(n)} (1) = \frac{{(- 1)}^{n} (n - 1)!}{5^{\frac{n}{2}}} \sin (n \arctan 2)$ $f^{(7)} (x) = \frac{{(- 1)}^{7} 6!}{{(x^{2} + 4)}^{\frac{7}{2}}} \sin (\frac{1}{7} \arctan \frac{2}{x})$ $f^{(7)} (x) = - \frac{720}{{(x^{2} + 4)}^{\frac{7}{2}}} \sin (\frac{1}{7} \arctan \frac{2}{x})$ $f^{(7)} (\frac{1}{7}) = - \frac{720}{{(\frac{1}{49} + 4)}^{\frac{7}{2}}} \sin (\frac{1}{7} \arctan 14)$ $f^{(7)} (\frac{1}{7}) = - \frac{720}{{(\frac{197}{49})}^{\frac{7}{2}}} \sin (\frac{1}{7} \arctan 14)$ $f^{(7)} (\frac{1}{7}) = - \frac{720 \times 7^{7}}{197^{3} \sqrt{197}} \sin (\frac{1}{7} \arctan 14)$ $f^{(7)} (\frac{1}{7}) = - \frac{823543}{38809 \sqrt{197}} \sin (\frac{1}{7} \arctan 14)$ $. . . maybe . . .$
	Commented by mathmax by abdo last updated on 30/Nov/20

	$thank you sir .$
	Answered by mathmax by abdo last updated on 04/Dec/20
	$we have f(x)=arctan((2/x))⇒f^((1)) (x)=−(2/(x^2 (1+(4/x^2 ))))=((−2)/(x^2 +4)) =((−2)/((x−2i)(x+2i))) =((−2)/(4i)){(1/(x−2i))−(1/(x+2i))} =(1/(2i)){(1/(x+2i))−(1/(x−2i))} ⇒ f^((n)) (x)=(1/(2i)){((1/(x+2i)))^()n−1)) −((1/(x−2i)))^((n−1)) } =(1/(2i)){(((−1)^(n−1) (n−1)!)/((x+2i)^n ))−(((−1)^(n−1) (n−1)!)/((x−2i)^n ))} ⇒ f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(2i)){(((x−2i)^n −(x+2i)^n )/((x^2 +4)^n ))} (n>0) f^((n)) (1) =(((−1)^(n−1) (n−1)!)/(2i)){(((1−2i)^n −(1+2i)^n )/5^n )} but 1−2i =(√5) e^(−iarctan(2)) and 1+2i =(√5)e^(iarctan(2)) ⇒ (1−2i)^n −(1+2i)^n =−(√5){ e^(iarctan(2)) −e^(−iarctan(2)) } =−(√5)(2isin(arctan(2))} ⇒ f^((n)) (1) =(((−1)^(n−1) (n−1)!)/(2i 5^n ))(−(√5)2isin(arctan(2)) =(1/5^(n−(1/2)) )(−1)^n (n−1)! sin(arctan2)$
	$we have f (x) = \arctan (\frac{2}{x}) \Rightarrow f^{(1)} (x) = - \frac{2}{x^{2} (1 + \frac{4}{x^{2}})} = \frac{- 2}{x^{2} + 4}$ $= \frac{- 2}{(x - 2 i) (x + 2 i)} = \frac{- 2}{4 i} {\frac{1}{x - 2 i} - \frac{1}{x + 2 i}} = \frac{1}{2 i} {\frac{1}{x + 2 i} - \frac{1}{x - 2 i}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{2 i} {{(\frac{1}{x + 2 i})}^{) n - 1)} - {(\frac{1}{x - 2 i})}^{(n - 1)}}$ $= \frac{1}{2 i} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + 2 i)}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - 2 i)}^{n}}} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{{(x - 2 i)}^{n} - {(x + 2 i)}^{n}}{{(x^{2} + 4)}^{n}}} (n > 0)$ $f^{(n)} (1) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{{(1 - 2 i)}^{n} - {(1 + 2 i)}^{n}}{5^{n}}}$ $but 1 - 2 i = \sqrt{5} e^{- iarctan (2)} and 1 + 2 i = \sqrt{5} e^{iarctan (2)} \Rightarrow$ ${(1 - 2 i)}^{n} - {(1 + 2 i)}^{n} = - \sqrt{5} {e^{iarctan (2)} - e^{- iarctan (2)}}$ $= - \sqrt{5} (2 isin (\arctan (2))} \Rightarrow$ $f^{(n)} (1) = \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i 5^{n}} (- \sqrt{5} 2 isin (\arctan (2))$ $= \frac{1}{5^{n - \frac{1}{2}}} {(- 1)}^{n} (n - 1)! \sin (\arctan 2)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum