Menu Close

let-f-x-arctan-2-x-calculate-f-n-x-and-f-n-1-find-f-7-1-7-

Tinku Tara 0 Comments
Pin




Question Number 124066 by Bird last updated on 30/Nov/20
let f(x)=arctan((2/x)) calculate f^((n)) (x) and f^((n)) (1) find f^((7)) ((1/7))
letf(x)=arctan(2x)calculatef(n)(x)andf(n)(1)findf(7)(17)
Answered by Olaf last updated on 30/Nov/20
f(x) = arctan((2/x)) f′(x) = −(2/x^2 )×(1/(1+(4/x^2 ))) = −(2/(x^2 +4)) f′(x) = (i/2)[(1/(x−2i))−(1/(x+2i))] f^((n)) (x) = (i/2)(−1)^(n−1) (n−1)![(1/((x−2i)^n ))−(1/((x+2i)^n ))] f^((n)) (x) = (i/2)(−1)^(n−1) (n−1)![(1/(((√(x^2 +4))e^(−iarctan(2/x)) )^n ))−(1/(((√(x^2 +4))e^(iarctan(2/x)) )^n ))] f^((n)) (x) = ((i(−1)^(n−1) (n−1)!)/(2(x^2 +4)^(n/2) ))[(1/e^(−inarctan(2/x)) )−(1/e^(inarctan(2/x)) )] f^((n)) (x) = ((i(−1)^(n−1) (n−1)!)/(2(x^2 +4)^(n/2) ))[e^(inarctan(2/x)) −e^(−inarctan(2/x)) ] f^((n)) (x) = −(((−1)^(n−1) (n−1)!)/((x^2 +4)^(n/2) ))sin(narctan(2/x)) f^((n)) (x) = (((−1)^n (n−1)!)/((x^2 +4)^(n/2) ))sin(narctan(2/x)) f^((n)) (1) = (((−1)^n (n−1)!)/5^(n/2) )sin(narctan2) f^((7)) (x) = (((−1)^7 6!)/((x^2 +4)^(7/2) ))sin((1/7)arctan(2/x)) f^((7)) (x) = −((720)/((x^2 +4)^(7/2) ))sin((1/7)arctan(2/x)) f^((7)) ((1/7)) = −((720)/(((1/(49))+4)^(7/2) ))sin((1/7)arctan14) f^((7)) ((1/7)) = −((720)/((((197)/(49)))^(7/2) ))sin((1/7)arctan14) f^((7)) ((1/7)) = −((720×7^7 )/(197^3 (√(197))))sin((1/7)arctan14) f^((7)) ((1/7)) = −((823543)/( 38809(√(197))))sin((1/7)arctan14) ...maybe...
f(x)=arctan(2x)f′(x)=−2x2×11+4x2=−2x2+4f′(x)=i2[1x−2i−1x+2i]f(n)(x)=i2(−1)n−1(n−1)![1(x−2i)n−1(x+2i)n]f(n)(x)=i2(−1)n−1(n−1)![1(x2+4e−iarctan2x)n−1(x2+4eiarctan2x)n]f(n)(x)=i(−1)n−1(n−1)!2(x2+4)n2[1e−inarctan2x−1einarctan2x]f(n)(x)=i(−1)n−1(n−1)!2(x2+4)n2[einarctan2x−e−inarctan2x]f(n)(x)=−(−1)n−1(n−1)!(x2+4)n2sin(narctan2x)f(n)(x)=(−1)n(n−1)!(x2+4)n2sin(narctan2x)f(n)(1)=(−1)n(n−1)!5n2sin(narctan2)f(7)(x)=(−1)76!(x2+4)72sin(17arctan2x)f(7)(x)=−720(x2+4)72sin(17arctan2x)f(7)(17)=−720(149+4)72sin(17arctan14)f(7)(17)=−720(19749)72sin(17arctan14)f(7)(17)=−720×771973197sin(17arctan14)f(7)(17)=−82354338809197sin(17arctan14)…maybe…
Commented by mathmax by abdo last updated on 30/Nov/20
thank you sir.
thankyousir.
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)
wehavef(x)=arctan(2x)⇒f(1)(x)=−2x2(1+4x2)=−2x2+4=−2(x−2i)(x+2i)=−24i{1x−2i−1x+2i}=12i{1x+2i−1x−2i}⇒f(n)(x)=12i{(1x+2i))n−1)−(1x−2i)(n−1)}=12i{(−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(x2+4)n}(n>0)f(n)(1)=(−1)n−1(n−1)!2i{(1−2i)n−(1+2i)n5n}but1−2i=5e−iarctan(2)and1+2i=5eiarctan(2)⇒(1−2i)n−(1+2i)n=−5{eiarctan(2)−e−iarctan(2)}=−5(2isin(arctan(2))}⇒f(n)(1)=(−1)n−1(n−1)!2i5n(−52isin(arctan(2))=15n−12(−1)n(n−1)!sin(arctan2)

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *