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let-f-x-0-pi-2-d-x-cos-2-with-x-gt-0-1-calculate-f-x-and-f-x-2-find-f-n-x-and-f-n-0-3-developp-f-at-integr-serie-




Question Number 40619 by math khazana by abdo last updated on 25/Jul/18
let  f(x)=∫_0 ^(π/2)    (dθ/(x  +cos^2 θ))  with x>0 .  1) calculate f(x) and f^′ (x)  2) find  f^((n)) (x) and f^((n)) (0)  3) developp f at integr serie.
letf(x)=0π2dθx+cos2θwithx>0.1)calculatef(x)andf(x)2)findf(n)(x)andf(n)(0)3)developpfatintegrserie.
Commented by abdo mathsup 649 cc last updated on 27/Jul/18
1) we have proved that f(x)=((π(√2))/(2(√x))) ⇒  f^′ (x)=((π(√2))/2)(x^(−(1/2)) )^′  =−(1/2) ((π(√2))/2) x^(−(3/2)) =−((π(√2))/(4x(√x)))  2) f^((n)) (x)=((π(√2))/2) (x^(−(1/2)) )^((n))    let find  (x^p )^((n))  with p ∈Q   (x^p )^((1)) =px^(p−1)  ,  (x^p )^((2)) =p(p−1)x^(p−2) ⇒  (x^p )^((n))  =p(p−1)...(p−n+1)x^(p−n)  ⇒  (x^(−(1/2)) )^((n)) =(−(1/2))(−(3/2))...(−(1/2)−n+1)x^(−(1/2)−n)   =(−(1/2))(−(3/2))...(((−1−2n+2)/2))x^(−(1/2)−n)   =(−(1/2))(−(3/2))....(−((2n−1)/2))(1/(x^n (√x))) ⇒  f^((n)) (x)=(π/( (√2))) (−(1/2))(−(3/2))...(−((2n−1)/2)) (1/(x^n (√x)))  f^((n)) (1)=(π/( (√2)))(−(1/2))(−(3/2))...(−((2n−1)/2))  3) f(x) =Σ_(n=0) ^∞   ((f^((n)) (1))/(n!))(x−1)^n   = Σ_(n=0) ^∞ (π/(n!(√2)))(−(1/2))(−(3/2))...(−((2n−1)/2))(x−1)^n .
1)wehaveprovedthatf(x)=π22xf(x)=π22(x12)=12π22x32=π24xx2)f(n)(x)=π22(x12)(n)letfind(xp)(n)withpQ(xp)(1)=pxp1,(xp)(2)=p(p1)xp2(xp)(n)=p(p1)(pn+1)xpn(x12)(n)=(12)(32)(12n+1)x12n=(12)(32)(12n+22)x12n=(12)(32).(2n12)1xnxf(n)(x)=π2(12)(32)(2n12)1xnxf(n)(1)=π2(12)(32)(2n12)3)f(x)=n=0f(n)(1)n!(x1)n=n=0πn!2(12)(32)(2n12)(x1)n.
Commented by abdo mathsup 649 cc last updated on 27/Jul/18
2) the Q is find f^((n)) (x) and f^((n)) (1)  3) the Q is developp f at integr serie at v(1).
2)theQisfindf(n)(x)andf(n)(1)3)theQisdeveloppfatintegrserieatv(1).

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