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let-f-x-2x-3-x-2-1-calculate-f-n-x-2-find-f-10-x-and-f-15-x-3-calculate-f-10-0-and-f-15-0-4-developp-f-at-integr-serie-5-let-g-x-0-x-f-t-dt-developp-g-at-in




Question Number 72260 by mathmax by abdo last updated on 26/Oct/19
let f(x)=((2x+3)/(x^2 +1))  calculate  f^((n)) (x)  2)find f^((10)) (x) and f^((15)) (x)  3)calculate f^((10)) (0) and f^((15)) (0)  4)developp f at integr serie  5)let g(x)=∫_0 ^x f(t)dt  developp g at integr serie.
letf(x)=2x+3x2+1calculatef(n)(x)2)findf(10)(x)andf(15)(x)3)calculatef(10)(0)andf(15)(0)4)developpfatintegrserie5)letg(x)=0xf(t)dtdeveloppgatintegrserie.
Commented by mathmax by abdo last updated on 27/Oct/19
1) f(x)=((2x+3)/(x^2 +1)) =((2x+3)/((x−i)(x+i))) =(a/(x−i)) +(b/(x+i))  a=((2i+3)/(2i)) =1+(3/(2i)) =1−((3i)/2) =((2−3i)/2)  b=((−2i+3)/((−2i))) =((2i−3)/(2i)) =1−(3/(2i)) =1+((3i)/2) =((2+3i)/2) ⇒  f(x)=((2−3i)/(2(x−i))) +((2+3i)/(2(x+i))) ⇒f^((n)) (x)=((2−3i)/2)((1/(x−i)))^((n)) +((2+3i)/2)((1/(x+i)))^((n))   =((2−3i)/2) (((−1)^n n!)/((x−i)^(n+1) )) +((2+3i)/2) (((−1)^n n!)/((x+i)^(n+1) ))  =(((−1)^n n!)/2){ ((2+3i)/((x+i)^(n+1) )) +((2−3i)/((x−i)^(n+1) ))}  =(((−1)^n n!)/2){(((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2  +1)^(n+1) ))}  =(((−1)^n n!)/((x^2  +1)^(n+1) ))Re{(2+3i)(x−i)^(n+1) } but  2+3i =(√(13))e^(iarctan((3/2)))    and x−i =(√(x^2 +1))e^(−iarctan((1/x)))  ⇒  (x−i)^(n+1) =(x^2 +1)^((n+1)/2)  e^(−i(n+1)arctan((1/x)))  ⇒  (2+3i)(x−i)^(n+1) =(√(13))(x^2  +1)^((n+1)/2)   e^(i(arctan((3/2))−(n+1)arctan((1/x))))  ⇒  f^((n)) (x)=(((−1)^n n!)/((x^2 +1)^(n+1) ))(√(13))(x^2  +1)^((n+1)/2)  cos(arctan((3/2))−(n+1)arctan((1/x)))  f^((n)) (x)=(√(13))(−1)^n n!(x^2  +1)^(−((n+1)/2)) cos(arctan((3/2))−(n+1)arctan((1/x)))
1)f(x)=2x+3x2+1=2x+3(xi)(x+i)=axi+bx+ia=2i+32i=1+32i=13i2=23i2b=2i+3(2i)=2i32i=132i=1+3i2=2+3i2f(x)=23i2(xi)+2+3i2(x+i)f(n)(x)=23i2(1xi)(n)+2+3i2(1x+i)(n)=23i2(1)nn!(xi)n+1+2+3i2(1)nn!(x+i)n+1=(1)nn!2{2+3i(x+i)n+1+23i(xi)n+1}=(1)nn!2{(2+3i)(xi)n+1+(23i)(x+i)n+1(x2+1)n+1}=(1)nn!(x2+1)n+1Re{(2+3i)(xi)n+1}but2+3i=13eiarctan(32)andxi=x2+1eiarctan(1x)(xi)n+1=(x2+1)n+12ei(n+1)arctan(1x)(2+3i)(xi)n+1=13(x2+1)n+12ei(arctan(32)(n+1)arctan(1x))f(n)(x)=(1)nn!(x2+1)n+113(x2+1)n+12cos(arctan(32)(n+1)arctan(1x))f(n)(x)=13(1)nn!(x2+1)n+12cos(arctan(32)(n+1)arctan(1x))
Commented by mathmax by abdo last updated on 27/Oct/19
2)f^((10)) (x) =(√(13))(10)!(x^2 +1)^(−((11)/2))  cos(arctan((3/2))−11arctan((1/x))  f^((15)) (x)=−(√(13))(15)! (x^2 +1)^(−8) cos(arctan((3/2))−16 arctan((1/x)))
2)f(10)(x)=13(10)!(x2+1)112cos(arctan(32)11arctan(1x)f(15)(x)=13(15)!(x2+1)8cos(arctan(32)16arctan(1x))
Commented by mathmax by abdo last updated on 27/Oct/19
4) we have f^((n)) (x)=  (√(13))(−1)^n n!(x^2 +1)^(−((n+1)/2))  cos(arctan((3/2))−(n+1)arctan((1/x))) ⇒  f^((n)) (0) =(√(13))(−1)^n n!cos(arctan((3/2))+^− (n+1)(π/2))  f(x)=Σ_(n=0) ^∞  ((f^((n)) (x))/(n!)) =Σ_(n=0) ^∞ (√(13))(−1)^n cos(arctan((3/2))+^− (n+1)(π/2))x^n
4)wehavef(n)(x)=13(1)nn!(x2+1)n+12cos(arctan(32)(n+1)arctan(1x))f(n)(0)=13(1)nn!cos(arctan(32)+(n+1)π2)f(x)=n=0f(n)(x)n!=n=013(1)ncos(arctan(32)+(n+1)π2)xn
Answered by mind is power last updated on 26/Oct/19
((2x+3)/(x^2 +1))=(a/(x+i))+(c/(x−i))  a+c=2,i(c−a)=3⇒c=((2−3i)/2),a=((2+3i)/2)  f(x)=((2+3i)/(2(x+i)))+((2−3i)/(2(x−i)))  f^n (x)=((2+3i)/2)(d^n x/dx^n ) (1/(x+i))+((2−3i)/2)(d^n x/dx^n ) (1/(x−i))  =((2+3i)/2)   (((−1)^n  n!)/((x+i)^(n+1) ))+((2−3i)/2).(((−1)^n n!)/((x−i)^(n+1) ))  =(((−1)^n n!)/2)((((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2 +1)^(n+1) )))  =(((−1)^n n! (2.Re{ (2+3i)(x+i)^(n+1) }))/((x^2 +1)^(n+1) ))  Re(2+3i)(x+i)^(n+1)   (x+i)^(n+1) =ΣC_(n+1) ^k .i^k .(x)^(n+1−k)   =Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^k i.X^(n−2k)   Re (2+3i).(x+i)^(n+1) =  2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k)   f^n (x)=(((−1)^n n!)/((x^2 +1)^(n+1) )).2(2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k) )    f^((10)) (x)=2.((10!)/((x^2 +1)^(11) )).(2Σ_(k=0) ^5 C_(11) ^(2k) .(−1)^k X^(11−2k) +3Σ_(k=0) ^5 C_(11) ^(2k+1) (−1)^(k+1) X^(10−2k) )  f^(15) (x) same idee  3)  f^(10) (x)=2.10!.(3.)=6.10!  f^(15) (0)=2(−1)^(15) .15!.(2.)=−4.15!  4) f^n (0) if n=2s  f^n (0)=(−1)^(2s) .(2s)!.2.(3.(−1)^(s+1) )=(((−1)^(s+1) .6.(2s)!)/)  n=2s+1⇒f^n (0)=(((−1)^(2s+1) .(2s+1)!.2(2.(−1)^(s+1) ))/)=(−1)^s .4(2s+1)!  f(x)=Σ_(n≥0) ((f^n (0).x^n )/(n!))=Σ((f^(2s) (0)x^(2s) )/((2s)!))+Σ((f^(2s+1) (0)x^(2s+1) )/((2s+1)!))  =Σ_(s=0) ^(+∞) (((f^(2s) (0)x^(2s) )/((2s)!))+((f^(2s) (0)x^(2s+1) )/((2s+1)!)))=Σ_(s=0) ^(+∞) (6(−1)^(s+1) x^(2s) +4(−1)^s x^(2s+1) )  =Σ_(s≥0) (−1)^s x^(2s) (−6+4x)  5)gx=∫_0 ^x f   g(x)=Σ_(s≥0) (∫_0 ^x 6(−1)^(s+1) t^(2s) +4(−1)^s t^(2s+1) )  =Σ_(s≥0) .(6(−1)^(s+1) .(x^(2s+1) /(2s+1))+((2(−1)^s x^(2s+2) )/(s+1)))
2x+3x2+1=ax+i+cxia+c=2,i(ca)=3c=23i2,a=2+3i2f(x)=2+3i2(x+i)+23i2(xi)fn(x)=2+3i2dnxdxn1x+i+23i2dnxdxn1xi=2+3i2(1)nn!(x+i)n+1+23i2.(1)nn!(xi)n+1=(1)nn!2((2+3i)(xi)n+1+(23i)(x+i)n+1(x2+1)n+1)=(1)nn!(2.Re{(2+3i)(x+i)n+1})(x2+1)n+1Re(2+3i)(x+i)n+1(x+i)n+1=ΣCn+1k.ik.(x)n+1k=E(n+12)k=0Cn+12k.(1)k.Xn+12k+E(n2)k=0Cn+12k+1(1)ki.Xn2kRe(2+3i).(x+i)n+1=2E(n+12)k=0Cn+12k.(1)k.Xn+12k+3E(n2)k=0Cn+12k+1(1)k+1.Xn2kfn(x)=(1)nn!(x2+1)n+1.2(2E(n+12)k=0Cn+12k.(1)k.Xn+12k+3E(n2)k=0Cn+12k+1(1)k+1.Xn2k)f(10)(x)=2.10!(x2+1)11.(25k=0C112k.(1)kX112k+35k=0C112k+1(1)k+1X102k)f15(x)sameidee3)f10(x)=2.10!.(3.)=6.10!f15(0)=2(1)15.15!.(2.)=4.15!4)fn(0)ifn=2sfn(0)=(1)2s.(2s)!.2.(3.(1)s+1)=(1)s+1.6.(2s)!n=2s+1fn(0)=(1)2s+1.(2s+1)!.2(2.(1)s+1)=(1)s.4(2s+1)!f(x)=n0fn(0).xnn!=Σf2s(0)x2s(2s)!+Σf2s+1(0)x2s+1(2s+1)!=+s=0(f2s(0)x2s(2s)!+f2s(0)x2s+1(2s+1)!)=+s=0(6(1)s+1x2s+4(1)sx2s+1)=s0(1)sx2s(6+4x)5)gx=0xfg(x)=s0(0x6(1)s+1t2s+4(1)st2s+1)=s0.(6(1)s+1.x2s+12s+1+2(1)sx2s+2s+1)
Commented by mathmax by abdo last updated on 27/Oct/19
thank you sir.
thankyousir.
Commented by mind is power last updated on 27/Oct/19
y′re welcom
yrewelcom
Commented by mind is power last updated on 27/Oct/19
y′re welcom
yrewelcom

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