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let-f-x-e-2x-x-1-1-calculate-f-n-x-and-f-n-0-2-develop-f-at-integr-serie-




Question Number 48668 by maxmathsup by imad last updated on 26/Nov/18
let f(x)=(e^(−2x) /(x+1))  1) calculate f^((n)) (x) and f^((n)) (0) .  2) develop f at integr serie .
letf(x)=e2xx+11)calculatef(n)(x)andf(n)(0).2)developfatintegrserie.
Commented by maxmathsup by imad last updated on 27/Nov/18
1) leibniz formula give f^((n)) (x) =Σ_(k=0) ^n  C_n ^k  ((1/(x+1)))^((k)) (e^(−2x) )^((n−k))  ⇒  f^((n)) (x) =Σ_(k=0) ^n  C_n ^k  (((−1)^k k!)/((x+1)^(k+1) )) (−2)^(n−k)  e^(−2x)   and   f^((n)) (0) =Σ_(k=0) ^n  C_n ^k   (−1)^k k! (−2)^(n−k) =Σ_(k=0) ^n  ((n!)/(k!(n−k)!))(−2)^n  k!  =(−2)^n n!Σ_(k=0) ^n   (1/((n−k)!))  2) we have f(x)=Σ_(n=0) ^∞   ((f^((n)) (0))/(n!)) x^n  ⇒f(x)=Σ_(n=0) ^∞ (−2)^n {Σ_(k=0) ^n  (1/((n−k)!))}x^n   ⇒f(x)=Σ_(n=0) ^∞ (Σ_(k=0) ^n    (((−2)^n )/((n−k)!)))x^n    .
1)leibnizformulagivef(n)(x)=k=0nCnk(1x+1)(k)(e2x)(nk)f(n)(x)=k=0nCnk(1)kk!(x+1)k+1(2)nke2xandf(n)(0)=k=0nCnk(1)kk!(2)nk=k=0nn!k!(nk)!(2)nk!=(2)nn!k=0n1(nk)!2)wehavef(x)=n=0f(n)(0)n!xnf(x)=n=0(2)n{k=0n1(nk)!}xnf(x)=n=0(k=0n(2)n(nk)!)xn.
Commented by Abdulhafeez Abu qatada last updated on 28/Nov/18
    very beautiful
verybeautiful
Commented by maxmathsup by imad last updated on 28/Nov/18
thank you sir.
thankyousir.
Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18
    Let y = f(x) = (e^(−2x) /(x + 1))  y′ = ((−2(x+1)e^(−2x)  − e^(−2x) )/((x + 1)^2 ))  y′ = −2y − (y/((x+1)))  y′(x + 1) + y = −2y(x + 1)       y′(x + 1) + y = −2y(x + 1)  y′(x + 1) + (1+2(x + 1))y = 0  y′(x + 1) + (2x + 3)y = 0    Applying Leibnitz theorem to obtain the nth derivative  y^((n+1)) (x + 1) + ny^((n))  + y^((n)) (2x + 3) + 2ny^((n−1))  = 0  at x = 0  (y^((n+1)) )_0  + n(y^((n)) )_0  + 3(y^((n)) )_0  + 2n(y^((n−1)) )_0  = 0  (y^((n+1)) )_0  + 4(y^((n)) )_0  + 2n(y^((n−1)) )_0  = 0    This is a recurrence relation, so we apply the Leibnitz−maclaurin method to obtain a series  (y)_0  = f(0) = (e^(−2(0)) /(0 + 1)) = 1  y′(x + 1) + (2x + 3)y = 0  (y′)_0  + 3(y)_0  = 0  (y′)_0  + 3(y)_0  = 0  (y′)_0  + 3 = 0  (y′)_0  = −3  (y′′)_0  + 4(−3) + 2 = 0   (y′′)_0   = 10  (y′′′)_0  + 4(y′′)_0  + 2(2)(y′)_0  = 0  (y′′′)_0  + 4(10) + 2(2)(−3) = 0  (y′′′)_0  = −28  (y^((iv)) )_0  = 52  (y^((v)) )_0  = 16  y = (y)_0  + x(y′)_0  + ((x^2 (y′′)_0 )/(2!)) + ((x^3 (y′′′)_0 )/(3!)) + ((x^4 (y^((iv)) )_0 )/(4!)) + ((x^5 (y^((v)) )_0 )/(5!)) +...  y = 1 − 3x + 5x^2  − ((14x^3 )/3) + ((13x^4 )/6) + ((2x^5 )/(15)) +...  y = f(x) = (e^(−2x) /(x + 1)) = 1 − 3x + 5x^2  − ((14x^3 )/3) + ((13x^4 )/6) + ((2x^5 )/(15)) +...    Hafeez Ayinde(Abu qatada)
Lety=f(x)=e2xx+1y=2(x+1)e2xe2x(x+1)2y=2yy(x+1)y(x+1)+y=2y(x+1)y(x+1)+y=2y(x+1)y(x+1)+(1+2(x+1))y=0y(x+1)+(2x+3)y=0ApplyingLeibnitztheoremtoobtainthenthderivativey(n+1)(x+1)+ny(n)+y(n)(2x+3)+2ny(n1)=0atx=0(y(n+1))0+n(y(n))0+3(y(n))0+2n(y(n1))0=0(y(n+1))0+4(y(n))0+2n(y(n1))0=0Thisisarecurrencerelation,soweapplytheLeibnitzmaclaurinmethodtoobtainaseries(y)0=f(0)=e2(0)0+1=1y(x+1)+(2x+3)y=0(y)0+3(y)0=0(y)0+3(y)0=0(y)0+3=0(y)0=3(y)0+4(3)+2=0(y)0=10(y)0+4(y)0+2(2)(y)0=0(y)0+4(10)+2(2)(3)=0(y)0=28(y(iv))0=52(y(v))0=16y=(y)0+x(y)0+x2(y)02!+x3(y)03!+x4(y(iv))04!+x5(y(v))05!+y=13x+5x214x33+13x46+2x515+y=f(x)=e2xx+1=13x+5x214x33+13x46+2x515+HafeezAyinde(Abuqatada)

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