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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)=e−2xx+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)(e−2x)(n−k)⇒f(n)(x)=∑k=0nCnk(−1)kk!(x+1)k+1(−2)n−ke−2xandf(n)(0)=∑k=0nCnk(−1)kk!(−2)n−k=∑k=0nn!k!(n−k)!(−2)nk!=(−2)nn!∑k=0n1(n−k)!2)wehavef(x)=∑n=0∞f(n)(0)n!xn⇒f(x)=∑n=0∞(−2)n{∑k=0n1(n−k)!}xn⇒f(x)=∑n=0∞(∑k=0n(−2)n(n−k)!)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)=e−2xx+1y′=−2(x+1)e−2x−e−2x(x+1)2y′=−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=0y′(x+1)+(2x+3)y=0ApplyingLeibnitztheoremtoobtainthenthderivativey(n+1)(x+1)+ny(n)+y(n)(2x+3)+2ny(n−1)=0atx=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=0Thisisarecurrencerelation,soweapplytheLeibnitz−maclaurinmethodtoobtainaseries(y)0=f(0)=e−2(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=1−3x+5x2−14x33+13x46+2x515+...y=f(x)=e−2xx+1=1−3x+5x2−14x33+13x46+2x515+...HafeezAyinde(Abuqatada)

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