Question Number 114185 by mohammad17 last updated on 17/Sep/20

Answered by Dwaipayan Shikari last updated on 17/Sep/20

f(x)=lim_(n→∞) Σ_(k=0) ^n ((f^((k)) (0))/(k!))x^k f(x)=e^(−7x) f′(x)=−7e^(−7x) f′′(x)=49e^(−7x) f′′′(x)=−343e^(−7x) f^((iv)) (x)=3401e^(−7x) f(x)=f(0)+f′(0)x+f^(′′) (0)(x^2 /2)+f′′′(0)(x^3 /6)+... f(x)=1−7x+7^2 (x^2 /2)−7^3 (x^3 /6)+... e^(−7x) =lim_(n→∞) Σ_(k=0) ^n (−1)^(k+1) .(7x)^k (1/(k!)) Derivation of Maclaurine series f(x)=a_0 +a_1 (x−x_0 )+a_2 (x−x_0 )^2 +a_3 (x−x_0 )^3 +.. f′(x_0 )=a_1 f′′(x_0 )=2a_2 f′′′(x_0 )=6a_3 f^n (x_0 )=n!a_n ((f^((n)) (x_0 ))/(n!))=a_n ((f^((n)) (x_0 ))/(n!))(x−x_0 )^n =a_n (x−x_0 )^n Σ^∞ ((f^((n)) (x_0 ))/(n!))(x−x_0 )^n =Σ^∞ a_n (x−x_0 )^n Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n =f(x) (take x_0 =0)

Answered by Dwaipayan Shikari last updated on 18/Sep/20

Σ_(n=1) ^∞ (((−3)^(n−1) )/2^(3n) ) =(1/2^3 )−(3/2^6 )+(3^2 /2^9 )+... =(1/2^3 )−((3/2^6 )−(3^2 /2^9 )+...) =(1/2^3 )−3(((1/2^6 )/(1+(3/2^3 ))))=(1/2^3 )−(3/(11.2^3 ))=(1/8)(1−(3/(11)))=(1/(11))

Answered by mathmax by abdo last updated on 17/Sep/20

1) S =Σ_(n=1) ^∞ (((−3)^(n−1) )/2^(3n) ) ⇒S =−3 Σ_(n=1) ^∞ (((−3)/8))^n =−(1/3){ Σ_(n=0) ^∞ (((−3)/8))^n −1} =(1/3)−(1/3)×(1/(1+(3/8))) =(1/3)−(1/3)×(8/(11)) =(1/3)(1−(8/(11))) =(1/3)×(3/(11)) =(1/(11))

Answered by mathmax by abdo last updated on 17/Sep/20

3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n we have f^((n)) (x) =(−7)^n e^(−7x) ⇒ f^((n)) (0) =(−7)^n ⇒f(x) =Σ_(n=0) ^∞ (((−7)^n )/(n!)) x^n =1−7x +((49)/2) x^2 −....with fadius R=+∞

Answered by mathmax by abdo last updated on 17/Sep/20

2) let u_n (x) =((2^n (x−2)^n )/((n+2)!)) ⇒∣((u_(n+1) (x))/(u_n (x)))∣ =∣(((2(x−2))^(n+1) )/((n+3)!))×(((n+2)!)/((2(x−2))^n ))∣ =∣(((n+2)!)/((n+3)!))×2(x−2)∣ =∣((2(x−2))/(n+3))∣→0 (n→+∞) ⇒the interval of cv is R (radius R =+∞)