Question Number 103205 by DGmichael last updated on 13/Jul/20

Answered by OlafThorendsen last updated on 13/Jul/20

f(λ) = Σ_(k=0) ^∞ (λ^k /(k!))  ⇒ f′(λ) = Σ_(k=1) ^∞ ((kλ^(k−1) )/(k!)) = Σ_(k=1) ^∞ (λ^(k−1) /((k−1)!))  ⇒ f′(λ) = Σ_(k=0) ^∞ (λ^k /(k!)) = f(λ)  f′(λ) = f(λ) ⇒ f(λ) = Ce^λ   But f(0) = Σ_(k=0) ^∞ (0^k /(k!)) = 1 = Ce^0  ⇒ C = 1  Then f(λ) = 1×e^λ  = e^λ

Answered by OlafThorendsen last updated on 13/Jul/20

f(x+h) = f(x)+Σ_(k=1) ^∞ ((f^((k)) (x))/(k!))h^n   If x = 0, h = λ, f = exp  e^λ  = e^0 +Σ_(k=1) ^∞ (e^0 /(k!))λ^k  = Σ_(k=0) ^∞ (λ^k /(k!))