Question

Expandir a função f(x) = 1 - $e^{2-2x}$ em série de taylor, nas vizinhanças do ponto x₀ = 1. Escreva os cinco primeiros termos e o termo geral da expansão obtida.

Answer 1

Olá


x₀ = 1


$\displaystyle\mathsf{f(x)=1-e^{2-2x}\qquad\Rightarrow\qquad f(1) = 1-e^{2-2\cdot1}\qquad\Rightarrow\qquad f(1) =0 }\\\\\boxed{{\mathsf{a_0=0}}}\\\\\\\\\mathsf{f'(x)=2e^{2-2x}\qquad\Rightarrow\qquad f'(1) = 2e^{2-2\cdot1}\qquad\Rightarrow\qquad f'(1) =2 }\\\\\boxed{\mathsf{a_1= \frac{2}{1!} }}\\\\\\\\\mathsf{f''(x)=-4e^{2-2x}\qquad\Rightarrow\qquad f''(1) = 2e^{2-2\cdot1}\qquad\Rightarrow\qquad f''(1) =-4 }\\\\\boxed{\mathsf{a_2= -\frac{4}{2!} }}$

$\displaystyle \mathsf{f'''(x)=8e^{2-2x}\qquad\Rightarrow\qquad f'''(1) = 8e^{2-2\cdot1}\qquad\Rightarrow\qquad f'''(1) =8 }\\\\\boxed{\mathsf{a_3= \frac{8}{3!} }}\\\\\\\\\\\mathsf{f^{iv}(1)=-16e^{2-2x}\qquad\Rightarrow\qquad f^{iv}(1) = -16e^{2-2\cdot1}\qquad\Rightarrow\quad f^{iv}(1) =-16 }\\\\\boxed{\mathsf{a_4= -\frac{16}{4!} }}$


$\displaystyle\mathsf{f(x)= \frac{2}{1!}(x-1)^1~-~ \frac{4}{2!}(x-1)^2~+~\frac{8}{3!}(x-1)^3~-~\frac{16}{4!}(x-1)^4~+~...}\\\\\\\\\text{Serie alternada, entao teremos um alternador de sinal}~(-1)^{n+1}\\\text{Perceba que o termo do numerador crescre exponencialmente}~2^n\\\text{E o denominador cresce de acordo com n fatorial}~n!\\\\\\\\ \boxed{\mathsf{f(x)=\sum\limits^\infty_{n=1} ~{ \frac{2^n\cdot (x-1)^n\cdot (-1)^{n+1}}{n!} } \, }}\qquad\qquad\Longleftarrow\quad\text{Termo geral}$