Question

Determina a Fórmula de McLaurin para os valores dados de n.

a) f(x) = Ln(x+1), n=4
b) f(x) = cos x, n=8

Answer 1

McLaurin é Taylor, desenvolvido em torno de $x_0=0:$
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a) $f(x)=\mathrm{\ell n}(x+1),~~n=4$

Como $n=4\,,$ precisamos calcular as derivadas até a ordem 4 de $f$ em $x_0=0.$

$\bullet\;\;f(x)=\mathrm{\ell n}(x+1)~~\Rightarrow~~f(0)=0\\\\ \bullet\;\;f'(x)=\dfrac{1}{x+1}~~\Rightarrow~~f'(0)=1\\\\\\ \bullet\;\;f''(x)=-\dfrac{1}{(x+1)^2}~~\Rightarrow~~f''(0)=-1\\\\\\ \bullet\;\;f'''(x)=\dfrac{2}{(x+1)^3}~~\Rightarrow~~f'''(0)=2\\\\\\ \bullet\;\;f^{(4)}(x)=-\,\dfrac{6}{(x+1)^4}~~\Rightarrow~~f^{(4)}(0)=-6$


O polinômio aproximador de $f$ é

$P_4(x)=f(0)+f'(0)\,x+\dfrac{f''(0)\,x^2}{2!}+\dfrac{f'''(0)\,x^3}{3!}+\dfrac{f^{(4)}(0)\,x^4}{4!}\\\\\\ P_4(x)=0+1\,x+\dfrac{(-1)\,x^2}{2!}+\dfrac{2\,x^3}{3!}+\dfrac{(-6)\,x^4}{4!}\\\\\\ P_4(x)=x-\dfrac{x^2}{2}+\dfrac{2x^3}{6}-\dfrac{6x^4}{24}\\\\\\ \boxed{\begin{array}{c}P_4(x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4} \end{array}}$


b) $f(x)=\cos x,~~n=8$

Como $n=8\,,$ precisamos calcular as derivadas até a ordem 8 de $f$ em $x_0=0.$


Mas um fato interessante acontece com a função cosseno.

$\bullet\;\;f(x)=\cos x~~\Rightarrow~~f(0)=1\\\\ \bullet\;\;f'(x)=-\mathrm{sen\,}x~~\Rightarrow~~f'(0)=0\\\\ \bullet\;\;f''(x)=-\cos x~~\Rightarrow~~f''(0)=-1\\\\ \bullet\;\;f'''(x)=\mathrm{sen\,}x~~\Rightarrow~~f'''(0)=0\\\\ \vdots$


A partir da derivada de ordem 4, o ciclo se repete. Dessa forma, todas as derivadas de ordem ímpar são nulas em $x_0=0\,,$ anulando os termos de graus ímpares.

E as derivadas de ordem par alternam entre $+1$ e $-1:$


Então, o polinômio aproximador de $f$ é

$P_8(x)=f(0)+f'(0)\,x+\dfrac{f''(0)\,x^2}{2!}+\dfrac{f'''(0)\,x^3}{3!}+\ldots+\dfrac{f^{(8)}(0)\,x^8}{8!}\\\\\\ P_8(x)=f(0)+\dfrac{f''(0)\,x^2}{2!}+\dfrac{f^{(4)}(0)\,x^4}{4!}+\dfrac{f^{(6)}(0)\,x^6}{6!}+\dfrac{f^{(8)}(0)\,x^8}{8!}\\\\\\ P_8(x)=1+\dfrac{(-1)\,x^2}{2!}+\dfrac{1\,x^4}{4!}+\dfrac{(-1)\,x^6}{6!}+\dfrac{1\,x^8}{8!}\\\\\\ P_8(x)=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}\\\\\\ \boxed{\begin{array}{c} P_8(x)=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dfrac{x^6}{720}+\dfrac{x^8}{40\,320} \end{array}}$


Bons estudos! :-)