Tentukan deret MacLaurin dari f(x) = e^(-x) cos⁡(x) 

Tentukan deret MacLaurin dari f(x) = e^(-x) cos⁡(x)

Jawab:

Deret MacLaurin;
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯

Maka:
f(x) = e^(-x) cos⁡x
f^1 (x)= – e^(-x) cos⁡x-e^(-x) sin⁡x
f^2 (x)= e^(-x) cos⁡x+e^(-x) sin⁡x+e^(-x) sin⁡x-e^(-x) cos⁡x =2e^(-x) sin⁡x
f^3 (x)=-2e^(-x) sin⁡x+2e^(-x) cos x
f^4 (x)= 2e^(-x) sin⁡x-2e^(-x) cos ⁡x-2e^(-x) cos⁡x – 2e^(-x) sin⁡x= -4e^(-x) cos⁡x

Maka:
f(0) = e^(-0) cos⁡0=1
f^1 (0)= -e^(-0) cos⁡0-e^(-0) sin⁡0=-1-0=-1
f^2 (0)=2e^(-0) sin⁡0=0
f^3 (0)= -2e^(-0) sin⁡0+2e^(-0) cos 0=0+2=2
f^4 (0)= -4e^(-0) cos⁡0=-4

Maka: deret macLaurinnya
f(x) = f(0) + f^1 (0)(x)+(f^2 (0))/2! x^2+(f^3 (0))/3! x^3+(f^4 (0))/4! x^4+⋯
f(x) = 1 + (-1)(x)+0/2! x^2+2/3! x^3+(-4)/4! x^4+⋯
f(x) = 1 -x+0+1/3 x^3-1/6 x^4+⋯
f(x) = 1 – x+1/3 x^3-1/6 x^4+⋯