Posted by andi telaumbanua on Jul 28, 2018 in
Matematika
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) cosx
f^1 (x)= – e^(-x) cosx-e^(-x) sinx
f^2 (x)= e^(-x) cosx+e^(-x) sinx+e^(-x) sinx-e^(-x) cosx =2e^(-x) sinx
f^3 (x)=-2e^(-x) sinx+2e^(-x) cos x
f^4 (x)= 2e^(-x) sinx-2e^(-x) cos x-2e^(-x) cosx – 2e^(-x) sinx= -4e^(-x) cosx
Maka:
f(0) = e^(-0) cos0=1
f^1 (0)= -e^(-0) cos0-e^(-0) sin0=-1-0=-1
f^2 (0)=2e^(-0) sin0=0
f^3 (0)= -2e^(-0) sin0+2e^(-0) cos 0=0+2=2
f^4 (0)= -4e^(-0) cos0=-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+⋯
Posted by andi telaumbanua on Jul 28, 2018 in
Matematika
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 cosx-e^x sinx-e^x sin(x) – e^x cos(x) =-2e^x sin(x)
f^3 (x)=-2e^x sinx-2e^x cos (x)
f^4 (x)= -2e^x sinx-2e^x cos (x) – 2e^x cos(x) + 2e^x sinx= -4e^x cos(x)
Maka:
f(0) = e^0 cos0=1
f^1 (0)= e^0 cos0-e^0 sin0 =1- 0 =1
f^2 (0)=-2e^0 sin0=0
f^3 (0)= -2e^0 sin0-2e^0 cos 0 = 0 – 2 =-2
f^4 (0)= -4e^0 cos0=-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+⋯
Posted by andi telaumbanua on Jul 28, 2018 in
Matematika
Tentukan deret MacLaurin dari f(x) = e^x sin(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 sinx
f^1 (x)= e^x sinx+e^x cosx
f^2 (x)= e^x sinx+e^x cosx+e^x cosx-e^x sinx=2e^x cosx
f^3 (x)= 2e^x cosx-2e^x sinx
f^4 (x)= 2e^x cosx-2e^x sinx -2e^x sinx-2e^x cosx= -4e^x sinx
Maka:
f(0) = e^0 sin0 = 0
f^1 (0)= e^0 sin0+e^0 cos0 = 0+1=1
f^2 (0)= 2e^0 cos0= 2
f^3 (0)= 2e^0 cos0-2e^0 sin0=2-0=2
f^4 (0)= -4e^0 sin0=0
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) = 0 + (1)(x)+2/2! x^2+2/3! x^3+0/4! x^4+⋯
f(x) = 0 + x+x^2+1/3 x^3+0+⋯
f(x) = x+x^2+1/3 x^3+⋯
Posted by andi telaumbanua on Jul 28, 2018 in
Matematika
Tentukan deret MacLaurin dari f(x) = e^(-x) sinx
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) sinx
f^1 (x)= -e^(-x) sinx+e^(-x) cosx
f^2 (x)= e^(-x) sinx – e^(-x) cosx -e^(-x) cosx – e^(-x) sinx =-2e^(-x) cosx
f^3 (x)= 2e^(-x) cos (x) + 2e^(-x) sinx
f^4 (x)= -2e^(-x) cos x -2e^(-x) sinx-2e^(-x) sin x+2e^(-x) cosx
Maka:
f(0) = e^(-0) sin 0=0
f^1 (0)= -e^(-0) sin 0 + e^(-0) cos0=0+1=1
f^2 (0)=-2e^(-0) cos 0 = -2
f^3 (0)= 2e^(-0) cos 0+2e^(-0) sin 0=2+0=2
f^4 (0)= -2e^(-0) cos 0-2e^(-0) sin 0 -2e^(-0) sin0+2e^(-0)cos0 = -2-0-0+2=0
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) = 0 + (1)(x)+(-2)/2! x^2+2/3! x^3+0/4! x^4+⋯
f(x) = 0 + x-x^2+1/3 x^3+0+⋯
f(x) = x-x^2+1/3 x^3+⋯
Posted by andi telaumbanua on Jul 28, 2018 in
Matematika
Tentukan turunan pertama dari : f(x) = 3^(2x+1)+ 2^sin2x
Jawab:
Note;
a^x =e^xlna
d/dx (a^x )= d/dx (e^xlna )=(e^xlna )(lna)= a^x lna
Maka: turunan pertama dari : f(x) = 3^(2x+1)+ 2^sin2x adalah
dy/dx = d/dx (3^(2x+1) )+ d/dx (2^sin2x )
= d/dx ( e^(2x+1)ln3 )+d/dx (e^((sin(2x)ln2) )
= e^(2x+1)ln3 d/dx [(2x+1)ln3] + e^((sin(2x)ln2) d/dx [sin(2x)ln2]
=e^(2x+1)ln3 d/dx [(2xln3+ln3) ]+e^((sin(2x)ln2) d/dx [sin(2x)ln2]
=e^(2x+1)ln3 (2ln3)+e^((sin(2x)ln2) (2cos(2x) ln2
Karena: e^(2x+1)ln3=3^(2x+1) dan e^((sin(2x)ln2)=2^sin2x
Sehingga:
=e^(2x+1)ln3 (2ln3)+e^((sin(2x)ln2) (2cos(2x) ln2)
=3^(2x+1) (2ln3)+2^sin2x (2 cos(2x) ln2)