Respuesta :
Answer and explanation:
To find : f(1) , f(2) , f(3) , f(4) , and f(5) if f(n) is defined recursively by f(0) = 3 and for n = 0 , 1 , 2 ,... ?
Solution :
We have given, f(0) = 3 for all parts.
a) [tex]f(n+1)=-2f(n)[/tex]
For f(1) put n=0,
[tex]f(0+1)=-2f(0)[/tex]
[tex]f(1)=-2(3)=-6[/tex]
For f(2) put n=1,
[tex]f(1+1)=-2f(1)[/tex]
[tex]f(2)=-2(-6)=12[/tex]
For f(3) put n=2,
[tex]f(2+1)=-2f(2)[/tex]
[tex]f(3)=-2(12)=-24[/tex]
For f(4) put n=3,
[tex]f(3+1)=-2f(3)[/tex]
[tex]f(4)=-2(-24)=48[/tex]
For f(5) put n=4,
[tex]f(4+1)=-2f(4)[/tex]
[tex]f(5)=-2(48)=-96[/tex]
b) [tex]f(n+1)=3f(n)+7[/tex]
For f(1) put n=0,
[tex]f(0+1)=3f(0)+7[/tex]
[tex]f(1)=3(3)+7=16[/tex]
For f(2) put n=1,
[tex]f(1+1)=3f(1)+7[/tex]
[tex]f(2)=3(16)+7=55[/tex]
For f(3) put n=2,
[tex]f(2+1)=3f(2)+7[/tex]
[tex]f(3)=3(55)+7=172[/tex]
For f(4) put n=3,
[tex]f(3+1)=3f(3)+7[/tex]
[tex]f(4)=3(172)+7=523[/tex]
For f(5) put n=4,
[tex]f(4+1)=3f(4)+7[/tex]
[tex]f(5)=3(523)+7=1576[/tex]
c) [tex]f(n+1)=f(n)^2-2f(n)-2[/tex]
For f(1) put n=0,
[tex]f(0+1)=f(0)^2-2f(0)-2[/tex]
[tex]f(1)=3^2-2(3)-2=1[/tex]
For f(2) put n=1,
[tex]f(1+1)=f(1)^2-2f(1)-2[/tex]
[tex]f(2)=1^2-2(1)-2=-3[/tex]
For f(3) put n=2,
[tex]f(2+1)=f(2)^2-2f(2)-2[/tex]
[tex]f(3)=-3^2-2(-3)-2=13[/tex]
For f(4) put n=3,
[tex]f(3+1)=f(3)^2-2f(3)-2[/tex]
[tex]f(4)=13^2-2(13)-2=141[/tex]
For f(5) put n=4,
[tex]f(4+1)=f(4)^2-2f(4)-2[/tex]
[tex]f(5)=141^2-2(141)-2=19597[/tex]
d) [tex]f(n+1)=3^{\frac{f(n)}{3}}[/tex]
For f(1) put n=0,
[tex]f(0+1)=3^{\frac{f(0)}{3}}[/tex]
[tex]f(1)=3^{\frac{3}{3}}=3[/tex]
For f(2) put n=1,
[tex]f(1+1)=3^{\frac{f(1)}{3}}[/tex]
[tex]f(2)=3^{\frac{3}{3}}=3[/tex]
For f(3) put n=2,
[tex]f(2+1)=3^{\frac{f(2)}{3}}[/tex]
[tex]f(3)=3^{\frac{3}{3}}=3[/tex]
For f(4) put n=3,
[tex]f(3+1)=3^{\frac{f(3)}{3}}[/tex]
[tex]f(4)=3^{\frac{3}{3}}=3[/tex]
For f(5) put n=4,
[tex]f(4+1)=3^{\frac{f(4)}{3}}[/tex]
[tex]f(5)=3^{\frac{3}{3}}=3[/tex]
