Question

4. Find f (2), f (3), f (4), and f (5) if f is defined recursively by f (0) = f (1) = 1 and for n = 1, 2, . . . a) f (n + 1) = f (n) − f (n − 1). b) f (n + 1) = f (n)f (n − 1). c) f (n + 1) = f (n)2 + f (n − 1)3. d) f (n + 1) = f (n)/f (n − 1).

Answer 1

a) f(2) = 0; f(3) = f(4) = -1; f(5) = 0

b) f(2)=f(3)=f(4)=f(5) = 1

c) f(2) = 2; f(3) = 5; f(4) = 27; f(5) = 734

d) f(2)=f(3)=f(4)=f(5) = 1

Explanation:

Answer 2

a) f(2) = 0; f(3) = f(4) = -1; f(5) = 0

b) f(2)=f(3)=f(4)=f(5) = 1

c) f(2) = 2; f(3) = 5; f(4) = 27; f(5) = 734

d) f(2)=f(3)=f(4)=f(5) = 1

Explanation:

In general we have f(0) = f(1) = 1

a) We have

f(n+1) = f(n)-f(n-1)

Thus,

f(2) = f(1)-f(0) = 1-1 = 0

f(3) = f(2)-f(1) = 0-1 = -1

f(4) = f(3)-f(2) = -1-0 = -1

f(5) = f(4)-f(3) = -1-(-1) = 0

b) Here instead we have

f(n+1) = f(n)-f(n-1)

If f(n) = f(n-1) = 1, then f(n+1) should be 1, thus

f(2) = 1*1 = 1

f(3) = 1*1 = 1

f(4) = 1*1 = 1

f(5) = 1*1 = 1.

We conclude that f(2) = f(3) = f(4) = f(5) = 1

c) In this one we have

f(n+1) = f(n)²+f(n-1)

f(2) = f(1)²+f(0) = 1²+1 = 2

f(3) = f(2)²+f(1) = 2²+1 = 5

f(4) = f(3)²+f(2) = 5²+2 = 27

f(5) = f(4)²+f(3) = 27²+5 = 734

d) Now

f(n+1) = f(n)/f(n-1)

f(2) = f(1)/f(0) = 1/1 = 1

f(3) = f(2)/f(1) = 1/1 = 1

f(4) = f(3)/f(2) = 1/1 = 1

f(5) = f(4)/f(3) = 1/1 = 1.

Thus, f(2) = f(3) = f(4) = f(5) = 1.