1. In this pattern, as the number of iterations grows, the numbers will keep doubling. This is because each iteration involves multiplying the previous value by 2. So, after 10 iterations, we would expect the number to be 2^10 = 1024. To verify this, let's conduct 10 iterations:
f(1) = 2^1 = 2
f(f(1)) = f(2) = 2^2 = 4
f(f(f(1))) = f(f(2)) = f(4) = 2^3 = 8
f(f(f(f(1)))) = f(f(f(2))) = f(f(8)) = f(16) = 2^4 = 16
f(f(f(f(f(1))))) = f(f(f(f(16)))) = f(f(32)) = f(64) = 2^5 = 32
f(f(f(f(f(f(1)))))) = f(f(f(f(32)))) = f(f(64)) = f(128) = 2^6 = 64
f(f(f(f(f(f(f(1))))))) = f(f(f(f(64)))) = f(f(128)) = f(256) = 2^7 = 128
f(f(f(f(f(f(f(f(1)))))))) = f(f(f(256))) = f(f(512)) = f(1024) = 2^8 = 256
f(f(f(f(f(f(f(f(f(1))))))))) = f(f(512)) = f(1024) = 2^9 = 512
f(f(f(f(f(f(f(f(f(f(1))))))))) = f(1024) = 2^10 = 1024
As you can see, after 10 iterations, the number is indeed 1024.
2. Let's repeat the process with an initial value of -1:
f(-1) = 2*(-1) = -2
f(f(-1)) = f(-2) = 2*(-2) = -4
f(f(f(-1))) = f(f(-2)) = f(-4) = 2*(-4) = -8
f(f(f(f(-1)))) = f(f(f(-4))) = f(f(-8)) = f(-16) = 2*(-16) = -32
f(f(f(f(f(-1))))) = f(f(f(-8))) = f(f(-16)) = f(-32) = 2*(-32) = -64
f(f(f(f(f(f(-1)))))) = f(f(f(-16))) = f(f(-32)) = f(-64) = 2*(-64) = -128
f(f(f(f(f(f(f(-1))))))) = f(f(f(-32))) = f(f(-64)) = f(-128) = 2*(-128) = -256
f(f(f(f(f(f(f(f(-1)))))))) = f(f(f(-64))) = f(f(-128)) = f(-256) = 2*(-256) = -512
f(f(f(f(f(f(f(f(f(-1))))))))) = f(f(-128)) = f(-256) = 2*(-512) = -1024
f(f(f(f(f(f(f(f(f(f(-1))))))))) = f(-256) = 2*(-1024) = -2048
As you can see, with an initial value of -1, as the number of iterations grows, the numbers keep doubling but with a negative sign.
Start with the basic function . If you have an initial value of 1, then you
end up with the following iterations.
fx x () 2 =
• f (1) 2 1 2 = ⋅=
• 2 f (1) 2 2 1 4 = ⋅⋅=
• 3 f (1) 2 2 2 1 8 = ⋅⋅⋅=
Questions
1. If you continue this pattern, what do you expect would happen to the
numbers as the number of iterations grows? Check your result by conducting
at least 10 iterations.
2. Repeat the process with an initial value of −1. What happens as the number
of iterations grows?
1 answer