Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations.
• f (1) = 2 ⋅1 =2
• f 2 (1) = 2 ⋅2 ⋅1 =4
• f 3 (1) = 2 ⋅ 2 ⋅ 2 ⋅1 = 8
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.
f (1) = 2 ⋅1 = 2
f 2 (1) = 2 ⋅2 ⋅1 = 4
f 3 (1) = 2 ⋅ 2 ⋅ 2 ⋅ 1 = 8
f4(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 16
f5(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 32
f6(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 64
f7(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 128
f8(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 256
f9(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 512
f10(1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 1 = 1024
2.Repeat the process with an initial value of −1. What happens as the number of iterations grows?
f (-1) = 2 ⋅(-1) = -2
f 2 (-1) = 2 ⋅2 ⋅(-1) = -4
f 3 (-1) = 2 ⋅ 2 ⋅ 2 ⋅ (-1) = -8
f4(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ (-1)= -16
f5(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ (-1) = -32
f6(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅(-1) = -64
f7(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ (-1) = -128
f8(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅(-1) = -256
f9(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅(-1) = -512
f10(-1) = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ (-1) = -1024
As you can see I've already done the conducting 10 iterations part for numbers 1 and 2, however my question would be : How do i explain to answer the question in number 2 "What happens as the number of iterations grows?" .
4 answers
As the number of iterations grows, so does the f value, either positive or negative.
And that's 2^x
2x means 2 times x: 2,4,6,8,...
f(-1)=2(-1)=-2
f^2(-1)=f(-1)*f(-1)=-2*-2=4
f^3(-1)=f^2(-1)*f(-1)=4*(-2)=-8
f^4(-1)=f^3(-1)*f(-1)=-8*(-2)=16
f^5(-1)=f^4(-1)*f(-1)=16*(-2)=-32
As you can see the magnitude of the number increases, by being multiplied by 2 every time. Also as you can see the signs are altered, from negative to positive.
1 answer