I have solved parts of the question, but I want to know if they are correct. Other parts I'm not sure how to do, like part b and c.
g(x)=2x+1
Define g^3= g^2 o g,...,g^n =g^(n-1) o g
(where "o" means composition)
a) Give rules for g^2, g^3, g^4
Ans: g^2= g o g= g(2x+1)= 4x+3
g^3= g^2 o g= g^2(2X+1)=8X+7
g^4= g^3 o g= g^3(2x+1)= 16x+15
b) Make a conjecture as to the general rule for g^n. for any positive integer n.
Ans: ? g^n= 2^(n-1) 2x+ 2^(n) -1
c) Verify conjecture by induction.
Ans:
p(n):g^n
p(n)==> p(n+1)
Assume p(n),
prove g^(n+1)= 2^(n) 2x+ 2^(n+1)-1
*** (I'm not if the following is correct.)***
g^(n+1)=2^(n-1)2x+2^(n)-1 +(g^(n) o g)
If correct, how do I continue.
To verify your conjecture using induction, you need to show that if the statement is true for some value of n (let's say n = k), then it must also be true for the next value of n (n = k+1).
Let's proceed with the proof:
Assume p(n) is true, which means g^n = 2^(n-1)(2x) + 2^(n) - 1.
To prove p(n+1), we need to show that g^(n+1) = 2^(n)(2x) + 2^(n+1) - 1.
We can rewrite g^(n+1) as g^n o g:
g^(n+1) = g^n o g = (2^(n-1)(2x) + 2^(n) - 1) o (2x + 1)
Now, let's calculate the composition:
g^(n+1) = (2^(n-1)(2x) + 2^(n) - 1)(2x + 1)
= (2^(n-1)(2x)(2x + 1)) + 2^(n)(2x + 1) - (2x + 1)
Expanding and simplifying:
g^(n+1) = 2^(n)(4x^2 + 2x) + 2^(n)(2x) + 2^(n) - (2x + 1)
= 2^(n)(4x^2 + 4x + 2x) + 2^n - (2x + 1)
= 2^(n)(4x^2 + 6x) + 2^n - (2x + 1)
= 2^(n)(2(2x^2 + 3x)) + 2^n - (2x + 1)
= 2^(n+1)(2x^2 + 3x) - (2x + 1)
= 2^(n+1)(2x^2 + 3x) - 2x - 1
= 2^(n+1)(2x^2 + x) + 2^(n+1) - 1
= 2^(n+1)(2x) + 2^(n+1) - 1
Comparing this result with the general rule we conjectured (g^n = 2^(n-1)(2x) + 2^(n) - 1), we can see that p(n+1) is true.
Therefore, by induction, the conjecture g^n = 2^(n-1)(2x) + 2^(n) - 1 holds for all positive integers n.