In this problem we will show that there is a positive number

In this problem we will show that there is a positive number x_0 R such that x_0^2 = 2. Let g: (0, infinity) rightarrow (0, infinity) be the function defined by g(x) = 2x + 1/x + 2. Let (a_n) and (b_n) be two sequences defined recursively as: a_1 = 1, b_1 = 2, a_n+l = g(a_n) and b_n+1 = g(b_n). Show the following. a) If 0

Solution

(a) Given:

g(x) = (2x+2)/(x+2)

g(y) = (2y+2)/(y+2)

g(y) - g(x) = (2y+2)/(y+2)   - (2x+2)/(x+2)

                = 2 (y-x)/((x+2)(y+2))

Given y > x and also x and y are positive.

So, from the above equation, wew note:

g(x) < g(y)

and also,

g(y) - g(x) < (y-x)/2.

(b) To prove: a_n < a_n+1 < b_n+1 < b_n:

Given: a₁ = 1,

          b₁₌ 2.

We get:

a₂ = 4/3.

b₂ = 3/2.

So, we note: 1 < (4/3) < (3/2) < 2.

So, a₁ < a₂ < b₂ < b_1.

In general,

a_n+1 = (2a_n+2)/(a_n + 2)

and b_{n+1 =} (2b_n + 2)/(b_n+2).

But, b₂>a_2., from the relults of a₁ and b_1.

So, by induction, we obtain the desired result.

Since it is long questiion, as per Directions for Answering, the first two questions are answered.