A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?

B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10?

C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
How do you find out the first five terms for it. then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?

D) How do you find out the exact limit of the sequence defined in part C? Are you supposed to square the recursive equation and take limits using limit theorems? If so, then which are these theorems?

Thank you very much.

I answered this below after I did the first two.

calculus - Damon, Saturday, April 3, 2010 at 7:26pm

.3
sqrt 1.3 = 1.14
sqrt 2.14 = 1.46
sqrt 2.46 = 1.57
sqrt 2.57 = 1.60

hmmm, not getting bigger very fast.
let's see what happens to the derivative for large n
.5/sqrt(x+1)
ah ha, look at that. When n gets big, the derivative goes to zero. So the function stops changing.

A) To prove that if 0 (≤) x (≤) 10, then 0 (≤) sqrt(x+1) (≤) 10, we can use the following steps:

Step 1: Start with the assumption that 0 (≤) x (≤) 10.

Step 2: Add 1 to both sides of the inequality to get 1 (≤) x + 1 (≤) 11.

Step 3: Take the square root of both sides of the inequality to get sqrt(1) (≤) sqrt(x + 1) (≤) sqrt(11).

Step 4: Simplify the square roots to get 1 (≤) sqrt(x + 1) (≤) sqrt(11).

Step 5: Since 1 (≤) sqrt(x + 1) (≤) sqrt(11), we can replace sqrt(11) with 10 since 10 is greater than or equal to sqrt(11).

Step 6: Therefore, we can conclude that if 0 (≤) x (≤) 10, then 0 (≤) sqrt(x + 1) (≤) 10.

B) To prove that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) sqrt(u + 1) (≤) sqrt(v + 1) (≤) 10, we can use a similar approach as in part A:

Step 1: Start with the assumption that 0 (≤) u (≤) v (≤) 10.

Step 2: Add 1 to all three parts of the inequality to get 1 (≤) u + 1 (≤) v + 1 (≤) 11.

Step 3: Take the square root of all three parts of the inequality to get sqrt(1) (≤) sqrt(u + 1) (≤) sqrt(v + 1) (≤) sqrt(11).

Step 4: Simplify the square roots to get 1 (≤) sqrt(u + 1) (≤) sqrt(v + 1) (≤) sqrt(11).

Step 5: Again, since sqrt(11) is less than or equal to 10, we can replace sqrt(11) with 10.

Step 6: Therefore, we can conclude that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) sqrt(u + 1) (≤) sqrt(v + 1) (≤) 10.

C) To find the first five terms of the recursively defined sequence a_n = sqrt(a_(n-1) + 1) with a_1 = 0.3, we can use the following steps:

Step 1: Start with the given initial term a_1 = 0.3.

Step 2: Use the recursive formula a_(n+1) = sqrt(a_n + 1) to calculate each subsequent term.

a_2 = sqrt(a_1 + 1) = sqrt(0.3 + 1) = sqrt(1.3) ≈ 1.14
a_3 = sqrt(a_2 + 1) = sqrt(1.14 + 1) ≈ 1.34
a_4 = sqrt(a_3 + 1) ≈ 1.45
a_5 = sqrt(a_4 + 1) ≈ 1.52

Therefore, the first five terms of the sequence are approximately 0.3, 1.14, 1.34, 1.45, and 1.52.

To prove that this sequence converges, observe that as n approaches infinity, the difference between consecutive terms becomes smaller and smaller. Additionally, the sequence is bounded above by 10 (as shown in part A). This suggests that the sequence is converging towards a finite limit.

One specific theorem that guarantees convergence for a recursively defined sequence is the Monotone Convergence Theorem. This theorem states that if a sequence is increasing (or decreasing) and bounded above (or below), then it must converge to a finite limit.

D) To find the exact limit of the sequence defined in part C, we can solve the recursive equation for the limit value. Let's denote the limit as L:

L = sqrt(L + 1)

To find L, we can square both sides of the equation:

L^2 = L + 1

Now we have a quadratic equation. Rearrange it to be in standard form:

L^2 - L - 1 = 0

We can solve this quadratic equation using the quadratic formula:

L = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -1, and c = -1. Plugging in these values, we get:

L = (-(-1) ± sqrt((-1)^2 - 4(1)(-1))) / (2(1))
= (1 ± sqrt(5)) / 2

Therefore, the exact limit of the sequence is given by the two possible solutions: (1 + sqrt(5))/2 and (1 - sqrt(5))/2.

To determine which value is the correct limit, we can check if each value satisfies the recursive equation L = sqrt(L + 1). Substitute each value into the equation and verify if the equation holds true. The correct value will satisfy the equation.

As for the theorems used in finding the limit, we utilize the algebraic manipulations rather than specific limit theorems. However, the limit theorems used to determine the convergence in part C still apply here. Specifically, the Monotone Convergence Theorem can be used to confirm that the sequence converges towards a finite limit.