Problem 33 - Precalculus: Mathematical for Calculus

Question

(a) A sequence is defined recursively by \(a_{1}=0\) and $$a_{n+1}=\sqrt{2+a_{n}}$$ Find the first ten terms of this sequence correct to eight decimal places. Does this sequence appear to be convergent? If so, guess the value of the limit. (b) Assuming the sequence in part (a) is convergent, let \(\lim _{n \rightarrow \infty} a_{n}=L .\) Explain why \(\lim _{n \rightarrow \infty} a_{n+1}=L\) also, and therefore $$L=\sqrt{2+L}$$ Solve this equation to find the exact value of \(L\)

Step-by-Step Solution

Answer

The sequence appears to converge to 2. The exact value of the limit is 2.

Step 1: Determine Parameters for Calculation

Identify the initial term and the recursive formula given in the problem. We know \(a_1 = 0\) and for each subsequent term \(a_{n+1} = \sqrt{2 + a_n}\). We need to calculate up to the first ten terms, each to eight decimal places.

Step 2: Calculate Terms 1 to 3

Begin calculating the sequence terms.1. \(a_1 = 0.00000000\) (as provided).2. Calculate \(a_2 = \sqrt{2 + a_1} = \sqrt{2 + 0} = \sqrt{2} = 1.41421356\).3. Calculate \(a_3 = \sqrt{2 + a_2} = \sqrt{2 + 1.41421356} = 1.84775907\).

Step 3: Calculate Terms 4 to 6

Continue calculating the sequence terms.4. Calculate \(a_4 = \sqrt{2 + a_3} = \sqrt{2 + 1.84775907} = 1.96157056\).5. Calculate \(a_5 = \sqrt{2 + a_4} = \sqrt{2 + 1.96157056} = 1.99036945\).6. Calculate \(a_6 = \sqrt{2 + a_5} = \sqrt{2 + 1.99036945} = 1.99759091\).

Step 4: Calculate Terms 7 to 10

Continue calculating the remaining sequence terms.7. Calculate \(a_7 = \sqrt{2 + a_6} = \sqrt{2 + 1.99759091} = 1.99939764\).8. Calculate \(a_8 = \sqrt{2 + a_7} = \sqrt{2 + 1.99939764} = 1.99984940\).9. Calculate \(a_9 = \sqrt{2 + a_8} = \sqrt{2 + 1.99984940} = 1.99996235\).10. Calculate \(a_{10} = \sqrt{2 + a_9} = \sqrt{2 + 1.99996235} = 1.99999059\).

Step 5: Examine for Convergence

Observe the calculated terms of the sequence. The numbers are getting closer to 2 as \(n\) increases, suggesting that the sequence is converging towards that number.

Step 6: Formulate the Limit Equation

Given the limit \(\lim_{n \rightarrow \infty} a_n = L\), due to the recursive property, \(\lim_{n \rightarrow \infty} a_{n+1} = L\). Therefore, we have:\[L = \sqrt{2 + L}\].

Step 7: Solve the Limit Equation

To find \(L\), square both sides to eliminate the square root:1. \(L^2 = 2 + L\).2. Rearrange to form a quadratic equation: \(L^2 - L - 2 = 0\).3. Solve using the quadratic formula: \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), \(c = -2\).This simplifies to \(L = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}\).4. The roots are \(L = 2\) and \(L = -1\). Since the terms of the sequence are non-negative, the limit \(L\) is \(2\).

Key Concepts

Convergence

When a sequence is convergent, its terms approach a specific value as the sequence goes on indefinitely. This behavior is referred to as convergence. In the context of our recursion problem, the sequence converges because each term gets closer and closer to a particular value. Over time, it seems to stabilize around this number.

Convergence is key in recursive sequences as it implies stability in the values. The sequence we've been working with shows convergence because each calculated term comes nearer to 2. We observe:

• The sequence starts with a small number.
• Each next term grows larger but at a decreasing rate.
• Eventually, the terms appear to stabilize near a single point.

Recognizing convergence in a sequence can help predict long-term behavior without calculating an infinite number of terms.

Limit of a Sequence

The limit of a sequence is the value that the terms of the sequence "settle down" to as they proceed toward infinity. In mathematical terms, if a sequence of numbers \(a_n\) converges, it approaches a unique value \(L\), known as the limit.

For our recursive sequence, the limit \(L\) is central to determining the behavior of \(a_n\) as \(n\) approaches infinity. Establishing \(\lim_{n \rightarrow \infty} a_n = L\) ensures that any of the subsequent terms \(a_{n+1}\) will also share this limit, due to the sequence's recursive nature. This brings us to the equation \[L = \sqrt{2 + L}\].

This equation must be solved to find the precise value that the sequence approaches. The resolution of this equation reveals that as \(n\) approaches infinity, \(a_n\) and \(a_{n+1}\) both stabilize to the same value.

Quadratic Equations

Quadratic equations arise when trying to solve limit equations for recursive sequences. When we set the limit equation from our recursive problem \[L = \sqrt{2 + L}\] into motion, we simplify it by squaring both sides resulting in a quadratic equation: \[L^2 = 2 + L\]. This becomes \[L^2 - L - 2 = 0\].

Solving this quadratic equation requires using the quadratic formula: \[L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. For our equation, \(a = 1\), \(b = -1\), and \(c = -2\), leading us to solve \[L = \frac{1 \pm 3}{2}\]. This provides roots, \(L = 2\) and \(L = -1\). Since our sequence started with non-negative terms, the reasonable solution for the limit, and thus the convergence point of the sequence, is \(L = 2\).

Understanding quadratic equations is crucial when one encounters non-linear relationships in sequences, as they can reveal the hidden properties of convergence and limit values.