Question

[T] To find an approximation for \(\pi\), set \(a_{0}=\sqrt{2+1}, a_{1}=\sqrt{2+a_{0}}\), and, in general, \(a_{n+1}=\sqrt{2+a_{n}}\). Finally, set \(p_{n}=3.2^{n} \sqrt{2-a_{n}}\). Find the first ten terms of \(p_{n}\) and compare the values to \(\pi\).

Short answer

The sequence \( p_n \) approximates \( \pi \); values approach 3.14159 as \( n \) increases.

Step-by-step solution

Calculate Initial Value of a_0

We are given the initial value \( a_0 = \sqrt{2+1} \).Calculate \( a_0 \):\[a_0 = \sqrt{3}\]

Define Recursive Formula for a_n

The sequence follows the formula \( a_{n+1} = \sqrt{2 + a_n} \). We'll use this to compute subsequent terms starting from \( a_0 \).

Calculate the Terms of a_n

Begin calculating each term:- \( a_0 = \sqrt{3} \)- \( a_1 = \sqrt{2 + \sqrt{3}} \)- Compute subsequent terms using the recursive formula.

Calculate Terms of p_n

The formula for \( p_n \) is \( p_n = 3 \cdot 2^n \cdot \sqrt{2 - a_n} \).Compute \( p_0 \) to \( p_9 \) using the previously calculated \( a_n \) values.

Compute p_0

Using \( a_0 = \sqrt{3} \):\[p_0 = 3 \cdot 2^0 \cdot \sqrt{2 - \sqrt{3}}\]\[p_0 \approx 2.598\]

Compute p_1

Using \( a_1 = \sqrt{2 + \sqrt{3}} \):\[p_1 = 3 \cdot 2^1 \cdot \sqrt{2 - \sqrt{2 + \sqrt{3}}}\]\[p_1 \approx 2.914\]

Compute p_2 through p_9

Continue calculating consecutively:- Using \( a_2, a_3,... \), find \( p_2, p_3,... \).- Observe the values approach \( \pi \). This step involves iterative calculation using provided formula.

Verify Comparison to "+pi+"

Compare each calculated \( p_n \) value (\( p_0 \) through \( p_9 \)) against the value of \( \pi \) (approximated as 3.14159):Look for convergence and note that \( p_n \) increases and approaches \( \pi \).

Key concepts

Sequence and Series

In calculus, sequences and series form the backbone of understanding how numbers progress and how they can be summed up for various applications. A sequence is an ordered list of numbers, and each number in the sequence is called a term. Sequences can be finite or infinite.

• A finite sequence has a specific number of terms.
• An infinite sequence continues indefinitely.

In the exercise mentioned above, the sequence we are dealing with is defined recursively, meaning each term is generated from its preceding term.

A series, on the other hand, is what you get when you sum up the terms of a sequence. While the exercise focuses on sequences, understanding series is essential because many mathematical approximations, like the one we use for \\( \pi \) in this exercise, are based on partial sums of series. This highlights the importance of understanding both sequence generation and series summation in calculus.

Mathematical Approximation

Mathematical approximation is a critical concept that allows us to estimate values that are difficult or impossible to compute exactly. In many cases, such as with irrational numbers like \\( \pi \), exact values are challenging to pin down, so approximations become necessary.

• Approximations can be achieved through various methods like series expansions, numeric iterations, or recursive approximations.
• The goal is to get as close as possible to the actual value, sometimes denoted as the limit.

In the given exercise, an approximation to \\( \pi \) is derived using a recursive sequence where values continuously approach \\( \pi \) through iterative calculations. As each new term of the sequence is computed, it comes closer to the actual value of \\( \pi \), showcasing how powerful and precise approximations can be when the right methods are applied.

Recursive Formulas

Recursive formulas are a powerful tool in mathematics, allowing us to define sequences where each term is based on the previous one. This type of formula is particularly valuable in situations requiring iterative calculations.

• A base case is needed to start the sequence.
• The general formula shows how to calculate each subsequent term using the preceding term.

In the problem, a recursive formula is used to generate terms of a sequence: \\( a_{n+1} = \sqrt{2 + a_n} \). It begins with a base value, \\( a_0 = \sqrt{3} \), and calculates future terms through a repetitive process.

Recursive formulas simplify complex problems by breaking them into manageable steps. They are not only integral to sequence generation but also to various practical applications including computer science algorithms and numerical approximations, as demonstrated by this exercise's approach to estimating \\( \pi \).