Problem 67 - Precalculus: Mathematical for Calculus

Question

Find a formula for the \(n\) th term of the sequence \(\sqrt{2}, \quad \sqrt{2 \sqrt{2}}, \quad \sqrt{2 \sqrt{2 \sqrt{2}}}, \quad \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, \ldots\)

Step-by-Step Solution

Answer

The formula for the \( n \)th term is \( 2^{1 - \frac{1}{2^n}} \).

Step 1: Identify the Pattern

Observe that each term in the sequence has a nested radical structure. The first term is \( \sqrt{2} \), the second term is \( \sqrt{2 \times \sqrt{2}} \), and the third term is \( \sqrt{2 \times \sqrt{2 \times \sqrt{2}}} \). Notice that this pattern can be expressed in terms of powers of 2.

Step 2: Express Radicals in Exponential Form

Rewrite the terms as exponents. The first term \( \sqrt{2} \) can be rewritten as \( (2^{1/2}) \). The second term \( \sqrt{2 \times \sqrt{2}} \) is equivalent to \( (2^{1/2})^{3/2} = 2^{3/4} \). Observe that each subsequent term builds upon the exponent from the previous term.

Step 3: Establish a Recurrence for Exponents

Define the exponent of \( 2 \) in the \( n \)th term as \( a_n \). Based on the pattern, we see \( a_1 = \frac{1}{2}, a_2 = \frac{3}{4} \), and \( a_3 = \frac{7}{8} \). Notice how \( a_n = 1 - \frac{1}{2^n} \).

Step 4: Derive the General Formula

Recognizing that the sequence follows \( a_n = 1 - \frac{1}{2^n} \), the \( n \)th term of the sequence is given by \( \sqrt[n]{2} = 2^{1 - \frac{1}{2^n}} \). This is equivalent to \( 2^{1 - \frac{1}{2^n}} = 2^{1} \times 2^{- \frac{1}{2^n}} \).

Step 5: Formulate the nth Term

The \( n \)th term of the sequence can be written as \( 2^{1 - \frac{1}{2^n}} \), which represents the nested radical expression in its exponential form.Therefore, the formula for the \( n \)th term of the sequence is \( 2^{1 - \frac{1}{2^n}} \).

Key Concepts

Exponential Form

The concept of exponential form is a powerful tool in mathematics, helping to simplify expressions with repeated multiplication. By using bases and exponents, complex calculations can be expressed in a more compact form. In our sequence, we transform endless products and square roots into simpler exponential terms. For instance:

• The number \( \sqrt{2} \) can be rewritten as \( 2^{1/2} \).
• The expression \( \sqrt{2 \times \sqrt{2}} \) becomes \( 2^{3/4} \) when expressed exponentially.

By rewriting numbers in exponential form, we uncover patterns and relationships that are not immediately obvious. This aids in understanding complex sequences, and helps in deriving general formulas for terms based on their position in the sequence.

Nested Radicals

Nested radicals are expressions containing a square root within another square root. These can often appear daunting and complex. In the given sequence:

• The first term has a simple radical: \( \sqrt{2} \).
• The second term adds another layer: \( \sqrt{2 \times \sqrt{2}} \).
• Each additional term introduces further nesting, like \( \sqrt{2 \times \sqrt{2 \times \sqrt{2}}} \).

Converting nested radicals into exponential form allows for simplification. It changes a potentially intimidating structure into a format where we can easily apply mathematical rules. Understanding nested radicals leads to discovering sequences and generating formulas to predict subsequent terms.

Recurrence Relation

Recurrence relations define sequences in a way that each term is based on the preceding ones. This method of expression is particularly useful for sequences where a simple pattern is detectable, as in our exercise. With the sequence provided, the exponents of 2 form a recurrence relation:

• The first exponent is \( a_1 = rac{1}{2} \).
• The second exponent is \( a_2 = rac{3}{4} \).
• The relation is modeled as \( a_n = 1 - rac{1}{2^n} \).

These exponents build from each prior exponent. This recurrence relation helps us to quickly find any term's exponent without evaluating all preceding terms. It's a shortcut that highlights the predictable pattern and regularity within the sequence. Using this relation, we can define the sequence formula, leading to an understanding of how terms evolve as the sequence progresses.