Problem 48 - Algebra and Trigonometry

Question

A sequence is defined recursively. List the first five terms. \(a_{1}=\sqrt{2} ; \quad a_{n}=\sqrt{\frac{a_{n-1}}{2}}\)

Step-by-Step Solution

Answer

The first five terms are \(\sqrt{2}\), \(2^{-1/4}\), \(2^{-5/8}\), \(2^{-13/16}\), and \(2^{-29/32}\).

Step 1 - Understand the Given

The sequence is defined recursively with the initial term given as \(a_1 = \sqrt{2}\). Each subsequent term is defined by the relationship \(a_n = \sqrt{\frac{a_{n-1}}{2}}\).

Step 2 - Find the First Term

The first term is provided directly: \(a_1 = \sqrt{2}\).

Step 3 - Find the Second Term

Use the recursive formula with \(a_1\) to find \(a_2\):

Step 4: Calculate

\[ a_2 = \sqrt{\frac{a_1}{2}} = \sqrt{\frac{\sqrt{2}}{2}} = \sqrt{\frac{2^{1/2}}{2}} = \sqrt{2^{-1/2}} = 2^{-1/4} \]

Step 4 - Find the Third Term

Use the recursive formula with \(a_2\) to find \(a_3\):

Step 6: Calculate

\[ a_3 = \sqrt{\frac{a_2}{2}} = \sqrt{\frac{2^{-1/4}}{2}} = \sqrt{2^{-1/4 - 1}} = \sqrt{2^{-5/4}} = 2^{-5/8} \]

Step 5 - Find the Fourth Term

Use the recursive formula with \(a_3\) to find \(a_4\):

Step 8: Calculate

\[ a_4 = \sqrt{\frac{a_3}{2}} = \sqrt{\frac{2^{-5/8}}{2}} = \sqrt{2^{-5/8 - 1}} = \sqrt{2^{-13/8}} = 2^{-13/16} \]

Step 6 - Find the Fifth Term

Use the recursive formula with \(a_4\) to find \(a_5\):

Step 10: Calculate

\[ a_5 = \sqrt{\frac{a_4}{2}} = \sqrt{\frac{2^{-13/16}}{2}} = \sqrt{2^{-13/16 - 1}} = \sqrt{2^{-29/16}} = 2^{-29/32} \]

Key Concepts

sequence and series

Sequences and series are fundamental concepts in mathematics. A sequence is an ordered list of numbers, while a series is the sum of the terms of a sequence.
Sequences are defined in various ways, including explicitly and recursively. An explicit definition provides a formula for any term given its position. A recursive definition, like the one in our exercise, defines each term based on previous terms.
This recursive method is often used because it simplifies the definition of complex sequences. In our exercise, we see the power of recursion:

• The initial term is given.
• Each following term relies on the previous one.

recursive formulas

Recursive formulas are a way to define the elements of a sequence based on preceding terms. This means each term is generated from the one before it.
In our exercise, the sequence starts with an initial term, and each following term is calculated using a recursive relationship. Specifically, we start with \(a_1 = \sqrt{2}\), and for \(n > 1\), the formula is \(a_n = \sqrt{\frac{a_{n-1}}{2}}\).
This recursive relationship helps to generate complex sequences with relatively simple rules. The key steps involve:

• Using the initial condition.
• Applying the recursive formula systematically.

Recursive formulas require careful handling, especially when dealing with square roots and fractions, to mitigate errors and ensure accuracy.

algebraic manipulation

Algebraic manipulation involves rearranging equations and expressions to simplify or solve them. For our recursive sequence, we need to manipulate the expression \(\sqrt{\frac{a_{n-1}}{2}}\) tailored to each subsequent term.
Let's break this down using the provided sequence steps:

• For \(a_2\):
  \(a_2 = \sqrt{\frac{a_1}{2}} = \sqrt{\frac{\sqrt{2}}{2}} = \sqrt{2^{-1/2}} = 2^{-1/4}\)
• For \(a_3\):
  \(a_3 = \sqrt{\frac{a_2}{2}} = \sqrt{2^{-1/4}}{2}} = \sqrt{2^{-5/4}} = 2^{-5/8}\)
• For \(a_4\):
  \(a_4 = \sqrt{\frac{a_3}{2}} = \sqrt{2^{-5/8}}{2}} = \sqrt{2^{-13/8}} = 2^{-13/16}\)
• For \(a_5\):
  \(a_5 = \sqrt{\frac{a_4}{2}} ={\sqrt{2^{-13/16}}{2}} = \sqrt{2^{-29/16}} = 2^{-29/32}\)

This cascade of steps displays the importance of careful algebraic manipulation in each recursion step.

exponent rules

Exponent rules are essential in simplifying expressions, especially in recursive sequences involving roots and fractional exponents. The rules used in our exercise include:

• Product rule: \(a^{m} \cdot a^{n} = a^{m+n}\)
• Quotient rule: \(\frac{a^{m}}{a^{n}} = a^{m-n}\)
• Power rule: \((a^{m})^n = a^{m \cdot n}\)
• Root notation: \(\sqrt{a} = a^{1/2}\)

In the steps of our sequence, these rules were extensively used to resolve the exponents:

• \(2^{-1/2} = 2^{-1/4}\)
• \(2^{-1/4 - 1} = 2^{-5/4}\)
• \(2^{-13/8} = 2^{-13/16}\)
• \(2^{-29/16}\)

Keeping these rules in mind helps in breaking down complex recursive sequences.