Problem 68 - Calculus: Early Transcendentals

Question

Find the first 40 terms of the sequence defined by \( a_{n + 1} =\left\\{ \begin{array}{ll} \frac{1}{2} a_n & \text{if } a_n \text{ is an even number} \\ 3a_n + 1 & \text{if } a_n \text{ is an odd number } \end{array} \right. \) and \( a_1 = 11. \) Do the same if \( a_1 = 25. \) Make a conjecture about this type of sequence.

Step-by-Step Solution

Answer

Both sequences eventually cycle through 4, 2, 1, repeatedly. This supports the Collatz conjecture that all positive integers will reach this cycle.

Step 1: First 10 Terms for a1 = 11

Start with the initial term, \( a_1 = 11 \). Since 11 is odd, the formula for the next term is \( 3a_n + 1 \). Calculate \( a_2 \): \( 3(11) + 1 = 34 \). Now, 34 is even, use the formula \( \frac{1}{2}a_n \). Continue this way:1. \( a_1 = 11 \) (odd) \( \Rightarrow a_2 = 34 \)2. \( a_2 = 34 \) (even) \( \Rightarrow a_3 = 17 \)3. \( a_3 = 17 \) (odd) \( \Rightarrow a_4 = 52 \)4. \( a_4 = 52 \) (even) \( \Rightarrow a_5 = 26 \)5. \( a_5 = 26 \) (even) \( \Rightarrow a_6 = 13 \)6. \( a_6 = 13 \) (odd) \( \Rightarrow a_7 = 40 \)7. \( a_7 = 40 \) (even) \( \Rightarrow a_8 = 20 \)8. \( a_8 = 20 \) (even) \( \Rightarrow a_9 = 10 \)9. \( a_9 = 10 \) (even) \( \Rightarrow a_{10} = 5 \)

Step 2: Complete to 40 Terms for a1 = 11

Continue calculating terms using the same rules until reaching 40 terms:10. \( a_{10} = 5 \) (odd) \( \Rightarrow a_{11} = 16 \)11. \( a_{11} = 16 \) (even) \( \Rightarrow a_{12} = 8 \)12. \( a_{12} = 8 \) (even) \( \Rightarrow a_{13} = 4 \)13. \( a_{13} = 4 \) (even) \( \Rightarrow a_{14} = 2 \)14. \( a_{14} = 2 \) (even) \( \Rightarrow a_{15} = 1 \)15. \( a_{15} = 1 \) (odd) \( \Rightarrow a_{16} = 4 \)16. Continue with this pattern: \( 2, 1, 4 \)... until 40 terms are reached.

Step 3: First 10 Terms for a1 = 25

Start with \( a_1 = 25 \): 1. \( a_1 = 25 \) (odd) \( \Rightarrow a_2 = 76 \)2. \( a_2 = 76 \) (even) \( \Rightarrow a_3 = 38 \)3. \( a_3 = 38 \) (even) \( \Rightarrow a_4 = 19 \)4. \( a_4 = 19 \) (odd) \( \Rightarrow a_5 = 58 \)5. \( a_5 = 58 \) (even) \( \Rightarrow a_6 = 29 \)6. \( a_6 = 29 \) (odd) \( \Rightarrow a_7 = 88 \)7. \( a_7 = 88 \) (even) \( \Rightarrow a_8 = 44 \)8. \( a_8 = 44 \) (even) \( \Rightarrow a_9 = 22 \)9. \( a_9 = 22 \) (even) \( \Rightarrow a_{10} = 11 \)

Step 4: Complete to 40 Terms for a1 = 25

Continue calculating terms using the same rules until reaching 40 terms:10. From \( a_{10} = 11 \), which loops back to starting case in Step 1 and continue.11. Without explicitly listing each term again, note the sequence will follow the pattern for \( a_1 = 11 \) since \( a_{10} = 11 \). Continue from Step 2 of \( a_1 = 11 \).

Step 5: Conjecture

Through calculations, notice that the sequence eventually falls into a repeating cycle of \( 4, 2, 1, \) and this cycle perpetuates. This behavior suggests that any positive integer will eventually enter this cycle within a finite number of steps, confirming the Collatz conjecture on which this sequence is based.

Key Concepts

Even and Odd Numbers

Understanding even and odd numbers is key to deciphering the Collatz conjecture sequence. In mathematics:

• Even numbers are integers divisible by 2 without a remainder, like 2, 4, 6, and so on.
• Odd numbers are not divisible by 2; when divided by 2, they leave a remainder of 1, such as 1, 3, 5, etc.

In the sequence defined by the Collatz conjecture, an even number leads to the following term by halving it, i.e., \( a_{n+1} = \frac{1}{2}a_n \). For odd numbers, the next term is calculated by multiplying it by 3 and then adding 1, i.e., \( a_{n+1} = 3a_n + 1 \).
This rule dictates how the sequence jumps backward and forwards based on the parity (whether a number is even or odd) of the current term. Understanding this parity is key to predicting and following the sequence's progression.

Sequence Patterns

The sequence patterns in a Collatz conjecture sequence are fascinating due to their structured unpredictability.
Once you start from any positive integer, the sequence shifts terms based on their positivity:

• If a number in the sequence is even, it shrinks by half.
• If the number is odd, it grows before falling back to an even number.

This alternation creates seemingly unpredictable patterns that draw attention. Yet, what's captivating is that no matter the initial number used to start the sequence, the numbers eventually shrink and fall into a repeating cycle of \( 4, 2, 1 \).
This repetitive pattern is the essence of the Collatz sequence, as demonstrated in the given exercise. By slicing any integer into this pattern eventually, it paints an interesting tapestry of numerical sequence behavior.

Mathematical Sequences

A mathematical sequence is essentially an ordered list of numbers following a particular rule or formula to determine the next term. In the realm of the Collatz conjecture, the formula is based on the parity of the current number.
Such sequences are vital in understanding numerical behaviors and predicting outcomes in number theory. Each step in the sequence involves calculating the next number based on the current number's even or odd status.

• Even terms: Transform by dividing by two.
• Odd terms: Transform by applying \( 3a_n + 1 \).

This transformation rule is what sequences this type of mathematical list, continually bringing it down to \( 4, 2, 1 \).
Seeing the sequence eventually terminating in a loop highlights that, no matter its initially unpredictable appearance, it adheres to a resilient set of mathematical rules, strongly hinting towards the validity of the Collatz conjecture.