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Magazine Chapter 11: Problem 33
Let \(F_{n}\) be the \(n\) th term of the Fibonacci sequence. Find and prove an inequality relating \(n\) and \(F_{n}\) for natural numbers \(n .\)
Short Answer
Expert verified
The inequality is \(F_n \leq 2^n\) for all natural numbers \(n\).
Step by step solution
01
Understanding the Fibonacci Sequence
The Fibonacci sequence is a series of numbers where each term, after the first two, is the sum of the two preceding terms. It starts with 0 and 1, so the sequence begins: 0, 1, 1, 2, 3, 5, 8,... Hence, each term can be defined as \( F_n = F_{n-1} + F_{n-2} \) for \( n \geq 2 \).
02
Recognizing the Inequality Form
The inequality we need relates the term \( F_n \) and its position \( n \). For large \( n \), it is known that \( F_n \sim \left( \frac{1+\sqrt{5}}{2} \right)^n \), but for simplicity, we can use the well-known fact that \( F_n \leq 2^n \), for all \( n \geq 0 \).
03
Inequality Base Case
Verify the base case to start the proof by induction. For \( n = 0 \), \( F_0 = 0 \leq 2^0 = 1 \). For \( n = 1 \), \( F_1 = 1 \leq 2^1 = 2 \). Both statements are true, so the base case holds.
04
Inductive Hypothesis
Assume that the inequality is true for some integer \( k \), such that \( F_k \leq 2^k \). Our goal is to prove it for \( k+1 \), i.e., \( F_{k+1} \leq 2^{k+1} \).
05
Inequality Inductive Step
Using the Fibonacci formula, \( F_{k+1} = F_k + F_{k-1} \). By the inductive hypothesis: \( F_k \leq 2^k \) and \( F_{k-1} \leq 2^{k-1} \). So, \( F_{k+1} = F_k + F_{k-1} \leq 2^k + 2^{k-1} = 2^{k-1}(2 + 1) = 2^{k-1} \cdot 3 \leq 2^{k+1} \) since \( 3 \leq 4 \). This completes the induction.
06
Conclusion of Proof
By mathematical induction, the inequality \( F_n \leq 2^n \) holds for all natural numbers \( n \). This shows the growth relationship between the Fibonacci sequence and exponential sequences.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mathematical Induction
Mathematical induction is a method used to prove a statement is true for all natural numbers. It's like a domino effect: if you prove the first domino falls and show that each domino knocks over the next, then they all fall. This technique has two main steps:
- Base Case: Show the statement is true for the initial value, often for the smallest number, say, n=0 or n=1.
- Inductive Step: Assume the statement is true for some arbitrary natural number, n=k. Then prove the statement for the next number, n=k+1, using this assumption.
Inequality
Inequalities are mathematical expressions which show the relationship between two values that are not exactly equal. They come in various forms, such as less than (<), greater than (>), less than or equal to (鈮�), and greater than or equal to (鈮�). In the context of the Fibonacci sequence, inequalities help us compare the size of a Fibonacci number to another function or number.
- For the Fibonacci sequence, the inequality is often used to compare Fibonacci numbers to exponential functions, recognizing that Fibonacci numbers grow quickly, but not as fast as certain exponential functions.
- In our case, it's shown that each Fibonacci number F_n is less than or equal to 2^n, which helps in understanding how Fibonacci numbers grow compared to powers of 2.
Exponential Growth
Exponential growth happens when the rate of change of a mathematical function is proportional to the current value. It's a constant multiplicative process, unlike linear growth, which is additive.
- In the Fibonacci sequence, even though each number is the sum of its two predecessors, the approximate growth can be compared to exponential growth due to its rapid increase.
- The authentic growth factor in the Fibonacci sequence approaches the golden ratio, \( \phi = \frac{1+\sqrt{5}}{2} \approx 1.618 \), but to simplify things, 2^n is often used as an upper bound for discussions on growth.
- Recognizing exponential growth helps in predicting future values quickly, crucial for applications in various fields from finance to biological science.
