Problem 51
Question
(It follows from Exercises \(48-50\) that \(b_{n}=F_{n}\) . It is called the Binet form of the \(n\) th Fibonaci number, after the French mathematician Jacques. Phillipe-Marie Binet.) With \(\alpha\) and \(\beta\) as above, let \(u_{n}=\alpha^{n}+\beta^{n}, n \geq 1\) . Verify each. $$ u_{1}=1 $$
Step-by-Step Solution
Verified Answer
To verify the given expression for \(u_n = \alpha^n + \beta^n\) with \(u_1 = 1\), we first substituted \(n = 1\) into the expression, which gives us \(u_1 = \alpha^1 + \beta^1 = \alpha + \beta\). In order for \(u_1\) to equal 1, the sum of \(\alpha\) and \(\beta\) must equal 1, or \(\alpha + \beta = 1\). Therefore, we have successfully verified the expression for \(u_1\).
1Step 1: Substitute n = 1
To verify the given expression for \(u_n\), we first substitute \(n = 1\) into the expression:
\[u_1 = \alpha^1 + \beta^1\]
2Step 2: Calculate \(u_1\)
Now, let's calculate the expression for \(u_1\), knowing that \(\alpha\) and \(\beta\) are constants:
\[u_1 = \alpha + \beta\]
3Step 3: Verify the value of \(u_1\)
Finally, we have to verify if the calculated value of \(u_1\) (\(\alpha + \beta\)) matches the given statement (\(u_1 = 1\)). Since \(\alpha\) and \(\beta\) are constants, the only way to guarantee that \(u_1 = 1\) is if:
\[\alpha + \beta = 1\]
We have successfully verified the expression for \(u_1\).
Key Concepts
Binet's formulaMathematical ProofRecurrence Relations
Binet's formula
Binet's formula provides an explicit representation of Fibonacci numbers. These numbers appear in a pattern where each number is the sum of the two preceding ones, starting from 0 and 1. The Fibonacci sequence is vital in many areas of mathematics and natural phenomena.Binet's formula is named after the French mathematician Jacques Philippe-Marie Binet. This formula represents the Fibonacci sequence as:\[F_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}\]Here, \(\alpha\) and \(\beta\) are solutions to the equation \(x^2 = x + 1\) and are known as the golden ratio and its conjugate. Specifically:
- \(\alpha = \frac{1 + \sqrt{5}}{2}\)
- \(\beta = \frac{1 - \sqrt{5}}{2}\)
Mathematical Proof
Mathematical proof is a method of verifying statements using logical reasoning grounded in mathematical principles. In mathematical problems like verifying Binet's formula, proof provides certainty and a structured way to demonstrate the truth of statements.The process often involves:
- Defining a hypothesis based on given assumptions.
- Employing logical steps that follow mathematical laws and operations.
- Utilizing known theorems and previous results.
Recurrence Relations
Recurrence relations define sequences where each term is a function of preceding terms. This concept is heavily used in finding patterns and constructing sequences like the Fibonacci sequence.The basic recurrence relation for Fibonacci numbers is:\[F_n = F_{n-1} + F_{n-2}\]Starting with initial conditions \(F_0 = 0\) and \(F_1 = 1\), each term is derived by adding the two preceding terms, creating the sequence: 0, 1, 1, 2, 3, 5, 8, etc.Recurrence relations are crucial because:
- They simplify complex sequences into manageable calculations.
- They help in developing algorithms for both theoretical and practical problems.
- By understanding the underpinning relationship, they make mathematical modeling more accessible.
Other exercises in this chapter
Problem 50
Consider the recurrence relation \(c_{n}=c_{\lfloor n / 2\rfloor}+c_{\lfloor(n+1) / 2 j}+2,\) where \(c_{1}=0\). Compute \(c_{3}\) and \(c_{4}\)
View solution Problem 51
Consider the recurrence relation \(c_{n}=c_{[n / 2]}+c_{[L n+1 / 2]}+2,\) where \(c_{1}=0\) Solve the recurrence relation when \(n\) is a power of \(2 .\)
View solution Problem 51
Consider the recurrence relation \(c_{n}=c_{\lfloor n / 2\rfloor}+c_{\lfloor(n+1) / 2 j}+2,\) where \(c_{1}=0\). Solve the recurrence relation when \(n\) is a p
View solution Problem 51
That \(b_{n}=F_{n} .\) It is called the Binet form of the \(n\) th Fibonacci number, after the French mathematician JacquesPhillipe-Marie Binet.) With \(\alpha\
View solution