Math Problem Statement

Use generating functions to find an explicit formula for the Fibonacci numbers.

Solution

To find an explicit formula for the Fibonacci numbers using generating functions, follow these steps:

Step 1: Define the Fibonacci Sequence

The Fibonacci sequence is defined by the recurrence relation: Fn=Fn1+Fn2,for n2F_n = F_{n-1} + F_{n-2}, \quad \text{for } n \geq 2 with initial conditions: F0=0,F1=1F_0 = 0, \quad F_1 = 1

Step 2: Define the Generating Function

Let G(x)G(x) be the generating function for the Fibonacci sequence: G(x)=n=0FnxnG(x) = \sum_{n=0}^{\infty} F_n x^n

Step 3: Express the Recurrence Relation in Terms of G(x)G(x)

Using the recurrence relation Fn=Fn1+Fn2F_n = F_{n-1} + F_{n-2}, we have: G(x)=F0+F1x+n=2FnxnG(x) = F_0 + F_1 x + \sum_{n=2}^{\infty} F_n x^n Substitute the recurrence relation into the sum: G(x)=0+x+n=2(Fn1+Fn2)xnG(x) = 0 + x + \sum_{n=2}^{\infty} (F_{n-1} + F_{n-2}) x^n This can be split into two sums: G(x)=x+n=2Fn1xn+n=2Fn2xnG(x) = x + \sum_{n=2}^{\infty} F_{n-1} x^n + \sum_{n=2}^{\infty} F_{n-2} x^n Re-index the sums: G(x)=x+n=1Fnxn+1+n=0Fnxn+2G(x) = x + \sum_{n=1}^{\infty} F_n x^{n+1} + \sum_{n=0}^{\infty} F_n x^{n+2} This simplifies to: G(x)=x+xn=1Fnxn+x2n=0FnxnG(x) = x + x \sum_{n=1}^{\infty} F_n x^n + x^2 \sum_{n=0}^{\infty} F_n x^n Thus, we can express G(x)G(x) as: G(x)=x+x(G(x)F0)+x2G(x)G(x) = x + x(G(x) - F_0) + x^2 G(x) Since F0=0F_0 = 0, this reduces to: G(x)=x+xG(x)+x2G(x)G(x) = x + xG(x) + x^2 G(x)

Step 4: Solve for G(x)G(x)

Now, collect terms involving G(x)G(x): G(x)xG(x)x2G(x)=xG(x) - xG(x) - x^2 G(x) = x Factor out G(x)G(x): G(x)(1xx2)=xG(x)(1 - x - x^2) = x So, we have: G(x)=x1xx2G(x) = \frac{x}{1 - x - x^2}

Step 5: Find the Explicit Formula

The generating function G(x)=x1xx2G(x) = \frac{x}{1 - x - x^2} can be expanded using partial fractions. First, find the roots of the denominator 1xx21 - x - x^2. The roots are: α=1+52,β=152\alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2} So, we can express G(x)G(x) as: G(x)=15(11αx11βx)G(x) = \frac{1}{\sqrt{5}} \left(\frac{1}{1 - \alpha x} - \frac{1}{1 - \beta x}\right) Using the geometric series expansion 11r=n=0rn\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n, we get: G(x)=15(n=0αnxnn=0βnxn)G(x) = \frac{1}{\sqrt{5}} \left(\sum_{n=0}^{\infty} \alpha^n x^n - \sum_{n=0}^{\infty} \beta^n x^n\right) Thus: G(x)=n=015(αnβn)xnG(x) = \sum_{n=0}^{\infty} \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n\right) x^n Comparing this with the original generating function G(x)=n=0FnxnG(x) = \sum_{n=0}^{\infty} F_n x^n, we have: Fn=15(αnβn)F_n = \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n\right)

Conclusion

The explicit formula for the Fibonacci numbers is: Fn=15((1+52)n(152)n)F_n = \frac{1}{\sqrt{5}} \left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right)

This is known as Binet's Formula.


Would you like more details or have any questions?

Here are five questions to consider:

  1. How is the generating function related to the Fibonacci sequence?
  2. What is the significance of the roots α\alpha and β\beta in the formula?
  3. How can partial fractions be used to solve generating functions?
  4. What are the applications of Binet's formula in combinatorics?
  5. How would the generating function approach change if the Fibonacci sequence had different initial conditions?

Tip: Understanding generating functions can simplify solving complex recurrence relations in sequences.

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Math Problem Analysis

Mathematical Concepts

Generating Functions
Fibonacci Sequence
Recurrence Relations
Geometric Series

Formulas

F_n = F_{n-1} + F_{n-2}
G(x) = \sum_{n=0}^{\infty} F_n x^n
G(x) = \frac{x}{1 - x - x^2}
F_n = \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n\right)

Theorems

Binet's Formula
Geometric Series Expansion

Suitable Grade Level

College or Advanced High School (Grades 11-12)