Q. Fibonacci generating function \( \frac{x}{1 – x – x^2} \).

Q: How do you derive the generating function \frac{x}{1-x-x^2} for the Fibonacci numbers?

Let F(x) = \sum_{n \ge 0} F_n x^n with F_0=0, F_1=1. Multiply by x and x^2 and subtract: F(x)-xF(x)-x^2F(x)=x. Solve for F(x) to get F(x) = \frac{x}{1-x-x^2}.

Q: How do you get Binet's formula from the generating function?

Do partial fractions using roots \phi = \frac{1+\sqrt{5}}{2}, \psi = \frac{1-\sqrt{5}}{2}. Decompose into geometric series to obtain F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}.

Q: How can I expand \frac{x}{1-x-x^2} as a power series?

Write \frac{x}{1-x-x^2} = \frac{1}{\sqrt{5}}\left(\frac{1}{1-\phi x} - \frac{1}{1-\psi x}\right) and expand each as \sum_{n \ge 0} \frac{(\phi^n - \psi^n)x^n}{\sqrt{5}} to recover F_n as coefficients.

Q: What is the radius of convergence of this generating function?

Singularities are at x=\phi and x=\psi; the nearest to 0 is \psi with |\psi| = \frac{1}{\phi} \approx 0.618. So the radius of convergence is \frac{1}{\phi}.

Q: How does the generating function give asymptotics for F_n?

The dominant pole at x = \psi implies exponential growth: F_n \sim \frac{\phi^n}{\sqrt{5}}. More precisely, F_n = (\phi^n - \psi^n)/\sqrt{5} so the \phi^n term dominates as n \to \infty.

Q: How to prove Fibonacci sum identities using the generating function?

Use algebraic manipulations: for example, \sum_{k=0}^n F_k = F_{n+2}-1 follows from (1-x)F(x)=\sum_{n \ge 0}(F_n-F_{n-1})x^n and comparing partial sums or extracting coefficients.

Q: Why is partial fraction decomposition useful here?

Because the denominator factors into linear terms over \mathbb{R}, partial fractions express F(x) as a sum of geometric series. That gives closed-form coefficients and simplifies proofs of identities and asymptotics.

Answer

Let \(G(x)=\sum_{n\ge 0}F_nx^n\) with \(F_0=0,\ F_1=1\). Then
\[
G(x)=x+\sum_{n\ge 2}F_nx^n
= x+\sum_{n\ge 2}(F_{n-1}+F_{n-2})x^n
= x + x\sum_{n\ge 2}F_{n-1}x^{\,n-1} + x^2\sum_{n\ge 2}F_{n-2}x^{\,n-2}
= x + xG(x) + x^2G(x),
\]
so \(G(x)(1-x-x^2)=x\) and hence
\[
G(x)=\frac{x}{1-x-x^2}.
\]

With \(\varphi=\dfrac{1+\sqrt{5}}{2}\) and \(\psi=\dfrac{1-\sqrt{5}}{2}\) one gets the partial fraction form
\[
G(x)=\frac{1}{\sqrt{5}}\!\left(\frac{1}{1-\varphi x}-\frac{1}{1-\psi x}\right),
\]
so the coefficients are the Fibonacci numbers
\[
F_n=\frac{\varphi^n-\psi^n}{\sqrt{5}}.
\]

Detailed Explanation

Define the Fibonacci sequence and its ordinary generating function. Let the Fibonacci numbers be defined by F_0 = 0, F_1 = 1, and for n ≥ 2, F_n = F_{n-1} + F_{n-2}. Define the generating function

\[F(x) \;=\; \sum_{n=0}^{\infty} F_n\,x^n.\]
Form the shifted series xF(x) and x^2F(x) so we can use the recurrence. Compute

\[xF(x) \;=\; \sum_{n=0}^{\infty} F_n\,x^{n+1} \;=\; \sum_{n=1}^{\infty} F_{n-1}\,x^{n},\]

\[x^2F(x) \;=\; \sum_{n=0}^{\infty} F_n\,x^{n+2} \;=\; \sum_{n=2}^{\infty} F_{n-2}\,x^{n}.\]
Subtract the shifted series from F(x) to incorporate the recurrence. Consider

\[F(x) – xF(x) – x^2F(x) \;=\; \sum_{n=0}^{\infty} \bigl(F_n – F_{n-1} – F_{n-2}\bigr)\,x^{n},\]

where by convention F_{-1} = 0 so the n = 0 and n = 1 terms are handled by the initial values. For n ≥ 2 the recurrence gives F_n – F_{n-1} – F_{n-2} = 0, so only the first two terms contribute.

Evaluate the left series directly from the first terms: F_0 = 0 and F_1 = 1, so the right-hand side reduces to x. Therefore

\[F(x) – xF(x) – x^2F(x) \;=\; x.\]
Solve for F(x). Factor the left-hand side:

\[F(x)\bigl(1 – x – x^2\bigr) \;=\; x.\]

Hence

\[F(x) \;=\; \dfrac{x}{1 – x – x^2}.\]

This shows the ordinary generating function of the Fibonacci sequence is x/(1 – x – x^2).
To extract an explicit formula for the coefficients (Binet’s formula), factor the denominator. Solve r^2 – r – 1 = 0 to find the roots

\[\varphi \;=\; \dfrac{1 + \sqrt{5}}{2}, \qquad \psi \;=\; \dfrac{1 – \sqrt{5}}{2}.\]

Then

\[1 – x – x^2 \;=\; (1 – \varphi x)(1 – \psi x),\]

because \varphi + \psi = 1 and \varphi\psi = -1.
Perform partial fraction decomposition. Write

\[\dfrac{x}{(1 – \varphi x)(1 – \psi x)} \;=\; \dfrac{A}{1 – \varphi x} \;+\; \dfrac{B}{1 – \psi x}.\]

Multiply through by the denominator to get

\[x \;=\; A(1 – \psi x) + B(1 – \varphi x) \;=\; (A + B) + \bigl(-A\psi – B\varphi\bigr)x.\]

Equate coefficients of powers of x: from the constant term, 0 = A + B, so B = -A. From the x-term, 1 = -A\psi – B\varphi. Substitute B = -A to obtain

\[1 \;=\; -A\psi + A\varphi \;=\; A(\varphi – \psi).\]

Since \varphi – \psi = \sqrt{5}, we get A = 1/\sqrt{5} and B = -1/\sqrt{5}.
Express F(x) using the geometric-series expansion for each partial fraction. We have

\[F(x) \;=\; \dfrac{1}{\sqrt{5}}\cdot\dfrac{1}{1 – \varphi x} \;-\; \dfrac{1}{\sqrt{5}}\cdot\dfrac{1}{1 – \psi x}.\]

Expand each as a power series (valid for |x| small enough):

\[\dfrac{1}{1 – \varphi x} \;=\; \sum_{n=0}^{\infty} \varphi^{\,n} x^{n}, \qquad
\dfrac{1}{1 – \psi x} \;=\; \sum_{n=0}^{\infty} \psi^{\,n} x^{n}.\]

Therefore

\[F(x) \;=\; \dfrac{1}{\sqrt{5}} \sum_{n=0}^{\infty} \bigl(\varphi^{\,n} – \psi^{\,n}\bigr) x^{n}.\]
Read off the coefficients to obtain the closed-form formula for Fibonacci numbers. For n ≥ 0,

\[F_n \;=\; \dfrac{\varphi^{\,n} – \psi^{\,n}}{\sqrt{5}}.\]

In particular F_0 = 0 and F_1 = 1, and the generating function is

\[F(x) \;=\; \sum_{n=0}^{\infty} F_n x^n \;=\; \dfrac{x}{1 – x – x^2}.\]

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FAQs

How do you derive the generating function \(\frac{x}{1-x-x^2}\) for the Fibonacci numbers?

Let \(F(x) = \sum_{n \ge 0} F_n x^n\) with \(F_0=0, F_1=1\). Multiply by \(x\) and \(x^2\) and subtract: \(F(x)-xF(x)-x^2F(x)=x\). Solve for \(F(x)\) to get \(F(x) = \frac{x}{1-x-x^2}\).

How do you get Binet's formula from the generating function?

Do partial fractions using roots \(\phi = \frac{1+\sqrt{5}}{2}, \psi = \frac{1-\sqrt{5}}{2}\). Decompose into geometric series to obtain \(F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}\).

How can I expand \(\frac{x}{1-x-x^2}\) as a power series?

Write \(\frac{x}{1-x-x^2} = \frac{1}{\sqrt{5}}\left(\frac{1}{1-\phi x} - \frac{1}{1-\psi x}\right)\) and expand each as \(\sum_{n \ge 0} \frac{(\phi^n - \psi^n)x^n}{\sqrt{5}}\) to recover \(F_n\) as coefficients.

What is the radius of convergence of this generating function?

Singularities are at \(x=\phi\) and \(x=\psi\); the nearest to 0 is \(\psi\) with \(|\psi| = \frac{1}{\phi} \approx 0.618\). So the radius of convergence is \(\frac{1}{\phi}\).

How does the generating function give asymptotics for F_n?

The dominant pole at \(x = \psi\) implies exponential growth: \(F_n \sim \frac{\phi^n}{\sqrt{5}}\). More precisely, \(F_n = (\phi^n - \psi^n)/\sqrt{5}\) so the \(\phi^n\) term dominates as \(n \to \infty\).

How do I get the generating function for shifted Fibonacci sequences F_{n+k}?

How to prove Fibonacci sum identities using the generating function?

Use algebraic manipulations: for example, \(\sum_{k=0}^n F_k = F_{n+2}-1\) follows from \((1-x)F(x)=\sum_{n \ge 0}(F_n-F_{n-1})x^n\) and comparing partial sums or extracting coefficients.

Why is partial fraction decomposition useful here?

Because the denominator factors into linear terms over \(\mathbb{R}\), partial fractions express \(F(x)\) as a sum of geometric series. That gives closed-form coefficients and simplifies proofs of identities and asymptotics.

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