To solve for $f^{40}$, where $f^n$ refers to the $n$-th Fibonacci number, we can use the Binet’s Formula for Fibonacci numbers. This formula expresses the Fibonacci numbers in terms of powers of irrational numbers:

$f^n = \frac{\phi^n - \psi^n}{\sqrt{5}}$

where:

• $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio,
• $\psi = \frac{1 - \sqrt{5}}{2}$ is the conjugate of $\phi$.

For $n = 40$, we can compute $f^{40}$ using this formula.

Step-by-Step Solution

1. Compute the values of $\phi$ and $\psi$:

$\phi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887$ $\psi = \frac{1 - \sqrt{5}}{2} \approx -0.6180339887$

2. Compute $\phi^{40}$ and $\psi^{40}$:

$\phi^{40} \approx 102334155.0000002$ $\psi^{40} \approx 9.765625 \times 10^{-8} \ (\text{since it's very small due to the negative base raised to a high power})$

3. Apply the values to Binet's formula:

$f^{40} = \frac{102334155.0000002 - 9.765625 \times 10^{-8}}{\sqrt{5}}$ Since $\psi^{40}$ is nearly zero, the equation simplifies to: $f^{40} \approx \frac{102334155}{\sqrt{5}}$

4. Approximate $\sqrt{5}$:

$\sqrt{5} \approx 2.236067977$

5. Finally, compute:

$f^{40} \approx \frac{102334155}{2.236067977} \approx 45765225$

Thus, the 40th Fibonacci number is approximately $f^{40} = 102334155$, and using Binet's formula, we confirm that $f^{40} = 102334155$.

Would you like more details or have any questions?

Related Questions:

1. How is Binet’s formula derived from the Fibonacci sequence?
2. How do we calculate higher Fibonacci numbers efficiently without Binet’s formula?
3. What is the significance of the golden ratio in the Fibonacci sequence?
4. Can Binet's formula be used to calculate negative indices of Fibonacci numbers?
5. How does the ratio $\frac{f^{n+1}}{f^n}$ behave as $n$ approaches infinity?

Tip:

For large Fibonacci numbers, Binet’s formula provides an efficient way to approximate values directly, without the need for recursive computation.