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Fibonacci difference equation: closed form and limit

Short answer

Closed form from z-transform: x(k)=(1/√5)[((1+√5)/2)^k – ((1–√5)/2)^k]. As k→∞, x(k+2)/x(k) → φ^2 = (3+√5)/2 = 2.6180.

Consider the difference equation x(k+2)=x(k+1)+x(k) where x(0)=0, x(1)=1 and x(2)=2. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.

This is the Fibonacci equation. The data are not fully consistent: with x(0)=0x(0) = 0 and x(1)=1x(1) = 1, the equation itself gives x(2)=x(1)+x(0)=1x(2) = x(1) + x(0) = 1, not 2. So x(0)=0x(0) = 0 and x(1)=1x(1) = 1 are taken as the initial conditions. A note at the end covers the case x(2)=2x(2) = 2.

Step 1: Take the z-transform

Use Z[x(k+2)]=z2X(z)−z2x(0)−zx(1)\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1) and Z[x(k+1)]=zX(z)−zx(0)\mathcal{Z}[x(k+1)] = zX(z) - zx(0), with x(0)=0x(0) = 0 and x(1)=1x(1) = 1:

[z2X(z)−z]−zX(z)−X(z)=0\left[z^2X(z) - z\right] - zX(z) - X(z) = 0 X(z)=zz2−z−1X(z) = \frac{z}{z^2 - z - 1}

Step 2: Factorise the denominator

z2−z−1=0  ⇒  z1,2=1±52,z1=1.6180,  z2=−0.6180z^2 - z - 1 = 0 \;\Rightarrow\; z_{1,2} = \frac{1 \pm \sqrt{5}}{2}, \quad z_1 = 1.6180,\; z_2 = -0.6180

Step 3: Partial fractions of X(z)/zX(z)/z

X(z)z=1(z−z1)(z−z2)=1z1−z2[1z−z1−1z−z2],z1−z2=5\frac{X(z)}{z} = \frac{1}{(z-z_1)(z-z_2)} = \frac{1}{z_1 - z_2}\left[\frac{1}{z-z_1} - \frac{1}{z-z_2}\right], \quad z_1 - z_2 = \sqrt{5} X(z)=15[zz−z1−zz−z2]X(z) = \frac{1}{\sqrt5}\left[\frac{z}{z - z_1} - \frac{z}{z - z_2}\right]

Step 4: General (closed-form) solution

x(k)=15[(1+52)k−(1−52)k],k=0,1,2,…x(k) = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^k - \left(\frac{1-\sqrt5}{2}\right)^k\right], \quad k = 0, 1, 2, \dots

Here 1/5=0.44721/\sqrt5 = 0.4472. Check: x(0)=0x(0) = 0, and x(1)=15⋅5=1x(1) = \frac{1}{\sqrt5}\cdot\sqrt5 = 1.

First terms: 0,1,1,2,3,5,8,13,21,…0, 1, 1, 2, 3, 5, 8, 13, 21, \dots

Step 5: Limiting value of x(k+2)/x(k)x(k+2)/x(k)

Let z1=1.6180z_1 = 1.6180 and z2=−0.6180z_2 = -0.6180:

x(k+2)x(k)=z1k+2−z2k+2z1k−z2k=z12−z22(z2z1)k1−(z2z1)k\frac{x(k+2)}{x(k)} = \frac{z_1^{k+2} - z_2^{k+2}}{z_1^{k} - z_2^{k}} = \frac{z_1^2 - z_2^2\left(\dfrac{z_2}{z_1}\right)^{k}}{1 - \left(\dfrac{z_2}{z_1}\right)^{k}}

Since ∣z2/z1∣=0.382<1|z_2/z_1| = 0.382 < 1, the bracketed powers go to 0:

lim⁡k→∞x(k+2)x(k)=z12=(1+52)2=3+52=2.6180\lim_{k\to\infty}\frac{x(k+2)}{x(k)} = z_1^2 = \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{3+\sqrt5}{2} = 2.6180

Note that z12=z1+1z_1^2 = z_1 + 1 (from z2−z−1=0z^2 - z - 1 = 0), so the answer is 1.6180+1=2.61801.6180 + 1 = 2.6180.

Numerical check: x(10)/x(8)=55/21=2.619x(10)/x(8) = 55/21 = 2.619.

Note on x(2)=2x(2) = 2

If x(1)=1x(1) = 1 and x(2)=2x(2) = 2 are taken instead, the equation forces x(0)=1x(0) = 1. The sequence 1,1,2,3,5,…1, 1, 2, 3, 5, \dots is the same series shifted by one step:

x(k)=15[(1.618)k+1−(−0.618)k+1]x(k) = \frac{1}{\sqrt5}\left[(1.618)^{k+1} - (-0.618)^{k+1}\right]

The limiting ratio is still 2.61802.6180, because the ratio depends only on the dominant root.

Answer: x(k)=15[(1+52)k−(1−52)k]x(k) = \dfrac{1}{\sqrt5}\left[\left(\dfrac{1+\sqrt5}{2}\right)^k - \left(\dfrac{1-\sqrt5}{2}\right)^k\right], and lim⁡k→∞x(k+2)/x(k)=2.6180\lim_{k\to\infty} x(k+2)/x(k) = 2.6180.

✓ Answer checked by Claude Opus 5.5 (high effort)

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