Problem 12 - Complex Analysis for Mathematics and Engineering

Question

(a) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=2\) and \(y[1]=-1\). (b) Solve \(y[n+2]+y[n+1]+y[n]=0\) with \(y[0]=0\) and \(y[1]=\sqrt{3}\).

Step-by-Step Solution

Answer

Both sequences are periodic with period 3: (a) 2, -1, -1; (b) 0, \(\sqrt{3}\), -\(\sqrt{3}\).

Step 1: Establish the Recursive Formula

The given difference equation is \( y[n+2] + y[n+1] + y[n] = 0 \). This implies each term can be expressed in terms of the previous two terms: \( y[n+2] = -y[n+1] - y[n] \).

Step 2: Solve for the First Sequence (Part a)

We have the initial conditions \( y[0] = 2 \) and \( y[1] = -1 \). To find \( y[2] \):\[ y[2] = -y[1] - y[0] = -(-1) - 2 = -1 \]Then find \( y[3] \):\[ y[3] = -y[2] - y[1] = -(-1) - (-1) = 2 \]And \( y[4] \):\[ y[4] = -y[3] - y[2] = -(2) - (-1) = -1 \]This cycle (2, -1, -1) continues, so the sequence is periodic with period 3.

Step 3: Solve for the Second Sequence (Part b)

For this sequence, we have the initial conditions \( y[0] = 0 \) and \( y[1] = \sqrt{3} \). Find \( y[2] \):\[ y[2] = -y[1] - y[0] = -\sqrt{3} - 0 = -\sqrt{3} \]Then find \( y[3] \):\[ y[3] = -y[2] - y[1] = -(-\sqrt{3}) - \sqrt{3} = 0 \]And \( y[4] \):\[ y[4] = -y[3] - y[2] = -0 - (-\sqrt{3}) = \sqrt{3} \]This cycle (0, \sqrt{3}, -\sqrt{3}) continues, so the sequence is periodic with period 3.

Key Concepts

Linear Recursive Sequences

A linear recursive sequence is one in which each term is a linear function of the previous terms. In simpler words, to find the next number, you use specific previous numbers in a set pattern. The equation given in the problem, \( y[n+2] + y[n+1] + y[n] = 0 \), is a classic example. Here, each term in the sequence depends on the two preceding terms. This is expressed as:

• \( y[n+2] = -y[n+1] - y[n] \)

This particular sequence is termed "second-order" because the next term hinges on the two prior terms. Linear recursive sequences are everywhere—from financial calculations to natural patterns. Understanding these helps in predicting future values and identifying repetitive cycles within sequences.

Initial Conditions in Recursive Sequences

For every recursive sequence, initial conditions are crucial because they set the starting point. They determine what numbers we start with and influence the entire sequence thereafter. In part (a) of the problem, our initial conditions are:

• \( y[0] = 2 \)
• \( y[1] = -1 \)

In part (b), they are:

• \( y[0] = 0 \)
• \( y[1] = \sqrt{3} \)

These initial conditions guide the sequence's path under the recursive formula. Without them, solving the sequence would be impossible. For any recursive sequence, these conditions are just as important as the formula itself—they're like the starting gear in a machine, getting everything in motion.

Periodic Sequences

Each sequence from the problem exhibits periodicity, meaning they repeat after a certain number of steps. For the sequence in part (a), after determining several terms, we find it repeats every three steps. The sequence pattern becomes:

• 2, -1, -1, 2, -1, -1...

This is termed as having a period of 3. The sequence repeats itself endlessly in this rhythm. Likewise, in part (b), the sequence:

• 0, \( \sqrt{3} \), \( -\sqrt{3} \), 0, \( \sqrt{3} \), \( -\sqrt{3} \)...

Also has a period of 3.
Understanding periodic sequences is important because they highlight predictable cycles. They are found in numerous areas like sound waves and even in day-to-day patterns. Recognizing and understanding these cycles can simplify complex problems by pointing out repeating trends.