Degenerated third order linear recurrences

Sep 3, 2002

Mario Catalani

In this blog, we'll delve into the topic of third order linear recurrences. When considering a third order linear recurrence, we follow the logic of Un=a1Un−1+a2Un−2+a3Un−3, U 0=u0, U1=u1, U2=u2. Now, let's consider a scenario where the roots of the characteristic polynomial are real, simple, but two of them are equal in absolute value. This scenario is denoted as degenerated. Suppose {λi} are the roots, we assume that λ3>0, λ1=−λ3 and λ1< λ2< λ3. In this case, the coefficients of the recurrence assume the value a1=λ2, a2=λ²3 , a3=−λ2λ²3.

1. Limits

We apply the generalized Binet’s formula (closed form) in this case: Un=c1(−λ3)ⁿ+c2λⁿ2+c3λⁿ3. The roots {λi} as well as the initial conditions {ui} determine the elements of {ci}. Depending on the parity of n, the limit of the ratio of two consecutive terms changes. For even n, the limit is specified in equation 1 while for odd n, the equation 2 applies.

By using the Binet’s formula, we can express c1 and c3 explicitly as a function of the initial conditions. This leads us to a matrix {A}. Multiplying the inverse of A with the column vector u (with elements {ui}) gives vector c (with elements {ci}). This solution further lets us calculate the parameters of limit L1 and L2.

2. Consequences

If we define γ as equation (3), we can state the limits L1 and L2 in terms of γ. Equation 4 then shows us that the limit defined does not depend on the initial conditions, whereas Equation 5 indicates that this limit indeed depends on initial conditions.

3. Existence of a Limit

In certain scenarios with particular initial conditions, our ratio may converge. The equality L1=L2 gives us an equation which in turn can help us derive the value for u2 as a function of u0 and u1. This eventually leads to a quadratic equation in u2. On simplifying this equation, we find two solutions which dictate the values of L1 and L2. If we choose the first solution, we find L1=L2= -λ3. However, if we choose the second solution, we get L1=L2= λ3.

Conclusion

Third-order linear recurrences are a vast and complex area of study. They provide a fascinating view into the mathematical world and reveal the intricate relationships within sequences, demonstrating how the properties of such equations can have far-reaching implications. This blog presents an introduction to the degenerated third-order linear recurrences and how the limits, consequences and the existence of a limit in such cases can be studied with examples.

Sign up to AI First Newsletter

Recommended

We use our own cookies as well as third-party cookies on our websites to enhance your experience, analyze our traffic, and for security and marketing. Select "Accept All" to allow them to be used. Read our Cookie Policy.