Log In Start studying!

Select your language

Suggested languages for you:
StudySmarter - The all-in-one study app.
4.8 • +11k Ratings
More than 3 Million Downloads
Free
|
|

Second Order Recurrence Relation

Second Order Recurrence Relation

The Characteristic Technique of solving second-order recurrence relations is similar to that of solving first-order recurrence relations. It involves deriving the complementary function then finding a suitable particular solution to solve for the closed-form of a given second-order recurrence relation. The Fibonacci sequence is a second order recurrence relation that can be solved using the Characteristic technique to find its closed form equation.

Meaning of Second order recurrence relations

Whenever you are describing a relation of events that require any information from different time positions, you are talking about recurrence relations. Now, second order recurrence relations are relations that require information two steps behind to get the information you want.

Second-order recurrence relations are recurrence relations of the form \[u_{n+2}=Au_{n+1}+B u_{n}+f(n),\] for all integers \(n\) greater than some fixed integer, \(A\) and \(B\) are constants and \(f(n)\) is a polynomial.

Examples of second order recurrence relations are,

  • \(u_{n+2}=2u_{n+1}-u_{n}+3\),
  • \(u_{n}=u_{n-1}-4u_{n-2}\),
  • \(u_{n+1}=-4u_{n}+7u_{n-1}+n^2\).

Second order recurrence relations are classified into homogeneous and non-homogeneous recurrence relations.

Homogeneous second order recurrence relations

Homogeneous second order relations are relations that only show a relation between the terms of the sequence at different iterations.

Homogeneous second-order recurrence relations are of the form\[u_{n+2}=Au_{n+1}+Bu_{n}\] for all integers \(n\) greater than some fixed integer, \(A\) and \(B\) are constants.

Examples of homogeneous second order recurrence relations are,

  • \(u_{n+2}=2u_{n+1}-u_{n}\),
  • \(u_{n}=u_{n-1}-4u_{n-2}\),
  • \(u_{n+1}=-4u_{n}+7u_{n-1}\).

Non-homogeneous second order recurrence relations

Non-homogeneous second order relations are relations that show a relation between the terms of the sequence at different iterations having a bit of extra information, generally a polynomial in terms of \(n\). In fact, this is the general definition introduced at the very first paragraph of this article.

Non- homogeneous second-order recurrence relations are recurrence relations of the form \[u_{n+2}=Au_{n+1}+B u_{n}+f(n),\] for all integers \(n\) greater than some fixed integer, \(A\) and \(B\) are constants and \(f(n)\) is a polynomial.

Examples of non-homogeneous second order recurrence relations are,

  • \(u_{n+2}=2u_{n+1}-u_{n}+3\),
  • \(u_{n}=u_{n-1}-4u_{n-2}+n+1\),
  • \(u_{n+1}=9u_{n}+3u_{n-1}-7n^2\)

Solving second order recurrence relations

When solving a second order recurrence relation of the form

\[u_{n+2}=Au_{n+1}+B u_{n}+f(n),\]

we look for an expression of the \(n^{\text{th}}\) term, that takes the form

\[u_{n}=c(n)+p(n),\]

where \(c(n)\) is the complementary function and \(p(n)\) is the particular function.

The very first step to solving second order recurrence relations, is to solve its homogeneous part, also called the reduced equation. Hiding \(f(n)\) will lead you to the homogenous part of a recurrence relation,

\[u_{n+2}=Au_{n+1}+B u_{n}, \]

and solving this part would require you to find what we call the complementary function \(c(n).\)

Now, to find the complementary function, we proceed as follow. We are looking for an expression of the form \(u_{n}=r^n\) where \(u_{n}\) satisfies

\[u_{n+2}=Au_{n+1}+B u_{n}. \]

Substituting in will lead to

\[\begin{align} u_{n+2}&=Au_{n+1}+Bu_{n} \\ r^{n+2}&=Ar^{n+1}+Br^{n}\\ r^2-Ar-B&=0\end{align}\]

\(r^2-Ar-B=0\) is called the characteristic equation and the number of solutions it has will determine the general form of the complementary function \(c(n).\)

We distinguish three cases for the characteristic equation \(r^2-Ar-B=0.\)

  • If \(r^2-Ar-B=0\) has two distinct real solutions \(r_1\) and \(r_2\), then \[u_n=Cr_{1}^n+Dr_{2}^n,\] for some constants \(C\) and \(D\).
  • If \(r^2-Ar-B=0\) has a double root \(r\), then \[u_n=Cr^n+Dnr^n.\]
  • If \(r^2-Ar-B=0\) has two complex roots \(z_1\) and \(z_2\), then \[u_n=C z_1^n+Dz_2^n,\] for some constants \(C\) and \(D\).

As for the particular solution \(p(n)\), it takes the form of the polynomial \(f(n).\)

Now, let's get the job done and recap the steps of calculation!

Step 1. Find the reduced equation by setting \(f(n)=0\).

For homogenous recurrence relations the reduced equation is the same as the recurrence relation equation. This gives you an equation of the form \(u_{n+2}=Au_{n+1}+Bu_{n}\).

Step 2. Find the characteristic equation and solve for \(r\).

Step 3. Find the complementary function using the values of the \(r\).

Real and distinct roots \(r_{1}\) and \(r_{2}\) \(c(n)=Cr_{1}^{n}+D r_{2}^{n}\)
Repeated real roots \(r_{1}=r_{2}=r\)\(c(n)=Cr^{n}+Dnr^{n}\)
Complex roots \(r_1=z_1\) and \(r_{2}=z_2\)\(c(n)=C z_1^n+D z_2^n \)

Step 4. Find the general form of particular solution and substitute in \(p(n)=u_{n}\) into the original equation and solve for unknowns.

Step 5. Using the initial value given in the question, find the value of \(C\) and \(D\).

Examples are always a good idea to understand the topic, so here we go!

Example of Solving a Second Order Non-Homogenous Recurrence Relation with Real Distinct Roots

Solve the recurrence relation \(u_{n+2}=2u_{n+1}+3u_{n}+4n+12\) with initial values \(u_{1}=-1\) and \(u_{2}=16\).

Solution

Step 1. Find the reduced equation by setting \(f(n)=0\), to get

$$u_{n+2}=2u_{n+1}+3u_n.$$

Step 2. Find the characteristic equation and solve for \(r\).

The characteristic equation is given by \(r^2-2r-3=0,\) solving it we get \(r_1=-1\) and \(r_2=3.\)

Step 3. Find the complementary function \(c(n).\)

\[c(n)=Cr_1^{n}+Dr_2^{n}=C(-1)^n+D 3^n\]

Step 4. Find the form of particular solution and substitute in \(p(n)=u_{n}\) into the original equation and solve for unknowns.

Since \(f(n)=4n+12\), the particular solution takes the form \(p(n)=an+b\).

Set \(p(n)=u_n=an+b\), therefore, \(p(n+1)=u_{n+1}=a(n+1)+b\) and \(p(n+2)=u_{n+2}=a(n+2)+b\).

Substituting these into the original equation gives us

\[\begin{align} u_{n+2}&=2u_{n+1}+3u_n+4n+12 \\ a(n+2)+b&=2(a(n+1)+b)+3(an+b)+4n+12 \\ an+2a+b&=2an+2a+2b+3an+3b+4n+12 \end{align}\]

To solve for \(a\) and \(b\), you compare coefficients.

Comparing coefficients of \(n\) gives,

\[\begin{align} a&=2a+3a+4 \\ a&=-1 \end{align}\]

Comparing constant terms gives,

\[\begin{align} 2a+b&=2a+2b+3b+12 \\ b&=-3 \end{align}\]

Therefore, the particular solution is \(p(n)=-n-3.\)

The general solution is thus \(u_n=C(-1)^n+D3^{n}-n-3.\)

Step 5. Using the initial values given in the question, find the values of \(C\) and \(D\).

Since the initial values are \(u_{1}=-1\) and \(u_{2}=16\), we have

\[\begin{align} u_1=-1&=C(-1)+D(3)-1-3 \\ -C+3D&=3\end{align}\]

\[\begin{align} u_2=16&=C(-1)^2+3^2D-2-3 \\ C+9D&=21 \end{align}\]

Solving the above equations simultaneously we get, \(C=3\) and \(D=2\).

Therefore the solution is the closed-form equation,

$$u_n=3\times (-1)^n+2\times 3^n -n-3.$$

Example of Solving a Second-Order Homogenous Recurrence Relation with Repeated Roots

Solve the recurrence relation \(u_{n+2}=6u_{n+1}-9u_{n}\) with initial values \(u_{1}=1\) and \(u_{2}=4\).

Solution

Step 1. Find the reduced equation.

Since this is a homogeneous equation, we have $$u_{n+2}=6u_{n+1}-9u_{n}.$$

Step 2. Find the characteristic equation and solve for \(r\).

The characteristic equation is given by,

\[r^2-6r-9=0\]

Therefore, \(r_1=r_2=r=3.\)

Step 3. Find the complementary function.

Since we have repeated roots, the complementary function is given by,

\[\begin{align} c(n)&= Cr^n+Dnr^n=C\times 3^n+D n\times 3^n \end{align}\]

Step 4. Since \(f(n)=0\) there is no particular solution.

Therefore, the general solution is \(u_{n}=(C+Dn)\times3^{n}\).

Step 5. Using the initial values given, we find the values of \(C\) and \(D\).

Since \(u_{1}=1\) and \(u_{2}=4\), we have

\[\begin{align} u_1&=1=3(C+D)=3C+3D\\ u_{2}&=4=9(C+2D)=9C+18D\end{align}\]

Solving simultaneously gives,

$$C=\frac{2}{9}, D=\frac{1}{9}.$$ Therefore,

$$ u_{n}=\left(\frac{2}{9}+\frac{n}{9}\right)\times3^{n}.$$

Example of Solving a Second-Order Homogenous Recurrence Relation with Complex Roots

Solve the recurrence relation \(u_{n+2}=8u_{n+1}-41u_{n}\) with initial values \(u_{1}=24\) and \(u_{2}=-54\).

Solution

Step 1. Find the reduced equation.

Since this is a homogeneous equation, we have $$u_{n+2}=8u_{n+1}-41u_n.$$

Step 2. Find the characteristic equation and solve for \(r\).

The characteristic equation is given by \[r^2-8r-41=0,\] thus \(r_1=z_1=4+5i\) and \(r_2=z_2=4-5i\).

Step 3. Find the complementary function.

\[\begin{align} c(n)&=Cz_{1}^{n}+Dz_{2}^{n} \\ &=C(4+5i)^n+D(4-5i)^n. \end{align}\]

Step 4. Find the particular solution.

Since \(f(n)=0\), there is no particular solution.

Now we have a general solution, \(u_n=C(4+5i)^n+D(4-5i)^n\).

Step 5. Using the initial values given in the question, find the values of \(C\) and \(D.\)

Since the initial values are \(u_{1}=24\) and \(u_{2}=-54\), we have

\[\begin{align}u_1&=24=C(4+5i)+D(4-5i)\\ u_2&=-54=C(4+5i)^2+D(4-5i)^2 \end{align}\]

Solving simultaneously, gives \(C=3\) and \(D=3\).

Therefore the solution is the closed-form equation,

$$u_n=3(4+5i)^n+3(4-5i)^n.$$

Second Order Recurrence Relation - Key takeaways

  • Second-order recurrence relations are ones in which each term of the sequence is a function of the two previous term and are of the form \(u_{n+2}=Au_{n+1}+Bu_{n}+f(n)\) where \(f(n)\) is a polynomial and \(A\) and \(B\) are constants.
  • Second-order recurrence relations are called homogenous if \(f(n)=0\) and non-homogenous otherwise.
  • Solving second-order recurrence relations involves finding the closed-form solution.
  • The method we use to solve these recurrence relations is called the Characteristic Technique and is summarized in the following steps,
    • Step 1. Find the reduced equation by setting \(f(n)=0\).
    • Step 2. Find the characteristic equation and solve for \(r\).
    • Step 3. Find the complementary function.
    • Step 4. Find the particular solution and substitute in \(p(n)=u_{n}\) into the original equation and solve for unknown.
    • Step 5. Now you have a general form for the solution, use the initial values given in the question to find the remaining unknowns.

Frequently Asked Questions about Second Order Recurrence Relation

We can solve second-order recurrence relations using the Characteristic Technique.

Second-order recurrence relations are used to model recursive relationships or to iteratively define a sequence.

The Fibonacci sequence is an example of a second order recurrence relation that can be written as: Fn = Fn-1 - Fn-2.

A second-order recurrence relation is a formula for the nth term in a sequence as a function of the previous two terms. They take the form u= Aun-1 + Aun-2 + f(n), where A and Bare constants, f(n) is a polynomial function of n, and uis the nth term of the sequence. 

The degree of a recurrence relation is the difference between the highest and lowest index of n. 

Final Second Order Recurrence Relation Quiz

Question

What would be the form for the particular solution of the recurrence relation \(u_{n+2}=u_{n+1}+5u_n+n^2+2n+1\)?

Show answer

Answer

\(p(n)=an^2+bn+c\).

Show question

Question

Which is the correct form for the particular solution of the following recurrence relation? \(u_n=3u_{n-1}+u_{n-2}+3\times 4^n\).

Show answer

Answer

\(p(n)=3a+b\).

Show question

Question

What is the method used to solve second-order recurrence relations?

Show answer

Answer

The Characteristic Technique

Show question

Question

How many initial values/boundary conditions do you need to solve a second-order recurrence relation?

Show answer

Answer

2

Show question

Question

How do you find the reduced equation of a recurrence relation of the form \(u_{n+2} =a_1u_{n+1}+a_2u_n+f(n)\)?

Show answer

Answer

Set \(f(n) = 0\).

Show question

Question

Which term best describes the following recurrence relation? \(u_{n+2}=2u_{n+1}+3u_n+4n+12\).

Show answer

Answer

Homogenous second-order recurrence relation.

Show question

Question

What form would the particular solution of the following recurrence relation take? \(u_{n+2}=2u_{n+1}+u_n+1\).

Show answer

Answer

\(p(n)=a\).

Show question

Question

Which term best describes the following recurrence relation? \(u_{n+2}=6u_{n+1}-9u_n\).

Show answer

Answer

Homogenous second-order recurrence relation.

Show question

Question

Which term best describes the following recurrence relation? \(u_{n+2}=2u_{n+1}+3u_n+n^2\).

Show answer

Answer

Homogenous first-order recurrence relation.

Show question

Question

Which of the following is the correct form for the particular solution of the recurrence relation \(u_{n+2}=6u_{n+1}-9u_n\)?

Show answer

Answer

There is no particular solution i.e. \(p(n) = 0\) since it is a homogenous recurrence relation.

Show question

More about Second Order Recurrence Relation
60%

of the users don't pass the Second Order Recurrence Relation quiz! Will you pass the quiz?

Start Quiz

Discover the right content for your subjects

No need to cheat if you have everything you need to succeed! Packed into one app!

Study Plan

Be perfectly prepared on time with an individual plan.

Quizzes

Test your knowledge with gamified quizzes.

Flashcards

Create and find flashcards in record time.

Notes

Create beautiful notes faster than ever before.

Study Sets

Have all your study materials in one place.

Documents

Upload unlimited documents and save them online.

Study Analytics

Identify your study strength and weaknesses.

Weekly Goals

Set individual study goals and earn points reaching them.

Smart Reminders

Stop procrastinating with our study reminders.

Rewards

Earn points, unlock badges and level up while studying.

Magic Marker

Create flashcards in notes completely automatically.

Smart Formatting

Create the most beautiful study materials using our templates.

Sign up to highlight and take notes. It’s 100% free.

Get FREE ACCESS to all of our study material, tailor-made!

Over 10 million students from across the world are already learning smarter.

Get Started for Free
Illustration