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.
Explore our app and discover over 50 million learning materials for free.
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Jetzt kostenlos anmeldenNie wieder prokastinieren mit unseren Lernerinnerungen.
Jetzt kostenlos anmeldenThe 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.
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,
Second order recurrence relations are classified into homogeneous and non-homogeneous 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,
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,
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 \(r^2-Ar-B=0.\)
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!
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,
$$u_n=3\times (-1)^n+2\times 3^n -n-3.$$
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,
$$ u_{n}=\left(\frac{2}{9}+\frac{n}{9}\right)\times3^{n}.$$
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,
$$u_n=3(4+5i)^n+3(4-5i)^n.$$
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 un = Aun-1 + Aun-2 + f(n), where A and Bare constants, f(n) is a polynomial function of n, and un is the nth term of the sequence.
The degree of a recurrence relation is the difference between the highest and lowest index of n.
What would be the form for the particular solution of the recurrence relation \(u_{n+2}=u_{n+1}+5u_n+n^2+2n+1\)?
\(p(n)=an^2+bn+c\).
Which is the correct form for the particular solution of the following recurrence relation? \(u_n=3u_{n-1}+u_{n-2}+3\times 4^n\).
\(p(n)=3a+b\).
What is the method used to solve second-order recurrence relations?
The Characteristic Technique
How many initial values/boundary conditions do you need to solve a second-order recurrence relation?
2
How do you find the reduced equation of a recurrence relation of the form \(u_{n+2} =a_1u_{n+1}+a_2u_n+f(n)\)?
Set \(f(n) = 0\).
Which term best describes the following recurrence relation? \(u_{n+2}=2u_{n+1}+3u_n+4n+12\).
Homogenous second-order recurrence relation.
Already have an account? Log in
Open in AppThe first learning app that truly has everything you need to ace your exams in one place
Sign up to highlight and take notes. It’s 100% free.
Save explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of StudySmarter.
Already have an account? Log in
Already have an account? Log in
The first learning app that truly has everything you need to ace your exams in one place
Already have an account? Log in