Prerequisite – Solving Recurrences, Different types of recurrence relations and their solutions, Practice Set for Recurrence Relations
The sequence which is defined by indicating a relation connecting its general term an with an-1, an-2, etc is called a recurrence relation for the sequence.
Types of recurrence relations
- First order Recurrence relation :- A recurrence relation of the form : an = can-1 + f(n) for n>=1
where c is a constant and f(n) is a known function is called linear recurrence relation of first order with constant coefficient. If f(n) = 0, the relation is homogeneous otherwise non-homogeneous.
Example :- xn = 2xn-1 – 1, an = nan-1 + 1, etc.
Question :- Solve the recurrence relation T(2k) = 3T(2k-1) + 1, T(1) = 1.
Let T(2k) = ak. Therefore, ak = 3ak-1 + 1
Multiplying by xk and then taking sum,
Σakxk = 3Σak-1xk + Σxk ——> (1)
Σak-1xk = [a0x + a1x2 + ……]
= x[a0 + a1x + ……] = x[G(x)]
G(x) – 3xG(x) – x/(1-x) = 0
G(x)(1-3x) – x/(1-x) = 0
G(x) = x/[(1-x)(1-3x)] = A/(1-x) + B/(1-3x)
–> A = -1/2 and B = 3/2
G(x) = (3/2)Σ(3x)k – (1/2)Σ(x)k
Coefficient of xk is, ak = (3/2)3k – (1/2)1k
So, ak = [3k+1 – 1]/2.
- Second order linear homogeneous Recurrence relation :- A recurrence relation of the form
cnan + cn-1an-1 + cn-2an-2 = 0 ——> (1)
for n>=2 where cn, cn-1 and cn-2 are real constants with cn != 0 is called a second order linear homogeneous recurrence relation with constant coefficients.
Solution to this is in form an = ckn where c, k!=0
Putting this in (1)
cnckn + cn-1ckn-1 + cn-2ckn-2 = 0
cnk2 + cn-1k + cn-2 = 0 —–> (2)
Thus, an = ckn is solution of (1) if k satisfies quadratic equation (2). This equation is called characteristic equation for relation (1).
Now three cases arises,
Case 1 : If the two roots k1, k2 of equation are real and distinct then, we take
an = A(k1)n + B(k2)n as general solution of (1) where A and B are arbitrary real constants.
Case 2 : If the two roots k1, k2 of equation are real and equal, with k as common value then, we take
an = (A + Bn)kn as general solution of (1) where A and B are arbitrary real constants.
Case 3 : If the two roots k1 and k2 of equation are complex then, k1 and k2 are complex conjugate of each other i.e k1 = p + iq and k2 = p – iq and we take
an = rn(Acosnθ + Bsinnθ) as general solution of (1) where A and B are arbitrary complex constants, r = |k1| = |k2| = √p2 + q2 and θ = tan-1(q/p).
Question :- Solve the recurrence relation an + an-1 – 6an-2 = 0 for n>=2 given that a0 = -1 and a1 = 8.
Here coefficients of an, an-1 and an-2 are cn = 1, cn-1 = 1 and cn-2 = -6 respectively. Hence, characteristic equation is
k2 + k – 6 or (k + 3)(k – 2) = 0 ——> (1)
The roots of (1) are k1 = -3 and k2 = 2 which are real and distinct. Therefore, general solution is
an = A(-3)n + B(2)n
where A and B are arbitrary constants. From above we get, a0 = A + B and a1 = -3A + 2B
A + B = -1
-3A + 2B = 8
Solving these we get A = -2 and B = 1
Therefore, an = -2(-3)n + (2)n
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
- Mathematics | Introduction to Propositional Logic | Set 2
- Different types of recurrence relations and their solutions
- Mathematics | Generating Functions – Set 2
- Last Minute Notes – Computer Networks
- Mathematics | Combinatorics Basics
- Mathematics | Rings, Integral domains and Fields
- Introduction to Mojette transform
- Univariate, Bivariate and Multivariate data and its analysis
- Proof that vertex cover is NP complete
- Graph measurements: length, distance, diameter, eccentricity, radius, center