Math Problem Statement

find the recurrence relation of following function- an+2 -3 a n+1 -2an=3" where-a0 = 0, a1=1

Solution

The given recurrence relation is:

an+23an+12an=3,a_{n+2} - 3a_{n+1} - 2a_n = 3,

with initial conditions a0=0a_0 = 0 and a1=1a_1 = 1.

Step 1: Solve the homogeneous recurrence relation

The homogeneous part of the recurrence is:

an+23an+12an=0.a_{n+2} - 3a_{n+1} - 2a_n = 0.

Assume a solution of the form an=rna_n = r^n. Substituting into the homogeneous equation:

rn+23rn+12rn=0.r^{n+2} - 3r^{n+1} - 2r^n = 0.

Divide through by rnr^n (for r0r \neq 0):

r23r2=0.r^2 - 3r - 2 = 0.

Factoring:

(r4)(r+1)=0.(r - 4)(r + 1) = 0.

Thus, the roots are r=4r = 4 and r=1r = -1. The general solution of the homogeneous recurrence is:

an(h)=C1(4n)+C2(1)n,a_n^{(h)} = C_1(4^n) + C_2(-1)^n,

where C1C_1 and C2C_2 are constants to be determined.

Step 2: Solve the non-homogeneous recurrence relation

The non-homogeneous equation is:

an+23an+12an=3.a_{n+2} - 3a_{n+1} - 2a_n = 3.

Assume a particular solution an(p)a_n^{(p)} is constant. Let an(p)=Aa_n^{(p)} = A. Substituting an(p)=Aa_n^{(p)} = A into the equation:

A3A2A=3.A - 3A - 2A = 3.

Simplify:

4A=3    A=34.-4A = 3 \quad \implies \quad A = -\frac{3}{4}.

Thus, the particular solution is:

an(p)=34.a_n^{(p)} = -\frac{3}{4}.

Step 3: General solution

The general solution is the sum of the homogeneous and particular solutions:

an=C1(4n)+C2(1)n34.a_n = C_1(4^n) + C_2(-1)^n - \frac{3}{4}.

Step 4: Apply initial conditions

Using a0=0a_0 = 0 and a1=1a_1 = 1:

  1. For n=0n = 0, a0=0a_0 = 0: C1(40)+C2(1)034=0.C_1(4^0) + C_2(-1)^0 - \frac{3}{4} = 0. Simplify: C1+C234=0.C_1 + C_2 - \frac{3}{4} = 0. C_1 + C_2 = \frac{3}{4}. \tag{1}

  2. For n=1n = 1, a1=1a_1 = 1: C1(41)+C2(1)134=1.C_1(4^1) + C_2(-1)^1 - \frac{3}{4} = 1. Simplify: 4C1C234=1.4C_1 - C_2 - \frac{3}{4} = 1. 4C_1 - C_2 = \frac{7}{4}. \tag{2}

Solve the system of equations (1)(1) and (2)(2):

  • Add equations (1)(1) and (2)(2): C1+C2+4C1C2=34+74.C_1 + C_2 + 4C_1 - C_2 = \frac{3}{4} + \frac{7}{4}. 5C1=104.5C_1 = \frac{10}{4}. C1=12.C_1 = \frac{1}{2}.

  • Substitute C1=12C_1 = \frac{1}{2} into (1)(1): 12+C2=34.\frac{1}{2} + C_2 = \frac{3}{4}. C2=14.C_2 = \frac{1}{4}.

Final Solution

The solution is:

an=12(4n)+14(1)n34.a_n = \frac{1}{2}(4^n) + \frac{1}{4}(-1)^n - \frac{3}{4}.

Want more details or have questions?

Related questions:

  1. How can we confirm the correctness of the solution by induction?
  2. What happens if the non-homogeneous term is a polynomial instead of a constant?
  3. Can we generalize the solution for different initial conditions?
  4. What role does the characteristic equation play in finding solutions?
  5. How do we handle cases where the roots of the characteristic equation are repeated?

Tip:

Always verify the solution by substituting it back into the original recurrence relation!

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Homogeneous and Non-Homogeneous Solutions
Characteristic Equations

Formulas

Characteristic equation: r^n for homogeneous recurrence relations
General solution: a_n = a_n^{(h)} + a_n^{(p)}
System of equations for constants based on initial conditions

Theorems

Linear Recurrence Relation Theorem
Superposition Principle for Recurrence Relations

Suitable Grade Level

Grades 10-12