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Recurrence Relations Simplified Revision Notes

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4.5.3 Recurrence Relations

A recurrence relation is an equation that defines a sequence based on previous terms. It expresses each term of the sequence as a function of its predecessors. Recurrence relations are commonly used in sequences, algorithm analysis, and mathematical modelling.

Key Methods to Solve Recurrence Relations:

  1. Substitution/Iteration: Expands the relation step-by-step to find a pattern.
  2. Characteristic Equation: Solves for linear recurrence relations by finding the roots of the characteristic equation.
  3. Generating Functions: Converts the relation into a power series for more complex cases.

Step-by-Step Method to Approach Recurrence Relations:

infoNote
  1. Identify the Recurrence Relation:
  • Determine the type of relation (linear, homogeneous, or non-homogeneous).
  • Identify the order of the relation (e.g., first-order, second-order).
  1. Determine the Initial Conditions:
  • Recurrence relations require initial terms to begin generating the sequence (e.g., a0a_0, a1a_1).
  1. Check for Patterns:
  • Substitution Method: Expand the recurrence for a few terms to observe any emerging patterns. This helps in forming a conjecture about the solution.
  • Iteration: Express an​ as a function of earlier terms and try to write it explicitly.
  1. Use the Characteristic Equation (if linear):
  • For linear recurrence relations (e.g., ana_n=c1an1+c2an2c_1a_{n−1}+c_2a_{n−2}), set up the characteristic equation: x2c1xc2=0x_2−c_1x−c_2=0
  • Solve for the roots r1r_1 and r2r_2 to write the general solution based on the type of roots (real, distinct, or repeated).
  1. Apply Initial Conditions:
  • Substitute the initial conditions into the general solution to determine any constants.
  1. For Non-Homogeneous Relations:
  • Solve the homogeneous part first.
  • Find a particular solution for the non-homogeneous part (e.g., using undetermined coefficients).
  • Combine the two solutions for the final form.

This definition requires a start value (usually called u0u_0 or u1u_1) and a term-to-term rule.


infoNote

Example:

u0=7oru1=7u_0 = 7 \quad \text{or} \quad u_1 = 7un+1=5unun2orun=5un1un12u_{n+1} = 5u_n - u_n^2\\ or\\u_n=5u_{n-1}-u^2_{n-1}

Both of the above say "NEXT=5PREVPREV2NEXT = 5PREV - PREV^2"

The first 5 terms of this sequence are:

u0=7u_0 = 7 u1=5(7)72=14u_1 = 5(7) - 7^2 = -14 u2=5(14)(14)2=266u_2 = 5(-14) - (-14)^2 = -266 u3=5(266)(266)2=72086u_3 = 5(-266) - (-266)^2 = -72086 u4=5(72086)(72086)2=5196756826u_4 = 5(-72086) - (-72086)^2 = -5196756826 u5=5(5196756826)(5196756826)2=2.7×1019u_5 = 5(-5196756826) - (-5196756826)^2 = -2.7 \times 10^{19}
infoNote

The nth term of a sequence, unu_n, is given by

un=c+3n2u_n = c + 3^{n-2}

Given that u3=11u_3 = 11,

a) Find the value of the constant cc.

b) Find the value of u6u_6.


Solution:

a)

u3=c+332=11u_3 = c + 3^{3-2} = 11 c+3=11c=:success[8]\Rightarrow c + 3 = 11 \Rightarrow c = :success[8]

b)

u6=8+362=8+34=:success[89]u_6 = 8 + 3^{6-2} = 8 + 3^4 = :success[89]
infoNote

The terms of a sequence u1,u2,u3,u_1, u_2, u_3, \ldots are given by

un=3(un1k),n>1,u_n = 3(u_{n-1} - k), \quad n > 1,

where kk is a constant. Given that u1=4u_1 = -4,

a) Find expressions for u2u_2 and u3u_3 in terms of kk.

b) Given also that u3=7u2+3u_3 = 7u_2 + 3, find the value of kk.

c) Find the value of u4u_4.


Solution:

a)

u1=4u_1 = -4 u2=3(u1k)=3(4k)=:success[123k]u_2 = 3(u_1 - k) = 3(-4 - k) = :success[-12 - 3k] u3=3(u2k)=3(123kk)=3(124k)=:success[3612k]u_3 = 3(u_2 - k) = 3(-12 - 3k - k) = 3(-12 - 4k) = :success[-36 - 12k]

b)

u3=7u2+3u_3 = 7u_2 + 3 3612k=7(123k)+3-36 - 12k = 7(-12 - 3k) + 3 3612k=8421k+3-36 - 12k = -84 - 21k + 3 36+843=9k-36 + 84 - 3 = -9k k=:success[5]k = :success[-5]

c)

u3=3612(5)=24u_3 = -36 - 12(-5) = 24 u4=3(24+5)=:success[87]u_4 = 3(24 + 5) = :success[87]
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