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Introduction to first order recurrence relations with a simple example, including homogenous and non-homogenous solutions.

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A recurrence relation is an equation which can be used to construct a sequence of terms $x_n$ from a given initial value. A first order recurrence relation is of the form

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$x_n + Cx_{n-1} = f(n)$

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where $C$ is a constant and $f(n)$ is a function of $n$. When $f(n) = 0$ the recurrence relation is called homogeneous. The concept of a homogeneous recurrence relation is both simple and fundamental.

The purpose of the first two parts is to present the idea that the solution to a first order recurrence relation is of the form

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$x_n = A\\times C^n$.

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In this example we explored $x_n - \\var{a}x_{n-1} = 0$, which is a homogeneous equation (because of the zero) and has homogeneous solution

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$h_n = \\var{x0}\\times \\var{a}^n$.

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This allows you to calculate the elements in the sequence $x_1 = \\var{x0*a}, x_2 = \\var{x0*a^2}, x_3 = \\var{x0*a^3}$ and $x_4 = \\var{x0*a^4}$.

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The rest of the question is about exploring the non-homogeneous recurrence relation $x_n - \\var{a}x_{n-1} = \\var{b}$. So you can see the terms of the sequence we begin by iteratively constructing the sequence starting with $x_0 = \\var{x0}$:

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$x_1 = \\var{a}x_0 + \\var{b} = \\var{a} \\times \\var{x0} + \\var{b} = \\var{x1}$.

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Similarly

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$x_2 = \\var{a}x_1 + \\var{b} = \\var{a} \\times \\var{x1} + \\var{b} = \\var{x2}$

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$x_3 = \\var{a}x_2 + \\var{b} = \\var{a} \\times \\var{x2} + \\var{b} = \\var{x3}$

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$x_4 = \\var{a}x_3 + \\var{b} = \\var{a} \\times \\var{x3} + \\var{b} = \\var{x4}$.

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The solution to the non-homogeneous recurrence relation is of the form $x_n = h_n + p_n$. So it is always necessary to find the homogeneous solution $h_n$ when solving a (non-homogeneous) recurrence relation. The particular solution in this case is just a constant (since $f(n) = \\var{b}$ is also constant, we should expect $p_n$ to also have the same 'shape'). In this case $p_n = \\dfrac{\\var{b}}{\\var{1-a}}=\\var{b/(1-a)}$ so the full solution is

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$x_n = h_n + p_n = A\\times \\var{a}^n + \\var{b/(1-a)}$.

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Use the initial condition $x_0 = \\var{x0}$ to evaluate $A = \\var{x0-b/(1-a)}$.

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This solution is very useful to find $x_n$ when $n$ is large. Such as $x_{\\var{n0}} = \\var{x0 - b/(1-a)}\\times \\var{a}^\\var{n0} + \\var{b/(1-a)} = \\var{xn0}$.

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Consider the homogeneous first order recurrence relation $x_n - \\var{a} x_{n-1}=0$ for $n>0$ with initial condition $x_0 = \\var{x0}$. Re-arrange the relation into the form $x_n = \\var{a}x_{n-1}$ to find

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• $x_1 =$ [[0]]
• \n
• $x_2 =$ [[1]]
• \n
• $x_3 =$ [[2]]
• \n
• $x_4 =$ [[3]].
• \n
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The $n$th term of a first order homogeneous recurrence relation can be expressed in the form

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$h_n = A\\times C^n$

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where $C$ is determined by the coefficients, and $A$ is determined by the initial conditions. This is also called the homogeneous solution. What is the value of $C$ in the homogeneous solution for the above sequence?

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Consider the non-homogeneous first order recurrence relation $x_n - \\var{a} x_{n-1} = \\var{b}$ for $n>0$ with initial condition $x_0 = \\var{x0}$. Re-arrange the relation into the form $x_n = \\var{a}x_{n - 1} - \\var{-1*b}$ to find

\n
\n
• $x_1 =$ [[0]]
• \n
• $x_2 =$ [[1]]
• \n
• $x_3 =$ [[2]]
• \n
• $x_4 =$ [[3]].
• \n

The $n$th term of a first order non-homogeneous recurrence relation can be expressed in the form

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$x_n = h_n + p_n$

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where $h_n$ is the homogeneous solution, and $p_n$ is a particular non-homogeneous solution. The sum of the homogeneous and particular solutions is the general solution to the non-homogeneous recurrence relation. Find the (general) solution to $x_n - \\var{a}x_{n-1} = \\var{b}$ subject to the initial condition $x_0 = \\var{x0}$.

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There is no simple guaranteed method to finding $p_n$, and so we must 'guess and check'. Suppose $p_n$ is a particular solution to the non-homogeneous equation:

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$p_n - \\var{a} p_{n-1} = \\var{b}$.

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Since the non-homogeneous part $f(n) = \\var{b}$ is constant, we 'guess' that $p_n = D$ for some constant $D$:

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$D - \\var{a} D = \\var{b}$.

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This guess must have worked since the above equation is solvable, and the value of $D$ is:

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The general solution is therefore $h_n + D$ where $h_n$ and $D$ are determined above. Now apply the initial condition

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$\\var{x0} = x_0 = h_0 + p_0=A\\var{a}^0+D=A+D$

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to determine the value of $A$, which is:

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The complete solution is therefore $h_n + p_n = A\\times \\var{a}^n + D$ where $A$ and $D$ have the values determined above.

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Use your solution to evaluate $x_{\\var{n0}}$.

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