MATH10001 Project 3 Difference Equations 1

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MATH10001 Project 3 Difference Equations 1 http://www.maths.manchester.ac.uk/undergraduate/ ugstudies/units/2009-10/level1/MATH10001/

A sequence (yn)n0 is an infinite list of terms y0, y1, y2,… A difference equation is a relationship between the terms in a sequence. Examples yn+1 = yn, for all n  0. yn+1 = r yn (1 – yn) for all n  0. Difference equations are particularly useful when using computers because they deal with discrete data.

Linear Difference Equations An m-th order linear difference equation with constant coefficients has the form am yn+m + am-1 yn+m-1 + … + a1 yn+1 + a0 yn = r(n) where a0,…,am constants and r(n) is some function of n. The solution is in two parts: (A) Solution to the homogeneous equation with r(n) = 0. (B) A particular solution ie. any function that satisfies the original difference equation. The general solution is the sum of solutions for (A) and (B).

(A) Homogeneous Equations We start with the 1st order case: a1 yn+1 + a0 yn = 0 where a1  0. We can rewrite this as yn+1=C yn for some constant C. So yn = Cyn-1 = C2 yn-2 = … = Cn y0. Example 2yn+1 – 3yn = 0, y0 = 1.

In the m-th order case, the homogeneous equation has the form am yn+m + am-1 yn+m-1 + … + a1 yn+1 + a0 yn = 0. Again look for solutions of the form yn+1 = C yn, for all n  0. The equation becomes am Cn+m + am-1 Cn+m-1 +…+ a1Cn+1 + a0 Cn= 0. Dividing by Cn gives am Cm + am-1 Cm-1 +…+ a1C + a0 = 0. a polynomial of degree m in C.

This equation has m roots C1, C2, … ,Cm. We get m solutions to the difference equation A1C1n, A2C2n, … , AmCmn. for some constants A1,…,Am. The general solution to the homogeneous difference equation is the sum of these solutions yn = A1C1n + A2C2n + … + AmCmn. Example The Fibonacci sequence 1,1,2,3,5,…

(A) Particular Solutions To solve a non-homogeneous difference equation we need to find an additional function of n to account for the right-hand side of the equation r(n). We look for a single solution to the equation. This will often be a function of the same type as r(n). Example Solve yn+1 – 2yn = n, where y0 = 1.

The Mandelbrot Set